Alekhya

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

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 3 Playing With Numbers

Directions: In questions 1 to 14, out of the four options, only one is correct. Write the correct answer.

1 Sum of the number of primes between 16 to 80 and 90 to 100 is

(1) 20
(2) 18
(4) 16
(3) 17

Solution: (3): Prime numbers between 16 to 80 are 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.

So, there are 16 prime numbers between 16 to 80.

Also, 97 is the only prime number between 90 to 100.

So, there is only a prime number between 90 to 100.

Required sum = 16 +1 = 17

Read and Learn More Class 6 Maths Exemplar Solutions

2. Which of the following statements is not true?

(1) The HCF of two distinct prime numbers is 1
(2) The HCF of two coprime numbers is 1
(3) The HCF oftwo consecutive even numbers is 2
(4) The HCF of an even and an odd number is even

Solution: (4): The HCF of an even and an odd number is always an odd number.

3. The number of distinct prime factors of the largest 4-digit number is

(1) 2
(3) 5
(2) 3
(4) 11

Solution: (2): The largest 4-digit number is 9999.

Now,

9999 = 3 x 3×11 x101

So, distinct prime factors of 9999 are 3, 11 and 101, i.e., 3 distinct prime factors.

4. The number of distinct prime factors of the smallest 5-digit number is

(1) 2
(2) 4
(4) 8
(3) 6

Solution: (1): The smallest 5-digit number is 10000.

Now,

10000 = 2x2x2x2x5x5x5x5

So, distinct prime factors of10000 are 2 and 5, i.e., 2 distinct prime factors.

5. If the number 7254*98 is divisible by 22, the digit at* is

(1) 1
(2) 2
(3) 6
(4) 0

Solution: (3): 7254*98 is divisible by 22, if it is divisible by both 2 and 11.

Given number is even, therefore it is divisible by 2.

7254*98 is divisible by 11, if (7 + 5 + * + 8) – (2 + 4 + 9) or (20 + *)- 15 or 5 + * is divisible by 11.

The digit at * should be filled by 6.

6. The largest number which always divides the sum of any pair of consecutive odd numbers is

(1) 2
(2) 4
(3) 6
(4) 8

Solution:(2): The sum of any pair of consecutive odd numbers comesin the form ofmultiple of 4.

The required largest number is 4.

7. A number is divisible by 5 and 6.It may not be divisible by

(1) 10
(3) 30
(2) 15
(4) 60

Solution: (4): The LCM of 5 and 6 is 30.

And 30 is divisible by 10, 15 and 30 but not by 60.

8. The sum of the prime factors of 1 729 is

(1) 13
(2) 19
(4) 39
(3) 32

Solution: (4): We have,

The prime factors of 1729 = 7 x 13 x 19.

The sum of the prime factors of 1729 = 7+ 13 +19 = 39

9. The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is

(1) 6
(2) 4
(3) 16
(4) 8

Solution: (2): Since the odd natural numbers other than1 are 3, 5, 7, 9 and so on.

Now, the predecessor and successor of 3 are 2 and 4 respectively, and their product is 2×4 = 8

Similarly, the predecessor and successor of 5 are 4 and 6 respectively and their product is 4 x 6 = 24 and so on.

Thus, the above shows that the greatest number which always divides the product of the predecessor and successor of an odd natural other than 1 is 4.

10. The number of common prime factors of 75, 60,105 is

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

Solution: (1): We have,

75 = 3 x 5 x 5

60 = 2 x 2 x 3 x 5

105 = 3 x 5 x 7

The common prime factors of 75, 60 and 105 are 3 and 5 i.e., 2 in number.

11. Which of the following pairs is not coprime?

(1) 8,10
(3) 1,3
(2) 11, 12
(4) 31,33

Solution: (1): 8 and 10 are not co-prime numbers.

Their common factor other than 1 is 2

12. Which of the following numbers is divisible by 11?

(1) 1011011
(2) 1111111
(3) 22222222
(4) 3333333

Solution: (3): The difference of the sum of digits of 22222222 at even and odd places is 0. It must be divisible by 11.

13. LCM of10, 15 and 20 is

(1) 30
(2) 60
(4) 180
(3) 90

Solution: (2): We have,

The LCM of 10, 15 and 20 is 2x2x3x5=60

14. LCM of two numbers is 1 80. Then which of the following is not the HCF of the numbers?

(1) 45
(2) 60
(3) 75
(4) 90

Solution:(3): We have,

The factors of 180 = 2 x 2 x 3 x 3 x 5

75 does not divide 180

75 can not be the HCF of the numbers whose LCM is 180.

Directions: In questions 15 to 32 state whether the given statements are true (T) orfalse (6).

15. Sum of two consecutive odd numbers is always divisible by 4.

Solution: True

16. If a number divides three numbers exactly, it must divide their sum exactly.

Solution: True

17. If a number exactly divides the sum of three numbers, it must exactly divide the numbers separately.

Solution: False

18. If a number is divisible both by 2 and 3, then it is divisible by 12.

Solution: False

Since, a number is divisible by both 2 and 3 implies that it is also divisible by 6.

19. A number with three or more digits is divisible by 6, if the number formed by its last two digits (i.e., ones and tens) is divisible by 6.

Solution: False

A number is divisible by 6 if the number is divisible by both 2 and 3.

20. A number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8.

Solution: True

21. If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 9.

Solution: False

If the sum of the digits of a number is divisible by 3, then the number it self is divisible by 3.

