Class 6 Mathematics

NCERT Notes For Class 6 Maths Chapter 2 Whole Numbers

Whole Numbers Introduction

As we know, we use 1, 2, 3, 4,… when we begin to count. They come naturally when Westart is counting. Hence, Mathematically call the counting numbers Natural numbers.

Predecessor And Successor

Given any natural number, you can add 1 to that number and get the next number i.e. you get its successor.

The successor of 16 is 16 + 1 = 17, that of 19 is 19 +1 = 20 and so on.

The number 16 comes before 17, we say that the predecessor of 17 is 17-1=16, the predecessor of 20 is 20 – 1 = 19, and so on.

The number 3 has a predecessor and a successor. What about 2? The successor is 3 and the predecessor is 1. Does 1 have both a successor and a predecessor?

We can count the number of children in our school; we can also count the number of people in a city; we can count the number of people in India. The number of people in India.

the number of people in the whole world can also be counted. We may not be able to count the number of stars in the sky or the number of hair on our head but if we are able, there would be a number for them also.

We can then add one more to such a number and get a larger number. In that case, we can even write the number of hair on two heads taken together.

It is now perhaps obvious that there is no largest number. Apart from these questions shared above, there are many others that can come to our mind when we work with natural numbers.

You can think of a few such questions and discuss them with your friends. You may not clearly know the answers to many of them!

NCERT Notes For Class 6 Maths Whole Numbers

We have seen that the number 1 has no predecessor in natural numbers. To the
collection of natural numbers we add zero as the predecessor for 1.

The natural numbers along with zero form the collection of whole numbers.

In your previous classes, you have learnt to perform all the basic operations like addition, subtraction, multiplication and division of numbers.

You also know how to apply them to problems. Let us try them on a number line. Before we proceed, let us find out what a number line is!

NCERT Notes For Class 6 Maths The Number Line

Draw a line. Mark a point on it. Label it 0. Mark a second point to the right of 0. Label it 1.

The distance between these points labelled as 0 and 1 is called unit distance.

On this line, mark a point to the right of 1 and at a unit distance from 1 and label it 2. In this way go on labelling points at unit distances as 3, 4, 5,… on the line. You can go to any whole number on the right in this manner.

This is a number line for all numbers.

What is the distance between the points 2 and 4? Certainly, it is 2 units. Can you tell the distance between the points 2 and 6, between 2 and 7?

On the number line, you will see that the number 7 is on the right of 4. his number 7 is greater than 4, i.e. 7 > 4. The number 8 lies on the right of 6 and 8 > 6.

These observations help us to say that, out of any two whole
numbers, the number on the right of the other number is the greater number.

We can also say that the whole number on the left is the smaller number.

For example, 4 < 9; 4 is on the left of 9. Similarly, 12 > 5; 12 is to the
right of 5.

What can you say about 10 and 20?

Mark 30, 12, and 18 on the number line. Which number is at the farthest left? Can you say which number from 1005 and 9756 would be on the right relative to the other number?

Place the successor of 12 and the predecessor of 7 on the number line.

Addition on the number line

The addition of whole numbers can be shown on the number line. Let us see the addition of 3 and 4.

Start from 3. Since we add 4 to this number we make 4 jumps to the right; from 3 to 4, 4 to 5, 5 to 6 and 6 to 7 as shown above. The tip of the last arrow in the fourth jump is at 7.

The sum of 3 and 4 is 7, i.e. 3 + 4 = 7.

Subtraction on the number line

The subtraction of two whole numbers can also be shown on the number line.

Let us find 7-5.

Start from 7. Since 5 is being subtracted, so move towards left with 1 jump of 1 unit. Make 5 such jumps. We reach the point 2. We get 7- 5 = 2.

Multiplication on the number line

We now see the multiplication of whole numbers on the number line.
Let us find 4 x 3.

Start from 0, move 3 units at a time to the right, and make such 4 moves. Where do you reach? You will reach 12. So, we say, 3 x 4 = 12.

NCERT Notes For Class 6 Maths Chapter 1 Knowing Our Numbers

Knowing Our Numbers Introduction

Counting things is easy for us now. We can count objects in large numbers, for example, the number of students in the school, and represent them through numerals. We can also communicate large numbers using suitable number names.

It is not as if we always knew how to convey large quantities in conversation or through symbols. Many thousands of years ago, people knew only small numbers. Gradually, they learned how to handle larger numbers.

They also learned how to express large numbers in symbols. All this came through the collective efforts of human beings. Their path was not easy, they struggled all along the way.

The development of the whole of Mathematics can be understood this way. As human beings progressed, there was a greater need for the development of Mathematics and as a result, Mathematics grew further and faster.

We use numbers and know many things about them. Numbers help us count concrete objects. They help us to say which collection of objects is bigger and arrange them for example, first, second, etc.

Numbers are used in many different contexts and in many ways. Think about various situations where we use numbers. List five distinct situations in which
numbers are used.

We enjoyed working with numbers in our previous classes. We have added, subtracted, multiplied, and divided them. We also looked for patterns in number sequences and did many other interesting things with numbers.

In this chapter, we shall move forward on such interesting things with a bit of review and revision as well.

Comparing Numbers

As we have done quite a lot of this earlier, let us see if we remember which is the greatest among these:

1. 92, 392, 4456, 89742

89742 I am the greatest

2. 1902, 1920, 9201, 9021, 9210

9210 I am the greatest

So, we know the answer

We just looked at the number of digits and found the answer. The greatest number has the most thousands and the smallest is only in hundreds or in tens.

Make five more problems of this kind and give them to your friends to solve.

Now, how do we compare 4875 and 3542?

This is also not very difficult. These two numbers have the same number of digits. They are both in thousands. But the digit at the thousands place in 4875 is greater than that in 3542. Therefore, 4875 is greater than 3542.

Next tell me which is greater, 4875 or 4542? Here too the numbers have the same number of digits. Further, the digits at the thousands place are the same in both.

What do we do then? We move to the next digit, that is to the digit at the hundreds place. The digit at the hundreds place is greater in 4875 than in 4542. Therefore, 4875 is greater than 4542.

If the digits at the hundreds place are also the same in the two numbers, then what do we do?

Compare 4875 and 4889; Also compare 4875 and 4879.

How Many Numbers Can You Make?

Suppose, we have four digits 7, 8, 3, 5. Using these digits we want to make different 4-digit numbers in such a way that no digit is repeated in them. Thus, 7835 is allowed, but 7735 is not. Make as many 4-digit numbers as you can.

Which is the greatest number you can get? Which is the smallest number?

The greatest number is 8753 and the smallest is 3578.

Think about the arrangement of the digits in both. Can you say how the largest number is formed? Write down your procedure.

Shifting Digits

Have you thought how fun it would be if the digits in a number could shift (move^frpm from one place to the other?

Thinlq about what would happen to 182. It could become as large as 821 and as small as 128. Try this with 391 as well.

Now think about this. Take any 3-digit number and exchange the digit at the hundreds place with the digit at the one’s place.

1. Is the new number greater than the former one?
2. Is the new number smaller than the former number?

Write the numbers formed in both ascending and descending order.

Before 7 9 5

Exchanging the 1st and the 3rd tiles.

After 5 9 7

If you exchange the 1st and the 3rd tiles (i.e. digits), in which case does the number become greater? In which case does it become smaller?

Try this with a 4-digit number.

Introducing 10,000

We know that beyond 99 there is no 2-digit number. 99 is the greatest 2-digit number. Similarly, the greatest 3-digit number is 999 and the greatest 4-digit number is 9999. What shall we get if we add 1 to 9999?

Look at the pattern:

9+1 = 10 = 10 x 1

99+1 = 100 = 10 x 10

999+1 = 1000 = 10 x 100

We observe that

Greatest single digit number + 1 = smallest 2-digit number

Greatest 2-digit number + 1 = smallest 3-digit number

Greatest 3-digit number + 1 = smallest 4-digit number

We should then expect that on adding 1 to the greatest 4-digit number, we would get the smallest 5-digit number, that is 9999 + 1 = 10000.

The new number which comes next to 9999 is 10000. It is called ten thousand. Further, 10000 = 10 x 1000.

Revisiting Place Value

You have done this quite earlier, and you will certainly remember the expansion of a 2-digit number like 78 as 78 =70 + 8 = 7vx 10 + 8

Similarly, you will remember the expansion of a 3-digit number like 278 as 278 = 200 + 70 + 8 = 2vx 100 + 7vx 10 + 8

We say, here, 8 is at one place, 7 is at the tens place, and 2 at the hundreds place.

Later on, we extended this idea to 4-digit numbers.

For example, the expansion of 5278 is 5278 = 5000 + 200 + 70 + 8

= 5 x 1000 +2 x 100 + 7 x 10 + 8

Here, 8 is at one place, 7 is at the tens place, 2 is at the hundreds place and 5 is at the thousands place.

With the number 10000 known to us, we may extend the idea further. We may write 5-digit numbers like 45278 = 4 x 10000 + 5 x 1000 + 2 x 100 + 7 x 10 + 8

We say that here 8 is at one place, 7 at the tens place, 2 at the hundreds place, 5 at the thousands place, and 4 at ten thousand places.

The number is read as forty-five thousand, two hundred seventy-eight. Can you now write the smallest and the greatest 5-digit numbers?

Introducing 1,00,000

Which is the greatest 5-digit number?

Adding 1 to the greatest 5-digit number should give the smallest 6-digit number: 99,999 + 1 = 1,00,000

This number is named one lakh. One lakh comes next to 99,999.

10 x 10,000 = 1,00,000

We may now write 6-digit numbers in the expanded form as

2,46,853 = 2 x 1,00,000 + 4x 10,000 + 6 x 1,000 + 8 x 100 + 5 x 10 + 3 x 1

This number is 3 in one place, 5 in the tens place, 8 in the hundreds place, 6 in the thousands place, 4 in ten thousand places, and 2 in lakh place. Its number name is two lakh forty-six thousand eight hundred fifty three.

Larger numbers

If we add one more to the greatest 6-digit number we get the smallest 7-digit number. It is called ten lakh.

Write down the greatest 6-digit number and the smallest 7-digit number. Write the greatest 7-digit number and the smallest 8-digit number. The smallest 8-digit number is called one crore.

Complete The Pattern:

Remember

1 hundred =10 tens

1 thousand = 10 hundreds = 100 tens

1 lakh = 100 thousands = 1000 hundreds

1 crore =100 lakhs = 10,000 thousands

9+1 = 10

99+1 = 100

999+1 = 1,000

9.999+1 = 10,000

99.999+1 = 1,00,000

9.99.999+1 = 10,00,000

99.99.999+1 = 1,00,00,000

We come across large numbers in many different situations. For example, while the number of children in your class would be a 2-digit number, the number of children in your school, would be a 3 or 4-digit number.

The number of people in the nearby town would be much larger.

NCERT Notes For Class 6 Maths Use Of Commas

You must have noticed that in writing large numbers in the sections above, we have often used commas.

Commas help us in reading and writing large numbers. In our Indian System of Numeration, we use ones, tens, hundreds, thousands, and then lakhs and crores.

Commas are used to mark thousands, lakhs, and crores. The first comma comes after hundreds place (three digits from the right) and marks thousands. The second comma comes two digits later (five digits from the right).

It comes after ten thousand places and marks lakh. The third comma comes after another two digits (seven digits from the right). It comes after ten lakh place and marks crore.

For example,

5,08,01,592

3,32,40,781

7,27,05,062

Try reading the numbers given above. Write five more numbers in this form and read them.

International System of Numeration

In the International System of Numeration, as it is being used we have ones, tens, hundreds, thousands, and then millions. One million is a thousand thousand. Commas are used to mark thousands and millions.

It comes after every three digits from the right. The first comma marks thousands and the next comma marks millions. For example, the number 50,801,592 is read in the International System as fifty million eight hundred one thousand five hundred ninety-two.

In the Indian System, it is five crore eight lakh one thousand five hundred ninety-two.

How many lakhs make a million?

How many millions make a crore?

Take three large numbers. Express them in both Indian and International Numeration systems.

Interesting Fact:

To express numbers larger than a million, a billion is used in the International System of Numeration: 1 billion = 1000 million.

NCERT Notes For Class 6 Maths Large Numbers In Practice

In earlier classes, we learned that we use centimeters (cm) as a unit of length. For measuring the length of a pencil, the width of a book or notebook, etc., we use centimeters. Our ruler has marks on each centimeter.

However, we find centimeters too big when measuring the thickness of a pencil. We use millimeters (mm) to show the thickness of a pencil.

1. 10 millimetres = 1 centimetre

To measure the length of the classroom or the school building, we shall find a centimeter too small. We use meters for this purpose.

2. 1 metre =100 centimetres = 1000 millimetres

Even meter is too small when we have to state distances between cities, say, Delhi and Mumbai, or Chennai and Kolkata. For this, we need kilometers (km).

3. 1 kilometre = 1000 metres

How many millimeters make 1 kilometer?

Since 1 m = 1000 mm

1 km = 1000 m = 1000 x 1000 mm = 10,00,000 mm

We go to the market to buy rice or wheat; we buy it in kilograms (kg). But items like ginger or chilies which we do not need in large quantities, we buy in grams (g). We know 1 kilogram = 1000 grams.

Have you noticed the weight of the medicine tablets given to the sick? It is very small. It is in milligrams (mg)

1 gram = 1000 milligrams.

What is the capacity of a bucket for holding water? It is usually 20 liters (l). Capacity is given in liters. But sometimes we need a smaller unit, the milliliters. A bottle of hair oil, a cleaning liquid, or a soft drink has labels that give the quantity of liquid inside in milliliters (ml).

1 litre = 1000 millilitres.

Note that in all these units we have some words in common like kilo, milli, and centi. You should remember that among these kilo is the greatest and milli is the smallest; the kilo shows 1000 times greater, and the milli shows 1000 times smaller, i.e. 1 kilogram = 1000 grams, 1 gram = 1000 milligrams.

Similarly, centi shows 100 times smaller, i.e. 1 metre = 100 centimetres.

We have done a lot of problems that have addition, subtraction, multiplication, and division. We will try solving some more here. Before starting, look at these examples and follow the methods used.

Example 1. The population of Sundamagar was 2,35,471 in the year 1991. In the year 2001, it was found to be increased by 72,958. What was the population of the city in 2001?

Solution:

Population of the city in 2001

= Population of the city in 1991 + Increase in population = 2,35,471 + 72,958

Now, \(\begin{array}{r}
235471 \\
+72958 \\
\hline 308429 \\
\hline
\end{array}\)

Salma added them by writing 235471 as 200000 + 35000 + 471 and 72958 as 72000+958. She got the addition as 200000 +107000 +1429 = 308429. Mary added it as 200000 + 35000 + 400 + 71 + 72000 + 900 + 58 = 308429

Answer: The population of the city in 2001 was 3,08,429

All three methods are correct.

Example 2. In one state, the number of bicycles sold in the year 2002-2003 was 7,43,000. In the year 2003-2004, the number of bicycles sold was 8,00,100. In which year were more bicycles sold? and how many more?

Solution:

Clearly, 8,00,100 is more than 7,43,000. So, in that state, more bicycles were sold in the year 2003-2004 than in 2002-2003.

Now, \(\begin{array}{r}
800100 \\
-743000 \\
\hline 057100 \\
\hline
\end{array}\)

Check the answer by adding \(\begin{array}{r}
743000 \\
+57100 \\
\hline 800100 \\
\hline
\end{array}\) (the answer is right)

Can you think of alternative ways of solving this problem?
Answer: 57,100 more bicycles were sold in the year 2003-2004.

Example 3. The town newspaper is published every day. One copy has 12 pages> every day 11,980 copies are printed. How many total pages are printed every day?

Solution:

Each copy has 12 pages. Hence, 11,980 copies. will have 12×11,980 pages. What would this number be? More than 1,00,000 or lesser.

Try to estimate.

Now, \(\begin{array}{r}
11980 \\
\times 12 \\
\hline 23960 \\
+\quad 119800 \\
\hline 143760 \\
\hline
\end{array}\)

Answer: Every day 1,43,760 pages are printed.

Example 4. The number of sheets of paper available for making notebooks is 75,000. Each sheet makes 8 pages of a notebook. Each notebook:(contains 200 pages. How many notebooks can be made from the paper available?

Solution:

Each sheet makes 8 pages.

Hence, 75,000 sheets make 8 x 75,000 pages,

Now, \(\begin{array}{r}
75000 \\
\times 8 \\
\hline 600000 \\
\hline
\end{array}\)

Thus, 6,00,000 pages are available for making notebooks.

Now, 200 pages make 1 notebook.

Hence, 6,00,000 pages make 6,00,000 -r 200 notebooks.

Now,

The answer is 3,000 notebooks.

Class 6 Maths Chapter 6 Integers

Exercise – 6.1

1. Write opposites of the following:

(1) Increase in weight
(2) 30 km north
(3) 80 m east
(4) Loss of? 700
(5) 1 00 m above sea level

Solution: (1) Decrease in weight

(2) 30 km south

(3) 80 m west

(4) Profit of? 700

(5) 100 m below sea level

Read and Learn More Class 6 Maths Exemplar Solutions

2. Represent the following numbers as integers with appropriate signs.

(1) An aeroplane is flying at a height, two thousand metre above the ground.
(2) A submarine is moving at a depth, eight hundred metre below the sea level.
(3) A deposit of rupees two hundred.
(4) Withdrawal of rupees seven hundred.

Solution: (1) Two thousand metre above the ground = + 2000

(2) Eight hundred metre below the sea level = -800

(3) Deposit of two hundred rupees = + 200

(4) Withdrawal of seven hundred rupees = -700

3. Represent the following numbers on a number line:

(1) +5
(2) -10
(3) + 8
(4) -1
(5) -6

Solution: (1)

(2)

(3)

(4)

(5)

4. Adjacent figure is a vertical number A line, representing integers. Observe it and locate the following points:

(1) If point D is + 8, then which point is -8?
(2) Is point G a negative integer or a positive integer?
(3) Write integers for points 8 and E.
(4) Which point marked on this number line has the least value?
(5) Arrange all the points in decreasing order of value.

Solution: (1) We have, point D is +8.

Therefore, 16 steps to the down from D is -8 i.e., the point F.

(2) Point G is a negative integer.

(3) Point B is four steps down from point D.

Value of point B = +8- 4 = +4

Point E is eighteen steps down from point D.

Value of point E = +8- 18 = -10

(4) Since, point E is located in the bottom.

So, point E has the least value.

(5) Decreasing order of all the points is,

D, C,B, A,0, H,G, F,E

5. Following Is the list of temperatures of five places In India on a particular day of the year.

Place                              Temperature

Siachin                           10°C below 0°C …………

Shimla                            2°C below 0°C …………

Ahmedabad                   30°C above 0°C …………

Delhi                              20°C above 0°C …………

Srinagar                          5°C below 0°C …………

(1) Write the temperatures of these places in the form of integers in the blank column.

(2) Following is the number line representing the temperature in degree Celsius

Plot the name of the city against its temperature.

(3) Which is the coolest place?

(4) Write the names of the places where temperatures are above 10°C.

Solution:

Place                           Temperature

(1) Siachin                  -10°C

Shimla                        -2°C

Ahmedabad               + 30°C

Delhi                          + 20°C

Srinagar                     -5°C

(3) Siachin is the coolest place.

(4) Ahmedabad and Delhi have temperature above 10°C.

6. In each of the following pairs, which number is to the right of the other on the number line?

(1) 2, 9
(2) -3,-8
(3) 0,-1
(4) -11,10
(5) -6,6
(6) 1,-100

Solution: (1) 9 is right to 2

(2) -3 is right to -8

(3) 0 is right to -1

(4) 10 is right to -11

(5) 6 is right to -6

(6) 1 is right to -100

7. Write all the integers between the given pairs (write them in the increasing order.)

(1) 0 and-7
(2) -4 and 4
(4) -30 and -23
(3) -8 and -15

Solution: (1) The integers between 0 and -7 are -6, -5, -4, -3, -2, -1

(2) The integers between -4 and 4 are -3, -2, -1,0, 1,2, 3

(3) The integers between-8 and -15 are -14, -13, -12, -11, -10, -9

(4) The integers between -30 and -23 are -29, -28, -27, -26, -25, -24

8. (1) Write four negative integers greater than -20.

(2) Write four integers less than- 1 0.

Solution: (1) There are 19 negative integers which are greater than -20. Four of them are -19, -18, -17, -16

(2) There are infinite integers which are less than -10. Four of them are -11, -12, -13, -14

9. For the following statements, write True (T) or False (6). If the statement is false, correct the statement.

(1) – 8 is to the right of- 1 0 on a number line.
(2) – 100 is to the right of – 50 on a number line.
(3) Smallest negative integer is -1.
(4) – 26 is greater than- 25

Solution: (1) True

(2) False

Since -100 is to the left of -50 on the number line.

(3) False

Since -1 is the greatest negative integer.

(4) False

Since -26 is less than -25.

10. Draw a number line and answer the following :

(1) Which number will we reach if we move 4 numbers to the right of- 2.
(2) Which number will we reach if we move 5 numbers to the left of 1.
(3) if wc arc at- 8 on the number line, in which direction should we move to reach – 1 3?
(4) If we are at- 6 on the number line, in which direction should we move to reach – 1 ?

Solution: (1)

Thus, we will reach 2 if we move 4 numbers to the right of -2.

(2)

 

Thus, we will reach -4 if we move 5 numbers to the left of 1.

(3)

Thus, we should move 5 numbers to the left of-8 to reach -13.

(4)

Thus, we should move 5 numbers to the right of -6 to reach -1.

Exercise – 6.2

1. Using the number line write the integer which is:

(1) 3 more than 5
(2) 5 more than -5
(3) 6 less than 2
(4) 3 less than -2

Solution: (1)

Thus, 3 more than 5 is 8.

(2)

Thus, 5 more than -5 is 0.

(3)

Thus, 6 less than 2 is -4.

(4)

Thus, 3 less than -2 is -5.

2. Use number line and add the following integers :

(1) 9 + (-6)
(2) 5 + (-11)
(3) (-1) + (-7)
(4) (- 5) + 10
(5) (- 1 ) + (- 2) + (- 3)
(6) (- 2) + 8 + (- 4)

Solution: (1)

Thus, 9 + (-6) = 3

(2)

Thus, 5 + (-11) = -6

(3)

Thus, (-1) + (-7) = -8

(4)

Thus, (-5) + 10 = 5

(5)

Thus, (-1) + (-2) + (-3) = -6

(6)

Thus, (-2) + 8 + (-4) = 2

3. Add without using number line:

(1) 1 1 + (- 7)
(2) (-13) + (+18)
(3) (-10) + (+19)
(4) (-250) + (+150)
(5) (- 380) + (- 270)
(6) (- 21 7) + (- 1 00)

Solution: (1) 11+ (-7) -11-7=4

(2) (-13) + (+18)= -13 + 18 =5

(3) (-10) + (+19)= -10 + 19 =9

(4) (-250) + (+150) = -250 + 150 = -100

(5) (-380) + (-270) = -380- 270 = -650

(6) (-217) + (-100) = -217 -100 = -317

4. Find the sum of:

(1) 137 and -354
(2) -52 and 52
(3) -31 2, 39 and 192
(4) -50, -200 and 300

Solution: (1) 137 + (-354) = 137- 354 = -217

(2) -52 + 52 = 0

(3) -312 + 39 + 192 = -312 + 231 =-81

(4) -50 + (-200) + 300 = -50 -200 + 300 = -250 + 300 = 50

5. Find the sum:

(1) (- 7) + (- 9) + 4 + 16
(2) (37) + (-2) + (-65) + (-8)

Solution: (1) (-7) + (-9) + 4 + 16

=-7-9 + 4 + 16

= -16 + 20 = 4

(2) (37) + (-2) + (-65) + (-8)

= 37-2-65-8

= 37- 75 =- 38

Exercise – 6.3

1. Find

(1) 35 -(20)
(2) 72 -(90)
(3) (-15) -(-18)
(4) (-20) -(13)
(5) 23 -(-12)
(6) (-32) -(-40)

Solution: (1) 35-20 = 15

(2) 72 -90 = -18

(3) (-15)- (-18) =- 15 + 18 = 3

(4) (-20) -(13) = -20 -13 = -33

(5) 23 -(-12) = 23 + 12 = 35

(6) (-32) -(-40) =-32 + 40 = 8

2. Fill in the blanks with >, < or = sign.

(1) (- 3) + (- 6) _________ (- 3)- (- 6)
(2) (-21) -(-10) ___________ (— 31) + (— 11)
(3) 45 – (- 11)___________ 57 + (-4)
(4) (-25) -(-42)__________ (- 42)- (- 25)

Solution: (1) < : (-3) + (-6) = -3- 6 = -9

(-3) -(-6) –3 + 6 = 3

Since, -9 < 3

(-3) + (-6) < (-3)- (-6)

(2) > : (-21)- (-10) = -21 + 10 = -11

(-31) + (-11) = -31 – 11 = -42

Since, -11 > -42

(-21) -(-10) >(-31) + (-11)

(3) > : 45- (-11) = 45 + 11 = 56

57 + (-4) = 57- 4 = 53

Since, 56 > 53

45 -(-11) >57 + (-4)

(4) > : (-25)- (-42) = -25 + 42 = 17

(-42)- (-25) = -42 + 25 = -17

Since, 17 >-17

(-25)- (-42) > (-42)- (-25)

3. Fill in the blanks

(1) (-8) +_______=0

(2) 13 +_________=0

(3) 12 + (-12) =_____

(4) (-4) +________=-12

(5) _________ -15 = -10

Solution: (1) 8 : (-8) + 8 = 0

(2) -13 : 13 + (-13) = 0

(3) 0:12 + (-12) = 0

(4) -8: (-4) + (-8) =-12

(5) 5: 5- 15 = -10

4. Find

(1) (-7)-8- (-25)
(2) (-13) + 32 -8-1
(3) (- 7) + (- 8) + (- 90)
(4) 50 -(-40) -(-2)

Solution: (1) (-7)- 8- (-25)

=-7-8 + 25

= -15 + 25 = 10

(2) (-13) +32-8-1

= -13 +32-8-1

= 32- 22 = 10

(3) (-7) + (-8) + (-90)

= -7- 8- 90 =- 105

(4) 50 -(-40) -(-2)

= 50 + 40 + 2 = 92

Section-2 NCERT Exemplar

Directions: In questions 1 to 17, only one of the four options is correct. Write the correct one.

1. Every integer less than 0 has the sign

(1) +
(2) –
(3) x
(4) ÷

Solution: (2): Every integer which is less than 0 has negative sign.

2. The integer ‘5 units to the right of 0 on the number line’ is

(1) +5
(2) -5
(3) +4
(4) -4

Solution: (1): The integer which is 5 units to the right of 0 on the number line is +5.

3. The predecessor of the integer -1 is

(1) 0
(2) 2
(3) -2
(4) 1

Solution: (3): The predecessor of the integer -1 is -2.

4. Number of integers lying between -1 and 1 is

(1) 1
(2) 2
(3) 3
(4) 0

Solution: (1): Only 1 integer lies between -1 and1 i.e., 0

5. Number of whole numbers lying between -5 and 5 is

(1) 10
(2) 3
(3) 4
(4) 5

Solution: (4): There are 5 whole numbers lying between -5 and 5 i.e., 0, 1, 2, 3 and 4.

6. The greatest integer lying between -10 and -15 is

(1) -10
(2) -11
(3) -15
(4) -14

Solution: (2): -11 is the greatest integer lying between -10 and -15

7. The least integer lying between -10 and -15 is

(1) -10
(2) -11
(3) -15
(4) -14

Solution: (4): -14 is the least integer lying between -10 and -15.

8. On the number line, the integer 5 is located

(1) to the left of 0
(2) to the right of 0
(3) to the left of 1
(4) to the left of-2

Solution: (2):

The above number line shows that the integer 5 is located to the right of 0

9. In which of the following pairs of integers, the first integer is not on the left of the other integer on the number line?

(1) (-1,10)
(2) (-3,-5)
(3) (-5,-3)
(4) (-6,0)

Solution: (2) :

On observing all the options by using a number line, we get that there is only one pair (-3, -5) in which the first integer is not on the left of the other integer.

10. The integer with negative sign (-) is always less than

(1) 0
(2) -3
(3) -1
(4) -2

Solution: (1): All the negative integers are less than 0.

11. An integer with positive sign (+) is always greater than

(1) 0
(2) 1
(3) 2
(4) 3

Solution: (1): All the positive integers are greater than 0.

12. The successor of the predecessor of -50 is

(1) -48
(2) -49
(3) -50
(4) -51

Solution: (3): The predecessor of -50 is -51 and the successor of -51 is -50.

13. The additive inverse of a negative integer

(1) is always negative
(2) is always positive
(3) is the same integer
(4) zero

Solution: (2): The additive inverse of a negative integer is always positive.

14. Amulya and Amar visited two places A and B respectively in Kashmir and recorded the minimum temperatures on a particular day as -4°C at A and -1°C at 6. Which of the following statement is true?

(1) A is cooler than B
(2) Bis cooler than A
(3) There is a difference of 2°C in the temperature
(4) The temperature at A is 4°C higher than that at B.

Solution: (1) :- 4°C < -1°C [ – 4 lies on the left of-1 on the number line]

Thus, A is cooler than B.

15. When a negative integer is subtracted from another negative integer, the sign of the result

(1) is always negative
(2) is always positive
(3) is never negative
(4) depends on the numerical value of the integers

Solution: (4): When a negative integer is subtracted from another negative integer, the sign of the result depends on the numerical
value of the integers.

16. The statement “When an integer is added to itself, the sum is greater than the integer” is

(1) always true
(2) never true
(3) true only when the integer is positive
(4) true for non-negative integers

Solution: (3): When an integer is added to itself, the sum is greater than the integer only when the integer is positive.