22. All numbers which are divisible by 4 may not be divisible by 8.

Solution: True

Since, 12 is divisible by 4 but not by 8.

23. The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple.

Solution: False

HCF of two or more numbers is smaller than their LCM.

24. LCM of two or more numbers is divisible by their HCF.

Solution: True

25. LCM of two numbers is 28 and their HCF is 8.

Solution: False

The LCM of two numbers i.e., 28 is not divisible by their HCF, i.e., 8.

Their HCF cannot be 8.

26. LCM of two or more numbers may be one of the numbers.

Solution: True

27. HCF of two or more numbers may be one of the numbers.

Solution: True

28. Any two consecutive numbers are coprime.

Solution: True

The common factor of any two consecutive numbers is 1.

29. If the HCF of two numbers is one of the numbers, then their LCM is the other number.

Solution: True

The product of two numbers =HCF * LCM.

30. The HCF of two numbers is smaller than the smaller ofthe numbers.

Solution: False

Since, HCF of two numbers is less than or equal to the smaller of the numbers.

31. The LCM of two numbers is greater than the larger ofthe numbers.

Solution: False

Since, LCM of two numbers is greater than or equal to the larger of the numbers.

32. The LCM of two coprime numbers is equal to the product ofthe numbers.

Solution: True

33. A number is a _________ of each ofits factor.

Solution: Multiple

34.__________ is a factor of every number.

Solution:1

35. The number of factors of a prime number is_______

Solution: 2

36. A number for which the sum of all its factors is equal to twice the number is called a___________ number.

Solution: Perfect

37. The numbers having more than two factors are called___________

Solution: Composite

38. 2 is the only__________

Solution: Prime

39. Two numbers having only 1 as a common factor are called___________numbers.

Solution: Co-prime

40. Number of primes between 1 to 100 is_______.

Solution: 25 : Prime numbers between I to 100,are 2. 3, 5. 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

So, there are 25 primes between 1 to 100.

41. If a number has ________in ones place, then it is divisible by 10.

Solution: 0

42. A number is divisible by 5, if it has____________ or in its ones place.

Solution: 0, 5

43. A number is divisible by________if it has any of the digits 0, 2, 4, 6, or 8 in its ones place. 1,1,1

Solution: 2

44. If the sum of the digits in a number is a __________ of 3, then the number is divisible by

Solution: Multiple

45. If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either 0 or divisible by_____________, then the number is divisible by 11.

Solution: 11

46. The LCM of two or more given numbers is the lowest of their common________.

Solution: Multiple

47. The HCF of two or more given numbers is the highest of their common____________.

Solution: Factors

48. Find the LCM of 80, 96, 125, 160.

Solution: We have,

The LCM of 80, 96, 125 and 160 is = 2x2x2x2x2x3x5x5x5 = 12000

49. Find the LCM of 160, 170 and 90.

Solution: We have,

The LCM of 160, 170 and 190

=2x2x2x2x2x3x3x5x17

= 24480

50. Determine the sum of the four numbers as given below:

(1) successor of 32
(2) predecessor of 49
(3) predecessor of the predecessor of 56
(4) successor of the successor of 67

Solution:

Since, successor of32is 33, predecessor of 49 is 48, predecessor of the predecessor of 56 is 54 and successor of the successor of 67 is 69.

The required sum = 33 + 48 + 54 + 69 = 204

51. Determine the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

Solution: We have,

Since, the LCM of 3, 4 and 5 is 2 x 2 x 3 x 5 = 60.

The required number is 60 + 2 = 62

So, 62 is the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

52. A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

Solution:

Given:

merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity.

A merchant has 3 kinds of oil with different quantities like 120 litres, 180 litres and 240 litres

Since, he wants lo soli the oil by filling the three kinds of oil in tins of equal capacity, so the greatest capacity of such a tin is the HCF of 120, 180 and 240.

Now, 120 = 2x2x2x3x5

180 = 2x2x3x3x5

240 = 2x2x2x2x3x5

The required greatest capacity of a tin

= (2 x 2 x 3 x 5) litres

= 60 litres

The greatest capacity of a tin = 60 litres

53. Find a 4-digit odd number using each of the digits 1, 2, 4 and 5 only once such that when the first and the last digits are interchanged, it is divisible by 4.

Solution: By using the digits 1, 2, 4 and 5 only once, we get a 4 digit odd number 4521.

When we interchanged its first and last digits we get a new number, i.e., 1524 which is divisible by 4.

Thus, the required number is 4521.

54. Using each of the digits 1, 2, 3 and 4 only once, determine the smallest 4-digit number divisible by 4.

Solution: By using the digits 1, 2, 3 and 4 only once, the smallest 4-digit number which is divisible by 4 is 1324.

55. Fatima wants to mail three parcels to three village schools. She finds that the postal charges are 20, 28 and 36, respectively. If she wants to buy stamps only of one denomination, what is the greatest denomination of stamps she must buy to mail the three parcels?

Solution:

Given

Fatima wants to mail three parcels to three village schools. She finds that the postal charges are 20, 28 and 36, respectively.

The postal charges to mail three parcels are 20, 28 and 36 respectively.

Also, Fatima wants to buy stamps only of one denomination.

So, to find the greatest denomination of stamps, we find the HCF of 20, 28 and 36.

Now, 20 = 2 x 2 x 5

28 = 2 x 2 x 7

36 = 2 x 2 x 3 x 3

The HCF of 20, 28 and 36 is 2×2 = 4

So, ? 4 is the greatest denomination of stamps, she must buy to mail the three parcels.

56. Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy an equal number of biscuits, of each brand, what is the minimum number of packets of each brand, he should buy?

Solution:

Given

Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy an equal number of biscuits, of each brand

A shopkeeper has three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively.
Also, a shopkeeper wants to buy equal number of biscuits of each brand for that we need to find the LCM of 12, 15 and 21.

The LCM of 12, 15 and 21 = 2x2x3x5x7 = 420

Thus, the required number of packets of

brand A =420/12= 35,

brand B =420/15 =28 and

brand C =420/21 = 20

57. The floor of a room is 8 m 96 cm long and 6 m 72 cm broad. Find the minimum number of square tiles of the same size needed to cover the entire floor.

Solution:

Given

The floor of a room is 8 m 96 cm long and 6 m 72 cm broad.

Length of floor of a room = 8 m 96 cm = 896 cm

Breadth of floor of the room = 6 m 72 cm = 672 cm

To find the minimum number of square tiles of same size needed to cover the entire floor, we find the LCM of 896 cm and 672 cm.

LCM of 672 and 896 is 2 X 2 X 2 x 2 X 2 X 2 X 2 X 3 X 7 = 2688

Number of square tiles =( 2688/672×2688/896)

= 4×3 = 12

58. In a school library, there are 780 books of English and 364 books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves such that each shelf should have the same number of books of each subject. What should be the minimum number of books in each shelf?

Solution:

Given

In a school library, there are 780 books of English and 364 books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves such that each shelf should have the same number of books of each subject.

Number of books of English = 780

Number of books of Science = 364

780 = 2x2x3x5x13

364 = 2 x 2 x 7 x 13

The HCF of 780 and 364 = 2 x 2 x 13 = 52

Thus, the minimum number of books in each shelf = 52

59. In a colony of 100 blocks of flats numbering 1 to 1 00, a school van stops at every sixth block while a school bus stops at every tenth block. On which stops will both of them stop if they start from the entrance of the colony?

Solution:

Given

In a colony of 100 blocks of flats numbering 1 to 1 00, a school van stops at every sixth block while a school bus stops at every tenth block.

There are 100 blocks in a colony numbering 1 to 100.

A school van stops at every sixth block and a school bus stops at every tenth block.

We have to find the common stops at which they both stop if they start from a same entrance.

We need to find the LCM of 6 and 10.

The LCM of 6 and 10 is 2 * 3 * 5 = 30. i.e.

Firstly both will stop at 30th block, then at 60th block and lastly at 90th block.

60. Test the divisibility of following numbers by 11

(1) 5335
(2) 9020814

Solution: (1) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number 5335 is (5 + 3) = (3 + 5) = 8- 8 = 0, which is divisible by 11.

5335 is divisible by 11.

(2) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from
the right) of the number 9020814 is (4 + 8 + 2 + 9) -(1 + 0 + 0) = 23-1 = 22, which is divisible by 11.

9020814 is divisible by 11.

61. Using divisibility tests, determine which of the following numbers are divisible by 4?

(1) 4096
(2) 21084
(3) 31795012

Solution: (1) We have, 4096

Since, the last two digits 96 is divisible by 4.

4096 must be divisible by 4.

(2) We have, 21084

Since, the last two digits 84 is divisible by 4.

21084 must be divisible by 4.

(3) We have, 31795012

Since, the last two digits 12 is divisible by 4.

31795012 must be divisible by 4.

62. Using divisibility test determine which of the following numbers are divisible by 9?

(1) 672
(2) 5652

Solution: (1) We have, 672

Since, the sum of all the digits of 672 is 15, which is not divisible by 9.

672 is not divisible by 9.

(2) We have, 5652

Since, the sum of all the digits of 5652 is 18, which is divisible by 9.

5652 must be divisible by 9.

Class 6 Maths Chapter 2 Whole Numbers

Directions: In questions 1 to 11, out of the four options, only one is correct. Write the correct answer.

1. The number of whole numbers between 38 and 68 is

(1) 31
(2) 30
(3) 29
(4) 28

Solution (3): There are 29 whole numbers between 38 and 68.

2. The product of the successor and predecessor of 999 is

(1) 999000
(2) 998000
(3) 989000
(4) 1998

Solution (2): Successor of 999 = 999 +1 = 1000

Predecessor of 999 = 999-1 = 998

Now, their product = 998 x 1000 = 998000

Read and Learn More Class 6 Maths Exemplar Solutions

3. The product of a non-zero whole number and its successor is always

(1) an even number
(2) an odd number
(3) a prime number
(4) divisible by 3

Solution (1): The product of a non-zero whole number (even/odd) and its successor (odd/ even) is always an even number.

4. A whole number is added to 25 and the same number is subtracted from 25. The sum of the resulting numbers is

(1) 0
(2) 25
(3) 50
(4) 75

Solution(3): Let the whole number be x.

According to question,

Required sum = (x + 25) + (25- x)

= x + 25 + 25- x = 50

5. Which of the following is not true?

(1) (7 + 8) + 9 = 7 + (8 + 9)
(2) (7 x 8) x 9 = 7 x (8 x 9)
(3) 7 + 8 x 9 = (7 + 8) x (7 + 9)
(4) 7 x (8 + 9) = (7 x 8) + (7 x 9)

Solution (3): 7 + 8 x 9 = 7 + 72 = 79,

(7 + 8) x (7 + 9)- 15 × 16 = 240 and 79 x 240

6. By using dot (•) patterns, which ofthe following numbers can be arranged in all the three ways namely a line, a triangle and a rectangle?