17. Which of the following shows the maximum rise in temperature?

(1) 0°C to 1 0°C
(2) -4°C to 8°C
(3) -15°C to -8°C
(4) -7°C to 0°C

Solution: (2) : (1) Rise in temperature = (10- 0)°C = 10°C

(2) Rise in temperature = (8- (-4))°C = (8 + 4)°C = 12°C

(3) Rise in temperature = (-8- (-15))°C = (-8 + 15)°C = 7°C

(4) Rise in temperature = (0- (-7))°C = (0 + 7)°C = 7°C

Thus, option (2) has maximum rise in temperature.

Directions: In questions 18 to 39, state whether the given statements are true (T) orfalse (6).

18. The smallest natural number is zero.

Solution: False

Since,1 is the smallest natural number.

19. Zero is not an integer as it is neither positive nor negative.

Solution: False

0 is neither positive nor negative, but it is an integer.

20. The sum of all the integers between -5 and -1 is -6.

Solution: False

-4, -3 and -2 lie between -5 and -1 and their sum is (-4) + (-3) + (-2) =-4-3-2 = -9

21. The successor of the integer 1 is 0.

Solution: False

0 is the predecessor of 1.

22. Every positive integer is larger than every negative integer.

Solution: True

Since positive integers lies on the right side of 0 and negative integers lies on the left side of 0 and the integers lying on the right are
always greater.

23. The sum of any two negative integers is always greater than both the integers.

Solution: False

Since, the sum of any two negative integers is always smaller than both the integers.

24. The sum ofany two negative integers is always smaller than both the integers.

Solution: True

25. The sum of any two positive integers is greater than both the integers.

Solution: True

26. All whole numbers are integers.

Solution: True

Integers are the collection of 0, positive integers and negative integers.

Whole numbers are the collection of 0 and positive integers.

All whole numbers are integers.

27. All integers are whole numbers.

Solution: False

Integers are the collection of 0, positive integers and negative integers.

Whole numbers are the collection of 0 and positive integers.

All integers are not whole numbers.

28. Since 5 >3, therefore -5 > -3.

Solution: False

Since,5 lies on the right of 3 on the number line 5>3

And-3 lies on the right of-5 on the number line, -3 >-5.

29. Zero is less than every positive integer.

Solution: True

Since, zero lies on the left side of every positive integer on the number line. Therefore, zero is less than every positive integer.

30. Zero is larger than every negative integer.

Solution: True

Since, zero lies on the right side of every negative integer on the number line.

Therefore, zero is larger than every negative integer.

31. Zero is neither positive nor negative.

Solution: True

32. On the number line, an integer on the right of a given integer is always larger than the integer.

Solution: True

33. -2 is to the left of-5 on the number line.

Solution: False

Since, -2 lies on the right of-5 on the number line.

34. The smallest integer is 0.

Solution: False

Since, zero is greater than all the negative integers.

0 is not the smallest integer.

35. 6 and -6 are at the same distance from 0 on the number line.

Solution: True

The integer 6 is 6 units to the right of 0 and the integer -6 is 6 units to the left of 0.

Thus, 6 and -6 are at the same distance from 0 on the number line.

36. The difference between an integer and its additive inverse is always even.

Solution: True

Let a be any integer and -a is its additive inverse.

Difference = a- (-a) = a + a = 2a, which is an even number.

37. The sum of an integer and its additive inverse is always zero.

Solution: True

Let a be any integer and -a is its additive inverse.

Sum = a + (-a) =a- a = 0.

38. The sum of two negative integers is a positive
integer.

Solution: False

Since, the sum of two negative integers is always negative.

39. The sum of three different integers can never be zero.

Solution: False

Let -3, 1, 2 are three different integers.

Sum = (-3) +1 + 2 =-3 + 3 = 0

Directions: In questions 40 to 49, fill in the blanks to make the statements true.

40. On the number line, -1 5 is to the zero.

Solution: Left

41. On the number line, 10 is to the of zero.

Solution: Right

42. The additive Inverse of 14 is_.

Solution: -14: Additive inverse of an integer is obtained by changing the sign of the integer.

Additive inverse of 14 is -14.

43. The additive inverse of-1 is

Solution: 1

44. The additive inverse of 0 is

Solution: 0

45. The number of integers lying between -5 and 5 is

Solution: 9: The integers lying between -5 and 5 are -4, -3, -2, -1, 0, 1, 2, 3, 4 i.e., 9 in number

46. (-11) + (-2) + (-1) =____________

Solution: -14: (-11) + (-2) + (-1) =-11- 2-1 = “14

47.___________ + (-11) + 111 = 130

Solution: 30

48. (-80) + 0 + (-90) =____________

Solution: -170: (-80) + 0 + (-90) =-80 + 0- 90 =-170

49. -3456 = -8910

Solution: -5454

Directions: In questions 50 to 58,fill in the blanks using <, = or >.

50. (-11) + (-15)___________ 11+15
Solution: < : (-11) + (-15) = -11-15 = -26

11 + 15 = 26 and -26 < 26

51. (-71) + (+9)___________ (-81) + (-9)
Solution: >: (-71) + (9) = -71 + 9 = -62

(-81) + (-9) = -81- 9 = -90 and -62 > -90

52. 0__________ 1

Solution: <:0<1

53. -60__________ 50

Solution: < : -60 < 50

54. -10__________ -11

Solution: >: —10 > —11

55. -101___________ -102

Solution: >: -101 >-102

56. (-2) + (-5) + (-6)__________ (-3) + (-4) + (-6)

Solution: = : (-2) + (-5) + (-6) = -2- 5- 6 = -13

(-3) + (-4) + (-6) =-3-4- 6 =-13

And -13 =-13

57. 0 __________ -2

Solution: >:0>-2

58. 1+2 + 3________ (-1 ) + (-2) + (-3)

Solution: >: 1 +2+3=6

(-1) + (-2) + (-3) =-l-2-3 =-6

And 6 > -6

59. Match the items of Column I with that of Column II:

Solution:

(i)–>(B), (ii)–> (E), (iii) —> (B), (iv) –> (A),(v) –> (B)

(i) The additive inverse of +2 is -2.
(ii) The greatest negative integer is -1.
(iii) The greatest negative even integer is -2.
(iv) The smallest integer 0 is greater than every negative integer.
(v) Predecessor and successor of -1 are -2 and 0 respectively.

∴ Sum = -2 + 0 = -2

60. Compute each of the following:

(1) 30 + (-25) + (-10)
(2) (-20) + (-5)
(3) 70 + (-20) + (-30)
(4) -50 + (-60) + 50
(5) 1 + (-2) + (-3) + (-4)
(6) 0 + (-5) + (-2)
(7) 0- (-6)- (+6)
(8) 0-2 -(-2)

Solution: (1) 30 + (-25) + (-10) = 30 + (-25- 10) = 30 + (-35) = 30- 35 = -5

(2) (-20) + (-5) = -20-5 = -25

(3) 70 + (-20) + (-30) = 70 + (-20- 30) = 70 + (-50) = 70- 50 = 20

(4) -50 + (-60) + 50 = (-50 -60) + 50 = -110 + 50 = -60

(5) 1 + (-2) + (-3) + (-4) =1 +(-2-3-4)=1+ (-9) =1- 9 = -8

(6) 0 + (-5) + (-2) 0 + (-5 -2) « 0 + (-7) = 0- 7 = -7

(7)0- (-6)- (+6) = 0 + 6- 6 = 6- 6 = 0

(8) 0- 2- (-2) = 0- 2 + 2 = -2 + 2 = 0

61. If we denote the height of a place above sea level by a positive integer and depth below the sea level by a negative integer, write the following using integers with the appropriate signs:

(1) 200 m above sea level
(2) 1 00 m below sea level
(3) 10m above sea level
(4) sea level

Solution: (1) 200 m above sea level = + 200

(2) 100 m below sea level =- 100

(3) 10 m above sea level = + 10

(4) Sea level = 0

62. Write the opposite of each of the following :

(1) Decrease in size
(2) Failure
(3) Profit of? 10
(4) 1000 AD
(5) Rise in water level
(6) 60 km south
(7)10m above the danger mark of river Ganga
(8) 20 m below the danger mark of the river Brahmaputra
(9) Winning by a margin of 2000 votes
(10)Depositing? 100 in the Bank account
(11) 20°C rise in temperature.

Solution: (1) Increase in size.

(2) Success.

(3) Loss of? 10

(4) 1000 BC

(5) Fall in water level.

(6) 60 km north.

(7)10 m below the danger mark of river Ganga.

(8) 20 m above the danger mark of the river Brahmaputra.

(9) Losing by a margin of 2000 votes.

(10)Withdrawing ? 100 from the Bank account.

(11) 20°C fall in temperature.

63. Temperature of a place at 12:00 noon was +5°C. Temperature increased by 3°C in first hour and decreased by 1°C in the second hour. What was the temperature at 2:00 pm?

Solution:

Given

Temperature of a place at 12:00 noon was +5°C. Temperature increased by 3°C in first hour and decreased by 1°C in the second hour.

Temperature at 12 : 00 noon = + 5°C

Temperature at 1 : 00 p.m. = 5°C + 3°C = 8°C

And temperature at 2 : 00 p.m. = 8°C- 1°C = 7°C

64. Write the digits 0, 1,2, 3, 9 in this order and insert ‘+’ or between them to get the result 3.

Solution: The digits can be written as 0-1-2-3-4-5-6+7+8+9=3

65. Write the integer which is its own additive inverse.

Solution: 0 is the integer which is its own additive inverse

66. Write six distinct integers whose sum is 7.

Solution: 1 + 2 + 3 + 6 + (-2) + (-3) = 7

The six distinct integers are 1, 2, 3, 6, -2 and -3.

67. Write the integer which is 4 more than its additive inverse.

Solution: Let x be the required integer.

According to question,

x = 4 + (-x), where (-x) is the additive inverse of x.

=> x = 4-x => x + x = 4

=> 2x = 4 => x = 2

The required integer is 2.

68. Write the integer which is 2 less than its additive inverse.

Solution: Let the required integer be x.

According to question,

x = (-x)- 2, where -x is the additive inverse of x.

x=-x-2 => x + x = -2

2x = -2 => x = —1

69. Write two integers whose sum is less than both the integers.

Solution: We can take any two negative integers, i.e., -2 and -4.

Sum = -2 + (-4) = -2- 4 = -6, which is less than both -2 and -4.

70. Write two distinct integers whose sum is equal to one of the integers.

Solution: Two distinct integers whose sum is equal to one of the integer, then one must be 0 in them.

Let us take 0 and 4.

Sum = 0 + 4 =4.

71. Using number line, how do you compare

(1) two negative integers?
(2) two positive integers?
(3) one positive and one negative integer?

Solution: Since, the integer lying on right is greater than the integer lying on left.

In all of the given cases (1), (2) and (3), we can compare by using the number line by observing which one of the given integers lie
on tine right or left.

72. Observe the following : 1 +2-3+4+5-6-7+8-9=-5 Change one’-‘sign as’+’sign to get the sum 9.

Solution: On observing the given expression, l + 2- 3 + 4 +5-6- 7 +8- 9 = -5, we noticed that (-7) should be replaced by (+7) to get a result of 9.

Thus, 1+2-3+4+5-6+7+8-9
= (l+2 + 4 + 5 + 7 + 8)-(3 + 6 + 9)
= 27-18 = 9

73. Arrange the following integers in the ascending order: -2, 1,0, -3, 4, -5

Solution: Ascending order of given integers is, -5, -3, -2, 0,1, 4

74. Arrange the following integers in the descending order: -3, 0, -1,-4, -6

Solution: Descending order of given integers is, 0, -1, -3, -4, -6

75. Write two integers whose sum is 6 and difference is also 6.

Solution: We have, 6 + 0 = 6,

6-0 = 6

The required two integers are 6 and 0.

76. Write five integers which are less than -100 but greater than -1 50.

Solution: The required five integers which are less than -100 but greater than -150 are -101,-102, -103, -104 and -105.

77. Write four pairs of integers which are at the same distance from 2 on the number line.

Solution: There are many pairs of integers which are at the same distance from 2 i.e., (1,3), (0,4), (-1, 5) and (-2, 6)

78. The sum of two integers is 30. If one of the integers is -42, then find the other.

Solution:

Given

The sum of two integers is 30. If one of the integers is -42

Let the required integer be x.

According to question,

x + (-42) = 30

=>x -42 = 30

=> x = 30 + 42 = 72

79. Sum of two integers is -80. If one of the integers is -90, then find the other.

Solution:

Given

Sum of two integers is -80. If one of the integers is -90

Let the required integer be x.

According to question,

x + (-90) = -80

=> x- 90 =- 80 => x = -80 + 90 = 10

80. If we are at 8 on the number line, in which direction should we move to reach the integer

(1) -5
(2) 11
(3) 0?

Solution: (1) If we are at 8 on the number line, then to reach the integer -5, we must move towards the left on the number line.

(2) If we are at 8 on the number line, then to reach the integer 11, we must move towards the right on the number line.

(3) If we are at 8 on the number line, then to reach the integer 0, we must move towards the left on the number line.

81. Using the number line, write the integer which is

(1) 4 more than -5
(2) 3 less than 2
(3) 2 less than -2

Solution: (1) We want to know an integer 4 more than -5.

So, we start from -5 and proceed 4 steps to the right, then we obtain -1 as shown below.

Hence, 4 more than -5 is -1.

(2) We want to know an integer 3 less than 2.

So, we start from 2 and proceed 3 steps to the left, then we obtain -1 as shown below.

Hence, 3 less than 2 is -1.

(3) We want to know an integer 2 less than-2.

So, we start from -2 and proceed 2 steps to the left, then we obtain -4 as shown below.

Hence, 2 less than -2 is -4

82. Find the value of 49 -(-40) -(-3) + 69

Solution: We have,

49- (-40)- (-3) + 69

= 49 + 40 + 3 + 69 = 161

The value of 49 -(-40) -(-3) + 69= 161

83. Subtract -5308 from the sum [(-2100) + (-2001)1]

Solution: We have, [(-2100) + (-2001)]

= [-2100-2001]

= -4101

Required difference = —4101- (-5308)

= -4101 + 5308 = 1207

Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise – 5.1

1. What is the disadvantage in comparing line segments by mere observation?

Solution: There may be chance of error due to improper viewing.

2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?

Solution: It is better to use a divider than a ruler, because the thickness of the ruler may cause difficulties in reading of the length of a line segment. However, a divider gives accurate measurement.

3. Draw any line segment, say AB. Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB=AC + CS? [Note : If A, B, C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B.]

Solution: AB = 4 cm, AC = 2 cm, CB = 2 cm

Hence, AC + CB = 2 cm + 2 cm = 4 cm = AB

Yes, AB = AC + CB.

Read and Learn More Class 6 Maths Exemplar Solutions

4. If A, B, C are three points on a line such that AB-5 cm, BC= 3 cm and AC=8 cm, which one of them lies between the other two?

Solution:

Given

A, B, C are three points on a line such that AB-5 cm, BC= 3 cm and AC=8 cm

Since, AC is the longest line segment.

Thus B is the point lying between A and C.

5. Verify, whether D is the mid point of AG.

Solution: AD= 3 units, DG = 3 units, AD = DG

Thus, D is Thw mid- point od AG.

6. If B is the mid point of AC and C is the mid point of BD, where A, B, C, D lie on a straight line, say why AB = CD?

Solution: B is the mid point of AC.

AB = BC …………..(1)

And C is the mid point of BD.

BC = CD ……………(2)

From (1) and (2), we get

AB = CD

7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.

Solution: Sum of the lengths of any two sides of a triangle can never be less than length of the third side.

Exercise – 5.2

1. What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from

(1) 3 to 9
(2) 4 to z7
(3) 7 to 10
(4) 12 to 9
(5) 1 to 10
(6) 6 to 3

Solution: (1)1/2 or two right angles

(2)1/4 or one right angle

(3)1/4 or one right angle

(4)3/4 or three right angles

(5)3/4 or three right angles

(6)3/4 or three right angles

2. Where will the hand of a clock stop if it

(1) starts at 12 and makes 1/2 of a revolution, clockwise?
(2) starts at 2 and makes 1/2 of a revolution, clockwise?
(3) starts at 5 and makes 1/4 of a revolution, clockwise?
(4) starts at 5 and makes 3/4 of a revolution, clockwise?

Solution: (1) At 6

(2) At 8

(3) At 8

(4) At 2

3. Which direction will you face if you start facing

(1) eastand make 1/2 ofa revolution clockwise?
(2) east and make 1×1/2 of a revolution clockwise?
(3) west and make 3/4 of a revolution anti clock wise?
(4)south and make one full revolution?
(Should we specify clockwise or anti-clockwise for this last question? Why not?)

Solution: (1) West

(2) West

(3) North

(4) South

No, it is not necessary to specify because whether we turn clockwise or anti-clockwise, one full revolution will bring us back to the
original position.

4. What part of a revolution have you turned through if you stand facing

(1) east and turn clockwise to face north?
(2) south and turn clockwise to face east?
(3) west and turn clockwise to face east7

Solution: (1) 3/4

(2) 3/4

(3) 1/2

5. Find the number of right angles turned through by the hour hand of a clock when it goes from

(1) 3 to 6
(2) 2 to 8
(3) 5 to 11
(4) 10 to 1
(5) 12 to 9
(6) 12 to 6

Solution: (1) One right angle

(2) Two right angles

(3) Two right angles

(4) One right angle

(5) Three right angles

(6) Two right angles

6. How many right angles do you make if you start facing

(1) south and turn clockwise to west?
(2) north and turn anti-clockwise to east?
(3) west and turn to west?
(4) south and turn to north?

Solution: (1) One right angle

(2) Three right angles

(3) Four right angles

(4) Two right angles

7. Where will the hour hand of a clock stop if it starts

(1) from 6 and turns through 1 right angle?
(2) from 8 and turns through 2 right angles?
(3) from 1 0 and turns through 3 right angles?
(4) from 7 and turns through 2 straight angles?

Solution: (1) At 9

(2) At 2

(3) At 7

(4) At 7

Exercise – 5.3

1. Match the following:

(i) Straight angle                (1) Less than one-fourth of a revolution

(ii) Right angle                   (2) More than half a revolution

(iii) Acute angle                 (3) Half of a revolution

(iv) Obtuse angle              (4) One-fourth of a revolution

(v) Reflex angle                 (5) Between 1/4 and 1/3 of a revolution

Solution: (i)—> (3); (ii)—> (4); (iii)—> (1);(iv)—>(5);(v)—>(2)

2. Classify each one of the following angles as right, straight, acute, obtuse or reflex :

Solution: (a) Acute angle

(b) Obtuse angle

(c) Right angle

(d) Reflex angle

(e) Straight angle

(f) Acute angle

Exercise – 5.4

1. What is the measure of (i) a right angle? (ii) a straight angle?

Solution: (1) 90°

(2) 180°

2. Say True or False:

(1) The measure of an acute angle < 90°.
(2) The measure of an obtuse angle < 90°.
(3) The measure of a reflex angle > 1 80°.
(4) The measure of one complete revolution = 360°.
(5) If m∠A = 53° and m∠B = 35°, then m∠A > m∠B.

Solution: (1) True

(2) False

Since, measure of an obtuse angle is greater than 90° and less than 180°.

(3) True

(4) True

(5) True

3. Write down the measures of

(1) some acute angles.
(2) some obtuse angles. (give at least two examples of each).

Solution: (1) Measures of 2 acute angles are 35°, 20″

(2) Measures of 2 obtuse angles are 110°,135°

4. Measure the angles given below using the Protractor and write down the measure.

Solution: (a) Near about 40°

(b) Near about 130°

(c) Near about 90°

(d) Near about 60°, 120°, 90°

Note: Students can measure the angles exactly with the help of protractor.

5. Which angle has a large measure ? First estimate and then measure.

Measure of Angle A =
Measure of Angle B =

Solution: By estimating, we observe that ZB has a large measure.

∠A = near about 40° and ∠B = near about 65°.

Note: Students can measure the angles exactly with the help of protractor.

6. From these two angles which has larger measure? Estimate and then confirm by measuring them

Solution: Second angle has larger measure.

First angle is near about .30″ and second angle is near about 70″.

Note: Students can measure the angles exactly with the help of protractor

7. Fill in the blanks with acute, obtuse, right or straight:

(1) An angle whose measure is less than that of a right angle is___________.

(2) An angle whose measure is greater than that of a right angle is____________.

(3) An angle whose measure is the sum of the measures of two right angles is__________.

(4) When the sum of the measures of two angles is that of a right angle, then each one of them is___________.

(5) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be_____________.

Solution: (1) Acute

(2) Obtuse

(3) Straight

(4) Acute

(5) Obtuse

8. Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).

Solution: (1) By estimating with our eyes, we came to know that the measure of angle is 30°.

(2) By estimating with our eyes, we came to know that the measure of angle is 120°.

(3) By estimating with our eyes, we came to know that the measure of angle is 60°.

(4) By estimating with our eyes, we came to know that the measure of angle is 150°.

Note: Students can measure the angles exactly with the help of protractor.

9. Find the angle measure between the hands of the clock in each figure :

Solution: (1) 90° (Right angle)

(2) 30° (Acute angle)

(3) 180° (Straight angle)

10. Investigate In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?

Solution: No, the measure of angle will be same.

11. Measure and classify each angle:

Solution:

Note: Students can measure the angles exactly with the help of protractor.

Exercise 5.5

1. Which of the following arc models for perpendicular lines :

(1) The adjacent edges of a table top.
(2) The lines of a railway track.
(3) The line segments forming the letter ‘L’.
(4) The letter V.

Solution: (1) Perpendicular

(2) Not perpendicular

(3) Perpendicular

(4) Not perpendicular

2. Let PQ be the perpendicular to the line segment XY. Let PQ and XY intersect in the point A. What is the measure of ∠PAY1

Solution: ∠PAY = 90°

3. There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?

Solution: One set-square has angles 45°, 90°, 45° and other set-square has angles 60°, 90°, 30°.

Yes, they have angle measure 90° as common.

4. Study the diagram. The line / is perpendicular to linem

(1) Is CE = EG?

(2) Does PE bisect CGI
(3) Identify any two line segments for which PE is the perpendicular bisector.
(4) Are these true?
(i) AC>FG
(ii) CD = GH
(iii) BC < EH.

Solution: (1) Yes, both measure 2 units.

(2) Yes, because CE = EG

(3) BH and DF are two line segments for which PE is the perpendicular bisector.

(4) (i) True

Since, AC = 2 units and FG =1 unit.
AC>FG

(ii) True
Since, CD = GH= 1 unit

(iii) True
Since, BC = 1 unit, EH = 3 units

BC < EH

Exercise – 5.6

1. Name the types of following triangles:

(1) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(2) ΔABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(3) ΔPQR such that PQ = QR = PR = 5 cm.
(4) ΔDEF with m∠D = 90°
(5) ΔXYZ with m∠Y = 90° and XY = YZ.
(6) ΔLMN with m∠L= 30°, m∠M = 70° and m∠N= 80°.

Solution: (1) Scalene triangle
(2) Scalene triangle
(3) Equilateral triangle
(4) Right-angled triangle
(5) Isosceles right-angled triangle
(6) Acute-angled triangle

2. Match the following :

Measures of Triangle                                                              Type of Triangle

(i) 3 sides of equal                                                               (1) Scalene length
(ii) 2 sides of equal length                                                   (2) Isosceles right
(iii) All sides are of different length                                     (3) Obtuse angled
(iv) 3 acute angles                                                               (4) Right angled
(v) 1 right angle                                                                   (5) Equilateral
(vi) 1 obtuse angle                                                              (6) Acute angled
(vii) 1 right angle with two sides of equal length               (7) Isosceles

(i) -> (5); (ii) -» (7); (iii) -> (1); (iv) -> (6); (v) —> (4); (vi) —> (3); (vii) —> (2)

3. Name each of the following triangles in two different ways: (you may judge the nature of the angle by observation)

Solution: (a) Acute angled triangle and Isosceles triangle

(b) Right-angled triangle and Scalene triangle

(c) Obtuse-angled triangle and Isosceles triangle

(d) Right-angled triangle and Isosceles triangle

(e) Acute angled triangle and Equilateral triangle

(f) Obtuse-angled triangle and Scalene triangle

4. Try to construct triangles using match sticks. Some are shown here. Can you make a triangle with

(1) 3 matchsticks?
(2) 4 matchsticks?
(4) 6 matchsticks?
(3) 5 matchsticks?
(Remember you have to use all the available matchsticks in each case)

Name the type of triangle in each case. If you cannot make a triangle, think of reasons for it.

Solution: (1) Yes, it is possible to make a triangle with 3 matchsticks because sum of lengths of two sides is greater than the length of third side.

(2) No, it is not possible to make a triangle with 4 matchsticks because sum of lengths of two sides is equal to the length of third side.

(3) Yes, it is possible to make a triangle with 5 matchsticks because sum of lengths of two sides is greater than the length of third side.

(4) Yes, it is possible to make a triangle with the help of 6 matchsticks because sum of lengths of two sides is greater than the length of third side.

Exercise – 5.7

1. Say True or False:

(1) Each angle of a rectangle is a right angle.
(2) The opposite sides of a rectangle are equal in length.
(3) The diagonals of a square are perpendicular to one another.
(4) All the sides of a rhombus are of equal length.
(5) All the sides of a parallelogram are of equal length.
(6) The opposite sides of a trapezium are parallel.

Solution: (1) True

(2) True

(3) True

(4) True

(5) False

Since, opposite sides of a parallelogram are of equal length.

(6) False

Since, only one pair of opposite sides of a trapezium is parallel.

2. Give reasons for the following:

(1) A square can be thought of as a special rectangle.
(2) A rectangle can be thought of as a special parallelogram.
(3) A square can be thought of as a special rhombus.
(4) Squares, rectangles, parallelograms are all quadrilaterals.
(5) Square is also a parallelogram.

Solution: (1) Because its all angles are right angle and opposite sides are equal.

(2) Because its opposite sides are equal and parallel.

(3) Because its all sides are equal and diagonals are perpendicular to each other.

(4) Because all of them have four sides’.

(5) Because its opposite sides are equal and parallel.

3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?

Solution:

Given

A figure is said to be regular if its sides are equal in length and angles are equal in measure.

A square is a regular quadrilateral.

Exercise – 5.8

1. Examine whether the following are polygons. If any one among them is not, say why?

Solution: (1) As it is not a closed figure, therefore, it is not a polygon.

(2) It is a polygon because it is closed by line segments.

(3) It is not a polygon because it is not made by line segments.

(4) It is not a polygon because it is not made by only line segments and also it has curved surface.

2. Name each polygon.

Make two more examples of each of these.

Solution:

(a) Quadrilateral

(b) Triangle

(c) Pentagon

(d) Octagon

Two more examples of each:

3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.

Solution:

ABCDEF is a regular hexagon and ΔAEF is a triangle formed by joining AE. Hence, ΔAEF is an isosceles triangle.

4. Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.

Solution:

ABCDEFGH is a regular octagon and CDGH is a rectangle formed by joining C and H; D and G.

5. A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.

Solution:

Given

A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon

ABODE is the required pentagon and its diagonals are AD, AC, BE, BD and CE.

Section-2 NCERT Exemplar

Directions: In each of the questions1 to 11, out of four options only one is correct. Write the correct answer.

1. Measures of the two angles between hour and minute hands of a clock at 9 O’ clock are

(1) 60°, 300°
(2) 270°, 90°
(4) 30°, 330°
(3) 75°, 285°

Solution: (2): We know that 1 minute = 6°

The number of minutes between hour and minute hands of a clock at 9 O’ clock is 15 and 45.

The required angles are 15 * 6° = 90° and 45×6° = 270°

2. If a bicycle wheel has 48 spokes, then the angle between a pair of two consecutive spokes is

\(\text { (A) }\left(5 \frac{1}{2}\right)\)

 

\(\text { (B) }\left(7 \frac{1}{2}\right)\)

 

\(\text { (C) }\left(\frac{2}{11}\right)\)

 

\(\text { (D) }\left(\frac{2}{15}\right)\)

Solution: A bicycle wheel has 48 spokes.

The angle formed in circle is 360°.

The angle between a pair of two consecutive spokesr is 360 + 48 = \(7 \frac{24}{48}=7 \frac{1}{2}\)

3. In the given figure, ∠XYZ cannot be written as

(1)∠Y
(2) ∠ZXY
(3) ∠ZYX
(4) ∠XYP

Solution: (2): In the given figure, the name of angles formed are ∠XYP, ∠XYZ, ∠PYX and ∠ZYX

Thus, ∠ZXY is not a correct option.

4. In the given figure, if point A is shifted to point B along the ray PX such that PB = 2PA, then the measure of ∠BPY is

(1) greater than 45°
(2) Y45°
(3) less than 45°
(d) 90°

Solution: (2) : Since, the increase and decrease in the length of arms of an angle does not affect the angle made by them.
∠BPY = 45°

5. The number of obtuse angles In figure Is

(1) 2
(2) 3
(3) 4
(4) 5

Solution: (3): The obtuse angles formed in the given figure are ∠AOD = 20° + 45° + 65° = 130°

∠BOD = 45° + 65° = 110°, ∠COE = 65° + 30° = 95° and ∠BOE = 45° + 65° + 30° = 140°

Thus, there are total  4 obtuse angles formed.

6. The number of triangles in the given figure is

(1) 10
(2) 12
(4) 14
(3) 13

Solution: (3): We have,

The names of triangles formed in the given figure are

ΔABC, ΔABD, ΔADC, ΔAFG, ΔAEG, ΔAFE, ΔFGD, ΔEGD, ΔFED, ΔFBD, ΔDEC, ΔAFD, ΔAED

7. If the sum of two angles is greater than 180°, then which of the following is not possible for the two angles?

(1) One obluso angle and one acute angle
(2) One reflex angle and one acute angle
(3) Two obtuse angles
(4) Two right angles

Solution: (4): Since, the sum of two rijÿht angles is 180°.