(1) 9
(2) 10
(3) 11
(4) 12

Solution (2): As we know that every number can be arranged as a line.

The number 10 can be shown as

Also, the number 10 can be shown as a triangle as given below :

And, the number 10 can also be shown as a rectangle, as given below :

7. Which of the following statements is not true?

(1) Both addition and multiplication are associative for whole numbers.
(2) Zero is the identity for multiplication of whole numbers.
(3) Addition and multiplication both are commutative for whole numbers.
(4) Multiplication is distributive over addition for whole numbers.

Solution (2): Zero is the identity for addition of whole numbers.

8. Which of the following statements is not true?

(1) 0 + 0 = 0
(3) 0x0 = 0
(2) 0-0 = 0
(4) 0-0 = 0

Solution (4): 0 + 0 is not defined.

9. The predecessor of 1 lakh is

(1) 99000
(2) 99999
(4) 100001
(3) 999999

Solution (2): 1 lakh = 100000

Predecessor of 100000 = 100000-1 = 99999

10. The successor of 1 million is

(1) 2 millions
(2) 1000001
(3) 100001
(4) 10001

Solution(2): 1 million = 1000000

Successor of 1000000 = 1000000 +1 = 1000001

11. Number of even numbers between 58 and 80 is

(1) 10
(2) 11
(3) 12
(4) 13

Solution (1): Even numbers between 58 and 80 are 60, 62, 64, 66, 68, 70, 72, 74, 76, 78.

So, here are 10 even numbers between 58 and 80.

Directions: In questions 12 to .37 stole whether the given statements are true (T) or false (F)

12. Successor of a one digit number is always a one digit number.

Solution: False

The successor of a one digit number 9 is a two digit number, i.c., 10.

13. Successor of a 3-digit number is always a 3-digit number.

Solution: False

The successor of a 3-digit number 999 is a 4-digit number, i.e., 1000.

14. Predecessor of a two digit number is always a two digit number.

Solution: False

The predecessor of a two digit number 10 is a one digit number, i.e., 9.

15. Every whole number has its successor.

Solution: True

16. Every whole number has its predecessor.

Solution: False

0 is a whole number and it does not have any predecessor.

17. Between any two natural numbers, there is one natural number.

Solution: False

Since, 1 and 2 are two natural numbers and there is no natural number between1 and 2.

18. The smallest 4-digit number is the successor of the largest 3-digit number.

Solution: True

The smallest 4-digit number is 1000, i.e., 999 +1 and the largest 3-digit number is 999.

19. Of the given two natural numbers, the one having more digits is greater.

Solution: True

20. Natural numbers are closed under addition.

Solution: True

The sum of two natural numbers is also a natural number.

21. Natural numbers are not closed under multiplication.

Solution: False

The multiplication of two natural numbers is also a natural number.

22. Natural numbers are closed under subtraction.

Solution: False

2 and 5 are natural numbers, but their subtraction, 2-5 =-3 isnot anatural number.

23. Addition is commutative for natural numbers.

Solution: True

Since, a + b = b + a, where ‘a’ and ‘V are natural numbers.

24. 1 is the identity for addition of whole numbers.

Solution: False

0 is the identity for addition of whole numbers.

As 0 + a = a + 0 = a, where a is any whole number.

25. 1 is the identity for multiplication of whole numbers.

Solution: True

a×1 = 1 x a = a, where ‘a’ is any whole number.

26. There is a whole number which when added to a whole number, gives the number itself.

Solution: True

27. There is a natural number which when added to a natural number, gives the number itself.

Solution: False

28. If a whole number is divided by another whole number, which is greater than the first one, the quotient is not equal to zero.

Solution: True

29. Any non-zero whole number divided by itself gives the quotient 1.

Solution: True

30. The product of two whole numbers need not be a whole number.

Solution: False

The product of any two whole numbers will always be a whole number.

31. A whole number divided by another whole number greater than 1 never gives the quotient equal to the former.

Solution : True

32. Every multiple of a number is greater than or equal to the number.

Solution: True

33. The number of multiples of a given number is finite.

Solution: False

Since, the number of multiples of a given number is infinite.

34. Every number is a multiple of itself.

Solution: True

35. Every whole number is the successor of another whole number.

Solution: False

0 is a whole number but it is not a successor of any whole number.

36. Sum of two whole numbers is always less than their product.

Solution: False

Product of two whole numbers may or may not be greater than their sum.

37. If the sum of two distinct whole numbers is odd, then their difference also must be odd.

Solution: True

Directions:In questions 38 to 59,fill in the blanks to make the statements true.

38. The smallest whole number is_________.

Solution 0 : 0 is the smallest whole number.

39. Successor of 1061 59 is_________.

Solution 106160: Successor of 106159 is 106159 +1, i.e., 106160

40. Predecessor of 1 00000 is_________.

Solution 99999: Predecessor of 100000 is 100000-l, i.e., 99999

41. 400 is the predecessor of__________.

Solution 401: 400 is the predecessor of 400 + 1, i.e., 401

42.________ is the successor of the largest 3 digit number.