8. If the sum of two angles is equal to an obtuse angle, then which of the following is not possible?

(1) One obtuse angle and one acute angle
(2) One right angle and one acute angle
(3) Two acute angles
(4) Two right angles

Solution: (4): Since, the sum of two right angles is 180°, which is a straight angle, not an obtuse angle.

9. In the given figure, AB = BC and AD = BD = DC. The number of isosceles triangles in the figure is

(1) 1
(2) 2
(3) 3
(4) 4

Solution: (3):

Given

In the given figure, AB = BC and AD = BD = DC.

We have, AB = BC and AD = BD = DC.

ΔABD, ΔBDC and ΔABC all are isosceles triangles.

There are 3 isosceles triangles formed in the given figure.

10. In the given figure, ∠BAC = 90° and AD ⊥ BC. The number of right triangles in the figure is

(1) 1
(2) 2
(3) 3
(4) 4

Solution: (3):

Given

In the given figure, ∠BAC = 90° and AD ⊥ BC.

We have, ∠BAC = 90° and AD ⊥ BC

∠BDA = ∠CDA = ∠BAC = 90°

There are 3 right triangles formed in the given figure.

11. In the given figure, PQ⊥ RQ, PQ = 5 cm and QR = 5 cm. Then ΔPQR is

(1) a right triangle but not isosceles
(2) an isosceles right triangle
(3) isosceles but not a right triangle
(4) neither isosceles nor right triangle

Solution: (13) :

Given

In the given figure, PQ⊥ RQ, PQ = 5 cm and QR = 5 cm.

We have, PQ⊥ RQ, PQ = 5 cm and QR = 5 cm, which shows that in a triangle, two sides are equal and their included angle is 90°.

ΔPQR is an isosceles right triangle

Directions: In questions 12 to 17,fill in the blanks to make the statements true.

12. An angle greater than 180° and less than a complete angle is called________.

Solution: Reflex angle

13. A pair of opposite sides of a trapezium are

Solution: Parallel

14. In the given figure,

(1) ∠AOD is a/an __________ angle
(2) ∠COA is a/an__________ angle.
(3) ∠AOE is a/an__________ angle.

Solution: (1) Right : ∠AOD = 30° + 20° + 40° = 90°

(2) Acute : ∠COA = 20° + 30° = 50°

(3) Obtuse : ∠AOE = 40° + 40° + 20° + 30° = 130°

15. The number of triangles in figure is Their names are________.

Solution: ΔAOB, ΔOC, ΔCOD, ΔCAB and ΔACD

16. Number of angles less than 180” in given figure is_______ and their names are_______.

Solution: 12, ∠ABO, ∠BAO, ∠AOB, ∠BOD, ∠COD, ∠ODC, ∠OCD, ∠OCA, ∠CAO, ∠AOC, ∠BAC and ∠ACD.

17. The number of right angles in a straight angle is__________ and that in a complete angle is___________

Solution: 2, 4 : The number of right angles in a straight angle is 2 and that in a complete angle is 4.

Directions: State whether the statements given in questions 18 to 20 are true (T) or false (F).

18. A horizontal line and a vertical line always intersect at right angles.

Solution: True

19. If the arms of an angle on the paper are increased, the angle increases.

Solution: False

If the arms of an angle on the paper are increasing or decreasing, it doesn’t affect the angle made by them.

20. If the arms of an angle on the paper are decreased, the angle decreases.

Solution: False

21. Write down fifteen angles (less than 180°) involved in given figure.

Solution: The names of fifteen angles (less than 180°) involved in figure are :
∠AEC, ∠ADB, ∠EAD, ∠EFD, ∠EFB, ∠DFC, ∠FBC, ∠FCB, ∠BFC, ∠ABC, ∠ACB, ∠DCF, ∠FDC, ∠EBF and ∠BEF.

22. Is it possible for the same

(1) line segment to have two different lengths?
(2) angle to have two different measures?

Solution: (1) No, it is not possible for the same line segment to have two different lengths.

(2) No, it is not possible for the same angle to have two different measures.

23. Will the measure of ∠ABC and of ∠CBD make measure of ∠ABD in given figure?

Solution: Yes, ∠ABD- ∠ABC + ∠CBD

=> ∠ABD is the sum of ∠ABC and ∠CBD.

24. Will the lengths of line segment AB and line segment BC make the length of line segment AC in given figure?

Solution: Yes, the length of line segment AC is the sum of the lengths of line segment AB and BC.

25. Look at a given figure. Mark a point

(1) A which is in the interior of both∠1 and ∠2.
(2) B which is in the interior of only ∠1.
(3) Point C in the interior of ∠1.
Now, state whether points B and C lie in the interior of ∠2 also.

Solution: Yes, the given figure shows that the points B and C lie in the interior of ∠2 also.

26. In which of the following figures,

(1) perpendicular bisector is shown?
(2) bisector Is shown?
(3) only bisector is shown?
(4) only perpendicular is shown?

Solution:

(1) Figure (ii) shows the perpendicular bisector.

(2) Figure (ii) and (iii) shows the bisector.

(3) Figure (iii) shows only the bisector.

(4) Figure (i) shows only the perpendicular.

27. In given figure,

(1) name any four angles that appear to be acute angles.
(2) name any two angles that appear to be obtuse angles.

Solution: (1) Acute angles: ∠ADE, ∠AEB,∠ABE and ∠ECD.

(2) Obtuse angles: ∠BCD and ∠BAD.

28. In given figure,

(1) is AC + CB = AB?
(2) is AB + AC =CB?
(3) is AB + BC = CA?

Solution: (1) Yes, AC + CB = AB

(2) No, AB- AC = CB

(3) No, AB-BC = CA

29. In given figure,

(1) What is AE + EC?
(2) What is AC-EC?
(3) What is BD-BE?
(4) What is BD-DE?

Solution: (1) AE+EC=AC

(2) AC-EC=AE

(3) BD-BE = ED

(4) BD- DE = BE

30. Using the information given, name the right angles in each part of given figures.

(1) BA⊥BD

(2) RT⊥ST

(3) AC⊥BD

(4) RS⊥RW

(5) AC⊥BD

(6) AE⊥CE

(7) AC⊥CD

(8) OP⊥AB

Solution: (1) BA ⊥ BD, The right angle is ∠ABD.

(2) RT ⊥ ST, The right angle is ∠RTS

(3) AC⊥BD, The right angles are ∠ACD and ∠ACB.

(4) RS⊥RW, The right angle is ∠SRW.

(5) AC ⊥ BD, The right angles are ∠AED, ∠AEB,∠BEC and ∠CED.

(6) AE ⊥CE, The right angle is ∠AEC.

(7) AC ⊥ CD, The right angle is ∠ACD.

(8) OP⊥AB, The right angles are ∠AKO, ∠AKP, ∠BKO and ∠BKP.

31. What conclusion can be drawn from each part of given figures, if

(1) DB is the bisector of ∠ADC?

(2) BD bisects ∠ABC?

(3) DC is the bisector of ∠ADB, CA ⊥ DA and CB ⊥ DB?

Solution: (1) DB is the bisector of ∠ADC.

∠ADB =∠CDB

(2) BD bisects ∠ABC.

∠ABD = ∠CBD

(3) DC is the bisector of ∠ADB, CA ⊥ DA and CB ⊥ DB

∠ADC = ∠BDC, ∠CAD = ∠CBD =90°

32. An angle is said to be trisected, if it is divided into three equal parts. If in a given figure, ∠BAC = ∠CAD = ∠DAE, how many trisectors are there for ∠BAE?

Solution:

Given

An angle is said to be trisected, if it is divided into three equal parts. If in a given figure, ∠BAC = ∠CAD = ∠DAE,

We have given, ∠BAC = ∠CAD = ∠DAE

There are two trisectors namely, AC and AD.

33. Can we have two acute angles whose sum is

(1) an acute angle? Why or why not?
(2) a right angle? Why or why not?
(3) an obtuse angle? Why or why not?
(4) a straight angle? Why or why not?
(5) a reflex angle? Why or why not?

Solution: (1) Yes, v the sum of two acute angles can be the acute angle.

E.g., 30° and 40° are two acute angles and their sum = 30° + 40° = 70°, which is also an acute angle.

(2) Yes, v the sum of two acute angles be a right angle.

E.g., 30° and 60° are two acute angles and their sum = 30° + 60° = 90°, which is a right angle.

(3) Yes, the sum of two acute angles can be an obtuse angle.

E.g., 45° and 60° are two acute angles and their sum = 45° + 60° = 105°, which is an obtuse angle.

(4) No, v the sum of two acute angles is always less than 180°.

(5) No, v the sum of two acute angles is always less than 180°.

34. Can we have two obtuse angles whose sum is

(1) a reflex angle? Why or why not?
(2) a complete angle? Why or why not?

Solution: (1) Yes, v the sum of two obtuse angles is always greater than 180°.

E.g., 135° and 100° are two obtuse angles and their sum = 135° + 100° – 235°, which is greater than 180°.

(2) No, v the sum of two obtuse angles is greater than 180° but less than 360°. In the above example, we can see that the sum of 135° and 100° i.e., 235° is greater than 180° but less than 360°.

Class 6 Maths Chapter 4 Basic Geometrical Ideas

Directions: In each of the questions 1 to 5, out of four options only one is correct. Write the correct answer.

1. The number of lines passing through five points such that no three of them are collinear is

(1) 10
(2) 5
(3) 20
(4) 8

Solution: (1) : Since, total number of points is 5 and we need two points to form a line.

Total number of lines passing through the points is 5 x 2 = 10

2. The number of diagonals in a septagon is

(1) 21
(2) 42
(3) 7
(4) 14

Solution: (4): Since, the number of diagonals in

a polygon =n(n-3)/2

Septagon has 7 sides, i.e., n = 7

The number of diagonals in a septagon

Read and Learn More Class 6 Maths Exemplar Solutions

3. Number of line segments in figure is

(1) 5
(2) 10
(3) 15
(4) 20

Solution: (2) : The adjacent figure shows the line segments;

AB, BC, CD, AC, AD, BD, AE, BE, CE, DE

Thus, there are 10 line segments.

4. The number of angles in the given figure is

(1) 3
(2) 4
(3) 5
(4) 6

Solution: (4): The names of angles formed in the given figure are ZAOB, ZAOC, ZAOD, ZBOC, ZBOD, and ZCOD.

There are a total of 6 angles formed.

5. A polygon has a prime number of sides. Its number of sides is equal to the sum of the two least consecutive primes. The number of diagonals of the polygon is

(1) 4
(2) 5
(3) 7
(4) 10

Solution: (2): We have given, the number of sides of a polygon

= Sum of the two least consecutive primes

= 2 +3 =5 [ ∴ 2 and 3 are the least consecutive prime numbers]

The number of diagonals = n(n- 3)/2

where n = 5

=5(5-3)/2=5×2/2=5

Directions: In questions 6 to 14, fill in the blanks to make the statements true

6. The number of diagonals in a hexagon is

Solution: 9 : Number of diagonals = n(n-3)/2

where n = 6

=6(6-3)/2=6×3/2=9

7. In the given figure, points lying in the interior of the triangle PQR are_________, that in the exterior are__________ and that on the triangle itself are__________.

Solution: O and S; T and N; P, Q, R and M

8. In the given figure, points A, B, C, D and E are collinear such that AB = BC = CD = DE. Then

(1) AD=AB +________
(2) AD=AC +_________
(3) midpoint of AE is_____________
(4) midpoint of CE is__________
(5) AE=_______x AB

Solution: (1) BD: AD=AB + BC + CD=AB + BD

(2) CD:AD = AB + BC+CD=AC + CD

(3) C: AB + BC = CD + DE

=> AC = CE :. Mid point of AE is C.

(4) D :  CD = DE

Mid point of CE is D.

(5) 4 : AB + BC + CD + DE = AE
=> AB+AB+AB + AB = AE ∴ 4AB = AE

9. The number of straight angles in given figure is____________.

Solution: 4: The number of straight angles in the given figure is 4.

10. The number of common points in the two angles marked in the given figure is_______.

Solution: Two: The two angles marked; ∠PAQ and ∠PDQ.

The number of common points are 2 and these are P and Q.

11. The number of common points in the two angles marked in a given figure is__________.

Solution: One: The two angles marked; ∠CAB and ∠DAE.

The number of common points is 1 and that is A.

12. The number of common points in the two angles marked in given figure _________.

Solution: Three: There are 3 common points in the two angles marked in the given figure and these are P, Q, and R.

13. The number of common points in the two angles marked in a given figure is_______.

Solution: Four: The number of common points in the two angles marked in the given figure is 4 and these are E, D, G and F.

14. The common part between the two angles BAC and DAB in given figure is_____.

Solution: Ray AB

Directions: State whether the statements given in questions 15 to 21 are true (T) or false (F).

15. If line PQ|| line m, then line segment PQ || m.

Solution: True

16. Two parallel lines meet each other at some point.

Solution: False

Two lines in a plane which do not meet even when produced indefinitely in either direction, are known as parallel lines.

17. Measures of ∠ABC and ∠CBA in given figure are the same.

Solution: True

ABC is same as ∠CBA.

18. Two line segments may intersect at two points.

Solution: False

The intersecting point of any two line segments is only one.

19. Many lines can pass through two given points.

Solution: False

There is only one line which passes through two given points.

20. Only one line can pass through a given point.

Solution: False

There are infinite number of lines which passes through a given point.

21. Two angles can have exactly five points in common.

Solution: False

It can have any number of points.

22. Name all the line segments in given figure.

Solution: The line segments are AB, BC, CD, DE, AC, AD, AE, BD, BE and CE

23. Name the line segments shown in given figure

Solution: The line segments are AB, BC, CD, DE and EA

24. State the mid points of all the sides of given figure.

Solution: X is a mid-point of AC,

Y is a mid-point of BC and

Z is a mid-point of AB.

25. Name the vertices and the line segments in given figure.

Solution: The vertices are : A, B, C, D and E.

The line segments are : AB, BC, CD, DE, EA, AC and AD.

26. Name the following angles of given figure,using three letters :

(1) ∠1
(2) ∠2
(3) ∠3
(4) ∠1+∠2
(5) ∠2 + ∠3
(6) ∠1+∠2 + ∠3
(7) ∠CBA-∠1

Solution:(1) ∠1 = ∠CBD

(2) ∠2 = ∠DBE

(3) ∠3 = ∠EBA

(4) ∠1 +∠2 = ∠CBD + ∠DBE = ∠CBE

(5) ∠2 + ∠3 = ∠DBE +vEBA = ∠DBA

(6) ∠1 + ∠2 + ∠3 = ∠CBD + ∠DBE + ∠EBA = ∠CBA

(7)∠CBA- ∠1 = ∠CBA- ∠CBD = ∠DBA

27. Name the points and then the line segments in each of the following figures:

Solution:(i) Name of the points -> A,B and C.

Name of the line segments —> AB, BC and CA.

(ii) Name of the points —>A,B,C and D.

Name of the line segments —> AB, BC, CD and DA.

(iii) Name of the points —> A, B, C, D and E.

Name of the line segments —> AB, BC, CD, DE, and EA.

(iv) Name of the points —> A, B, C, D, E and F.

Name of the line segments —> AB, CD and EF.

28. Which points in given figures, appear to be mid-points of the line segments? When you locate a mid-point, name the two equal line segments formed by it.

Solution:(i) The given figure shows there is no mid-point.

(ii) The given figure shows that O is the mid-point of AB and the name of the two equal line segments are AO and OB.

(iii) The given figure shows that D is the mid-point of BC and the name of the two equal line segments are BD and DC.

29. Find out the incorrect statement, if any, in the following : An angle is formed when we have

(1) two rays with a common end-point
(2) two line segments with a common endpoint
(3) a ray and a line segment with a common end-point

Solution: All the three statements (1), (2) and (3) are incorrect.

The common initial point of two rays forms an angle.

30. What is common in the following figures (i) and (ii)? Is figure (i) that of triangle? if not, why?

Solution: Both the figures (i) and (ii) have 3 line segments.

No, Fig. (i) is not a triangle since the three line segments does not form a closed figure.

31. If two rays intersect, will their point of intersection be the vertex ofan angle of which the rays are the two sides?

Solution: Yes

32. How many points are marked in given figure?

Solution: Two points A and B are marked

33. How many line segments are there in given figure?

Solution: Only one line segment, AB is there.

34. In given figure, how many points are marked? Name them.

Solution: From the given figure, Three points A, B, and C are marked.

35. How many line segments are there in the given figure? Name them.

Solution: From the given figure, Three line segments, namely AB, BC, and AC are there.

36. In the given figure, how many points are marked? Name them.

Solution: From the given figure, Four points A, B, C, and D are marked.

37. In the given figure how many line segments are there? Name them.

Solution: Six line segments, namely AB, AC, AD, BC, BD, and CD.

38. In the given figure, how many points are marked? Name them

Solution: From the given figure, Five points are marked, namely A, B, D, E and C.

39. In given figure how many line segments are there? Name them

Solution: From the given figure, Ten line segments, namely AB, AD, AE, AC, BD, BE, BC, DE, DC and EC.

40. In given figure,O is the centre of the circle.

(1) Name all chords of the circle.
(2) Name all radii of the circle.
(3) Name a chord, which is not the diameter of the circle.
(4) Shade sectors OAC and OPB.
(5) Shade the smaller segment of the circle formed by CP.

Solution:(1) Name of chords : PC and BA.

(2) Name of radii : PO, OC, OB and OA.

(3) PC is a chord which is not the diameter of the circle

(4)

(5)

41. Write the name of

(1) vertices
(2) edges, and
(3) faces of the prism shown In given figure.

Solution: (1) Vertices: A, B, C, D, E and F.

(2) Edges: AB, BC, AC, DF, FC, BD. EF, ED and AE.

(3) Faces: EACF, EDBA, ABC, DEF and DBCF.

42. How many edges, faces and vertices are there in a sphere?

Solution: In a sphere, edges – 0, faces – 0 and vertices- 0.

43. Draw all the diagonals of a pentagon ABCDE and name them.

Solution: The diagonals of a pentagon ABCDE are AC, AD, BE, BD and EC.

Class 6 Maths Chapter 12 Ratio And Proportion

Question 1. The ratio of books to 20 books is

1. 2:5
2. 5:2
3. 4:5
4. 5:4

Solution: (1): The ratio of 8 books to 20 books

= 8 books: 20 books

⇒ \(=\frac{8}{20}=\frac{2}{5}=2: 5\)

Question 2. The ratio of the number of sides of a square to the number of edges of a cube is

1. 1:2
2. 3:2
3. 4:1
4. 1 :3

Solution: (4) : Required ratio = \(\frac{\text { Number of sides of a square }}{\text { Number of edges of a cube }}=\frac{4}{12}\)

⇒ \(=\frac{1}{3}=1: 3\)

Read and Learn More Class 6 Maths Exemplar Solutions

Question 3. A picture is 60 cm wide and 1.8 m long. The ratio of its width to its perimeter in the lowest form is

1. 1:2
2. 1:3
3. 1:4
4. 1:8

Solution:

(4) : Width of the picture = 60 cm and length of the picture = 1.8 m = 1.8 x 100 cm = 180 cm

Primetor of the picture = 2* (180 + 60) cm

= 2 x 240 cm = 480 cm

The required ratio \(=\frac{\text { Width of the picture }}{\text { Perimeter of the picture }}\)

⇒ \(\frac{60 \mathrm{~cm}}{480 \mathrm{~cm}}=\frac{1}{8}=1: 8\)

Question 4. Neelam’s annual income is ₹  288000. Her annual savings amount to ₹  36000. The ratio of her savings to her expenditure is
Solution:

1. 1:8
2. 1:7
3. 1:6
4. 1:5

Solution: (2):

Given

Neelam’s annual income =₹ 288000

Her savings =₹  36000

Her expenditure = Annual income – Savings

= ₹  288000- ₹  36000 =₹  252000

The required ratio = \(=\frac{\text { Savings }}{\text { Expenditure }}\)

⇒ \(=\frac{₹ 36000}{₹ 252000}=\frac{1}{7}=1: 7\)

Question 5. The mathematics textbook for Class VI has 320 pages. The chapter’symmetry’runs from page 261 to page 272. The ratio of the number of pages of this chapter to the total number of pages of the book is

1. 11: 320
2. 3: 40
3. 3: 80
4. 272: 320

Solution: (3):

Given

The mathematics textbook for Class VI has 320 pages. The chapter’symmetry’runs from page 261 to page 272.

Total number of pages = 320

The number of pages of the chapter ‘symmetry’ = 272- 261 +1 = 12

The required ratio \(=\frac{\text { Number of pages of chapter ‘symmetry’ }}{\text { Total number of pages of the book }}\)

⇒ \(=\frac{12}{320}=\frac{3}{80}=3: 80\)

Question 6. In a box, the ratio of red marbles to blue marbles is 7: 4. Which of the following could be the total number of marbles in the box?

1. 18
2. 19
3. 21
4. 22

Solution: (4):

Given

In a box, the ratio of red marbles to blue marbles is 7: 4.

Since the sum of the given ratio is 7 + 4 = 11 and from the given options, only 22 is Divisible by 11.

Option (D) is correct.

Question 7. On a shelf, books with green covers and those with brown covers are in the ratio 2 : 3. If there are 18 books with green covers, then the number of books with brown covers is

1. 12
2. 24
3. 27
4. 36

Solution:

Given

On a shelf, books with green covers and those with brown covers are in the ratio 2 : 3. If there are 18 books with green covers

(3): Let the number of green cover and brown cover books be 2x and 3x respectively. Since, the number of green cover books = 18

⇒  2x = 18

⇒ \( x=\frac{18}{2}=9\)

The number of brown cover books = 3×9 = 27

Question 8. The greatest ratio among the ratios 2 : 3, 5: 8, 75:121 and 40: 25 is

1. 2 : 3
2. 5: 8
3. 75: 121
4. 40: 25

Solution: (4): we have \(2: 3=\frac{2}{3}, 5: 8=\frac{5}{8}, 75: 121=\frac{75}{121}\) and \(40: 25=\frac{40}{25}\)

The LCM of 3, 8, 121 and 25 is 2 x 2 x 2 x 3 x 5 x 5x11x11 = 72600

Making the denominator of each ratio equal to 72600, we get

⇒ \(\frac{2}{3}=\frac{2 \times 24200}{3 \times 24200}=\frac{48400}{72600}\)

⇒ \(\frac{5}{8}=\frac{5 \times 9075}{8 \times 9075}=\frac{45375}{72600}\)

⇒ \(\frac{75}{121}=\frac{75 \times 600}{121 \times 600}=\frac{45000}{72600}\)

⇒ \(\frac{40}{25}=\frac{40 \times 2904}{25 \times 2904}=\frac{116160}{72600}\)

On comparing the numerator of the above ratios, we get \(\frac{40}{25}\) is the greatest ratio, i.e., 40: 25 is the greatest ratio among the given ratios.

Question 9. There are ‘b’ boys and ‘g’ girls in a class. The ratio of the number of boys to the total number of students in the class is: 20

1. \(\frac{b}{b+g}\)
2. \(\frac{g}{b+g}\)
3. \(\frac{b}{g}\)
4. \(\frac{b+g}{b}\)

Solution: (1) :

Given

There are ‘b’ boys and ‘g’ girls in a class. The ratio of the number of boys to the total number of students in the class is: 20

Total number of students = number of boys + number of girls = b+g

The required ratio \(=\frac{\text { Number of boys }}{\text { Total number of students }}\)

⇒ \(\frac{b}{b+g}\).

Question 10. If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then the ratio of the distance travelled by them in one hour is
Solution:

1. 1: 2
2. 4: 5
3. 5: 8
4. 8: 5

Solution: (3):

Given

If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds

Distance travelled by bus in 4 hours = 160 km

Distance travelled by bus in 1 hour \(=\frac{160}{4} \mathrm{~km}=40 \mathrm{~km}\)

Distance travelled by train in 5 hours = 320 km

Distance travelled by train in1 hour \(=\frac{320}{5} \mathrm{~km}=64 \mathrm{~km}\)

The required ratio = \(\frac{40 \mathrm{~km}}{64 \mathrm{~km}}=\frac{5}{8}=5: 8\)

Question 11.
Solution:

To get the missing number, we consider the fact that 4 * 5 = 20, i.e., we get 20 when we multiply 5 by 4. This indicates that to get the missing number, 3 must also be multiplied by 4.

When we multiply, we have, 3 x 4 = 12

Hence, the ratio is \(\frac{12}{20}.\)

Question 12.
Solution:

To get the missing number, we consider the fact that 9 * 2 = 18, i.e., when we multiply 9 by 2 we get 18. This indicates that to get the missing number, 2 must also be multiplied by 2.

When we multiply, we have, 2×2 = 4

Hence, the ratio is \(\frac{4}{18}.\)

Question 13.
Solution:

To get the missing number, we consider the fact that 3.2 x 2.5 = 8, i.e., when we multiply 3.2 by 2.5 we get 8. This indicates that to get the missing number, 4 must also be multiplied by 2.5.

When we multiply, we have, 4 x 2.5 = 10

Hence, the ratio is \(\frac{8}{10}.\)

Question 14.
Solution:

To get the missing number, we consider the fact that 40 x 1.125 = 45, i.e., when we multiply 40 by 1.125 we get 45. This indicates that to get the missing number of the first ratio, 16 must also be multiplied by 1.125.

When we multiply, we have, 16 * 1.125 = 18.

Hence, the first ratio is \(\frac{18}{45}\)

Similarly, to get the third ratio we multiply both terms of the second ratio by 1.5.

Hence, tine third ratio is \(\frac{24}{60}\)

Question 15.
Solution:

To get the missing number, we consider the fact that 36 x 1.75 = 63, i.e., we get 63 when we multiply 36 by 1.75. This indicates that to get the missing number of the second ratio, 16 must also be multiplied by 1.75.

When we multiply, we have, 16 * 1.75 = 28

Hence, the second ratio is \(\frac{28}{63} \text {. }\)

Similarly, to get the third ratio we multiply both terms of the first ratio by 2.25

Hence, the third ratio is \(\frac{36}{81} \text {. }\)

And to get the fourth ratio we multiply both terms of the first ratio by 3.25.

Hence, the fourth ratio is \(\frac{52}{117}\)

Question 16. \(\frac{3}{8}=\frac{15}{40}\)
Solution: True

We have, \(\frac{15}{40}=\frac{3}{8}\)

Question 17. \(4: 7=20: 35\)
Solution: True

We have, \(20: 35=\frac{20}{35}=\frac{4}{7}=4: 7\)

Question 18. 0.2:5 = 2:0.5
Solution: False

⇒ \(0.2: 5=\frac{0.2}{5}=\frac{2}{50}=\frac{1}{25}\)

⇒ \(2: 0.5=\frac{2}{0.5}=\frac{20}{5}=\frac{4}{1}\)

Since, \(\frac{1}{25} \neq \frac{4}{1}\)

0.2: 5* 2: 0.5

Question 19. 3:33 = 33:333
Solution: False

We have \(3: 33=\frac{3}{33}=\frac{1}{11}\)

⇒ \(\text { and } 33: 333=\frac{33}{333}=\frac{11}{111}\)

Since, \(\frac{1}{11} \neq \frac{11}{111}\)

3: 33 x 33: 333

Question 20. 1 5 m : 40 m = 35 m: 65 m
Solution: False

We have \(15 \mathrm{~m}: 40 \mathrm{~m}=\frac{15}{40}=\frac{3}{8}\)

⇒ \(\text { and } 35 \mathrm{~m}: 65 \mathrm{~m}=\frac{35}{65}=\frac{7}{13}\)

Since,\(\frac{3}{8} \neq \frac{7}{13}\)

15 m : 40 m* 35 m : 65 m

Question 21. 27 cm2: 57 cm2 = 18 cm: 38 cm
Solution: True

We have, 27 cm2: 57 cm2

⇒ \(=\frac{27}{57}=\frac{9}{19}\)

and 18 cm : 38 cm \(=\frac{18}{38}=\frac{9}{19}\)

27 cm2: 57 cm2 = 18 cm : 38 cm

Question 22. 5 kg: 7.5 kg = 7.50:? 5
Solution: False

We have, 5 kg: 7.5 kg

⇒ \(=\frac{5}{7.5}=\frac{50}{75}=\frac{2}{3}\)

and ? 7.50:? 5 =\(\frac{7.50}{5}=\frac{750}{500}=\frac{3}{2}\)

5 kg: 7.5 kg*? 7.50:? 5

Question 23. 20 g : 100 g = 1 metre: 500 cm
Solution: True

We have, \(20 \mathrm{~g}: 100 \mathrm{~g}=\frac{20}{100}=\frac{1}{5}=1: 5\) and 1 metre : 500 cm = 100 cm : 500 cm

⇒ \(=\frac{100}{500}=\frac{1}{5}=1: 5\)

20 g : 100 g =1 metre : 500 cm

Question 24. 1 2 hours: 30 hours = 8 km: 20 km
Solution: True

We have 12 hours : 30 hours = \(=\frac{12}{30}\)

⇒ \(=\frac{2}{5}=2: 5\)

⇒ \(\text { and } 8 \mathrm{~km}: 20 \mathrm{~km}=\frac{8}{20}=\frac{2}{5}=2: 5\)

12 hours: 30 hours- 8 km: 20 km

Question 25. The ratio of 0 kg to 1 00 kg is 1: 1 0
Solution: True

The ratio of 10 kg to 100 kg \(=\frac{10}{100}=\frac{1}{10}=1: 10\)

Question 26. The ratio of 1 50 cm to 1 metre is 1: 1 .5
Solution: False

The ratio of 150 cm to 1 metre = 150 cm : 1 metre = 150 cm : 100 cm

⇒ \(=\frac{150}{100}=\frac{3}{2}=3: 2\)

⇒ \(\text { and } 1: 1.5=\frac{1}{1.5}=\frac{10}{15}=\frac{2}{3}=2: 3\)

Since, 3:2*2:3

150 cm : 1 metre*1 : 1.5

Question 27. 25 kg: 20 g = 50 kg : 40 g
Solution: True

We have, 25 kg : 20 g = 25 * 1000 g : 20 g

⇒ \(=\frac{25000}{20}=\frac{1250}{1}=1250: 1\)

and 50 kg : 40 g = 50 x 1000 g : 40 g

⇒ \(=\frac{50000}{40}=\frac{1250}{1}=1250: 1\)

25 kg : 20 g = 50 kg : 40 g

Question 28. The ratio of 1 hour to one day is 1: 1.
Solution: False

The ratio of1 hour to 1 day =1 hour : 1 day

= 1 hour : 24 hours = \(=\frac{1}{24}=1: 24 \neq 1: 1\)

Question 29. The ratio of 4: 16 is in its lowest form
Solution: False

We have, 4:16 =\(\frac{4}{16}=\frac{1}{4}=1: 4\)

4: 16 is not in its lowest form.