Solution 1000: Largest 3 digit number = 999

Successor of 999 is 999 + 1, i.e., 1000

43. If 0 is subtracted from a whole number, then the result is the_________ itself.

Solution: Number

44. Whole numbers are closed under and under________.

Solution: Addition, multiplication

45. Natural numbers are closed under and under_______.

Solution: Addition, multiplication

46. Division of a whole number by ___________ is not defined.

Solution: 0

47. Multiplication is distributive over__________ for whole numbers.

Solution: Addition

48. 2395 x_________ = 6195×2395

Solution 6195: Since, multiplication is commutative for whole numbers

49. 1001 x 2002 = 1001 x (1001 +_________)

Solution: 1001

1001 x 2002 = 1001 x (1001 +1001)

50. 10001 x 0 = __________

Solution: 0.

10001 x 0 = 0

51. 2916 x_________ = 0

Solution: 0.

2916 x 0 = 0

52. 9128 x_______= 9128

Solution 1 : Since,1 is the multiplicative identity for whole numbers.

53. 125 + (68 + 17) = (125 +_______)+ 17

Solution 68: Since, addition is associative for whole numbers.

54. 8925x 1 =_____

Solution: 8925

8925x 1= 8925

55. 19 x 12 + 19 = 19 x (12 +_______)

Solution 1: Since multiplication is distributive over addition for whole numbers.

56. 24 x 35 = 24 x 1 8 + 24 X_______

Solution: 17

24 x 35 = 24 x 1 8 + 24 X 17

57. 32x(27×19) = (32x__________)x19

Solution 27: Since, multiplication is associative for whole numbers.

58. 786×3 + 786×7 =_______

Solution: 7860 : 786 x 3 + 786 x 7 = 786 * (3 + 7)

= 786 x 10 = 7860

786×3 + 786×7 =7860

59. 24 x 25 = 24 X □÷4= 600

Solution: 100

60. Given below are two columns – Column I and Column II. Match each item of Column I with the corresponding item of Column II.

Class 6 Maths Chapter 1 Knowing Our Numbers

Directions: In questions 1 to 13, out of the four options, only one is correct. Write the correct answer.

1. The product of the place values of two 2’s in 428721 is

(1) 4
(2) 40000
(3) 400000
(4) 40000000

Solution : (3) Place values of 2’s in 428721 are 20000 and 20

The required product = 20000 x 20 = 400000

2. 3x 1 0000 + 7x 1 000 + 9×1 00 + 0x10 + 4 is the same as

(1) 3794
(2) 37940
(3) 37904
(4) 379409

Read and Learn More Class 6 Maths Exemplar Solutions

Solution (3) : 3 x 10000 + 7 x 1000 + 9 x 100 + 0 x 10 + 4

= 30000 + 7000 + 900 + 4 = 37904

3. If 1 is added to the greatest 7-digit number, it will be equal to

(1) 10 thousand
(2) 1 lakh
(3) 10 lakh
(4)1,00,00,000

Solution (4): The greatest 7-digit number = 99,99,999

Now, 99,99,999 +1 = 1,00,00,000

4. The expanded form of the number 9578 is

(1) 9×10000 + 5×1000 + 7×10 + 8×1
(2) 9x 1000 + 5x 100 + 7x 10 + 8×1
(3) 9x 1000 + 57x 10 + 8×1
(4) 9x 100 + 5x 100 + 7x 10 + 8×1

Solution (2) : Expanded form of 9578 = 9 x 1000 + 5 x 100 + 7 x 10 + 8×1

5. When rounded off to the nearest thousands, the number 85642 is

(1) 85600
(2) 85700
(3) 85000
(4) 86000

Solution (4): Round off 85642 to the nearest thousands = 86000

6. The largest 4-digit number, using any one digit twice, from digits 5, 9, 2 and 6 is

(1) 9652
(2) 9562
(3) 9659
(4) 9965

Solution: (4): The largest 4-digitnumber formed by 5, 9, 2 and 6 using digit ‘9’ twice = 9965

7. In Indian System of Numeration, the number 58695376 is written as

(1) 58,69,53,76
(2) 58,695,376
(3) 5,86,95,376
(4) 586,95,376

Solution: (3): In Indian System of Numeration, 58695376 can be written as 5,86,95,376

8. One million is equal to

(1) 1 lakh
(3) 1 crore
(2) 10 lakh
(4) 10 crore

Solution: (2): One million = 1,000,000

Also, 10,00,000 = 10 lakh

9. The greatest number which on rounding off to nearest thousands gives 5000, is

(1) 5001
(2) 5559
(3) 5999
(4) 5499

Solution: (4) : (1) Rounding off 5001 to nearest thousands = 5000

(2) Rounding off 5559 to nearest thousands = 6000

(3) Rounding off 5999 to nearest thousands = 6000

(4) Rounding off 5499 to nearest thousands = 5000

And 5499 > 5001

10. Keeping the place of 6 in the number 6350947 same, the smallest number obtained by rearranging other digits is

(1) 6975430
(2) 6043579
(3) 6034579
(4) 6034759

Solution (3): The new number formed = 6034579

11. Which of the following numbers in Roman numerals is incorrect?

(1) LXXX
(2) LXX
(3) LX
(4) LLX

Solution: (4) : LLX is incorrect because L can be used exactly once in Roman numerals.

12. The largest 5-digit number having three different digits is

(1) 98978
(2) 99897
(3) 99987
(4) 98799

Solution: (3) : The largest 5-digit number with three different digits is 99987

13. The smallest 4-digit number having three different digits is

(1) 1102
(2) 1012
(3) 1020
(4) 1002

Solution: (4) : The smallest 4-digit number with three different digits is 1002.

Directions: In questions 14 to 29 state whether the given statements are true (T) or false (F)

14. In Roman numeration, a symbol is not repeated more than three times.

Solution: True

15. In Roman numeration, if a symbol is repeated, its value is multiplied as many times as it occurs.

Solution: False

If a symbol is repeated, then its value is added as many times as it occurs in Roman numeration

16. 5555 = 5×1000 + 5×100 + 5×10 + 5×1

Solution: True

17. 39746 = 3 × 10000 + 9 × 1000 + 7 × 100 + 4×10 + 6

Solution: True

18. 82546 = 8 x 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6

Solution: False

Since, 82546 = 8 × 10000 + 2× 1000 + 5×100 + 4×10 + 6

19. 532235 = 5 X 100000 + 3 x 10000 + 2 x 1000 + 2X 100 + 3X 10 + 5

Solution: True

20. XXIX = 31

Solution: False

Since, XXIX = 29

21. LXXIV = 74

Solution: True

22. The number LIV is greater than LVI.

Solution: False

Since, LIV = 54 and LVI = 56

LVI is greater than LTV

23. The numbers 4578, 4587, 5478, 5487 are in descending order.

Solution: False

Since, the numbers 4578, 4587, 5478, 5487 are in ascending order.

24. The number 85764 rounded off to nearest hundreds is written as 85700.

Solution: False

The number 85764 rounded off to nearest hundreds is 85800.

25. Estimated sum of 7826 and 1 2469 rounded off to hundreds is 20,000.

Solution: False

Round off 7826 to hundreds is 7800.

Round off 12469 to hundreds is 12500.

Required sum = 7800 + 12500 = 20300

26. The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875403.

Solution: False

The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875430.

27. The number 81652318 will be read as eighty one crore six lakh fifty two thousand three hundred eighteen.

Solution: False

The number 81652318 will be read as eight crore sixteen lakh fifty two thousand three hundred eighteen.

28. The largest 4-diglt number formed by the digits 6, 7, 0, 9 using each digit only once is 9760.

Solution: True

29. Among kilo, milli and centi, the smallest is centi.

Solution: False

Among kilo, milli and centi, the smallest is milli.

Directions: In questions 30 to 44, fill in the blanks to make the statements true

30. (1) 10 million =__________ crore

Solution:1

(2) 10 lakh =_______ million.

Solution: 1

31. (1) 1 metre = ____________ millimetres

Solution: 1000

(2) 1 centimetre =___________ millimetres

Solution: 10

(3) 1 kilometre =__________millimetres

Solution: 10,00,000

33. 100 thousands =________ lakh.

Solution: 1

34. Height of a person is 1 m 65 cm. His height in millimetres is_________.

Solution: 1650 : 1 m 65 cm = (1000 + 650) mm = 1650mm

35. Length of river ‘Narmada’ is about 1290 km. Its length in metres is___________.

Solution: 1290000 : 1290 km = (1290 x 1000) m = 1290000m

36. The distance between Srinagar and Leh is 422 km. The same distance in metres is___________

Solution: 422000: 422 km= (422 x 1000) m = 422000 m

37. Writing of numbers from the greatest to the smallest is called an arrangement in__________ order.

Solution: Descending

38. By reversing the order of digits of the greatest number made by five different non zero digits, the new number Is the________ number of five digits.

Solution: Smallest*

By reversing the order of digits of the greatest number made by five different non zero digits, the new number is the smallest number of these digits.

39. By adding 1 to greatest_______ digt number, we get ten lakh.

Solution: 6: Greatest 6-digit number = 999999 By adding1 to 999999, we get 1000000.

40. The number five crore twenty three lakh seventy eight thousand four hundred one can be written, using commas, in the Indian
System of Numeration as_______.

Solution: 5,23,78,401

41. In Roman Numeration, the symbol X can be subtracted from ________, M and C only.

Solution: L

42. The number 66 in Roman numerals is__________.

Solution: LXVI: 66 =LXVI

43. The population of Pune was 2,538,473 in 2001. Rounded off to nearest thousands, the population was____________.

Solution: 2,538,000

44. The smallest 6 digit natural number ending in 5 is__________

Solution: 100005

45. Arrange the following numbersin descending order: 8435, 4835, 13584, 5348, 25843

Solution: Descending order of given numbers is, 25843, 13584, 8435, 5348, 4835

46. Of the following numbers which is the greatest? Which is the smallest? 38051425, 30040700, 67205602

Solution: The greatest number is 67205602 and the smallest number is 30040700.

47. Write In expanded form :

(1) 74836
(2) 574021
(3) 8907010

Solution:(1) 74836- 7 x10000) x 4 x 1000 + 8 x 100 +3x 10+6×1

(2) 574021 =5x 1 00000 + 7x 1 0000 + 4x 1000 + 0x 100 + 2x 10+ 1 x1

(3) 8907010 = 8 x 1000000 + 9x 1 00000 + 0 x 10000 + 7 x 1000 + 0 x 100 + 1 x 10 + 0 x 1

48. As per the census of 2001, the population of four states are given below. Arrange the states in ascending and descending order of their population.

(1) Maharashtra 96878627
(2) Andhra Pradesh 76210007
(3) Bihar 82998509
(4) Uttar Pradesh 166197921

Solution: Ascending order -» (2), (3), (1), (4)

Descending order —» (4), (1), (3), (2)

49. The diameter of Jupiter is 142800000 metres. Insert commas suitably and write the diameter according to International System of Numeration.