Question 30. The ratio 5: 4 is different from the ratio 4: 5.
Solution: True

We have, 5 : 4 = \(\frac{5}{4} \text { and } 4: 5=\frac{4}{5}\)

5: 4 x 4: 5

Question 31. A ratio will always be more than 1.
Solution: False

A ratio may be less than 1.

Question 32. A ratio can be equal to 1.
Solution: True

Question 33. If b: a = c: d, then a, b, c, d are in proportion.
Solution: False

If b: a = c : d, then b, a, c, d are in proportion.

34. The two terms of a ratio can be in two different units_____.
Solution: False

Question 35. A ratio is a form of comparison by________.
Solution: Division

Question 36. 20 m : 70 m = ?8: ?________.
Solution:

28: We have, 20 m:70 m \(=\frac{20}{70}=\frac{2}{7}=2: 7\)

20 m : 70 m = ? 8 : ? 28

Question 37. There is a number in the box such that, 24, 9, and 12 are in proportion. The number in the box is _______.
Solution:

18: Since, 24, 9, and 12 are in proportion.

Question 38. If two ratios are equal, then they are in______.
Solution: Proportion

Question 39. The ratio of the perimeter of the boundary of the shaded portion to the perimeter of the whole figure is______.
Solution:

3:7: Since each square is of unit length. The perimeter of the shaded portion is 6 units and the perimeter of the whole figure is 14 units.

The required ratio = \(\frac{6}{14}=\frac{3}{7}=3: 7\)

Question 40. The ratio of the area of the shaded portion to that of the whole figure is________.
Solution:

1 : 6 : Area of shaded portion = (1 unit) x (2 units) = 2 sq units

and area of whole figure = (3 units) x (4 units) = 12 sq units

The required ratio = \(\frac{2}{12}=\frac{1}{6}=1: 6\)

Question 41. The sleeping time of a python in a 24-hour clock is represented by the shaded portion in the given figure.

The ratio of sleeping time to awaking time is______.

Solution: 3:1: Sleeping time = 18 hours and awaking time = (24- 18) hours = 6 hours

The required ratio = \(\frac{18}{6}=\frac{3}{1}=3: 1\)

Question 42. A ratio expressed in the lowest form has no common factor other than _____ in its terms.
Solution: One

43. To find the ratio of two quantities, they must be expressed in______ units.
Solution: Same

Question 44. The ratio of 5 paise to 25 paise is the same as the ratio of 20 paise to_____.
Solution: 100 paise or 1 rupee : We have, 5 paise : 25 paise = 20 paise :

5 paise : 25 paise = 20 paise : 100 paise

Question 45. Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes. The ratio of the time taken by Saturn and Jupiter in the lowest form is_______.
Solution:

Given

Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes.

149 : 160: Time taken by Saturn

= 9 hours 56 minutes

= (9 x 60 + 56) minutes

= (540 + 56) minutes

= 596 minutes

Time taken by Jupiter = 10 hours 40 minutes

= (10 x 60 + 40) minutes

= 640 minutes

The required ratio = \(=\frac{\text { Time taken by Saturn }}{\text { Time taken by Jupiter }}\)

⇒ \(=\frac{596 \text { minutes }}{640 \text { minutes }}=\frac{149}{160}=149: 160\)

Question 46. 10 g of caustic soda dissolved in 100 mL of water makes a solution of caustic soda. The amount of caustic soda needed for 1 litre of water to make the same type of solution is______.
Solution:

Given

10 g of caustic soda dissolved in 100 mL of water makes a solution of caustic soda.

100 g: Let the required amount of caustic soda be x g.

According to the question,

10 g: x g = 100 mL : 1 L

=> 10 g : .v g = 100 mL : 1000 mL

⇒ \(\frac{10}{x}=\frac{100}{1000} \Rightarrow \frac{10}{x}=\frac{1}{10}\)

X 1-10×10 ⇒ x — 100

The required amount of caustic soda is 100 g.

Question 47. The marked price of a table is ₹ 625 and its sale price is ₹ 500. What is the ratio of the sale price to the marked price?
Solution:

Given

The marked price of a table is ₹625 and its sale price is ₹500.

4:5: We have given the marked price = ₹625 and selling price = ₹500

The required ratio \(=\frac{\text { Selling price }}{\text { Marked price }}\)

⇒ \(=\frac{₹ 500}{₹ 625}=\frac{4}{5}=4: 5\)

Question 48. Which pair of ratios are equal? And why?

1. \(\frac{2}{3}, \frac{4}{6}\)
2. \(\frac{8}{4}, \frac{2}{1}\)
3. \(\frac{4}{5}, \frac{12}{20}\)

Solution:

1. We Have, \(\frac{2}{3} \text { and } \frac{4}{6}\)

The simplest form of \(\frac{4}{6} \text { is } \frac{2}{3}\)

⇒ \(\frac{2}{3}=\frac{4}{6}\)

2. We have, \(\frac{8}{4} \text { and } \frac{2}{1}\)

The simplest form of\(\)

⇒ \(\frac{8}{4}=\frac{2}{1}\)

3. We have, \(\frac{4}{5} \text { and } \frac{12}{20}\)

The simplest form of \(\frac{12}{20} \text { is } \frac{3}{5}\)

⇒ \(\frac{4}{5} \neq \frac{12}{20}\)

Question 49. Which ratio is larger 1 0: 21 or 21: 93?
Solution:

We have, 10: 21 \(=\frac{10}{21} \text { and } 21: 93=\frac{21}{93}\)

Now, the I, CM of 21; and 93 is 3 * 7 x 31 – 651 Mnking the denominator of each ratio equal to 651, we have

⇒ \(\frac{10}{21}=\frac{10 \times 31}{21 \times 31}=\frac{310}{651}\)

⇒ \(\text { and } \frac{21}{93}=\frac{21 \times 7}{93 \times 7}=\frac{147}{651}\)

⇒ \(\frac{310}{651}>\frac{147}{651}\)

⇒ \(\frac{10}{21}>\frac{21}{93}\)

Thus, the ratio 10: 21 is larger.

Question 50. Reshma prepared 18 kg of Burfi by mixing Khoya with sugar in the ratio of 7: 2. How much Khoya did she use?
Solution:

Given

Reshma prepared 18 kg of Burfi by mixing Khoya with sugar in the ratio of 7: 2.

The ratio of Khoya to sugar = 7:2

Let the quantity of Khoya be 7x kg and sugar be 2x kg.

7x + 2x =18 => 9x =18 => x = \(\frac{18}{9}=2\)

The quantity of Khoya be (7 x 2) kg = 14 kg

Question 51. A line segment 56 cm long is to be divided into two parts in the ratio of 2: 5. Find the length of each part.
Solution:

Given

A line segment 56 cm long is to be divided into two parts in the ratio of 2: 5.

The length of the line segment = 56 cm

Let the lengths of the two parts be 2 cm and 5 cm

2x + 5x = 56

7x = 56 ⇒ x \(=\frac{56}{7}=8\)

The length of one part is (2 x 8) cm =16 cm and length of other part is (5 x 8) cm = 40 cm

Question 52. The number of milk teeth in human beings is 20 and the number of permanent teeth is 32. Find the ratio of the number of milk teeth to the number of permanent teeth.
Solution:

Given

The number of milk teeth = 20

and the number of permanent teeth = 32

The required ratio \(=\frac{\text { Number of milk teeth }}{\text { Number of permanent teeth }}\)

⇒ \(\frac{20}{32}=\frac{5}{8}=5: 8\)

Question 53. The sex ratio is defined as the number of females per 1,000 males in the population. Find the sex ratio if there are 3732 females per 4000 males in a town.
Solution:

Given

The sex ratio is defined as the number of females per 1,000 males in the population.

Number of females = 3732

Number of males = 4000

The required sex ratio = \(\frac{3732}{4000}=\frac{933}{1000}\)

Thus, there are 933 females per 1000 males.

Question 54. In a year, Ravi earns? 360000 and paid? 24000 as income tax. Find the ratio of his

1. Income to income tax.
2. Income tax to income after paying income tax.

Solution: Ravi’s annual income =? 360000

Income tax paid by him = ? 24000

1. The required ratio \(=\frac{\text { The amount of income }}{\text { The amount of income tax }}\)

⇒ \(=\frac{₹ 360000}{₹ 24000}=\frac{15}{1}=15: 1\)

After paying income tax, the remaining amount of income =? (360000- 24000) = ? 336000

The required ratio \(\begin{aligned}
& =\frac{\text { The amount of income tax }}{\text { The remaining amount of income }} \\
& \text { after paying the income tax }
\end{aligned}\)

⇒ \(=\frac{₹ 24000}{₹ 336000}=\frac{1}{14}=1: 14\)

Question 55. Ramesh earns₹ 28000 per month. His wife Rama earns₹ 36000 per month. Find the ratio of

Ramesh’s earnings to their total earnings

Rama’s earnings to their total earnings.

Ramesh’s monthly earnings =₹ 28000

Rama’s monthly earnings =₹ 36000

Their total earnings =₹ (28000 + 36000) = ₹ 64000

1. Required ratio = \(=\frac{\text { Ramesh’s monthly earnings }}{\text { Their total monthly earnings }}\)

⇒ \(=\frac{₹ 28000}{₹ 64000}=\frac{7}{16}=7: 16\)

Question 56. Required ratio

⇒ \(=\frac{\text { Rama’s monthly earnings }}{\text { Their total monthly earnings }}\)

⇒ \(=\frac{₹ 36000}{₹ 64000}=\frac{9}{16}=9: 16\)

Question 56. Of the 288 persons working in a company, 11 2 are men and the remaining are women. Find the ratio of the number of

1. men to that of women.
2. men to the total number of persons.
3. women to the total number of persons

Solution:

Total number of persons working in a company = 288

Number of men = 112

Number of women = 288 -112 = 176

Required ratio=

⇒ \(=\frac{\text { Number of men }}{\text { Number of women }}=\frac{112}{176}\)

⇒ \(=\frac{7}{11}=7: 11\)

Required ratio \(=\frac{\text { Number of men }}{\text { Total number of persons }}\)

⇒ \(=\frac{112}{288}=\frac{7}{18}=7: 18\)

Required ratio \(=\frac{\text { Number of women }}{\text { Total number of persons }}\)

⇒ \(=\frac{176}{288}=\frac{11}{18}=11: 18\)

Question 57. A rectangular sheet of paper is of length 1.2 m and width 21 cm. Find the ratio of the width of the paper to its length.
Solution:

Given

A rectangular sheet of paper is of length 1.2 m and width 21 cm.

Length of paper = 1.2 m = 1.2 x 100 cm = 120 cm

And the width of the paper = is 21 cm

The required ratio = \(=\frac{\text { width of paper }}{\text { length of paper }}\)

⇒ \(=\frac{21 \mathrm{~cm}}{120 \mathrm{~cm}}=\frac{7}{40}=7: 40\)

Question 58. A scooter travels 1 20 km in 3 hours and a train travels 1 20 km in 2 hours. Find the ratio of their speeds.

\(\left(\text { Hint: Speed }=\frac{\text { distance travelled }}{\text { time taken }}\right)\)

Solution:

Given

A scooter travels 1 20 km in 3 hours and a train travels 1 20 km in 2 hours.

We know that Speed = \(\frac{\text { Distance }}{\text { Time }}\)

The speed of the scooter = \(\frac{120 \mathrm{~km}}{3 \text { hours }}\)

= 40 km/hour

And the speed of the train = \(\frac{120 \mathrm{~km}}{2 \text { hours }}\)

Thus, the required ratio \(=\frac{\text { Speed of the scooter }}{\text { Speed of the train }}\)

⇒ \(=\frac{40 \mathrm{~km} / \text { hour }}{60 \mathrm{~km} / \text { hour }}=\frac{2}{3}=2: 3\)

Question 59. An office opens at 9 a.m. and closes at 5.30 p.m. with a lunch break of 30 minutes. What is the ratio of lunch breaks to the total period in the office?
Solution:

Given

An office opens at 9 a.m. and closes at 5.30 p.m. with a lunch break of 30 minutes.

Opening time of office = 9 a.m.

Closing time of office = 5.30 p.m.

Total period in office = 8 hours 30 minutes

= (8 x 60 + 30) minutes = 510 minutes

The duration of lunch break = 30 minutes

The required ratio \(=\frac{\text { Duration of lunch break }}{\text { Total time period in office }}\)

⇒ \(=\frac{30 \text { minutes }}{510 \text { minutes }}=\frac{1}{17}=1: 17\)

Question 60. The shadow of a 3 m long stick is 4 m long. At the same time of the day, if the shadow of a flagstaff is 24 m long, how tall is the flagstaff?
Solution:

Given

The shadow of a 3 m long stick is 4 m long. At the same time of the day, if the shadow of a flagstaff is 24 m long

Let the length of the flagstaff be x m.

According to the given question,

3m : xm:: 4m: 24m

⇒ \(\frac{3}{x}=\frac{4}{24} \Rightarrow 4 \times x=3 \times 24\)

⇒ \(x=\frac{3 \times 24}{4}=18\)

Thus, the length of the flagstaff is 18 m.

Question 61. A recipe calls for 1 cup of milk for every 2– cups of flour to make a cake that would feed 6 persons. How many cups of both flour and milk will be needed to make a similar cake for 8 people?
Solution:

Given

A recipe calls for 1 cup of milk for every 2– cups of flour to make a cake that would feed 6 persons.

Quantity of both milk and flour to make a cake for 6 persons

⇒ \(=\left(1+2 \frac{1}{2}\right) \mathrm{cups}\)

⇒ \(=\left(1+\frac{5}{2}\right) \text { cups }=\left(\frac{2+5}{2}\right) \mathrm{cups}=\frac{7}{2} \text { cups }\)

Let the quantity of both milk and flour to make a cake for 8 persons be x cups.

According to the given question,

⇒ \(\frac{7}{2} \text { cups : } 6 \text { persons }:: x \text { cups }: 8 \text { persons }\)

⇒ \(\frac{\frac{7}{2}}{6}=\frac{x}{8} \Rightarrow 6 \times x=8 \times \frac{7}{2}\)

⇒ \(x=\frac{28}{6} \Rightarrow x=\frac{14}{3}\)

Thus, \(\frac{14}{3}=4 \frac{2}{3} \text { cups }\) of both flour and milk will be needed to make a cake for 8 people.

Question 62. In a school, the ratio of the number of large classrooms to small classrooms is 3: 4. If the number of small rooms is 20, then find the number of large rooms.
Solution:

Given

In a school, the ratio of the number of large classrooms to small classrooms is 3: 4. If the number of small rooms is 20,

Let the number of large classrooms is 3x and the number of small classrooms be 4x.

We have given, the number of small rooms = 20

⇒ \(4 x=20 \Rightarrow x=\frac{20}{4} \Rightarrow x=5\)

The number of large classrooms is 3 x 5 = 15

Question 63. Samira sells newspapers at Janpath CCrossing daily. On a particular day, she had 312 newspapers out of which 216 were in English and the remaining in Hindi. Find the ratio of

1. the number of English newspapers to the number of Hindi newspapers.
2. the number of Hindi newspapers to the total number of newspapers.

Solution:

Given

Samira sells newspapers at Janpath CCrossing daily. On a particular day, she had 312 newspapers out of which 216 were in English and the remaining in Hindi.

Total number of newspapers = 312.

Number of English newspapers = 216

Number of Hindi newspapers- 31 2- 216

= 96

The required ratio

⇒ \(=\frac{\text { Number of English newspapers }}{\text { Number of Hindi newspapers }}\)

⇒ \(=\frac{216}{96}=\frac{9}{4}=9: 4\)

The required ratio

⇒ \(=\frac{\text { Number of Hindi newspapers }}{\text { Total number of newspapers }}\)

⇒ \(=\frac{96}{312}=\frac{4}{13}=4: 13\)

Question 64. The students of a school belong to different religious backgrounds. The number of Hindu students is 288, the number of Muslim students is 252, the number of Sikh students is 1 44 and the number of Christian students is 72. Find the ratio of

Solution: Given

The students of a school belong to different religious backgrounds. The number of Hindu students is 288, the number of Muslim students is 252, the number of Sikh students is 1 44 and the number of Christian students is 72.

The number of Hindu students to the number of Christian students.

The number of Muslim students to the total number of students.

Number of Hindu students = 288

Number of Muslim students = 252

Number of Sikh students = 144

Number of Christian students = 72

Total number of students in the school = 288 + 252 + 144 + 72 = 756

The required ratio

⇒ \(=\frac{\text { Number of Hindu students }}{\text { Number of Christian students }}\)

⇒ \(=\frac{288}{72}=\frac{4}{1}=4: 1\)

The required ratio \(=\frac{\text { Number of Muslim students }}{\text { Total number of students }}\)

⇒ \(=\frac{252}{756}=\frac{1}{3}=1: 3\)

Question 65. When Chinmay visited Chowpati in Mumbai on a holiday, he observed that the ratio of North Indian food stalls to South Indian food stalls is 5: 4. If the total number of food stalls is 117, find the number of each type of food stalls.
Solution:

Given

When Chinmay visited Chowpati in Mumbai on a holiday, he observed that the ratio of North Indian food stalls to South Indian food stalls is 5: 4. If the total number of food stalls is 117,

Let the number of North Indian food stalls be 5x and the number of South Indian food stalls be 4x.

The total number of food stalls = 117

5x + 4x = 117 => 9x = 117

⇒ \(x=\frac{117}{9} \Rightarrow x=13\)

Thus, the number of North Indian food stalls is 5 x 13 = 65

And the number of South Indian food stalls is 4×13 = 52

Question 66. At the parking stand of Ramleela ground, Kartik counted that there were 115 cycles, 75 scooters and 45 bikes. Find the ratio of the number of cycles to the total number of vehicles.
Solution:

Given

At the parking stand of Ramleela ground, Kartik counted that there were 115 cycles, 75 scooters and 45 bikes.

Number of cycles = 115

Number of scooters = 75

Number of bikes = 45

Total number of vehicles = 115 + 75 + 45 = 235

The required ratio \(=\frac{\text { Number of cycles }}{\text { Total number of vehicles }}\)

⇒ \(=\frac{115}{235}=\frac{23}{47}=23: 47\)

Question 67. A train takes 2 hours to travel from Ajmer to Jaipur, which are 130 km apart. How much time will it take to travel from Delhi to Bhopal which are 780 km apart if the train is travelling at a uniform speed?
Solution:

Given

A train takes 2 hours to travel from Ajmer to Jaipur, which are 130 km apart.

A train covers a distance of 130 km in 2 hours and it covers a distance of 780 km in x hours.

According to the question, 130 km: 2 hours:: 780 km: x hours

⇒ \(\frac{130}{2}=\frac{780}{x} \Rightarrow 130 \times x=2 \times 780\)

⇒ \(x=\frac{2 \times 780}{130}=12\)

Thus, the required time is 12 hours.

Question 68. The length and breadth of a school ground are 150 m and 90 m respectively, while the length and breadth of a mela ground are 210 m and 126 m, respectively. Are these measurements in proportion?
Solution:

Given

The length and breadth of a school ground are 150 m and 90 m respectively, while the length and breadth of a mela ground are 210 m and 126 m, respectively.

The ratio of length to breadth of school ground = 150 m : 90 m \(=\frac{150}{90}=\frac{5}{3}=5: 3\)

The ratio of length to breadth of mela ground = 210 m: 126 m

⇒ \(=\frac{210}{126}=\frac{5}{3}=5: 3\)

Hence, both ratios are equal.

The given measurements are in proportion.

Question 69. In the given figure, the comparative areas of the continents are given :

1. What is the ratio of the areas of
2. Africa to Europe
3. Australia to Asia
4. Antarctica to a Combined area of North America and South America.

Solution:

Let each square be of unit length.

The required ratio = \(=\frac{26 \text { sq units }}{10 \text { sq units }}\)

⇒ \(=\frac{13}{5}=13: 5\)

The required ratio = \(=\frac{\text { Area of Australia }}{\text { Area of Asia }}\)

⇒ \(=\frac{8 \mathrm{sq} \text { units }}{44 \text { sq units }}\)

⇒ \(=\frac{2}{11}:=2: 11\)

The required ratio = \(=\frac{\text { Area of Antarctica }}{\text { Area of (North America }+ \text { South America) })}\)

⇒ \(=\frac{13 \mathrm{sq} \text { units }}{(17+18) \mathrm{sq} \text { units }}=\frac{13 \mathrm{sq} \text { units }}{35 \mathrm{sq} \text { units }}\)

⇒ \(=\frac{13}{35}=13: 35\)

Question 70. A tea merchant blends two varieties of tea costing her? 234 and 1 30 per kg in the ratio of their costs. If the weight of the mixture is 84 kg, then find the weight of each variety of tea.
Solution:

Given

A tea merchant blends two varieties of tea costing her? 234 and 1 30 per kg in the ratio of their costs. If the weight of the mixture is 84 kg,

The ratio of the costs of two varieties of tea is? 234 per kg? 130 per kg.

Sum of ratios = ?(234 + 130) per kg = ? 364 per kg.

Total weight of the mixture = 84 kg

The weights of each variety of tea are given by

⇒ \(\frac{234 \text { per } \mathrm{kg}}{364 \text { per } \mathrm{kg}} \times 84 \mathrm{~kg}=54 \mathrm{~kg}\)

⇒ \(\text { and } \frac{130 \text { per } \mathrm{kg}}{364 \text { per } \mathrm{kg}} \times 84 \mathrm{~kg}=30 \mathrm{~kg}\)

Question 71. An alloy contains only zinc and copper and they are in the ratio of 7: 9. If the weight of the alloy is 8 kg, then find the weight of copper in the alloy.
Solution:

Given

An alloy contains only zinc and copper and they are in the ratio of 7: 9. If the weight of the alloy is 8 kg

The ratio of the weight of zinc and copper is 7: 9.

Let the weight of zinc be lx kg and the weight of copper be 9x kg.

The weight of the alloy = 8 kg

7x + 9x = S => 16x = 8 => x \(=\frac{8}{16}=\frac{1}{2}\)

So, the weight of copper = \(9 \times \frac{1}{2} \mathrm{~kg}=\frac{9}{2} \mathrm{~kg}\)

⇒ \(=4 \frac{1}{2} \mathrm{~kg}\)

Question 72. In the following figure, each division represents 1 cm:

Express numerically the ratios of the following distances:

1. AC: AF
2. AG: AD
3. BF: AI
4. CE: DI

Solution:

1. AC:AF \(=\frac{A C}{A F}=\frac{2 \mathrm{~cm}}{5 \mathrm{~cm}}=\frac{2}{5}=2: 5\)

2. \(A G: A D=\frac{A G}{A D}=\frac{6 \mathrm{~cm}}{3 \mathrm{~cm}}=\frac{2}{1}=2: 1\)

3. \(B F: A I=\frac{B F}{A I}=\frac{4 \mathrm{~cm}}{8 \mathrm{~cm}}=\frac{1}{2}=1: 2\)

4. \(C E: D I=\frac{C E}{D I}=\frac{2 \mathrm{~cm}}{5 \mathrm{~cm}}=\frac{2}{5}=2: 5\)

Question 73. Find two numbers whose sum is 100 and whose ratio is 9: 16.
Solution:

The ratio of the two numbers is 9: 16.

Let the numbers be 9x and 16. v.

The sum of numbers = 100

⇒ 9.v + 16.v = 100 => 25.v = 100

\(x=\frac{100}{25} \Rightarrow x=4\)

Thus, the required numbers are 9 x 4 = 36 and 16×4 = 64

Question 74. In Figure (1) and Figure (2), find the ratio of the area of the shaded portion to that of the whole figure:

Solution:

Required ratio \(=\frac{\text { Area of the shaded portion }}{\text { Area of the whole figure }}\)

⇒ \(=\frac{8 \text { sq units }}{16 \text { sq units }}=\frac{1}{2}=1: 2\)

Lei the area of each small triangle is 1 sq unit.

Required ratio

⇒ \(=\frac{\text { Area of shaded portion }}{\text { Area of the whole figure }}\)

⇒ \(=\frac{16 \mathrm{sq} \text { units }}{32 \mathrm{sq} \text { units }}=\frac{1}{2}=1: 2\)

Question 75. A typist has to type a manuscript of 40 pages. She has typed 30 pages of the manuscript. What is the ratio of the number of pages typed to the number of pages left?
Solution:

Given

A typist has to type a manuscript of 40 pages. She has typed 30 pages of the manuscript.

Total number of pages = 40

Number of pages typed = 30

The number of pages left = 40- 30 = 10

The required ratio \(=\frac{\text { Number of pages typed }}{\text { Number of pages left }}\)

⇒ \(=\frac{30}{10}=\frac{3}{1}=3: 1\)

Question 76. In a floral design made from tiles each of dimensions 40 cm by 60 cm (see figure), find the ratios of:

1. The perimeter of the shaded portion to the perimeter of the whole design.
2. The area of the shaded portion to the area of the unshaded portion.

Solution:

Length of one tile = 60 cm

And the breadth of one tile = 40 cm

Required ratio \(=\frac{\text { Perimeter of shaded portion }}{\text { Perimeter of whole design }}\)

⇒ \(=\frac{(4 \times 60+6 \times 40) \mathrm{cm}}{(8 \times 60+10 \times 40) \mathrm{cm}}=\frac{480 \mathrm{~m}}{880 \mathrm{~m}}\)

⇒ \(=\frac{6}{11}=6: 11\)

Area ofone tile- 60 an x 40 an = 2400 cm2 Required ratio

⇒ \(=\frac{\text { Area of the shaded portion }}{\text { Area of the unshaded portion }}\)

⇒ \(=\frac{6 \times \text { area of } 1 \text { tile }}{14 \times \text { area of } 1 \text { tile }}\)

⇒ \(=\frac{6 \times 2400 \mathrm{~cm}^2}{14 \times 2400 \mathrm{~cm}^2}=\frac{6}{14}=\frac{3}{7}=3: 7\)

Question 77. In the given figure, what is the ratio of the areas of

1. Shaded portion 1 to shaded portion 2?

2. shaded portion 2 to shaded portion 3?

3. shaded portions 1 and 2 taken together and shaded portion 3?

Solution:

Area of shaded portion I = 5 x 5 sq units = 25 sq units

Area of shaded portion HI = 7 x 5 sq units = 35 sq units

Area of shaded portion II = Area of the whole figure- Area of a shaded portion (I + III)

= [10 x 10- (25 + 35)] sq units

= [100- 60] sq units = 40 sq units

1.  Required ratio

⇒ \(=\frac{\text { Area of shaded portion I }}{\text { Area of shaded portion II }}\)

⇒ \(=\frac{25 \mathrm{sq} \text { units }}{40 \mathrm{sq} \text { units }}=\frac{5}{8}=5: 8\)

Required ratio \(=\frac{\text { Area of shaded portion II }}{\text { Area of shaded portion III }}\)

\(=\frac{40 \text { sq units }}{35 \text { sq units }}=\frac{8}{7}=8: 7\)

Required ratio \(=\frac{\text { Area of shaded portion (I + II) }}{\text { Area of shaded portion III }}\)

\(=\frac{(25+40) \text { sq units }}{35 \text { sq units }}=\frac{65 \text { sq units }}{35 \text { sq units }}\) \(=\frac{13}{7}=13: 7\)

Question 78. A car can travel 240 km in 15 litres of petrol. How much distance will it travel in 25 litres of petrol?
Solution:

Given

A car can travel 240 km in 15 litres of petrol.

The distance can be travelled in 15 litres of petrol = 240 km

The distance can be travelled in 1 litre of petrol \(=\frac{240}{15} \mathrm{~km}=16 \mathrm{~km}\)

Distance can be travelled in 25 litres of petrol = (25 x 16) km = 400 km

Question 79. Bachhu Manjhi earns ₹ 24000 in 8 months. At this rate,

How much does he earn in one year?

In how many months does he earn₹ 42000?

Solution:

Given

Bachhu Manjhi earns ₹ 24000 in 8 months.

In 8 months, money earned by Bachhu Manjhi = ₹ 24000

In 1 month, money earned by him

\(=₹ \frac{24000}{8}=₹ 3000\)

In 1 year, he earns =? (12 x 3000)

= ₹ 36000

Number of months for earning₹ 3000 = 1

Number of months for earning₹ 42000

\(=\frac{4200}{300} = 14 \)

Question 80. The yield of wheat from 8 hectares of land is 360 quintals. Find the number of hectares of land required for a yield of 540 quintals.
Solution:

Given

The yield of wheat from 8 hectares of land is 360 quintals.

Land required for a yield of 360 quintals of wheat \(=\frac{8}{360} \text { hectares }=\frac{1}{45} \text { hectares }\)

Land required for yield of 540 quintals of wheat = \(=\frac{1}{45} \times 540 \text { hectares }=12 \text { hectares }\)

Question 81. The earth rotates 360° about its axis in about 24 hours. By how many degrees will it rotate in 2 hours?
Solution:

Given

The earth rotates 360° about its axis in about 24 hours.