Solution:

Given

The diameter of Jupiter is 142800000 metres.

The diameter of Jupiter is 142,800,000 metres.

50. India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in Indian System of Numeration, using commas suitably.

Solution:

Given

India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001.

Total increaseinpopulation from 1961 to 2001= 1028 millions- 439 millions = 589 millions

According toIndian System of Numeration, the increase in population = 58,90,00,000

51. Radius of the Earth is 6400 km and that of Mars is 4300000 m. Whose radius is bigger and by how much?

Solution:

Given

Radius of the Earth is 6400 km and that of Mars is 4300000 m.

Radius of the Earth = 6400 km = 6400 x 1000 m = 6400000 m

And radius of the Mars = 4300000 m

Radius of the Earth is greater than the radius of the Mars by (6400(X)0- 4300000) m = 2100000 m.

52. In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively. Write the populations of these two states in words.

Solution:

Given

In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively.

The population of Tripura = 3,199,203 i.e., Three million one hundred ninety-nine thousand two hundred three.

And the population of Meghalaya = 2,318,822, i.e., Two million three hundred eighteen thousand eight hundred twenty two.

53. In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month. Find the difference of the number of children getting polio drops in the two months.

Solution:

Given

In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month.

Number of children getting polio drops in March 2008 = 2,12,583

Number of children getting polio drops in April 2008 = 2,16,813

The required difference of the number of children getting polio drops in the two months = 2,16,813- 2,12,583 = 4,230

54. A person had ? 1000000 with him. He purchased a colour T.V. for ? 16580, a motor cycle for ? 45890 and a flat for ? 870000. How much money was left with him?

Solution:

Given :

A person had 1000000 with him. He purchased a colour T.V. for 16580, a motor cycle for  45890 and a flat for  870000.

Total amount a personhad = 1000000

The amounthe spent on a colour T.V. = 16580

The amount he spent on a motorcycle = 45890

The amount he spent on a flat =870000

Total amount he spent = (16580 + 45890 + 870000) = 932470

Thus, the amount left with him = 1000000- 932470 = 67530

55. Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district. Find the number of the remaining vitamin tablets.

Solution:

Given:

Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district.

Total tablets of Vitamin A= 180000

Number of tablets distributed among the students in a district = 18734

The number of remaining vitamin tablets -180000-18734-161266

56. Chinmay had ? 610000. He gave 87500 to Jyoti, 126380 to Javed and 350000 to John. How much money was left with him?

Solution:

Chinmay had total amount = 610000

The amount he gave to Jyoti = 87500

The amount he gave to Javed = 126380

The amount he gave to John = 350000

Total amount given by Chinmay = (87500 + 126380+ 350000) = 563880

Thus, the amount left with him = 610000- 563880 = 46120

57. Find the difference between the largest number of seven digits and the smallest number of eight digits.

Solution: The smallest number of eight digits = 10000000

The largest number of seven digits = 9999999

The required difference = 10000000-9999999 =1

58. A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number. What are these digits?

Solution:

Given

A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number.

A mobile number consists of 10 digits.

If the first four digits of the number are 9, 9, 8 and 7 and the last three digits of the number are 3, 5 and 5.

Thus, for the greatest possible number, the remaining distinct digits are 6, 4 and 2.

59. A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0.

Solution:

Given

A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9.

A mobile number consists of 10 digits.

If the first four digits are 9, 9, 7 and 9.

Thus, the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0 is 9979003568.

60. In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place. Write the number.

Solution:

Given

In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place.

A number consists of 5 digits.

Now, the digit at ten’s place = 4,

the digit at unit’s place = 1/4×4=1,

tine digit at hundred’s place = 0,

the digit at thousand’s place = 5 x1 = 5

the digit at ten thousand’s place = 2×4 =8

Therefore, the number is 85041.

61. Find the sum of the greatest and the least six digit numbers formed by the digits 2, 0, 4, 7,6, 5 using each digit only once.

Solution: By using the digits 2, 0, 4, 7, 6, 5

The greatest number formed = 765420,

and the least number formed = 204567

The required sum= 765420 + 204567= 969987

62. A factory has a container filled with 35874 litres of cold drink. In how many bottles of 200 ml capacity each can it be filled?

Solution:

Given:

A factory has a container filled with 35874 litres of cold drink.

Quantity of cold drink in a container = 35874 litres = 35874 x 1000 ml = 35874000 ml

The capacity of one bottle = 200 ml

The required number of bottles = 35874000 + 200 = 179370

Therefore, 179370 bottles can be filledby cold drink.

63. The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate. In all how many illiterate persons are there in the town?

Solution:

Given:

The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate.

Total population of a town = 450772

Since, one out of every 14 personsis illiterate.

The number of illiterate persons in the town = 450772 + 14 = 32198

Therefore, 32198 persons are illiterate in the town.

64. Make the greatest and the smallest 5-digit numbers using different digits in which 5 appears at ten’s place.

Solution: By using the digit 5 at ten’s place, the greatest 5-digit number is 98756, and the smallest 5-digitnumber is 10253

65. How many grams should be added to 2 kg 300 g to make it 5 kg 68 g?

Solution: 5 kg 68 g = (5 x 1000 + 68) g = 5068 g and 2 kg 300 g = (2 x 1000 + 300) g = 2300 g

The required number of grams should be added = 5068 g- 2300 g = 2768 g or 2kg 768 g

2768 grams should be added to 2 kg 300 g to make it 5 kg 68 g

66. A box contains 50 packets of biscuits each weighing 1 20 g. How many such boxes can be loaded in a van which cannot carry beyond 900 kg?