Rotation of earth in 24 hours = 360°

Rotation of earth in1 hour \(=\frac{360^{\circ}}{24}\) = 15°

Rotation of earth in 2 hours = 2 x 15° = 30°

Question 82. Shivangi is suffering from anaemia as the haemoglobin level in her blood is lower than the normal range. The doctor advised her to take one iron tablet two times a day. If the cost of 10 tablets₹ 17, then what amount will she be required to pay for her medical bill for 15 days?
Solution:

Given

Shivangi is suffering from anaemia as the haemoglobin level in her blood is lower than the normal range. The doctor advised her to take one iron tablet two times a day. If the cost of 10 tablets ₹ 17

Shivangi will consume 2 x 15 tablets i.e., 30 tablets in 15 days.

Cost of 10 tablets = ₹17

Thus, the cost of 30 tablets = \(₹ \frac{17}{10} \times 30\)

= ₹ 51

Question 83. The quarterly school fee in Kendriya Vidyalaya for Class VI is₹ 540. What will be the fee for seven months?
Solution:

Given

The quarterly (3 months) fee =₹ 540

The fee for 1 month = \(₹ \frac{540}{3}=₹ 180\)

Thus, the fee for 7 months =? (180 x 7)

= ₹ 1260

The fee for seven months= ₹ 1260

Question 84. In an electron, the votes cast for two of the candidates were in the ratio 5::7. If the successful candidate received 20734 votes, how many votes did his opponent receive?
Solution:

Given

In an electron, the votes cast for two of the candidates were in the ratio 5::7. If the successful candidate received 20734 votes

Let the number of votes cast for the candidates be 5x and 7x.

According to the question,

7x = 20734 => \(\)

⇒ \(x=\frac{20734}{7}=2962\)

⇒ \(x=\frac{20734}{7}=2962\)

Number of votes received by opponent candidate = 5×2962 = 14810

5. A metal pipe 3 metres long was found to weigh 7.6 kg. What would be the weight of the same kind of 7.8 m long pipe?
Solution:

Given

A metal pipe 3 metres long was found to weigh 7.6 kg.

The weight of 3 metre long pipe = 7.6 kg

The weight of1 metre long pipe = \(=\frac{7.6}{3} \mathrm{~kg}\)

Thus, the weight o7.8-metreetre long pipe \(=\frac{7.6}{3} \times 7.8 \mathrm{~kg}=19.76 \mathrm{~kg}\)

Question 86. A recipe for raspberry jelly calls for 5 cups of raspberry juice and \(2 \frac{1}{2}\) cups of sugar. Find the amount of sugar needed for 6 cups of the juice?
Solution:

Given

A recipe for raspberry jelly calls for 5 cups of raspberry juice and \(2 \frac{1}{2}\) cups of sugar.

The requirement of sugar for 5 cups of raspberry juice \(=2 \frac{1}{2} \text { cups }=\frac{5}{2} \text { cups }\)

The requirement of sugar for 1 cup of raspberry juice \(=\left(\frac{5}{2} \div 5\right) \text { cups }\)

⇒ \(\left(\frac{5}{2} \times \frac{1}{5}\right) \text { cups }=\frac{1}{2} \text { cups }\)

Thus, the requirement of sugar for 6 cups of the raspberry juice \(=6 \times \frac{1}{2} \mathrm{cups}=3 \mathrm{cups}\)

Thus, the requirement of sugar for 6 cups of raspberry juice = \(=6 \times \frac{1}{2} \text { cups }=3 \text { cups }\)

Question 87. A farmer planted 1890 tomato plants in a field in rows each having 63 plants. A certain type of worm destroyed 18 plants in each row. How many plants did the worm destroy in the whole field?
Solution:

Given

A farmer planted 1890 tomato plants in a field in rows each having 63 plants. A certain type of worm destroyed 18 plants in each row.

Total number of plants = number of rows x number of plants in each row

1890 = number of rows x 63

⇒m\(\frac{1890}{63}=\text { number of rows }\)

number of rows = 30

Now, the number of plants destroyed worms in 1 row = IS

Total number of plants destroyed by worm in 30 rows = 18 x 30 = 540

Question 88. The length and breadth of the floor of a room are 5 m and 3 m, respectively. Forty tiles, each witan h area are used to cover the floor partially. Find the ratio of the tiled and the non tiled portion of the floor.
Solution:

Given

The length and breadth of the floor of a room are 5 m and 3 m, respectively. Forty tiles, each witan h area are used to cover the floor partially.

Tire total area of the floor = 5 m x 3 m = 15 m2

Tire area covered with tiles \(=40 \times \frac{1}{16} \mathrm{~m}^2\)

⇒ \(=\frac{5}{2} \mathrm{~m}^2\)

The non tiled area = \(\left[15-\frac{5}{2}\right] \mathrm{m}^2\)

⇒ \(=\left[\frac{30-5}{2}\right] \mathrm{m}^2=\frac{25}{2} \mathrm{~m}^2\)

The required ratio \(=\frac{\text { Tiled area }}{\text { Non tiled area }}\)

⇒ \(=\frac{\frac{5}{2} \mathrm{~m}^2}{\frac{25}{2} \mathrm{~m}^2}=\frac{5}{2} \times \frac{2}{25}=\frac{1}{5}=1: 5\)

Question 89. A carpenter had a board which measured 3 m x 2m. She cut out a rectangular piece of 250 cm x 90 cm. What is the ratio of the area of the cutout piece and the remaining piece?
Solution:

Given

A carpenter had a board which measured 3 m x 2m. She cut out a rectangular piece of 250 cm x 90 cm.

The area of the board = 3 m x 2 m = 6 m2 = 6 x 10000 cm2 = 60000 cm2

And the area of cut out piece = 250 cm x 90 cm = 22500 cm2

The area of remaining piece = (60000-22500) cm2

= 37500 cm2

Thus, the required ratio \(=\frac{\text { The area of cut out piece }}{\text { The area of remaining piece }}\)

⇒ \(=\frac{22500 \mathrm{~cm}^2}{37500 \mathrm{~cm}^2}=\frac{3}{5}=3: 5\)

Class 6 Maths Chapter 11 Algebra

Question 1. If each matchbox contains 50 matchsticks, the number of matchsticks required to fill n such boxes is

1. 50 + n
2. 50n
3. 50 ÷ n
4. 50 – n

Solution: (2): Number of matchsticks in one box = 50

The number of matchsticks required to fill n boxes = 50 x n = 50n

Question 2. Amulya is x years of age now. 5 years ago her age was

1. (5 -x) years
2. (5 +x) years
3. (x- 5) years
4. (5 ÷ X) years

Solution: (3): The present age of Amulya is x years.

Question 3. Which of the following represents 6 x x

1. 6x
2. \(\frac{x}{6}\)
3. 6+x
4. 6-x

Solution: (1): 6 x x can be represented as 6x.

Read and Learn More Class 6 Maths Exemplar Solutions

Question 4.  Which of the following is an equation?

1. x + 1
2. x – 1
3. x – 1=0
4. x+1 >0

Solution:(3): Equation means an expression with a variable, constants and the sign of equality (=).

x- 1 = 0 is an equation.

Question 5. If x takes the value 2, then the value of + 1 0 is

1. 20
2. 12
3. 5
4. 8

Solution:(2) : When we put x = 2 in the given expression x + 10, we get

x + 10 = 2 + 10 = 12.

Question 6. If the perimeter of a regular hexagon is x metres, then the length of each of its sides is

1. (x + 6) metres
2. (x ÷ 6 ) meters
3. (x- 6) metres
4. (6 ÷ x) metres

Solution: (2): Perimeter of a regular hexagon

= 6 x Length of each side

⇒ x m = 6 x Length of each side

⇒ Length of each side = (x + 6) m

Question 7. Which of the following equations has = 2 as a Solution?

1. x + 2 = 5
2. x – 2 = 0
3. 2x + 1 =0
4. x + 3 = 6

Solution: (2) : (A) Putting x = 2 in x + 2, we get 2+2=4*5

Putting x = 2 in x- 2, we get 2- 2 = 0

Putting x = 2 in 2x + 1, we get 2×2+l=5*0

Putting x = 2 in x + 3, we get 2 + 3 = 5* 6

Thus, the above conditions shows that x = 2 is the solution of x- 2 = 0

Question 8. For any two integers x and y, which of the following suggests that the operation of addition is commutative?

1. x + y = y + x
2. x + y > x
3. x – y = y – x
4. x x y = y x x

Solution: (A): x + y- y + x shows that addition is commutative.

Question 9. Which of the following equations does not have a solution in integers?

1. x + 1 =1
2. x – 1 =3
3. 2x+ 1 =6
4. 1 -x = 5

Solution: (3) : (A) x +1 =1

x + 1- 1 = 1 – 1

[Subtracting1 from both sides]

⇒ x = 0, which is an integer

x-1 =3

⇒ x-l + l= 3 + 1 [Adding1 to both sides]

⇒  x = 4, which is an integer.

2x +1 = 6

⇒ 2x + 1- 1 =6- 1

[Subtracting1 from both sides]

2x = 5

⇒ \(\frac{2 x}{2}=\frac{5}{2}\) [Dividing both sides by 2]

⇒ \(x=\frac{5}{2} \text {, }\) which is not an integer.

1 -x = 5

1-x-l =5-1

[Subtracting1 from both sides]

⇒ -x = 4

⇒ -(-x) =-4

[Multiplying both sides by (-1)]

⇒ x = -4, which is an integer

Thus, the above conditions show that equation 2x + 1 = 6 does not have a solution in integers.

Question 10. In algebra, an x b means ab, but in arithmetic 3×5 is

1. 35
2. 53
3. 15
4. 8

Solution: (3) : 3 x 5 means 15.

Question 11. In algebra, letters may stand for

1. known quantities
2. Unknown quantities
3. Fixed numbers
4. None of these

Solution: (2): In algebra, letters may stand for unknown quantities.

Question 12. “Variable” means that it

1. Can take different values
2. Has a fixed value
3. Can take only 2 values
4. Can take only three values

Solution: (1): “Variable” means that it can take different values.

Question 13. 10 -x means

1. 10 is subtracted x times
2. x is subtracted 10 times
3. x is subtracted from 1 0
4. 10 is subtracted from x

Solution: (3): 10- x means x is subtracted from 10.

Question 14. Savitri has a sum of ₹ x. She spent ₹ 1000 on groceries,₹ 500 on clothes and ₹ 400 on education, and received ₹ 200 as a gift. How much money (in ₹) is left with her?

1. x – 1700
2. x – 1900
3. x + 200
4. x – 2100

Solution: (1):

Given

Savitri has a sum of ₹ x. She spent ₹ 1000 on groceries,₹ 500 on clothes and ₹ 400 on education, and received ₹ 200 as a gift.

Amount of money Savitri has = ₹ x.

Amount of money spent by her =₹ (1000 + 500 + 400) = ₹ 1900

Amount of money received by her as a gift = ₹ 200

Amount of money left with her

= ₹(x- 1900 + 200)

= ₹ (x- 1700)

Question 15. The perimeter of the triangle shown in the given figure is

1. 2x+y
2. x + 2y
3. x + y
4. 2x-y

Solution: (1): The perimeter of the triangle = Sum of all sides

=x+x+y

= 2x + y

Question 16. The area of a square with each side x is

1. xxx
2. 4x
3. x + x
4. 4 = x + x

Solution: (1) : The area of a square = (side) x (side) =x*x

Question 17. The expression obtained when x is multiplied by 2 and then subtracted from 3 is

1. 2x- 3
2. 2x + 3
3. 2x- 3
4. 3x- 2

Solution : x is multiplied by 2 and then subtracted from 3 =3-2 x * x = 3-2x

Question 18. \(\frac{q}{2}=3\) has a solution

1. 6
2. 8
3. 3
4. 2
5. Solution: (1): We have, \(\frac{q}{2}=3\) Multiplying both sides by 2, we get

⇒ \(\Rightarrow \quad \frac{q}{2} \times 2=3 \times 2\)

q = 6

Question 19. x-4 =- 2 has a solution

1. 6
2. 2
3. -6
4. -2

Solution:(2) : We have, x- 4 =-2

⇒  x -4 + 4 = -2 + 4

[Adding 4 to both sides]

⇒  x = 2

Question 20. \(\frac{4}{2}=2\) denotes a V

1. Numerical equation
2. Algebraic expression
3. Equation with a variable
4. False statement

Solution: (1): \(\frac{4}{2}=2\) denotes a numerical equation.

Question 21. Kanta has p pencils in her box. She puts q more pencils in the box. The total number of pencils with her is.

1. p+q
2. PQ
3. p-q
4. \(\frac{p}{q}\)

Solution: (1): Kanta has p pencils in her box. After putting q more pencils in the box, she has a total number of pencils = p + q.

Question 22. The equation 4x = 16 is satisfied by the following value of x

1. 4
2. 2
3. 12
4. -12

Solution:(1) : We have, 4x = 16

⇒ \(\frac{4 x}{4}=\frac{16}{4}\)

x = 4

Question 23. I think of a number and on adding 13 to it, I get 27. The equation for this is

1. x – 27 = 13
2. x – 13 = 27
3. x + 27 = 13
4. x + 13 = 27

Solution: (D): Let the number be x.

According to the question, x + 13 = 27

Question 24. The distance (in km) travelled in h hours at a constant speed of 40 km per hour is_______.
Solution: 40 h : Distance = Speed * Time = (40 x h) km = 40 h km

Question 25. p kg of potatoes are bought for ₹70. The cost of 1 kg of potatoes (in ₹) is_______.
Solution:

⇒ \(\frac{70}{p}\) : The cost of p kg of potatoes = ₹ 70

The cost of1 kg of potatoes = \(₹ \frac{70}{p}\)

Question 26. An auto rickshaw charges ₹ 10 for the first kilometre then ₹ 8 for each such subsequent kilometre. The total charge (in ₹) for d kilometres is_______.
Solution:

Given

An auto rickshaw charges ₹ 10 for the first kilometre then ₹ 8 for each such subsequent kilometre.

2 + 8d: For the first kilometre, rickshaw charges = ₹ 10.

And the charges for subsequent kilometres = ₹ 8.

The charges for d kilometres

= 10 + (d- 1)8

= 10 +8d-8 = 2 + 8d

Question 27. lf 7x + 4 = 25, then the value of x is_______.
Solution: 3:

We have given, 7x + 4 = 25

⇒  7x + 4- 4 = 25- 4

[Subtracting 4 from both sides]

⇒ \(\frac{7 x}{7}=\frac{21}{7}\)

⇒  x = 21

⇒  x = 3

Question 28. The solution of equation 3x + 7 = -20 is _______.
Solution:

-9 : We have, 3x + 7 =- 20

3x + 7- 7 =- 20- 7

[Subtracting 7 from both sides]

⇒  3x = -27

[Dividing both sides by 3]

⇒  x = -9

Question 29. ‘exceeds by 7’ca can be expressed as_______.
Solution: x = y + 7: x exceeds y by 7 can be expressed as = y + 7 or x-y = 7.

Question 30. ‘8 more than three times the number x7 can be written as_______.
Solution: 3x + 8 : 8 more than three times the number x can be expressed as 3x + 8

Question 31. Many pencils bought for? x at the rate of ₹ 2 per pencil is_______.
Solution: \(\frac{x}{2}\) Number of pencils bought for ₹ 2 =1

Number of pencils bought for ₹ x = \(\frac{1}{2} \times x=\frac{x}{2}\)

Question 32. The number of days in w weeks is_______.
Solution: 7W : Number of days in a week = 7

Number of days in w weeks = 7 * w = 7w

Question 33. Annual salary at r rupees per month along with a festival bonus of? 2000 is_______.
Solution: ₹ (12r + 2000) : We have given, monthly salary = ₹ r

Annual salary = ₹ (12 * r) = ₹ 12r

Now, total annual salary with a festive bonus = ₹ (12r + 2000)

Question 34. The two-digit number whose ten digit is Y and unit’s digit is’ is_______.
Solution: + u: The digit at the unit’s place is V.

And the digit at ten’s place is Y.

The number can be expressed as 10 x t + u = lOf + u

Question 35. The variable used in the equation 2p +8 = 18 is _______.
Solution: p : p is the variable in the equation Ip + 8 = 18

Question 36. x metres = _______centimeters.
Solution:

x : 1 metre = 100 cm

⇒  x metres = lOOx cm

Question 37. p litres = _______.millilitres.
Solution:

lOOOp : 1 litre = 1000 millilitres

⇒ p litres = lOOOp millilitres

Question 38. r rupees = _______ paise.
Solution:

Floor : 1 rupee = 100 paise

⇒  r rupees = lOOr paise

Question 39. If their present age of Ramandeep is n years, then her age after 7 years will be_______.
Solution:

(n + 7) years: Present age of Ramandeep = n years

Her age after 7 years = (n + 7) years

Question 40. If I spend f rupees from 100 rupees, the money left with me is_______rupees.
Solution:

100 – f: I have 100 rupees and spend f rupees.

Money left with me = (100 -f) rupees 40.

Question 41. 0 is a solution of the equation x+ 1 = 0
Solution: False

We have given, x +1 = 0

⇒  — 1 – 0- 1

[Subtracting 1 from both sides]

⇒  x –1, which is the solution of the given equation.

Question 42. The equations x + 1 = 0 and 2x + 2 = 0 have the same solution.
Solution: True

We have given, x + 1 = 0

⇒  x +1 -1 =0-1

[Subtracting 1 from both sides]

⇒  x = —1 …(1)

and 2x + 2 = 0

2x + 2-2 = 0-2

[Subtracting 2 from both sides]

⇒  2x = -2

⇒ \(\frac{2 x}{2}=\frac{-2}{2}\)

[Dividing both sides by 2]

⇒  x =-1 …(2)

Thus, (1) and (2) imply that both equations have the same solution.

Question 43. If m is a whole number, then 2m denotes a multiple of 2.
Solution: True

Question 44. The additive inverse of an integer x is 2x.
Solution: False

Since the additive inverse of x is -x.

Question 45. If x is a negative integer, – x is a positive integer.
Solution: True

The negative of a negative integer is always a positive integer.

Question 46. 2x- 5 > 1 1 is an equation.
Solution: False

Since an equation includes a sign of equality(=)

Question 47. In an equation, the LHS is equal to the RHS.
Solution: True

Question 48. In the equation 7k- 7 = 7, the variable is 7.
Solution: False

Since, in the equation 7k-7 = 7, the variable is k.

Question 49. a = 3 Is a solution of the equation 2a- 1 = 5
Solution: True

We have, 2a- 1 = 5

⇒  2n- 1 +1 = 5 + 1

[Adding 1 to both sides]

⇒  2a = 6

⇒  \(\frac{2 a}{2}=\frac{6}{2}\) [Dividing both sides by 2]

⇒  a = 3, which is the solution of the given equation

Question 50. The distance between New Delhi and Bhopal is not a variable.
Solution: True

Question 51. t minutes are equal to 60f seconds.
Solution: True

Since, 1 minute = 60 seconds

t minutes = 60 x f seconds = 60f seconds

Question 52. x = 5 is the solution of the equation 3x + 2 = 20
Solution: False

⇒  3x + 2- 2 = 20- 2

[Subtracting 2 from both sides]

⇒  3x = 18

⇒  \(\quad \frac{3 x}{3}=\frac{18}{3}\) [Dividing both sides by 3]

⇒  x = 6, which is the solution of the given equation.

Question 53. ‘One-third of a number added to itself gives 8’, can be expressed as\(\frac{x}{3}+8=x\)
Solution: False

Let the number be x.

One third of the number = \(\frac{x}{3} \text {. }\)

According to the given question, \(\frac{x}{3}+x=8\)

Question 54. The difference between the ages of the two sisters Leela and Yamini is a variable.
Solution: False

The difference between the ages of the two sisters will be a fixed number.

Question 55. The number of lines that can be drawn through a point is a variable. ,
Solution: False

Question 56. One more than twice the number.
Solution:

Let the number be x.

Twice the number = 2x.

Now, 1 more than 2x can be expressed as 2x + l.

Question 57. 20°C less than the present temperature.
Solution:

Let the present temperature be t°C.

Now, 20°C less than t = (t- 20)°C

Question 58. The successor of an integer.
Solution:

Let the integer be n.

Now, the successor of the integer n = n +1.

Question 59. The perimeter of an equilateral triangle, if the side of the triangle is m.
Solution: We have given, the side of the equilateral triangle = m.

The perimeter of an equilateral triangle

= 3 x side

= 3 m

Question 60. Area of the rectangle with length k units and breadth n units.
Solution:

We have given, a rectangle of length = k units and breadth = n units.

Area of the rectangle = length * breadth = (k x n) sq units = kn sq units

Question 61. Omar helps his mother 1 hour more than his sister does.
Solution:

Let Omar’s sister help her mother for x hours.

Omar helps his mother for (x +1) hours.

Question 62. Two consecutive odd integers.
Solution: Two consecutive odd integers are (2n + 1) and (2n + 3), where n is any integer.

Question 63. Two consecutive even integers.
Solution: Two consecutive even integers are 2m and 2m + 2, where m is any integer.

Question 64. Multiple of 5.
Solution:

A multiple of 5 can be expressed as 5n, where n is an integer.

Question 65. The denominator of a fraction is 1 more than its numerator.
Solution:

Let the numerator of the fraction be x.

The denominator of the fraction can be expressed as x + 1.

The fraction \(=\frac{x}{x+1}\)

Question 66. The height of Mount Everest is 20 times the height of the Empire State Building.
Solution:

Let the height of the Empire State Building be y.

Height of Mount Everest = 20y.

Question 67. If a notebook costs ₹ p and a pencil costs ₹ 3, then the total cost (in ₹) of two notebooks and one pencil.
Solution:

The cost of a notebook = ₹ p

Cost of 2 notebooks = ₹ 2p.

The cost of a pencil =₹ 3.

Cost of two notebooks and one pencil = ₹(2p + 3)

Question 68. Z is multiplied by 3 and the result is subtracted from 1 3.
Solution:

z multiplied by -3 = -3z

– 3z subtracted from 13 =13- (-3z) = 13 + 3z

Question 69. p is divided by 11 and the result is added to 1 0.
Solution:

p divided by 11 = \(\frac{p}{11} .\)

\(\frac{p}{11}\) added to 10 = \(\frac{p}{11}+10 .\)

Question 70. x times 3 is added to the smallest natural number.
Solution: x times of 3 = 3x

And the smallest natural number =1.

So, 3x added to1 = 3x + 1.

Question 71. 6 times q is subtracted from the smallest two-digit number.
Solution:

6 times q = 6q.

The smallest two-digit number = 10

6q subtracted from 10 = 10- 6q.

Question 72. Write two equations for which 2 Is the solution.
Solution: The required equations are 3i/ + 4 = 10 and 2. V- 3 = 1, i.e., for both equations, 2 is the solution.

Question 73. Write an equation for which 0 is a solution.
Solution: The required equation is 2f + 3 = 3, which has solution t = 0.

Question 74. Write an equation whose solution is not a whole number.
Solution: The required equation is x + 1 = 0, and its solution is x = -1, which is not a whole number.

Question 75. A pencil costs? p and a pen cost 5p.
Solution: A pen costs 5 times the cost of a pencil.

Question 76. Leela contributed ₹ y towards the Prime Minister’s Relief Fund. Leela is now left with ₹ (y+ 10000).
Solution: The amount left with Leela is? 10,000 more than the amount she contributed towards the Prime Minister’s Relief Fund.

Question 77. Kartik is n years old. His father is In years old.
Solution: The age of Kartik’s father is seven times the age of Kartik.

Question 78. The maximum temperature on a day in Delhi was p°C. The minimum temperature was (P-10)°C.
Solution: The minimum temperature on a day in Delhi was 10°C less than the maximum temperature.

Question 79. John planted two plants last year. His friend Jay planted 2f + 1 0 plants that year.
Solution: Last year Jay planted 10 more plants than twice the plants planted by his friend John.

Question 80. Sharad used to take p cups of tea a day. After having some health problems, he takes p – 5 cups of tea a day.
Solution: Sharad reduced the consumption of tea per day by 5 cups after having some health problems.

Question 81. The number of students dropping out of school last year was m. Number of students dropping out of school this year is m- 30.
Solution: The number of students dropping out this year is 30 less than the number of students who dropped out last year.

Question 82. The price of petrol was ₹ p per litre last month. The price of petrol now is ₹ (p- 5) per litre.
Solution: The price of petrol per litre decreased this month by 5 than its price last month.

Question 83. Khader’s monthly salary was P in the year 2005. His salary in 2006 was ₹ (P + 1 000).
Solution: Khader’s monthly salary increased by? 1000 in the year 2006 than that of 2005.

Question 84. The number of girls enrolled in a school last year was g. The number of girls enrolled this year in the school is 3g- 1 0.
Solution: The number of girls enrolled this year is 10 less than 3 times the girls enrolled last year.

Question 85. Translate each of the following statements into an equation, using x as the variable:

1. 13 subtracted from twice a number gives 3.
2. One-fifth of a number is 5 less than that number.
3. Two-thirds of the number is 12.
4. 9 added to twice a number gives 1 3.
5. 1 subtracted from one-third of a number gives 1.

Solution:

Let the number be x.

Twice of the number x = 2x

According to question, 2x- 13 = 3

Let the number be x.

One-fifth of the number x = \(\frac{x}{5}.\)

5 less than the number x = x- 5

According to question, \(\frac{x}{5}=x-5\)

Let the number be x.

Two-thin! of the number x \(=\frac{2}{3} x\)

According to question, \(\frac{2}{3} x=12\)

Let the number be x.

Twice of the number x = 2x.

Now, 9 is added to 2x = 9 + 2x

According to the question, 9 + 2x = 13.

Let the number be x.

One-third of the number x = \(\frac{x}{3}\)

Now, 1 is subtracted from \(\frac{x}{3}=\frac{x}{3}-1\)

According to question, “\(\frac{x}{3}-1=1 \text {. }\)

Question 86. Translate each of the following statements into an equation :

• The perimeter (p) of an equilateral triangle is three times its side (a).
• The diameter (d) of a circle is twice its radius (r).
• The selling price (s) of an item is equal to the sum of the cost price (c) of an item and the profit (p) earned.
• Amount (a) is equal to the sum of principal (p) and interest (/).

Solution:

We have given the perimeter of an equilateral triangle

= 3 (the side of an equilateral triangle)

⇒  p = 3a

We have given the diameter of a circle

= 2 (the radius of the circle)

⇒  d = 2r

We have given, Selling price = cost price + profit

⇒  s = c + p

We have given, Amount = principal + interest

⇒  a = p + i

Question 87. Let Kanika’s present age be x years. Complete the following table, showing the ages of her relatives:

Solution:

We have given the present age of Kanika = x years.

Her brother’s age = (x- 2) years

Her father’s age = (x + 35) years

Her mother’s age = (x + 35- 3) years = (x + 32) years

Her grandfather’s age = 8x years

Question 88. If m is a whole number less than 5, complete the table and by inspection of the table, find the solution of the equation 2m- 5 = -1 :

Solution:

Given

If m is a whole number less than 5, complete the table and by inspection of the table

For m = 0, 2m-5=2×0-5=0-5=-5

For m = 1, 2m-5=2xl-5=2-5=-3

For m = 3, 2m-5=2×3-5=6-5=l

For m = 4, 2m-5=2×4-5=8-5=3

Thus, the solution of the equation 2m- 5 =-1 is m = 2.

Question 89. A class with p students has planned a picnic. 50 per student is collected, out of which ₹ 1800 is paid in advance for transport. How much money is left with them to spend on other items?
Solution:

Given

A class with p students has planned a picnic. 50 per student is collected, out of which ₹ 1800 is paid in advance for transport.

Number of students in the class = p

The total amount collected from p students =₹ 50p

Amount paid in advance for transport =₹ 1800

The amount left with them = ₹ (50p- 1800)

Question 90. In a village, there are 8 water tanks to collect rainwater. On a particular day, x litres of rainwater is collected per tank. If 100 litres of water was already there in one of the tanks, what was the total amount of water in the tanks on that day?
Solution:

Given

In a village, there are 8 water tanks to collect rainwater. On a particular day, x litres of rainwater is collected per tank. If 100 litres of water was already there in one of the tanks

Number of water tanks = 8

Rainwater collected by each tank = x litres

Rainwater collected by 8 tanks = 8x litres

But 100 litres of water was already there in one of the tanks.

Total amount of water in the tanks = (8x + 100) litres.

Question 91. What is the area of a square whose side is m cm?
Solution:

We have given, the side of a square = m cm

Area of the square = side x side = m x m sq cm

Question 92. The perimeter of a triangle is found by using the formula P = a + b + c, where a, b and c are the sides of the triangle. Write the rule that is expressed by this formula in words.
Solution:

Given

The perimeter of a triangle is found by using the formula P = a + b + c, where a, b and c are the sides of the triangle.

The perimeter of a triangle is the sum of all its sides.

Question 93. The perimeter of a rectangle is found by using the formula P = 2 (I + w), where l and w are respectively the length and breadth of the rectangle. Write the rule that is expressed by this formula in words.
Solution:

Gien

The perimeter of a rectangle is found by using the formula P = 2 (I + w), where l and w are respectively the length and breadth of the rectangle.

The perimeter of a rectangle is twice the sum of its length and breadth.

Question 94. On my last birthday, I weighed 40 kg. If I put on m kg of weight after a year, what is my present weight?
Solution:

Given

On my last birthday, I weighed 40 kg. If I put on m kg of weight after a year

Present weight = weight on last birthday + weight put on after a year.

D 40 kg + tn kg

= (40 + m) kg

Question 95. Length and breadth of rcm and tern, respectively.

1. What will be the length (in cm) of the aluminium strip required to frame the board, if 1 0 cm extra strip is required to fix it properly?
2. Ifx nails are used to repair one board, how many nails will be required to repair 15 such boards?
3. If 500 sq cm extra cloth per board is required to cover the edges, what will be the total area of the cloth required to cover 8 such boards?
4. What will be the expenditure for making 23 boards if the carpenter charges? x per board.

Solution:

We have given, the length of the board = r cm and breadth = t cm.

The length of the aluminium strip to frame the board = Perimeter of the board

= 2(length + breadth)

= 2(r + t) cm

But a 10 cm extra strip is required to fix it properly.

The total length of the aluminium strip = 2(r + t) cm + 10 cm.

Number of nails required to repair 1 board = x.

Number of nails required to repair 15 boards = 15 x  x = 15 x.

The area of the cloth required for the board

= Area of the board

= length x breadth

= r cm x t cm

= (rt) sq cm

The area of the cloth required for 8 boards

= 8 x (rt) sq cm

= 8rt sq cm

But 500 sq cm extra cloth per board is required to cover the edges.

For 8 boards we need 8 x 500 sq cm = 4000 sq cm extra cloth.

Thus, the total area of the cloth required = (8rt + 4000) sq cm

The carpenter charges for 1 board = %x.

The carpenter charges for 23 boards = ?(23*x) = ?23x

Question 96. Sunita is half the age of her mother Geeta. Find their ages

1. After 4 years.
2. Before 3 years.

Solution:

Let the present age of Sunita be x years.

The present age of her mother Geeta = 2(Sunita’s present age) = 2x years.

1.  After 4 years,

Sunita’s age = (x + 4) years

Geeta’s age = (2x + 4) years

2.  Before 3 years,

Sunita’s age = (x- 3) years

Geeta’s age = (2x- 3) years.

Question 97. Match the items of Column I with that of Column 2:
Solution:

(1)⇒ (B), (2) ⇒(E), (3) ⇒ (C), (4) ⇒ (C), (5) ⇒ (A)

The number of comers of a quadrilateral is 4, i.e., a constant.

The variable in the equation 2p + 3 =5 is p.

We have given, x + 2 = 3

⇒ x+2-2=3-2

[Subtracting 2 from both sides]

⇒ x = 1, which is the solution of the given equation.

We have given, 2p + 3 = 5

⇒ 2p + 3 — 3 =5- 3

[Subtracting 3 from both sides]

⇒ 2p = 2

⇒ \(\frac{2 p}{2}=\frac{2}{2}\)

[Dividing both sides by 2]

p = 1, which is the solution of the given equation.

The equality sign (=) is used in an equation

Class 6 Maths Chapter 7 Fractions

Question 1. The fraction which is not equal to\(\frac{4}{5}\)

1. \(\frac{40}{50}\)
2. \(\frac{12}{15}\)
3. \(\frac{16}{20}\)
4. \(\frac{9}{15}\)

Solution:

4. Since, \(\frac{9}{15}=\frac{9 \div 3}{15 \div 3}=\frac{3}{5} \text {, }\) which is not equal to \(\frac{4}{5}\)

\(\frac{9}{15}\) is the required fraction.

Question 2. The two consecutive integers between which the fraction \(\frac{5}{7}\) lies are

1. 5 and 6
2. 0 and 1
3. 5 and 7
4. 6 and 7

Solution:

Read and Learn More Class 6 Maths Exemplar Solutions

2. Since we know that a proper fraction whose numerator is less than its denominator always lies between 0 and 1.

∴ \(\frac{5}{7},\) which is also a proper fraction, must
lie between 0 and 1.

Question 3. when \(\frac{1}{4}\) is written with denominator as 12, its numerator is

1. 3
2. 8
3. 24
4. 12

Solution:

1. Since, \(\frac{1}{4}=\frac{1}{4} \times \frac{3}{3}\) [By multiplying the numerator and denominator by 3 to get 12 as the denominator] \(=\frac{3}{12},\) which shows that the required numerator is 3.

Question 4. Which of the following is not in the lowest form?

1. \(\frac{7}{5}\)
2. \(\frac{15}{20}\)
3. \(\frac{13}{33}\)
4. \(\frac{27}{28}\)

Solution:

In the case of \(\frac{7}{5}\) the common factor of 7 and 5 is

i.e. already given in the simplest form.

Similarly,

In \(\frac{15}{20}\) the common factor of 15 and 20 is 4

\(\frac{13}{33}\) the common factor of 13 and 33 is 1

i.e., already given in the simplest form

\(\frac{27}{28}\) the common factor of 27 and 28 is 1

i.e., already given in the simplest form.

So, the above cases show that \(\frac{15}{20}\) is not in the lowest form.

Question 5. If \(\frac{5}{8}=\frac{20}{p}\) then value of P is

1. 23
2. 2
3. 32
4. 16

Solution:

3. We have given, \(\frac{5}{8}=\frac{20}{p}\)

To find the value of p, we must multiply the numerator and denominator of \(\frac{5}{8} \text { by } 4.\)

∴ \(\frac{5}{8}=\frac{5 \times 4}{8 \times 4}=\frac{20}{32}\)

∴ \(\frac{20}{32}=\frac{20}{p}\)

∴ The value of p is 32.

Question 6. Which of the following is not equal to the others?

1. \(\frac{6}{8}\)
2. \(\frac{12}{16}\)
3. \(\frac{15}{25}\)
4. \(\frac{18}{24}\)

Solution:

3. Firstly, we will simplify all the fractions into their lowest form, we get,

\(\frac{6}{8}=\frac{6 \div 2}{8 \div 2}=\frac{3}{4}, \frac{12}{16}=\frac{12 \div 4}{16 \div 4}=\frac{3}{4}, \frac{15}{25}=\frac{15 \div 5}{25 \div 5}=\frac{3}{5}\) and \(\frac{18}{24}=\frac{18 \div 6}{24 \div 6}=\frac{3}{4}\)

Thus, on comparing the above cases, observe that \(\frac{3}{5}=\frac{15}{25}\) is not equal to the above-given fractions.

Question 7. Which of the following fractions is the greatest?

1. \(\frac{5}{7}\)
2. \(\frac{5}{6}\)
3. \(\frac{5}{9}\)
4. \(\frac{5}{8}\)

Solution:

2. We have given, \(\frac{5}{7}, \frac{5}{6}, \frac{5}{9} \text { and } \frac{5}{8}\) Since we know that, among all the fractions with the same numerator, the one with a smaller denominator will be the greatest

\(\frac{5}{6}\) is the required fraction.

Question 8. Which of the following fractions is the smallest?

1. \(\frac{7}{8}\)
2. \(\frac{9}{8}\)
3. \(\frac{3}{8}\)
4. \(\frac{5}{8}\)

Solution:

3. We have given \(\frac{7}{8}, \frac{9}{8}, \frac{3}{8} \text { and } \frac{5}{8}\)

Since we know that, among all the fractions with the same denominator, the one with a smaller numerator will be the smallest.

∴ \(\frac{3}{8}\) is the required fraction.

QueSomeon 9. Sum of \(\frac{4}{17} \text { and } \frac{15}{17} \text { is }\)

1. \(\frac{19}{17}\)
2. \(\frac{11}{17}\)
3. \(\frac{19}{34}\)
4. \(\frac{2}{17}\)

Solution:

1.  We have, \(\frac{4}{17}+\frac{15}{17}\)

⇒ \(=\frac{4+15}{17}\)

⇒ \(\left[\begin{array}{l}
\text { Sum of like fractions } \\
=\frac{\text { Sum of their numerators }}{\text { common denominator }}
\end{array}\right]\)

⇒ \(=\frac{19}{17}\)

Question 10. On subtracting \(\frac{5}{9} \text { from } \frac{19}{9} \text {, }\) the result is

1. \(\frac{24}{9}\)
2. \(\frac{14}{9}\)
3. \(\frac{14}{18}\)
4. \(\frac{14}{0}\)

Solution:

According to the question, \(\frac{19}{9}-\frac{5}{9}\)

⇒ \(\left[\begin{array}{l}\text { Difference of like fraction } \\
=\frac{\text { Difference of their numerator }}{\text { Common denominator }}
\end{array}\right]\)

∴ \(=\frac{14}{9}\)

Question 11. \(\frac{11}{7}\) can be expressed in the form

1. \(\frac{33}{7}\)
2. \(\frac{39}{7}\)
3. \(\frac{33}{4}\)
4. \(\frac{39}{4}\)

Solution:

We have, \(5 \frac{4}{7}=\frac{(5 \times 7)+4}{7}=\frac{35+4}{7}=\frac{39}{7}\)

Question 13. A number representing a part of a ______ is called a fraction.
Solution: whole

Question 14. A fraction with a denominator greater than the numerator is called a______fraction.
Solution: Proper

Question 15. Fractions with the same denominator are called _____ fractions.
Solution: Like

Question 16. \(13 \frac{5}{18}\) is a _____ fraction.
Solution: Mixed

Question 17. \(\frac{18}{5}\) is an _____ fraction.
Solution: Improper

Question 18. \(\frac{7}{19}\) is an _____ fraction.
Solution: Proper

Question 19. \(\frac{5}{8} \text { and } \frac{3}{8} \text { are }\) proper _____ fraction.
Solution: Like

Question 20. \(\frac{6}{11} \text { and } \frac{6}{13}\) proper _____ fraction.
Solution: Unlike

Question 21. The fraction \(\frac{6}{15}\) in simplest form is_____.
Solution:

\(\frac{2}{5}: \text { We have, } \frac{6}{15}=\frac{6 \div 3}{15 \div 3}=\frac{2}{5}\)

Question 22. The fraction \(\frac{17}{34}\)  in simplest form is _____.
Solution:

\(\frac{1}{2} \text { : We have, } \frac{17}{34}=\frac{17 \div 17}{34 \div 17}=\frac{1}{2}\)

Question 23. \(\frac{18}{135} \text { and } \frac{90}{675}\)  are proper, unlike and _____ fractions.
Solution:  Equivalent

Question 24. \(8 \frac{2}{7}\)  is equal to the improper fraction______.
Solution:

\(\frac{58}{7}: 8 \frac{2}{7}=\frac{(8 \times 7)+2}{7}=\frac{56+2}{7}=\frac{58}{7}\)

Question 25. \(\frac{87}{7}\)  is equal to the mixed fraction______.
Solution:

\(12 \frac{3}{7}:\)

∴ \(\frac{87}{7}=12 \frac{3}{7}\)

Question 26. \(\frac{17}{9}+\frac{41}{9}=\)______.
Solution:

⇒ \(\frac{58}{9}: \frac{17}{9}+\frac{41}{9}=\frac{17+41}{9}=\frac{58}{9}\)

Question 27. \(\frac{67}{14}-\frac{24}{14}=\)_______.
Solution:

\(\frac{43}{14}: \frac{67}{14}-\frac{24}{14}=\frac{67-24}{14}=\frac{43}{14}\)

Question 28. \(\frac{17}{2}+3 \frac{1}{2}=\)______.
Solution:

\(\text { 12: } \frac{17}{2}+3 \frac{1}{2}=\frac{17}{2}+\frac{7}{2}=\frac{17+7}{2}=\frac{24}{2}=12\)

Question 29. \(9 \frac{1}{4}-\frac{5}{4}=\)______.
Solution:

\(8: 9 \frac{1}{4}-\frac{5}{4}=\frac{37}{4}-\frac{5}{4}=\frac{37-5}{4}=\frac{32}{4}=8\)

Question 30. Fractions with the same numerator are called fractions.
Solution: False

Because fractions with the same denominators are called fractions.

Question 31. Fraction \(\frac{18}{39}\) is in its lowest form
Solution: False

Since the common factor of 18 and 39 is 3 and its simplest form is

⇒ \(\frac{18 \div 3}{39 \div 3}=\frac{6}{13}\)

Question 32. Fractions  \(\frac{15}{39} \text { and } \frac{45}{117}\) are equivalent fractions.
Solution:  True

⇒\(\text { Since, } \frac{45 \div 3}{117 \div 3}=\frac{15}{39}\)

∴ Both are equivalent

Question 33. The sum of two fractions is always a fraction.
Solution: True

Let, \(\frac{2}{3} \text { and } \frac{4}{3}\) are two fractions.

When we add them, we get \(\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}\)

\(=\frac{6}{3}=\frac{2}{1} \text {, }\) which is a fraction.

Question 34. The result obtained by subtracting a fraction from another fraction is necessarily a fraction.
Solution:  False

Let \(\frac{1}{2} \text { and } \frac{2}{3}\) are two fractions.

When we subtract \(\frac{2}{3} \text { from } \frac{1}{2} \text {, }\)

⇒ \(\frac{1}{2}-\frac{2}{3}=\frac{3-4}{6}=\frac{-1}{6}\)

Which is not a fraction.

Question 35. If a whole or an object is divided into several equal parts, then each part represents a fraction.
Solution: True

Question 36. The fraction represented by the shaded portion in the adjoining figure is \(\frac{3}{8}.\)

Solution: True

The total number of parts in a given figure is 8 and the shaded parts are 3.

∴ The required fraction is \(\frac{3}{8}\)

Question 37. The fraction represented by the unshaded portion in the adjoining figure is\(\frac{5}{9}.\)

Solution: False

The total number of parts in a given figure is 9 of which unshaded parts are 4.

∴ The required fraction is \(\frac{4}{9}.\)

Question 38. \(\text { 38. } \frac{25}{19}+\frac{6}{19}=\frac{31}{38}\)
Solution: False

⇒ \(\frac{25}{19}+\frac{6}{19}=\frac{25+6}{19}=\frac{31}{19} \neq \frac{31}{38}\)

∴ The above condition is false.

Question 39. \(\frac{8}{18}-\frac{8}{15}=\frac{8}{3}\)
Solution: False

The L.C.M. of 18 and 15 is 2 * 3 x 3 x 5- 90

∴ \(\frac{8}{18}-\frac{8}{15}=\frac{8 \times 5}{90}-\frac{8 \times 6}{90}\)

⇒\(=\frac{40}{90}-\frac{48}{90}=-\frac{8}{90} \neq \frac{8}{3}\)

∴ The above condition is false.

Question 40. \(\frac{7}{12}+\frac{11}{12}=\frac{3}{2}\)
Solution: True

⇒ \(\frac{7}{12}+\frac{11}{12}=\frac{7+11}{12}=\frac{18}{12}=\frac{18 \div 6}{12 \div 6}=\frac{3}{2}\)

Question 41. \(\frac{16}{25}>\frac{13}{25}\)
Solution: True

The given fractions are like fractions. On comparing the numerators, we get

⇒ \(\frac{16}{25}>\frac{13}{25}\)

Question 42. \(\frac{11}{16} \ldots \frac{14}{15}\)
Solution:

< : The L.C.M. of 16 and 15 is 2x2x2x2x3x5 = 240

⇒ \(\text { Thus, } \frac{11}{16}=\frac{11 \times 15}{16 \times 15}=\frac{165}{240}\)

⇒ \(\text { and } \frac{14}{15}=\frac{14 \times 16}{15 \times 16}=\frac{224}{240}\)

On comparing, we observe that

∴ \(\frac{224}{240}>\frac{165}{240} \text {, i.e., } \frac{11}{16}<\frac{14}{15}\)

Question 43. \(\frac{8}{15} \ldots \frac{95}{14}\)
Solution:

< : The L.C.M. of 15 and 14 is 2 x 3 x 5 x 7 = 210

⇒ \(\text { Thus, } \frac{8}{15}=\frac{8 \times 14}{15 \times 14}=\frac{112}{210}\)

⇒ \(\text { and } \frac{95}{14}=\frac{95 \times 15}{14 \times 15}=\frac{1425}{210}\)

On comparing, we observed that

⇒ \(\frac{1425}{210}>\frac{112}{210} \text { i.e., } \frac{8}{15}<\frac{95}{14}\)

Question 44. \(\frac{12}{75} \ldots \frac{32}{200}\)
Solution:

= : The L.C.M. of 75 and 200 is 2x2x2x3x5x5 = 600

⇒ \(\text { Thus, } \frac{12}{75}=\frac{12 \times 8}{75 \times 8}=\frac{96}{600}\)

⇒ \(\text { and } \frac{32}{200}=\frac{32 \times 3}{200 \times 3}=\frac{96}{600}\)

On comparing, we observe that

⇒ \(\frac{12}{75}=\frac{32}{200}\)

Question 45. \(\frac{18}{15} \ldots 1.3\)
Solution:

< : \(1.3=\frac{13}{10}\)

The L.C.M. of 15 and 10 is 2 x 3 x 5 = 30

⇒ \(\text { Now, } \frac{18}{15}=\frac{18 \times 2}{15 \times 2}=\frac{36}{30} \text { and } \frac{13}{10}=\frac{13 \times 3}{10 \times 3}=\frac{39}{30}\)

⇒ \(\frac{36}{30}<\frac{39}{30}\)

∴ \(\text { Thus, } \frac{18}{15}<1.3\)

Question 46. Write the fraction represented by the shaded portion of the adjoining figure:

Solution:

In the given figure, the total parts in which the figure has been divided is 8 out of which 7 parts are shaded.

∴ The required fraction is \(\frac{7}{8}.\)

Question 47. Write the fraction represented by the unshaded portion of the adjoining figure:

Solution:

In the given figure, the total number of parts in which the 1444 figure has been divided is 15, out of which 4 are unshaded.

∴ The required fraction is \(\frac{4}{15}.\)

Question 48. Ali divided one fruit cake equally among six persons. What part of the cake he gave to each person?
Solution:  Since Ali has to divide one fruit cake equally among 6 persons

Each person will get \(\frac{1}{6}\) part.

Question 49. Express \(6 \frac{2}{3}\) as an improper fraction.
Solution:

We have \(6 \frac{2}{3}=\frac{(6 \times 3)+2}{3}=\frac{18+2}{3}=\frac{20}{3}\)

Question 50. Arrange the fractions \(\frac{2}{3}, \frac{3}{4}, \frac{1}{2} \text { and } \frac{5}{6} \text { in }\) in ascending order
Solution:

⇒ \(\text { We have, } 6 \frac{2}{3}=\frac{(6 \times 3)+2}{3}=\frac{18+2}{3}=\frac{20}{3}\)

Firstly find the L.C.M. of 3, 4, 2, and 6.

∴ The L.C.M. of 3, 4, 2 and 6 is 2 x 2 x 3 = 12

⇒ \(\text { Now, } \frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}, \frac{3}{4}=\frac{3 \times 3}{4 \times 3}=\frac{9}{12} \text {, }\)

⇒ \(\frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12} \text { and } \frac{5}{6}=\frac{5 \times 2}{6 \times 2}=\frac{10}{12}\)

On comparing the above, we get

⇒ \(\frac{6}{12}<\frac{8}{12}<\frac{9}{12}<\frac{10}{12} \text { i.e., } \frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{5}{6}\)

Question 51. Arrange the fractions \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4} \text { in }\) descending order.
Solution:

⇒ We have given, \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4} \text {. }\)

The L.C.M. of 7, 8, 5 and 4 is 2x2x2x5x7 = 280

⇒ \(\text { Now, } \frac{6}{7}=\frac{6 \times 40}{7 \times 40}=\frac{240}{280}\)

⇒  \(\frac{7}{8}=\frac{7 \times 35}{8 \times 35}=\frac{245}{280}\)

⇒  \(\frac{4}{5}=\frac{4 \times 56}{5 \times 56}=\frac{224}{280} \text { and } \frac{3}{4}=\frac{3 \times 70}{4 \times 70}=\frac{210}{280}\)

On comparing the above, we get;

⇒  \(\frac{245}{280}>\frac{240}{280}>\frac{224}{280}>\frac{210}{280}\)

⇒ \(\text { i.e., } \frac{7}{8}>\frac{6}{7}>\frac{4}{5}>\frac{3}{4}\)

Question 52. Write \(\frac{3}{4}\) as a fraction with denominator 44.
Solution:

⇒ \(\text { Let } \frac{3}{4}=\frac{?}{44}\)

Then, we have to find the missing numeral. To get 44 in the denominator, we multiply by 4 by 11.

So, we multiply the numerator and denominator by 11.

∴ \(\frac{3}{4}=\frac{3 \times 11}{4 \times 11}=\frac{33}{44}\)

\(\text { Hence, } \frac{3}{4} \text { and } \frac{33}{44}\) are equivalent fractions.

Question 53. Write \(\frac{5}{6}\) as a fraction with the numerator 60.
Solution:

Let, \(\frac{5}{6}=\frac{60}{?}\)

Then, we have to find the missing numeral. To get 60 in the numerator, we multiply 5 by

So, we multiply the numerator and denominator by 12.

∴ \(\frac{5 \times 12}{6 \times 12}=\frac{60}{72}\)

⇒ \(\text { Hence, } \frac{5}{6} \text { and } \frac{60}{72}\) are equivalent fractions.

Question 54. Write \(\frac{129}{8}\) as a mixed fraction
Solution:

We have, \(\frac{129}{8}=16 \frac{1}{8}\)

Question 55. Add the fractions \(\frac{3}{8} \text { and } \frac{2}{3} \text {. }\)
Solution:

L.C.M. 0f 8 and 3 is 2x2x2x3 = 24

Now, \(\frac{3}{8}+\frac{2}{3}=\frac{3 \times 3}{8 \times 3}+\frac{2 \times 8}{3 \times 8}\)

∴  \(=\frac{9}{24}+\frac{16}{24}=\frac{9+16}{24}=\frac{25}{24}\)

Question 56. Add the fractions \(\frac{3}{8} \text { and } 6 \frac{3}{4} \text {. }\)
Solution:

L.C.M. of 8 and 4 is 2 x 2 x 2 = 8

Now, \(\frac{3}{8}+6 \frac{3}{4}=\frac{3}{8}+\frac{27}{4}\)

⇒  \(\frac{3}{8}+\frac{27 \times 2}{4 \times 2}\)

⇒  \(\frac{3}{8}+\frac{54}{8}=\frac{3+54}{8}=\frac{57}{8}=7 \frac{1}{8}\)

Question 57. Subtract \(\frac{1}{6} \text { from } \frac{1}{2} \text {. }\)
Solution:

The L.C.M. of 6 and 2 = 6

Now, \(\frac{1}{2}-\frac{1}{6}=\frac{1 \times 3}{2 \times 3}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}\)

⇒ \(\frac{3-1}{6}=\frac{2}{6}=\frac{2 \div 2}{6 \div 2}\)

[H.C.F. of 2 and 6 is 2]

⇒ \(\frac{1}{3}\)

Question 58. Subtract \(8 \frac{1}{3} \text { from } \frac{100}{9}\).
Solution:

The L.C.M. of 3 and 9 = 9

⇒ \(\frac{100}{9}-8 \frac{1}{3}=\frac{100}{9}-\frac{25}{3}\)

⇒ \(=\frac{100}{9}-\frac{25 \times 3}{3 \times 3}=\frac{100}{9}-\frac{75}{9}=\frac{100-75}{9}=\frac{25}{9}\)

⇒ \(2 \frac{7}{9}\)

Question 59. Subtract \(1 \frac{1}{4} \text { from } 6 \frac{1}{2}\)
Solution:

The L.C.M. of 4 and 2 = 4

Now, \(6 \frac{1}{2}-1 \frac{1}{4}=\frac{13}{2}-\frac{5}{4}\)

⇒ \(\frac{13 \times 2}{2 \times 2}-\frac{5}{4}=\frac{26}{4}-\frac{5}{4}=\frac{26-5}{4}\)

⇒ \(\frac{21}{4}=5 \frac{1}{4}\)

Question 60. Add \(1 \frac{1}{4} \text { and } 6 \frac{1}{2}\)
Solution:

The L.C.M. of 4 and 2 = 4

⇒ \(\text { Now, } 1 \frac{1}{4}+6 \frac{1}{2}=\frac{5}{4}+\frac{13}{2}\)

⇒ \(\frac{5}{4}+\frac{13 \times 2}{2 \times 2}=\frac{5}{4}+\frac{26}{4}=\frac{5+26}{4}=\frac{31}{4}\)

⇒ \(7 \frac{3}{4}\)

⇒ \(\begin{array}{l|l}
2 & 2,4 \\
\hline 2 & 1,2 \\
\hline & 1,1
\end{array}\)

Question 61. Katrina rode her bicycle \(6 \frac{1}{2}\) km in the morning and \(8 \frac{3}{4}\) km in the evening. Find the distance traveled by her altogether on that day.
Solution:

Given

Katrina rode her bicycle \(6 \frac{1}{2}\) km in the morning and \(8 \frac{3}{4}\) km in the evening.

The distance travelled by Katrina in the morning  \(=6 \frac{1}{2} \mathrm{~km}=\frac{13}{2} \mathrm{~km}\)

The distance traveled by Katrina in the evening \(=8 \frac{3}{4} \mathrm{~km}=\frac{35}{4} \mathrm{~km}\)

Total =distance traveled by her

⇒ \(\frac{13}{2} \mathrm{~km}+\frac{35}{4} \mathrm{~km}\)

⇒ \(\frac{13 \times 2}{2 \times 2} \mathrm{~km}+\frac{35}{4} \mathrm{~km}\)

∴ [ L.C.M. of 2 and 4 = 4]

⇒ \(\frac{26}{4} \mathrm{~km}+\frac{35}{4} \mathrm{~km}=\frac{(26+35) \mathrm{km}}{4}\)

⇒ \(\frac{61}{4} \mathrm{~km}=15 \frac{1}{4} \mathrm{~km}\)

Question 62. A rectangle is divided Into a certain number of equal parts. If 16 of the parts so formed represent the fraction \(\frac{1}{4}\) find the number of parts in which the rectangle has been divided.
Solution:

Given

A rectangle is divided Into a certain number of equal parts. If 16 of the parts so formed represent the fraction \(\frac{1}{4}\)

Let the number of parts in which the rectangle has been divided be x.

According to the question, \(\frac{1}{4}=\frac{16}{x}\)

By cross-multiplication, x = 16 Χ 4 = 64

∴ The required number of parts is 64.

Question 63. The grip size of a tennis racquet is \(11 \frac{9}{80} \mathrm{~cm}\). Express the size as an improper fraction.
Solution:

Given

The grip size of a tennis racquet is \(11 \frac{9}{80} \mathrm{~cm}\).

We have given, you the grip size of a tennis racquet \( =11 \frac{9}{80} \mathrm{~cm}=\frac{889}{80} \mathrm{~cm}\), which is the required improper fraction.

Question 64. On average \(\frac{1}{10}\) the food eaten is turned into an organism’s own body and is available for the next level of consumer in a food chain. What fraction of the food eaten is not available for the next level?
Solution:

Given

On average \(\frac{1}{10}\) the food eaten is turned into an organism’s own body and is available for the next level of consumer in a food chain.

We have given, \(\frac{1}{10}\) of the food eaten is turned into the organism’s own body.

The required fraction of the food eaten not available for the next level is

⇒ \(1-\frac{1}{10}=\frac{10-1}{10}\)

[ L.C.M. of1 and 10 is 10]

⇒ \(\frac{9}{10}\)

Question 65. Mr. Rajan got a job at the age of 24 years and he got retired from the job at the age of 60 years. What fraction of his age till retirement was he in the job?
Solution:

Given

Mr. Rajan got a job at the age of 24 years and he got retired from the job at the age of 60 years.

Mr. Rajan got a job at the age of 24 years.

He retired at the age of 60 years.

He worked for (60- 24) years = 36 years

∴ The required fraction \(=\frac{36}{60}=\frac{36 \div 12}{60 \div 12}=\frac{3}{5}\)

[ H.C.F. of 36 and 60 is 12]

Question 66. The food we eat remains in the stomach for a maximum of 4 hours. For what fraction of a day, does it remain there? 
Solution:

Given

The food we eat remains in the stomach for a maximum of 4 hours.

Total number of hours in a day 24 hours

The required fraction \(=\frac{4}{24}= \div {4+4}{24+4}= \div {1}{6}\)

[ H.C.F. of 4 and 24 is 4]

Question 67. It was estimated that because of people switching to Metro trains, about 33000 tonnes of CNG, 3300 tonnes of diesel, and 21000 tonnes of petrol were saved by the end of the year 2007. Find the fraction of:

1. The quantity of diesel saved to the quantity of petrol saved.
2. The quantity of diesel saved to the quantity of CNG saved.

Solution:

Given

It was estimated that because of people switching to Metro trains, about 33000 tonnes of CNG, 3300 tonnes of diesel, and 21000 tonnes of petrol were saved by the end of the year 2007.

By the end of the year 2007,

The quantity of CNG saved 33000 tonnes

The quantity of diesel saved 3300 tonnes and

The quantity of petrol saved 21000 tonnes

⇒ \(\frac{\text { The quantity of diesel saved }}{\text { The quantity of petrol saved }}=\frac{3300}{21000}\)

⇒ \(\frac{3300 \div 300}{21000 \div 300}\)

[H.C.F. of 3300 and 21000 is 300]

\(=\frac{11}{70}\)

⇒ \(\frac{\text { The quantity of diesel saved }}{\text { The quantity of CNG saved }}=\frac{3300}{33000}\)

⇒ \(\frac{3300 \div 3300}{33000 \div 3300}\)

[H.C.F. of 3300 and 33000 is 3300]

∴ \(=\frac{1}{10}\)

Question 68. A cup is \(\frac{1}{3}\) full of milk. What part of the cup is still to be filled with milk to make it full?
Solution:

Given

A cup is \(\frac{1}{3}\) full of milk.