Solution:

Given

A box contains 50 packets of biscuits each weighing 1 20 g.

The total weight of a box containing 50 packets ofbiscuits each weighing 120 g = 50 x 120 g = 6000 g

The capacity of a van = 900 kg = 900 x 1000 g = 900000 g

The required number of boxes = 900000/6000 = 150

Therefore, 150 boxes can be loaded in the van.

67. How many lakhs make five billions?

Solution: 50000 lakhs make 5 billions.

68. How many millions make 3 crores?

Solution: 30 millions make 3 crores.

69.Estimate each of the following by rounding off each number to nearest hundreds:

(1) 874 + 478
(2) 793 + 397
(3) 11244 + 3507
(4) 17677 + 13589

Solution:(1) 874 rounded off to The nearest hundreds-900

478 rounded off to the nearest hundreds -500

Estimated sum = 900 + 500 – 1400

(2) 793 rounded off to the nearest hundreds = 800

397 rounded off to the nearest hundreds = 400

Estimated sum = 800 + 400 = 1200

(3) 11244 rounded off to the nearest hundreds= 11200

3507 rounded off to the nearest hundreds = 3500

Estimated sum = 11200 + 3500 = 14700

(4) 17677roundedofftothenearesthundreds = 17700

13589 rounded off to the nearest hundreds = 13600

Estimated sum = 17700 + 13600 = 31300

70. Estimate each of the following by rounding off each number to nearest tens:

(1) 11963-9369
(2) 76877-7783
(3) 10732-4354
(4) 78203-16407

Solution: (1) 11963 rounded off to the nearest tens- 11960

9369 rounded off to the nearest tens = 9370

Estimated difference=11960- 9370= 2590

(2) 76877 rounded off to the nearest tens = 76880

7783 rounded off to the nearest tens = 7780

Estimated difference= 76880- 7780= 69100

(3) 10732 rounded off to the nearest tens = 10730

4354 rounded off to the nearest tens = 4350

Estimated difference = 10730- 4350 = 6380

(4) 78203 rounded off to the nearest tens = 78200

16407 rounded off to the nearest tens = 16410

Estimated difference = 78200 – 16410 = 61790

71. Estimate each of the following products by rounding off each number to nearest tens:

(1) 87×32
(2) 311 × 113
(3) 3239 × 28
(4) 1385×789

Solution: (1) 87 rounded off to the nearest tens = 90

32 rounded off to the nearest tens = 30

Estimated product = 90 * 30 = 2700

(2) 311 rounded off to the nearest tens = 310

113 rounded off to the nearest tens = 110

Estimated product = 310 * 110 = 34100

(3) 3239 rounded off to the nearest tens= 3240

28 rounded off to the nearest tens = 30

Estimated product = 3240 * 30 = 97200

(4) 1385 rounded off to the nearest tens=1390

789 rounded off to the nearest tens = 790

Estimated product= 1390 * 790= 1098100

72. The population of a town was 78787 in the year 1991 and 95833 in the year 2001. Estimate the increase in population by
rounding off each population to nearest hundreds.

Solution:

Given

The population of a town was 78787 in the year 1991 and 95833 in the year 2001.

78787 rounded off to the nearest hundreds = 78800

95833 rounded off to the nearest hundreds = 95800

The estimated increase in population = 95800-78800 = 17000

73. Estimate the product 758 x 6784 using the general rule.

Solution: 758 can be rounded off to 800 and 6784 can be rounded off to 7000 Estimated product= 800 × 7000 = 5600000

The product 758 x 6784 using the general rule = 5600000

74. A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year. What is the total production of all the three items in that year?

Solution:

Given

A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year.

Number of shirts produced by the factory = 216315

Number of trousers produced by the factory = 182736

Number of jackets produced by the factory = 58704

Total production of the factory = 216315 + 182736 + 58704 = 457755

Total production of the factory = 457755

75. A vessel has 13 litres 200 mL of fruit juice. In how many glasses each of capacity 60 mL can it be filled?

Solution:

Given:

A vessel has 13 litres 200 mL of fruit juice.

Quantity of fruit juice in a vessel = 13 L 200 mL

= (13 x 1000 + 200) mL = 13200 mL

Capacity of one glass = 60 mL

The required number of glasses = 13200 ÷ 60 = 220

Therefore, 220 glasses can be filled by fruit juice.

76. A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight. Find the total weight that can be carried by both the vehicles.

Solution:

Given

A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight.

Total weight can be carried by a tempo= (482×15) kg = 7230kg

and the total weight can be carried by a van = (518×15) kg = 7770 kg

Thus, the total weight that can be carried by both the vehicles = (7230 + 7770) kg = 15000 kg

The total weight that can be carried by both the vehicles = 15000 kg

77. In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items. Find the total amount spent by her on the above items.

Solution:

Given :

In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items.

Amount spent by Leela on food and decoration = 216766

Amount spent by her on jewellery = 122322

Amount spent by her on furniture = 88234

Amount spent by her on kitchen items = 26780

Total amount spent by her = (216766 + 122322 + 88234 + 26780) = 454102

Total amount spent by her = 454102

78. A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule. Find the total weight in grams of medicine in 32 such boxes.

Solution:

Given :

A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule.

Quantity of medicine in one capsule = 500 mg

Quantity ofmedicinein 12 capsules or 1 strip = (500 x 12) mg = 6000 