The remaining part of the cup which is still to be filled with milk  \(=1-\frac{1}{3}\)

⇒ \(\frac{3-1}{3}\)

⇒  \(\frac{2}{3}\)

Question 69. Mary bought \(3 \frac{1}{2} \mathrm{~m}\) of lace. She used \(1 \frac{3}{4} \mathrm{~m}\) of Hires lace for her new dress. How much lace is left with her?
Solution:

Given

Mary bought the lace = \(3 \frac{1}{2} \mathrm{~m}\) = \(\frac{7}{2} \mathrm{~m}\)

Lace used by Mary \(=1 \frac{3}{4} \mathrm{~m}=\frac{7}{4} \mathrm{~m}\)

She is left with \(\frac{7}{2} m-\frac{7}{4} m\)

⇒\(\frac{7 \times 2}{2 \times 2} m-\frac{7}{4} m=\frac{14 m-7 m}{4}\)

[L.C.M. of 2 and 4 is 4]

⇒ \(\frac{7}{4} m=1 \frac{3}{4} m \text { of lace }\)

Question 70. When Sunita weighed herself on Monday, she found that she had gained \( \frac{1}{4} \mathrm{~kg}\) Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}\)  What was her weight on Monday?
Solution:

Given

Sunita weighed herself on Monday, she found that she had gained \( \frac{1}{4} \mathrm{~kg}\) Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}\)

Sunita had gained \(=1 \frac{1}{4} \mathrm{~kg}=\frac{5}{4} \mathrm{~kg}\)

Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}=\frac{371}{8} \mathrm{~kg}\)

Her total weight on Monday

⇒ \(\frac{371}{8} \mathrm{~kg}+\frac{5}{4} \mathrm{~kg}=\frac{371}{8} \mathrm{~kg}+\frac{5 \times 2}{4 \times 2} \mathrm{~kg}\)

[L.C.M. of 8 and 4 is 8]

⇒ \(\frac{371}{8} \mathrm{~kg}+\frac{10}{8} \mathrm{~kg}\)

⇒ \(\frac{(371+10)}{8} \mathrm{~kg}=\frac{381}{8} \mathrm{~kg}=47 \frac{5}{8} \mathrm{~kg}\)

Question 71. Sunil purchased \(12 \frac{1}{2}\)liters of juice on Monday and \(14 \frac{3}{4}\) liters of juice on Tuesday. How many liters of juice did he purchase together in two days?
Solution:

Given

Sunil purchased \(12 \frac{1}{2}\)liters of juice on Monday and \(14 \frac{3}{4}\) liters of juice on Tuesday.

Sunil purchased juice on Monday = \(12 \frac{1}{2}\)liters \(=\frac{25}{2} \text { litres }\)

and on Tuesday \(14 \frac{3}{4}\)liters  \(=\frac{59}{4} \text { litres }\)

The total quantity of juice Sunil purchased in two days  \(=\left(\frac{25}{2}+\frac{59}{4}\right) \text {liters}\)

⇒ \(=\left(\frac{25 \times 2}{2 \times 2}+\frac{59}{4}\right) \text { litres }=\left(\frac{50}{4}+\frac{59}{4}\right) \text { litres }\)

⇒ \(=\left(\frac{50+59}{4}\right) \text { litres }=\frac{109}{4} \text { litres }=27 \frac{1}{4} \text { litres }\)

Question 72. Nazima gave \(2 \frac{3}{4}\) liters out of the \(5 \frac{1}{2}\) liters of juice she purchased for her friends. How many liters of juice is left with her?
Solution:

Given

Nazima gave \(2 \frac{3}{4}\) liters out of the \(5 \frac{1}{2}\) liters of juice she purchased for her friends.

Total quantity of juice = \(5 \frac{1}{2}\)liters

= \(\frac{11}{2} \text { litres }\)

Nazima gave to her friends \(=2 \frac{3}{4} \text { litres }\)

⇒ \(\frac{11}{4} \text { litres }\)

The required quantity of juice she is left \(=\left(\frac{11}{2}-\frac{11}{4}\right) \text { litres }=\left(\frac{11 \times 2}{2 \times 2}-\frac{11}{4}\right) \text { litres }\)

⇒ \(\left(\frac{22-11}{4}\right) \text { litres }=\frac{11}{4} \text { litres }=2 \frac{3}{4} \text { litres }\)

Question 73. Roma gave a wooden board of length \(150 \frac{1}{4} \mathrm{~cm}\) to a carpenter for making a shelf. The Carpenter sawed off a piece of \(40 \frac{1}{5} \mathrm{~cm}\) cm from it. What is the length of the remaining piece?
Solution:

Given

Roma gave a wooden board of length \(150 \frac{1}{4} \mathrm{~cm}\) to a carpenter for making a shelf. The Carpenter sawed off a piece of \(40 \frac{1}{5} \mathrm{~cm}\) cm from it.

Total length of a wooden board \(=150 \frac{1}{4} \mathrm{~cm}=\frac{601}{4} \mathrm{~cm}\)

The carpenter sawed off a piece of length \(=40 \frac{1}{5} \mathrm{~cm}=\frac{201}{5} \mathrm{~cm}\)

The length of the remaining piece \(=\left(\frac{601}{4}-\frac{201}{5}\right) \mathrm{cm}=\left(\frac{601 \times 5}{4 \times 5}-\frac{201 \times 4}{5 \times 4}\right) \mathrm{cm}\)

⇒\(\left(\frac{3005}{20}-\frac{804}{20}\right) \mathrm{cm}=\frac{2201}{20} \mathrm{~cm} \text { or } 110 \frac{1}{20} \mathrm{~cm}\)

Question 74. Naslr travelled \(3 \frac{1}{2} \mathrm{~km}\) in a bus and then walked \(1 \frac{1}{8} \mathrm{~km}\) to reach a town. How much did he travel to reach the town?
Answer:

Given

Naslr travelled \(3 \frac{1}{2} \mathrm{~km}\) in a bus and then walked \(1 \frac{1}{8} \mathrm{~km}\) to reach a town.

Nasir traveled by a bus = \(3 \frac{1}{2} \mathrm{~km}\)

⇒ \(\frac{7}{2} \mathrm{~km}\)

Nasir walked \(=1 \frac{1}{8} \mathrm{~km}=\frac{9}{8} \mathrm{~km}\)

Total distance traveled by him latex]=\frac{7}{2} \mathrm{~km}+\frac{9}{8} \mathrm{~km}=\frac{7 \times 4}{2 \times 4} \mathrm{~km}+\frac{9}{8} \mathrm{~km}[/latex]

⇒ \(\left(\frac{28}{8}+\frac{9}{8}\right) \mathrm{km}\)

∴ \(\left(\frac{28+9}{8}\right) \mathrm{km}=\frac{37}{8} \mathrm{~km} \text { or } 4 \frac{5}{8} \mathrm{~km}\)

Question 75. The fish caught by Neetu was of weight \(3 \frac{3}{4} \mathrm{~kg}\) kg and the fish caught by Narendra was of weight 2 \frac{1}{2} \mathrm{~kg} kg. How much more did Neetu’s fish weigh than that of Narendra?
Solution:

Given

The fish caught by Neetu was of weight \(3 \frac{3}{4} \mathrm{~kg}\) kg and the fish caught by Narendra was of weight 2 \frac{1}{2} \mathrm{~kg} kg.

The weight of fish caught by Neetu \(=3 \frac{3}{4} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}\)

The weight of fish caught by Narendra \(=2 \frac{1}{2} \mathrm{~kg}=\frac{5}{2} \mathrm{~kg}\)

∴ Neetu’s fish weigh more than that of Narendra by \(\frac{15}{4} \mathrm{~kg}-\frac{5}{2} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}-\frac{5 \times 2}{2 \times 2} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}-\frac{10}{4} \mathrm{~kg}\)

⇒ \(\frac{(15-10)}{4} \mathrm{~kg}=\frac{5}{4} \mathrm{~kg} \text { or } 1 \frac{1}{4} \mathrm{~kg}\)

Question 76. Neelam’s father needs \(1 \frac{3}{4} \mathrm{~m}\) of cloth for the skirt of Neelam’s new dress and \(\frac{1}{2} m\) for the scarf. How much cloth must he buy in all?
Solution:

Given

Neelam’s father needs \(1 \frac{3}{4} \mathrm{~m}\) of cloth for the skirt of Neelam’s new dress and \(\frac{1}{2} m\) for the scarf.

Neelam’s father purchased the length of cloth for the skirt \(=1 \frac{3}{4} \mathrm{~m}=\frac{7}{4} \mathrm{~m}\) and for the scarf \(=\frac{1}{2} \mathrm{~m}\)

Total length he buys in all = \(=\frac{7}{4} m+\frac{1}{2} m\)

⇒ \(\frac{7}{4} m+\frac{1 \times 2}{2 \times 2} m=\frac{7}{4} m+\frac{2}{4} m\)

⇒ \(\frac{(7+2)}{4} m=\frac{9}{4} m \text { or } 2 \frac{1}{4} m\)

Question 77. What is wrong with the following additions?

Solution:

1. Equal denominators are added.
2. Numerators and denominators are added

Question 78. Match the fractions of Column I with the shaded or marked portion of figures of Column II:

Solution:

1→(D); 2→(A); 3→(E); 4→(B);

Marked portion in (A) \(=\frac{6}{10}\)

Shaded fraction in (B) \(=\frac{6}{16}\)

Shaded fraction in (C) \(=\frac{6}{7}\)

Shaded fraction in (D) \(=\frac{4}{4}+\frac{2}{4}=\frac{6}{4}\)

Shaded fraction in (E) \(=\frac{6}{6}\)

Question 79. Find the fraction that represents the number of natural numbers to total numbers in the collection 0, 1, 2, 3, 4, 5. What fraction will it be for whole numbers?
Solution:

Out of 0, 1, 2, 3, 4, 5 → 1, 2, 3, 4 and 5 are the natural numbers.

The fraction that represents the number of natural numbers to the total numbers \(=\frac{5}{6}\) and the whole numbers are 0, 1, 2, 3, 4, and 5.

∴ The fraction that represents the number of whole numbers to the total numbers \(=\frac{6}{6}\)

Question 80. Write the fraction representing the total number of natural numbers in the collection of numbers -3, -2, -1, 0, 1, 2, 3. What fraction will it be for whole numbers? What fraction will it be for integers?
Solution:

Out of -3, -2, -1, 0, 1, 2, 3 → 1, 2, and 3 are the natural numbers, 0, 1, 2, and 3 are the whole numbers and -3, -2, -l, 0, 1, 2, 3 are integers.

∴ The fraction representing the natural numbers to the total numbers \(=\frac{3}{7}\)

The fraction representing the whole numbers to the total numbers \(=\frac{4}{7}\)

The required fraction representing the integers to the total numbers  \(=\frac{7}{7}\)

Question 81. Write a pair of fractions whose sum is \(\frac{7}{11}\) and the difference is \(\frac{2}{11}\)
Solution: Let one fraction be x

Another fraction be \(\frac{7}{11}-x\)

Now, according to the question,

⇒ \(x-\left(\frac{7}{11}-x\right)=\frac{2}{11}\)

⇒ \(\Rightarrow x-\frac{7}{11}+x=\frac{2}{11}\)

⇒ \(\frac{11 x}{11}-\frac{7}{11}+\frac{11 x}{11}=\frac{2}{11}\)

⇒ \(\frac{11 x-7+11 x}{11}=\frac{2}{11}\)

⇒ \(22 x-7=\frac{2}{11} \times 11\)

⇒ \(22 x-7=2 \Rightarrow 22 x=2+7=9\)

⇒ \(x=\frac{9}{22}\)

Thus, one fraction is \(\frac{9}{22}\) and another fraction is \(\frac{7}{11}-\frac{9}{22}=\frac{7 \times 2}{11 \times 2}-\frac{9}{22}=\frac{14-9}{22}=\frac{5}{22}\)

Question 82. What fraction of a straight angle is a right angle?
Solution: Since we know that the measurement of a straight angle is 180° and a right angle is 90°

∴ The required fraction is \(\frac{90^{\circ}}{180^{\circ}}=\frac{1}{2} .\)

Question 83. Put the right card in the right bag.

Solution:

We know that if the numerator is less than the denominator, then the fraction is less than 1.

If the numerator is equal to the denominator, then the fraction is equal to 1 and if the numerator is greater than the denominator, then the fraction is greater than 1.

The cards in Bag 1 are

1.  \(\frac{3}{7},\)

4. \(\frac{8}{9}\)

5.  \(\frac{5}{6},\)

6. \(\frac{6}{11},\)

8. \(\frac{19}{25}\)

9. \(\frac{2}{3}\) and

10. \(\frac{13}{17}\)

The cards in Bag 2 are

2.  \(\frac{4}{4}\)

7. \(\frac{18}{18}\)

And cards in B ag 3 are

3. \(\frac{9}{8} .\)

Class 6 Maths Chapter 10 Mensuration

Question 1. The following figures are formed by joining six unit squares. Which figure has the smallest perimeter?

1. 2
2. 3
3. 4
4. 1

Solution: (4): Perimeter of the figure (1) = 10 units

Perimeter of figure (2) = 12 units

Perimeter of figure (3) = 14 units

Perimeter of figure (4) = 14 units

So, figure (1) has the smallest perimeter

Read and Learn More Class 6 Maths Exemplar Solutions

Question 2. A square-shaped park ABCD of side 100 m has two rectangular flower beds each of size 10 m x 5 m (see figure). The length of the boundary of the remaining park is

1. 360 m
2. 400 m
3. 340 m
4. 460 m

Solution: (2): Length of required boundary

= Perimeter of the remaining park

= 90m + 95m + 10m + 5m + 90m + 95m + 10m + 5m

= 400 m

Question 3. The side of a square is 1 0 cm. How many times will the new perimeter become if the side of the square is doubled?

1. 2 times
2. 4 times
3. 6 times
4. 8 times

Solution: (1): Side of square = 10 cm

Perimeter of square = 4 x 10 cm = 40 cm

Now, new side of square = 2 x 10 cm = 20 cm

Perimeter of new square = 4 x 20 cm = 80 cm = 2 x 40 cm …(ii)

(1) and (2) show that the new perimeter will be 2 times the perimeter of a given square.

Question 4. The length and breadth of a rectangular sheet column II: of paper are 20 cm and 10 cm, respectively. A rectangular piece is cut from the sheet as shown in the figure. Which of the following statements is correct for the remaining sheet?

1. The perimeter remains the same but the area changes.
2. The area remains the same but the perimeter changes.
3. Both area and perimeter are changing.
4. Both the area and perimeter remain the same.

Solution: (1): The given figure shows that the perimeter will remain the same but the area changes

Question 5. Two regular hexagons of perimeter 30 cm each are joined as shown in the figure. The perimeter of the new figure is

1. 65 cm
2. 60 cm
3. 55 cm
4. 50 cm

Solution: (4): Perimeter of a regular hexagon = 6 x side

30 cm= 6 x side => side \(\frac{30}{6} \mathrm{~cm}=5 \mathrm{~cm}\)

Each side of a regular hexagon = 5 cm

The perimeter of the new figure = 10 x side- 10 x 5 cm = 50 cm

Question 6. In the given figure, which of the following is a regular polygon? All have equal sides except (i)

1. 1
2. 2
3. 3
4. 4

Solution: (2): Since we know that in a regular polygon, all angles and all sides are equal. Thus figure (2) satisfies the above condition.

Question 7. Match the shapes (each side measures 2 cm) in column 1 with the corresponding perimeters in column 2.

Solution: (A) → (4); (B) → (1); (C) → (2); (D) → (3)

1.  The given figure has 14 sides.

Each side measures 2 cm.

∴ The perimeter of the given figure = 14 x 2 cm = 28 cm

2.  The given figure has 8 sides.

∴ The perimeter of the given figure = 8 x 2 cm = 16 cm

3. The given figure has 10 sides.

∴ The perimeter of the given figure = 10 x 2 cm = 20 cm

4.  The given figure has 12 sides.

∴  The perimeter of the given figure = 12 x 2 cm = 24 cm

Question 8. Match the following:

Solution: (A) → (3); (B) → (3); (C) → (2); (D) → (1)

1.  The perimeter of a rectangle

= 2 (length + breadth)

= 2(6 + 4) = 2(10) = 20

2. Perimeter of a square = 4 x side

= 4×5 = 20

3. The perimeter of an equilateral triangle

= 3 x side = 3×6 = 18

4. Perimeter of an isosceles triangle

= Sum of all sides = 4 + 2 + 4 = 10

Question 9. The perimeter of the shaded portion in the given figure is

AB+ _ + _ + _ + _ + _ + _ + HA

Solution:

EM, MD, DE, EK NG, GH: Perimeter of the shaded portion of the given figure

= AB + BM +MD + DE + EN + NG + GH +HA

Question 10. The amount of region enclosed by a plane closed figure is called its____.
Solution: Area

11. The area of a rectangle with a length of 5 cm and breadth of 3 cm is____.
Solution:

15 cm2: Area of a rectangle

= length x breadth = (5×3) cm2= 15 cm2

Question 12. A rectangle and a square have the same perimeter (see figure).

1. The area of the rectangle is____.
   The area of the square is____.

Solution:

12 sq units: Area of a rectangle

= length x breadth

= (6×2) sq units = 12 sq units

16 sq units: We have given,

Perimeter of a rectangle = Perimeter of a square

⇒  2(length + breadth) = 4 x side

⇒  2(6 + 2) = 4 x side ⇒  2(8) = 4 x side

⇒  \(\frac{16}{4}=\text { side }\)

Side = 4 units

∴ The area of square = side x side = (4×4) sq units = 16 sq units

Question 13.

1. 1 m = ____ cm.
2. 1 sq cm = ____ cm x 1 cm.
3. 1 sq m = 1 m x ____ m = 100 cm x ____ cm.
4. 1 sq m = ____ sq cm

Solution:

1. 100: 1 ni = 100 cm
2. 1 : 1 sq cm =1 cm x 1 cm
3. 1 and 100: 1 sq m =1 m x1 m = 100 cm x 100 cm
4. 10000: 1 sq m = 10000 sq cm

Question 14. If the length of a rectangle is halved and the breadth is doubled then the area of the rectangle obtained remains the same.
Solution: True

Let l and b are the dimensions of a given rectangle, whereas l’ and b’ are the dimensions of a new rectangle.

According to question, \(l^{\prime}=\frac{1}{2} l \text { and } b^{\prime}=2 b\)

Area of new rectangle = \(l^{\prime} b^{\prime}=\frac{1}{2} l \times 2 b=l b \text {, }\)

which is the area of the given rectangle

Question 15. The area of the square is doubled if the side of the square is doubled.
Solution: False

Let the side of a given square be a.

∴ Area of the square = a2

The side of a new square = 2a

∴ Area of the new square = (2a)2 = 4a2

which concludes that the area of the new square is four times the area of the given square.

Question 16. The perimeter of a regular octagon of side 6 cm is 36 cm.
Solution: False

The side of a regular octagon = 6 cm

The perimeter of a regular octagon = 8 x side

= 8x6cm = 48cm

Question 17. A farmer who wants to fence his field must find the perimeter of the field.
Solution: True

Question 18. An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
Solution: False

To build a compound wall on all sides of a house, the engineer must find the perimeter of the compound

Question 19. To find the cost of painting a wall we need to find the perimeter of the wall.
Solution: False

To find the cost of painting a wall we need to find the area of the wall.

Question 20. To find the cost of a frame of a picture, we need to find the perimeter of the picture.
Solution: True

Question 21. Four regular hexagons are drawn to form the design as shown in the figure. If the perimeter of the design is 28 cm, find the length of each side of the hexagon.

Solution:

The perimeter of the design = 28 cm

⇒  14 x side = 28 cm

⇒  \(\text { side }=\frac{28}{14} \mathrm{~cm}=2 \mathrm{~cm}\)

Question 22. The perimeter of an isosceles triangle is 50 cm. If one of the two equal sides is 18 cm, find the third side.
Solution:

Given

The perimeter of an isosceles triangle is 50 cm. If one of the two equal sides is 18 cm

Let AABC be the given isosceles triangle, where AB = CA = 18 cm.

Perimeter of the AABC = 50 cm

AB + BC + CA = 50 cm A

⇒  18 cm + BC + 18 cm = 50 cm

⇒  36 cm + BC = 50 cm

⇒  BC = 50 cm- 36 cm

⇒  BC = 14 cm

The third side BC = 14 cm

Question 23. The length of a rectangle is three times its breadth. The perimeter of the rectangle is 40 cm. Find its length and width.
Solution:

Given

The length of a rectangle is three times its breadth. The perimeter of the rectangle is 40 cm.

Let l and b be the length and breadth respectively of the given rectangle.

According to question,l = 3b

Perimeter of the rectangle = 2(1 + b)

⇒  40 cm = 2(3b + b)

⇒  40 cm = 2 x 4b => 40 cm = 8b

⇒  \(\frac{40}{8} \mathrm{~cm}=b \Rightarrow b=5 \mathrm{~cm}\)

The required length = 3x5cm = 15cm and breadth = 5 cm

Question 24. There is a rectangular lawn 10 m long and 4 m wide in front of Meena’s house (see figure). It is fenced along the two smaller sides and one longer side leaving a gap of 1 m for the entrance. Find the length of the fencing.

Solution:

Given

There is a rectangular lawn 10 m long and 4 m wide in front of Meena’s house (see figure). It is fenced along the two smaller sides and one longer side leaving a gap of 1 m for the entrance.

Length of rectangular lawn = 10 m

And the breadth of rectangular lawn = 4 m

The length of fencing along the two smaller

sides of lawn = 4m + 4m = 8 m

Andthelengthoffencingalongthelongerside of lawn leaving a gap of1 m for the entrance =10m-lm = 9m

Total length of fencing=8m +9m=17m

Question 25. The region given in the figure is measured by taking it as a unit. What is the area of the region?

Solution:

Given

he region given in the figure is measured by taking it as a unit.

Since, in the given figure there are 13 rectangular shapes.

Area of the region = 13 sq units.

Question 26. Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was 140 cm. Find the side of this field in metres.
Solution:

Given

Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was 140 cm.

The perimeter of a square field = the distance around a square field

⇒ 4 x side = length of 200 rods

⇒ 4 x side = 200 x 140 cm

[ The length of the rod = 140 cm]

Side = \(\frac{200 \times 140}{4} \mathrm{~cm}\)

⇒ Side = 7000 cm \(=\frac{7000}{100} \mathrm{~m}=70 \mathrm{~m}\)

∴ The side of the field = 70 m

Question 27. The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6 km. What is the length of the field?
Solution:

Given

The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6 km.

We have given,

The length of a rectangular field = 2 (breadth)

The perimeter of field = 2(length + breadth)

Jamal jogged around it 4 times and covered a distance of 6 km.

4[2(length + breadth)] = 6 km

⇒  8(length + breadth) = 6 x 1000 m

(2 breadth + breadth) \(\frac{6 \times 1000}{8} \mathrm{~m}\)

⇒  3 breadth = 3 x 250 m

⇒  breadth = \(\frac{3 \times 250}{3} \mathrm{~m}=250 \mathrm{~m}\)

∴ Length of the field = 2 x 250 m = 500 m

Question 28. Three squares are joined together as shown in the figure. Their sides are 4 cm, 10 cm and 3 cm. Find the perimeter of the figure.

Solution:

Given

Three squares are joined together as shown in the figure. Their sides are 4 cm, 10 cm and 3 cm.

We have

The perimeter of the given figure

= (10 + 3 + 3 + 3 + 7 + 10 + 6 + 4 + 4 + 4) cm

= 54 cm

Question 29. In the given figure all triangles are equilateral and AB = 8 units. Other triangles have been formed by taking the midpoints of the sides. What is the perimeter of the figure?

Solution:

Given

In the given figure all triangles are equilateral and AB = 8 units. Other triangles have been formed by taking the midpoints of the sides.

We have given, all the triangles are equilateral and AB = 8 units

Now, the perimeter of the given figure

= AE + EF + FG + GH +HI+ID + DC + CK + KN + NO + OM +ML + LJ + JB + BU + US + SR +RT+TQ + QP + PA

=(4+2+1 +1+1+2+4+4+2+1+1 +1+2 + 4 + 4 + 2 +1 +1 +1 + 2 + 4) units

= 45 units

The perimeter of the figure = 45 units

Question 30. The length of a rectangular field is 250 m and the width is 150 m. Anuradha runs around this field 3 times. How far did she run? How many times she should run around the field to cover a distance of 4 km?
Solution:

Given

The length of a rectangular field is 250 m and the width is 150 m. Anuradha runs around this field 3 times.

Let ABCD be the given rectangular field where length = AB = 250 m and breadth = AD = 150 m

The perimeter of the rectangle ABCD

= 2(AB+AD)

= 2(250 + 150) m

= 2(400) m = 800 m

Since Anuradha runs around the field 3 times.

∴ She covers a distance of 3 x 800 m = 2400 m = 2 km 400 m [ 1 km = 1000 m]

Now, the number of times she will run to cover the distance of 4 km or 4000 m \(=\frac{4000}{800}=5\)

Question 31. Bajinder runs ten times around a square track and covers 4 km. Find the length of the track.
Solution:

Given

Bajinder runs ten times around a square track and covers 4 km.

Let a be the side of the square track.

The perimeter of the square track = 4a

Bajinder runs ten times around the square track and covers 4 km.

10 x 4a = 4 km

40a = 4 x 1000 m

⇒ \(a=\frac{4 \times 1000}{40} \mathrm{~m}=100 \mathrm{~m}\)

∴ Length of the track = 4a = 4 x 100 m = 400 m

Question 32. The lawn in front of Molly’s house Is 12 m x 8 m, whereas the lawn in front of Dolly’s house is 1 5 m x 5 m. A bamboo fencing is built around both the lawns. How much fencing is required for both?
Solution:

Given

The lawn in front of Molly’s house Is 12 m x 8 m, whereas the lawn in front of Dolly’s house is 1 5 m x 5 m. A bamboo fencing is built around both the lawns.

The perimeter of the lawn in front of Molly’s house= 2(length + breadth) = 2(12 + 8) m = 2(20) m = 40 m

The perimeter of the lawn in front of Dolly’s house = 2(length + breadth) = 2(15 + 5) m = 2 x 20 m = 40 m

∴ Total length of fencing for both the lawns = (40 + 40) m = 80 m

Question 33. The perimeter of a regular pentagon is 1540 cm. How long is it on each side?
Solution:

The perimeter of the regular pentagon = 1540 cm

⇒ 5 x side = 1540 cm

⇒ Side = \(\frac{1540}{5} \mathrm{~cm}=308 \mathrm{~cm}\)

Each side of the regular pentagon is 308 cm.

Question 34. The perimeter of a triangle is 28 cm. One of its sides is 8 cm. Write all the sides of the possible isosceles triangles with these measurements.
Solution:

Given

The perimeter of a triangle is 28 cm. One of its sides is 8 cm.

Let AABC be an isosceles triangle,

where AB = 8 cm

Case 1: If AB = BC

AB = BC = 8 cm

The perimeter of the triangle ABC = 28 cm

⇒  AB + BC+ CA = 28 cm

⇒  8 cm + 8 cm + CA = 28 cm

⇒  16 cm + CA = 28 cm

⇒  CA = 28 cm- 16 cm = 12 cm

∴ Sides are 8 cm, 8 cm and 12 cm

Case 2: If BC = CA

The perimeter of the triangle ABC = 28 cm

⇒  AB + BC + CA = 28 cm

⇒  8 cm + 2BC = 28 cm

⇒  2BC = 28 cm- 8cm = 20 cm

⇒  BC = 10 cm

∴ Sides are 10 cm, 10 cm and 8 cm

Question 35. The length of an aluminium strip is 40 cm. If the lengths in cm are measured in natural numbers, write the measurement of all the possible rectangular frames which can be made out of it. (For example, a rectangular frame with 1 5 cm length and 5 cm breadth can be made from this strip.)
Solution:

Given

The length of an aluminium strip is 40 cm. If the lengths in cm are measured in natural numbers,

Perimeter of rectangular frame = Length of the aluminium strip

2(length + breadth) = 40 cm

length + breadth = \(\frac{40}{2} \mathrm{~cm}=20 \mathrm{~cm}\)

The possible measurement of rectangular frames are 1 cm x 19 cm, 2 cm x 18 cm, 3 cm x 17 cm, 4 cm x 16 cm, 5 cm x 15 cm, 6 cm x 14 cm, 7 cm x 13 cm, 8 cm x 12 cm, 9 cm x 11 cm, 10 cm x 10 cm

Question 36. The base of a tent is a regular hexagon with a perimeter of 60 cm. What is the length of each side of the base?
Solution:

Given

The base of a tent is a regular hexagon with a perimeter of 60 cm.

The perimeter of the regular hexagon= 60 cm

⇒  6 x side = 60 cm

⇒  \(=\frac{60}{6} \mathrm{~cm}=10 \mathrm{~cm}\)

∴ Each side of the base is 10 cm.

Question 37. In an exhibition hall, there are 24 display boards each of length 1 m 50 cm and breadth 1 m. There is a 100 m long aluminium strip, which is used to frame these boards. How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.
Solution:

Given

In an exhibition hall, there are 24 display boards each of length 1 m 50 cm and breadth 1 m. There is a 100 m long aluminium strip, which is used to frame these boards.

Length of display board =1 m 50 cm

= 1.50 m

Breadth of display board =1 m

Perimeter of one display board

= 2(length + breadth) = 2(1.50 + 1) m

= 2(2.50) m = 5 m

∴ Perimeter of 24 display boards

= (24 x 5) m = 120 m

The number of boards will be framed using the aluminium strip of 100 m long = \(\frac{100}{5}=20\)

The length of the aluminium strip required to frame the remaining boards = 120 m- 100 m = 20 m

Question 38. In the above question, how many square metres of cloth is required to cover all the display boards? What will be the length in metres of the cloth used, if its breadth is 120 cm?
Solution:

Length of display board =1 m 50 cm

= 1.5 m

Breadth of display board =1 m

∴ Area of 24 display boards = 24(/ x b)

= 24(1.5 x 1) m2= 24 x 1.5 m2= 36 m²

Therefore, 36 m2 of cloth is required to cover all the display boards.

Now, breadth of cloth = 120 cm = \(\frac{120}{100} \mathrm{~m}\)

∴ Length of cloth = \(\frac{\text { area }}{\text { breadth }}\)

⇒  \(\frac{36 \times 100}{120} \mathrm{~m}=30 \mathrm{~m}\)

Question 39. What is the length of the outer boundary of the park shown in the figure? What will be the total cost of fencing it at the rate of? 20 per metre? There is a rectangular flower bed in the centre of the park. Find the cost of manuring the flower bed at the rate of? 50 per square metre.

Solution:

The length of outer boundary of the park = (200 + 300 + 80 + 300 + 200 + 260) m = 1340 m

The cost of fencing the park at the rate of? 20 per metre = ₹ (20 x 1340) = ₹ 26800 Area of the flowerbed = (100 x 80) m2 = 8000 m2 Now, the cost of manuring the flower bed at the rate of? 50 per square metre

= ₹(50×8000)

= ₹ 400000

Question 40. The total cost of fencing the park shown in the figure is 55000. Find the cost of fencing per metre.

Solution:

Given

The total cost of fencing the park shown in the figure is 55000.

The perimeter of the park

= (150 + 100 + 120 + 180 + 270 + 280) m

= 1100 m

∴ Cost of fencing the park per metre \(=\frac{\text { Total cost of fencing }}{\text { Perimeter of the park }}\)

⇒ \(=₹\left(\frac{55000}{1100}\right)=₹ 50\)

Question 41. In the given figure each square is a length

1. What is the perimeter of the rectangle ABCD?
2. What is the area of the rectangle ABCD?
3. Divide this rectangle into ten parts of equal area by shading squares.
   (Two parts of equal area are shown here)
4. Find the perimeter of each part which you have divided. Are they all equal?

Solution:

We have given each square is of unit length, where length = 10 units and breadth = 6 units

1.  Perimeter of ABCD = 2 (length +breadth)

= 2(10 + 6) units

= 2 x 16 units = 32 units

2.  Area of ABCD = length x breadth

= (10 x 6) sq units = 60 sq units

3.  Since, there are several ways to divide this rectangle into 10 parts, from which one of the ways is shown below :

So, the above figure shows that there are 10 parts of equal area i.e., E, F, G, H, I, /, K, L, M and N.

The perimeter of each part is 12 units. Yes, the perimeter of each part is equal.

Question 42. The rectangular wall MNOP of the kitchen is covered with square tiles 15 cm in length (see figure). Find the area of the wall.

Solution:

Given

The rectangular wall MNOP of the kitchen is covered with square tiles 15 cm in length (see figure).

Area of one square tile with side 15 cm = (15 x 15) cm2 = 225 cm²

Since the walls are covered with 28 square tiles.

∴Area of the wall = (225 x 28) cm2 = 6300 cm²

Question 43. The length of a rectangular field is 6 times its breadth. If the length of the field is 120 cm, find the breadth and perimeter of the field.
Solution:

Given

The length of a rectangular field is 6 times its breadth. If the length of the field is 120 cm

Length of rectangular field = 120 cm

According to the question,

⇒ 6(breadth) = 120 cm

⇒ breadth = \(\frac{120}{6} \mathrm{~cm}=20 \mathrm{~cm}\)

∴ Perimeter of the field = 2(length +breadth)

= 2(120 + 20) cm

= 2(140) cm = 280 cm

Perimeter of the field = 280 cm

Question 44. Anmol has a chart paper of measures 90 cm x 40 cm, whereas Abhishek has one which measures 50 cm x 70 cm. Which will cover more area on the table and by how much?
Solution:

Given

Anmol has a chart paper of measures 90 cm x 40 cm, whereas Abhishek has one which measures 50 cm x 70 cm.

The area of Anmol’s chart paper = (90×40) cm² = 3600 cm²

And the area of Abhishek’s chart paper = (50 x 70) cm² = 3500 cm²

The chart paper of Anmol will cover more area on the table than that of Abhishek by (3600- 3500) cm2 = 100 cm²

Question 45. A rectangular path of 60 m in length and 3 m in width is covered by square tiles of side 25 cm. How many tiles will there be in one row along its width? How many such rows will be there? Find the number of tiles used to make this path.
Solution:

Given

A rectangular path of 60 m in length and 3 m in width is covered by square tiles of side 25 cm.

Length of rectangular path = 60 m = 6000 cm

Width of rectangular path = 3 m = 300 cm

The side of a square tile = 25 cm

⇒ The number of tiles will be in one row along the width of the path = \(=\frac{\text { width of the path }}{\text { side of a tile }}=\frac{300}{25}=12\)

⇒ Number of rows = \(\frac{\text { length of the path }}{\text { side of a tile }}\)

⇒ \(=\frac{6000}{25}=240\)

∴ Total number of tiles required = 12 x 240

= 2880

Total number of tiles required = 2880

Question 46. How many square slabs each with side 90 cm are needed to cover a floor of area 81 sq m?
Solution:

Area of the floor = 81 m² = 81 x 10000 cm²

= 810000 cm²

Area of one square slab = (90 x 90) cm²

= 8100 cm²

⇒ The required number of square slabs \(=\frac{\text { Area of the floor }}{\text { Area of a square slab }}=\frac{810000}{8100}=100\)

Question 47. The length of a rectangular field is 8 m and the breadth is 2 m. If a square field has the same perimeter as this rectangular field, find which field has the greater area.
Solution:

Given

The length of a rectangular field is 8 m and the breadth is 2 m. If a square field has the same perimeter as this rectangular field,

We have given, the perimeter of the rectangular field = perimeter of the square field

⇒ 2(length + breadth) – 4 * side

⇒ 2(8 + 2) m = 4 x side

⇒ 2(10) m = 4 x side

⇒ 20 m = 4 x side

⇒ \(\frac{20}{4} \mathrm{~m}=\text { side } \Rightarrow \text { side }=5 \mathrm{~m}\)

Now, area of the rectangular field = length x breadth = 8 m x 2 m = 16 m²

Area of the square field= side x side = 5mx5m = 25m²

∴ The square field has a greater area than that of the rectangular field.

Question 48. Parmindar walks around a square park once and covers 800 m. What will be the area of this park?
Solution:

Given

Parmindar walks around a square park once and covers 800 m.

The perimeter of the square park = 800m

⇒ 4 x side = 800 m

⇒ \(\text { side }=\frac{800}{4} m=200 \mathrm{~m}\)

∴ Area of the square park = side x side

= (200 x 200) m²

= 40000 m²

Question 49. The side of a square is 5 cm. How many times does the area increase, if the side of the square is doubled?
Solution:

Given

Side of the square = 5 cm

∴ Area of the square = side x side = 5 cm x 5 cm = 25 cm²…..(1)

Side of a new square

= 2 (side of the given square)

= 2 x 5 cm = 10 cm

∴ Area of the new square = side x side

= 10 cm x 10 cm

= 100 cm2 …(2)

Thus, (1) and (2) show that the area will be increased by 4 times.

Question 50. Amita wants to make rectangular cards measuring 8 cm x 5 cm. She has a square chart paper of side 60 cm. How many complete cards can she make from this chart? What area of the chart paper will be left?
Solution:

Given

Amita wants to make rectangular cards measuring 8 cm x 5 cm. She has a square chart paper of side 60 cm.

The square chart of length 60 cm can be divided into 7 parts each of length 8 cm along one side of chart paper and exactly in 12 parts each of length 5 cm along the other side of chart paper.

The figure shows that the number of rectangular cards which can be made from the square chart paper is 7 x 12 = 84

Now, area of the square chart paper = (60 x 60) cm²

= 3600 cm²

Area of one rectangular card = (8 x 5) cm2 = 40 cm²

∴ Area of 84 rectangular cards = 84 x 40 cm2= 3360 cm²

∴ Area of the remaining chart paper = (3600- 3360) cm²= 240 cm²

Question 51. A magazine charges ₹ 300 per 10 sq cm area for advertising. A company decided to order a half-page advertisement. If each page of the magazine is 15 cm x 24 cm, what amount will the company have to pay for it?
Solution:

Given

A magazine charges ₹ 300 per 10 sq cm area for advertising. A company decided to order a half-page advertisement. If each page of the magazine is 15 cm x 24 cm

The area of one page of magazine = length x breadth

= (15 x 24) cm2 = 360 cm2

∴ Area of the half page of magazine \(=\frac{360}{2} \mathrm{~cm}^2=180 \mathrm{~cm}^2\)

Cost of advertising for 10 cm2 = ₹ 300

∴ Cost of advertising for 180 cm2 \(=₹ \frac{300}{10} \times 180=₹ 5400\)

Question 52. The perimeter of a square garden is 48 m. A small flower bed covers an 18 sq m area inside this garden. What is the area of the garden that is not covered by the flower bed? What fractional part of the garden is covered by a flower bed? Find the ratio of the area covered by the flower bed and the remaining area.
Solution:

Given

The perimeter of a square garden is 48 m. A small flower bed covers an 18 sq m area inside this garden.

The perimeter of the square garden= 48m

⇒ \(4 \times \text { side }=48 \mathrm{~m} \Rightarrow \text { side }=\frac{48}{4} \mathrm{~m}=12 \mathrm{~m}\)

Area of the square garden = side * side

= 12 m x 12 m

= 144 m²

The area of the small flower bed = is 18 m²

∴ The area of the garden that is not covered by the flower bed = 144 m2- 18 m2= 126 m2 The required fractional part of the garden which is covered by the flower bed

⇒ \(=\frac{\text { Area of the flower bed }}{\text { Area of the square garden }}\)

⇒ \(=\frac{18}{144}=\frac{1}{8}\)

The ratio of the area covered by the flower bed and the remaining area

⇒ \(\frac{18}{126}=\frac{1}{7} \text { i.e., } 1: 7\)

Question 53. The perimeter of a square and a rectangle is the same. If a side of the square is 15 cm and one side of the rectangle is 18 cm, find the area of the rectangle.
Solution:

Given

The perimeter of a square and a rectangle is the same. If a side of the square is 15 cm and one side of the rectangle is 18 cm

We have given,

Perimeter of the square= Perimeter of the rectangle

⇒ 4 x side = 2(length + breadth)

⇒ 4 x 15 cm = 2(18 cm + breadth)

⇒ \(\frac{4 \times 15}{2} \mathrm{~cm}=18 \mathrm{~cm}+\text { breadth }\)

30 cm = 18 cm + breadth

⇒ (30- 18) cm = breadth

=» breadth = 12 cm

Now, the area of the rectangle= length x breadth

= (18 x 12) cm²

= 216 cm²

Question 54. A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2 cm. If the total area of the small squares is 28 square cm, what was the original length of the wire?
Solution:

Given

A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2 cm. If the total area of the small squares is 28 square cm

Length of one piece = Perimeter of the square

= 4 x 2 cm = 8 cm

Let the number of pieces be n.

We have given, the total area of n squares = 28 cm²

⇒ n (side)2 = 28

⇒ n (2)2 = 28

n x 4 = 28

⇒ \(n=\frac{28}{4}=7\)

The original length of the wire

=Number of pieces x length of one piece

=7×8 cm- 56 cm

Question 55. Divide the park shown in the figure of question 40 into two rectangles. Find the total area of this park. If one packet of fertilizer is used for 300 sq m, how many packets of fertilizer are required for the whole park?
Solution:

Area of rectangle ABCD=length x breadth

= (150 x 100) m²

= 15000 m²

And area of rectangle AGFE = length x breadth = (270×180) m²= 48600 m²

Total area of the park = 15000 m2+ 48600 m² = 63600 m²

Number of packets of fertilizer used for 300 m²=1

Number of packets of fertilizer used for \(63600 \mathrm{~m}^2=\frac{63600}{300}=212\)

Question 56. The area of a rectangular field is 1600 sqm. If the length of the field is 80 m, find the perimeter of the field.
Solution:

Given

The area of a rectangular field is 1600 sqm. If the length of the field is 80 m,

Area of the rectangular field = 1600 m²

⇒ length x breadth= 1600 m²

⇒ 80 m x breadth = 1600 m²

⇒ breadth = \(\frac{1600 \mathrm{~m}^2}{80 \mathrm{~m}}=20 \mathrm{~m}\)

Now, the perimeter of the rectangular field

= 2(length + breadth)

= 2(80 + 20) m

= 2 x 100 m = 200 m

Question 57. The area of each square on a chessboard is cm. Find the area of the board.

1. At the beginning of the game when all the chessmen are put on the board, write the area of the squares left unoccupied.
2. Find the area of the squares occupied by chessmen.

Solution:

We have given, The area of one square on a chess board = 4 sq cm

Since there are 64 squares on a chessboard.

Area of the chess board = 64 x 4 sq cm = 256 sq cm

1.  At the beginning of the game when all the chessmen are put on the board, 32 squares are left unoccupied.

∴ The area of the squares left unoccupied

= 32 x area of one square

= 32 x 4 sq cm = 128 sq cm

2.  Since 32 squares are occupied by chessmen.

∴ Area of the squares occupied by chessmen = 32 x (area of one square)

= 32 x 4 sq cm = 128 sq cm

Question 58.

Find all the possible dimensions (in natural numbers) of a rectangle with a perimeter of 36 cm and find their areas.

Find all the possible dimensions (in natural numbers) of a rectangle with an area of 36 sq cm, and find their perimeters.
Solution:

The perimeter of the rectangle = 36 cm

⇒ 2(length + breadth) = 36 cm

⇒ length + breadth = \(\frac{36}{2} \mathrm{~cm}=18 \mathrm{~cm}\)

The possible dimensions of the rectangles and their areas are as follows:

The area of the rectangle = 36 sq cm

⇒ length x breadth = 36 sq cm

The possible dimensions of the rectangle and their perimeters are as follows:

Question 59. Find the area and perimeter of each of the following figures, if the area of each small square is 1 sq cm

Solution:

In the given figure (i), there are 11 small squares with an area of sq cm each.

Area of the given = (11 x 1) sq cm = 11 sq cm

Perimeter of the given  = (18 x 1) cm = 18 cm

In the given, there are 13 small squares with an area of 1 sq cm each.

Area of the given= (13 x 1) sq cm = 13 sq cm

Perimeter of the given  = (28 x 1) cm = 28 cm

In the given, there are 13 small squares with an area of 1 sq cm each.

Area of the given figure = (13 x 1) sq cm = 13 sq cm

Perimeter of the given figure = (28 x 1) cm = 28 cm

Question 60. What is the area of each small square in the given figure if the area of the entire figure is 96 sq cm? Find the perimeter of the figure.

Solution:

We have 24 small squares in the figure.

Area of the entire figure = 96 sq cm

The area of one small square

⇒ \(\frac{96}{24} \mathrm{sq} \mathrm{cm}\)

= 4 sqcm

Thus, the area of each small square = 4 sq cm

(side) x (side) = 4 sq cm

side = √4 cm = 2 cm

Now, the perimeter of the given figure

= 34 x 2 cm

= 68 cm

Class 6 Maths Chapter 9 Data Handling

Question 1. Using tally marks, which one of the following represents the number eight:

Solution: (4): 8 can be represented as an image

Question 2. The marks (out of 0) obtained by 28 students in a Mathematics test are listed below:

8, 1, 2, 6, 5, 5, 5, 0, 1, 9, 7, 8, 0, 5, 8, 3, 0, 8, 1 0, 1 0, 3, 4, 8, 7, 8, 9, 2,0

The number of students who obtained marks more than or equal to 5 is

1. 13
2. 15
3. 16
4. 17

Solution: (4): The number of students who obtained marks more than or equal to 5 is 17

Question 3. In question 2 above, the number of students who scored marks less than 4 is

1. 15
2. 13
3. 12
4. 10

Solution: (4): The number of students who scored marks less than 4 is 10

Read and Learn More Class 6 Maths Exemplar Solutions

Question 4. The choices of the fruits of 42 students in a class are as follows:

A,0,B,M,A,G,B,G,A,G, B,M,A,G,M,A,B,G,M,B, A,0,M,0,G,B,0,M,G,A, A,B,M,0,M,G,B,A,M,0,M,0,

where A, B, G, M, and O stand for the fruits Apple, Banana, Grapes, Mango, and Orange respectively.

Which two fruits are liked by an equal number of students?

1. A and M
2. M and B
3. B and O
4. B and G

Solution: (4): We have given two fruits Banana and Grapes, which are liked by an equal number of students i.e., 8.

Question 5. According to data from question 4, which fruit Is liked by most of the students?

1. O
2. G
3. M
4. A

Solution: (3): Mango is the only fruit liked by most students.

Question 6. In a bar graph, the width of bars may be unequal.
Solution: False

In a bar graph, bars have equal width.

Question 7. In a bar graph, bars of uniform width are drawn vertically only.
Solution: False

In a bar graph, bars of uniform width can be drawn vertically as well as horizontally.

Question 8. In a bar graph, the gap between two consecutive bars may not be the same.
Solution: False

The gap between two consecutive bars should be the same in a bar graph.

Question 9. In a bar graph, each bar (rectangle) represents only one value of the numerical data.
Solution: True

Question 10. To represent the population of different towns using a bar graph, it is convenient to take one unit length to represent one person.
Solution: False

Question 11. Pictographs and bar graphs are pictorial representations of the numerical data.
Solution: True

Question 12. An observation occurring five times in data is recorded as an image using tally marks.
Solution: False

Using tally marks, the occurrence of any observation five times can be represented as an image

Question 13. In a pictograph, if a symbol represents 50 books on a library shelf, then the symbol represents 25 books.

Solution: True

Question 14. A _______is a collection of numbers gathered to give some meaningful information.
Solution: Data

Question 15. The data can be arranged in a tabular form using_______marks.
Solution: Tally

Question 16. A_______ represents data through pictures of objects
Soluction: Pictograph

17. In a bar graph, _______ can be drawn horizontally or vertically.
Solution: Bars

Question 18. In a bar graph, bars of _______ width can be drawn horizontally or vertically with _______ spacing between them.
Solution: Uniform, Equal

Question 19. An observation occurring seven times in a data is represented as _______ using tally marks.
Solution: 

Question 20. In a pictograph, if a symbol represents 20 flowers in a basket then stands for _______ flowers.
Solution: 60

Question 21. On the scale of 1 unit length = 10 crores, the bar of length 6 units will represent _______ crore, and of _______units will represent 75 crores.
Solution: 60, 7.5

Question 22. In an examination, the grades achieved by 30 students of a class are given below. Arrange these grades in a table using tally marks:

B, C, C, E, A, C, B, B, D, D, D, D, B, C, C, C, A, C, B, E, A, D, C, B, E, C, B, E, C, D

Solution:

Question 23. The number of two-wheelers owned individually by each of the 50 families is listed below. Make a table using tally marks.

1,1, 2, .1,1, 1,2, 1,2, 1,0, 1,1, 2, 3,1,2, 1,1,2, 1, 2, 3, 1, 0, 2, 1, 0, 2, 1, 2, 1, 2, 1, 1, 4, 1, 3, 1, 1, 2, 1,1, 1,1, 2, 3, 2,1,1

Find the number of families having two or more, two-wheelers

Solution:

So, 14 + 4 + 1 = 19 families have two or more, two-wheelers

Question 24. The lengths in centimeters (to the nearest centimeter) of 30 carrots are given as follows:

15, 22, 21, 20, 22,15, 15, 20, 20,15, 20, 18, 20, 22, 21, 20, 21, 18, 21, 18, 20, 18, 21, 18, 22, 20, 15,21,18, 20

Arrange the data given above in a table using tally marks and answer the following questions.

1. What is the number of carrots which have a length of more than 20 cm?
2. Which length of the carrots occur the maximum number of times? The minimum number of times?

Solution:

1. The number of carrots that have a length of more than 20 cm is 6 + 4 = 10.
2. The carrots of length 20 cm occur a maximum number of times and the carrots of length 22 cm occur a minimum number of times

Question 25. Thirty students were interviewed to find out what they wanted to be in the future. Their responses are listed below:

doctor, engineer, doctor, pilot, officer, doctor, engineer, doctor, pilot, officer, pilot, engineer, officer, pilot, doctor, engineer, pilot, officer, doctor, officer, doctor, pilot, engineer, doctor, pilot, officer, doctor, pilot, doctor, engineer.

Arrange the data in a table using tally marks.

Solution:

Question 26. Following are the choices of games for 40 students of Class 6:

football, cricket, football, kho-kho, hockey, cricket, hockey, kho-kho, tennis, tennis, cricket, football, football, hockey, kho-kho, football, cricket, tennis, football, hockey, khokho, football, cricket, cricket, football, hockey, kho-kho, tennis, football, hockey, cricket, football, hockey, cricket, football, kho-kho, football, cricket, hockey, football.

Arrange the choices of games in a table using tally marks.

1. Which game is liked by most of the students?
2. Which game is liked by the minimum number of students?

Solution: 1.

1. Football is liked by most of the students.
2. Tennis is liked by a minimum number of students.

Question 27. Fill in the blanks in the following table which represents the shirt size of 40 students of a school.

Solution:

Question 28. The following pictograph represents some surnames of people listed in the telephone directory of a city.

Observe the pictograph and answer the following questions:

1. How many people have the surname ‘Roy’?
2. Which surname appears the maximum number of times in the telephone directory?
3. Which surname appears the least number of times in the directory?
4. Which two surnames appear an equal number of times?

Solution:

1. 400 people have the surname ‘Roy’.
2. The surname ‘Patel’ appears the maximum number of times in the telephone directory.
3. The surname ‘Saikia’ appears the least number of times in the telephone directory.
4. The surnames ‘Rao’ and ‘Roy’ appear an equal number of times.

Question 29. Students of Class 6 in a school were given a task to count the number of articles made of different materials in the school. The information collected by them is represented as follows:

Observe the pictograph and answer the following questions:

1. Which material is used in a maximum number of articles?
2. Which material is used in a minimum number of articles?
3. Which material is used in exactly half the number of articles as those made up of metal?
4. What is the total number of articles counted by the students?

Solution:

1. Metal is used in a maximum number of articles.
2. Glass is used in the minimum number of articles.
3. Rubber is used in exactly half the number of articles as those made up of metal.
4. The total number of articles counted by the students is 160.

Question 30. The number of scouts in a school is depicted by the following pictograph:

Observe the pictograph and answer the following questions:

1. Which class has the minimum number of scouts?
2. Which class has the maximum number of scouts?
3. How many scouts are there in Class 6?
4. Which class has exactly four times the scouts as that of Class X?
5. What is the total number of scouts in the Classes 6 to 10?

Solution:

1. Class 10 has a minimum number of scouts.
2. Class 8 has a maximum number of scouts.
3. There are 40 scouts in Class 6.
4. Class 6 has exactly four times the scouts as that of Class 10.
5. Total number of scouts in the Classes 6 to 10 is 160.

Question 31. A survey was carried out in a certain school to find out the popular school subjects among students of Classes VI to VIII. The data in this regard is displayed as a pictograph given below:

1. Which subject is most popular among the students?
2. How many students like Mathematics?
3. Find the number of students who like subjects other than Mathematics and Science.

Solution:

1. Hindi is the most popular subject among the students.
2. 175 students like Mathematics.
3. A total of 200 + 150 + 75 = 425 students are there who like subjects other than Mathematics and Science.

Question 32. The following pictograph depicts the information about the areas in sq km (to the nearest hundred) of some districts of Chhattisgarh State:

1. What is the area of the Koria district?
2. Which two districts have the same area?
3. How many districts have an area of more than 5000 square kilometers?

Solution:

1. The area of Koria district is 6000 sq km.
2. Raigarh and Jashpur districts have the same area of 6500 sq km.
3. 4 districts have an area of more than 5000 square kilometers and they are Raigarh, Rajnandgaon, Koria, and Jashpur

Question 33. The number of bottles of cold drinks sold by a shopkeeper on six consecutive days is as follows:

Prepare a pictograph of the data using one symbol to represent 50 bottles.

Solution:

Question 34. The following table gives information about the circulation of newspapers (dailies) in a town in five languages.

Prepare a pictograph of the above data, using a symbol of your choice, each representing 1000 newspapers.

Solution:

Question 35. Annual expenditure of a company in the year 2007-2008 is given below:

Prepare a pictograph of the above data using an appropriate symbol to represent ₹ 10 lakh.

Solution:

Question 36. The houses following (out ofbar100) graphing a showdown using the number of different types of fuels for cooking.

Read the bar graph and answer the following questions:

1. Which fuel is used in the maximum number of houses?
2. How many houses are using coal as fuel?
3. Suppose that the total number of houses in the town is ₹ 1 lakh. From the above graph estimate the number of houses using electricity.

Solution:

1. LPG is used in a maximum number of houses.
2. 10 houses are using coal as fuel.
3. Since there are 5 houses using electricity out of 100. Therefore, out of ₹ 1 lakh, the electricity will be used by \(\frac{5}{100} \times 100000\) = 5000 houses.

Question 37. The following bar graph represents the data for different sizes of shoes worn by the students in a school. Read the graph and answer the following questions.

1. Find the number of students whose shoe sizes have been collected.
2. What is the number of students wearing shoe size 6?
3. What are the different sizes of the shoes worn by the students?
4. Which shoe size is worn by the maximum number of students?
5. Which shoe size is worn by the minimum number of students?
6. State whether true or false: The total number of students wearing shoe sizes 5 and 8 is the same as the number of students wearing shoe size 6.

Solution:

1. The total number of students whose shoe sizes have been collected = 250 + 200 + 300 + 400 + 150 = 1300
2. The number of students wearing shoe size 6 is 300.
3. The different sizes of the shoes worn by the students are 4, 5, 6, 7, and 8.
4. The shoe size 7 is worn by the maximum number of students.
5. The shoe size 8 is worn by the minimum number of students.
6. False: Since the number of students wearing shoe sizes 5 and 8 is 350 and the number of students wearing shoe size 6 is 300.

Question 38. The following graph gives information about the number of railway tickets sold for different cities on a railway ticket counter between 6.00 am to 10.00 am. Read the bar graph and answer the following questions.

1. How many tickets were sold in all?
2. For which city were the maximum number of tickets sold?
3. For which city were the minimum number of tickets sold?
4. Name the cities for which the number of tickets sold is more than 20.
5. Fill in the blanks: The number of tickets sold for Delhi and Jaipur together exceeds the total number of tickets sold for Patna and Chennai by________.

Solution:

1. Total number of tickets sold = 80 + 50 + 100 + 20 + 45 = 295
2. The maximum number of tickets were sold for Delhi.
3. The minimum number of tickets were sold for Chennai.
4. The cities for which the number of tickets sold is more than 20 are as follows: Patna, Jaipur, Delhi, and Guwahati.
5. 50: Number of tickets sold for Delhi and Jaipur together = 100 + 50 = 150.

• Number of tickets sold for Patna and Chennai together = 80 + 20 = 100
• Therefore, tickets sold for Delhi and Jaipur exceed the tickets sold for Patna and Chennai by 150- 100 = 50.

Question 39. The bar graph given below represents the approximate length (in kilometers) of some National Highways in India. Study the bar graph and answer the following questions:

1. Which National Highway (N.H.) is the longest among the above?
2. Which National Highway is the shortest among the above?
3. What is the length of National Highway 9?
4. Length of which National Highway is about three times the National Highway l0?

Solution:

1. The longest National Highway is N.H. 2.
2. The shortest National Highway is N.H. 10.
3. The length of National Highway 9 is 900 km.
4. Length ofNational Highway 10 = 450km Length ofNational Highway 8 = 1400km

So, the length of National Highway 8 is about three times the of National Highway 10

Question 40. The bar graph given below represents the circulation of newspapers in different languages in a town. Study the bar graph and answer the following questions:

What is the circulation of English newspapers?

Name the two languages in which the circulation of newspapers is the same.

By how much is the circulation of newspapers in Hindi more than the newspaper in Bengali?

Solution:

1. The circulation of English newspapers is 1000.
2. In Bengali and Marathi languages the circulation of newspapers is the same.
3. The circulation of newspapers in Hindi is more than the newspaper in Bengali by 1400-600 = 800.

Question 41. Read the bar graph given below and answer the following questions:

1. What information is given by the bar graph?
2. In which year is the number of students maximum?
3. In which year is the number of students twice that of 2001 -02?
4. In which year did the number of students decrease as compared to the previous year?
5. In which year is the increase in number of students maximum as compared to the previous year?

Solution:

1. The given bar graph shows the number of students in different academic years.
2. In the year 2005-06, the number of students is maximum.
3. Since number of students in 2001-02 is150 and number of students in 2004-05 is 300. So, in the year 2004-05, the number of students is twice that of 2001-02.
4. In the year 2003-04, the number of students decreased as compared to the previous year.
5. The year is 2004-05in which the increase in number of students is maximum as compared to the previous year

Question 42. The lengths in km (rounded to the nearest hundred) of some major rivers of India are given below:

Draw a bar graph to represent the above information.

Solution:

Question 43. The number of ATMs of different banks in a city is shown below:

Draw a bar graph to represent the above information by choosing the scale of your choice.

Solution:

Question 44. Number of mobile phone users in various age groups in a city is listed below:

Draw a bar graph to represent the above information.

Solution:

Question 45. The following table gives the number of vehicles passing through a toll gate, every hour from 8.00 am. to 1.00 pm:

Draw a bar graph representing the above data.

Solution:

Question 46. The following table represents the income of a Gram Panchayat from different sources in a particular year:

Draw a bar graph to represent the above information.

Solution:

Question 47. The following table gives the data on several schools (stage-wise) in a country in the year 2002.

Draw a bar graph to represent the above data:

Solution:

Question 48. Home appliances sold by a shop in one month are given as below:

Draw a bar graph to represent the above information.

Solution:

Question 49. In a botanical garden, the number of different types of plants are found as follows:

Draw a bar graph to represent the above information and answer the following questions:

1. Which type of plant is the maximum in number in the garden?
2. Which type of plant is the minimum in number in the garden?

Solution:

1. The tree is the maximum in number in the garden.
2. Creeper is the minimum in number in the garden.

Question 50. Prepare a bar graph of the data given in question 28.

Solution:

Question 51. Refer to question 39. Prepare a pictograph of the data by taking a suitable symbol to represent 200 kilometers.
Solution:

Question 52. Prepare a pictograph of the information given in question 38.
Solution:

Question 53. Refer to question 23. Prepare a bar graph of the data.
Solution:

Question 54. The following table shows the area of the land on which different crops were grown.

Prepare a pictograph by choosing a suitable symbol to represent 10 million hectares.

Solution:

Question 55. Refer to question 54. Prepare a bar graph of the data.
Solution: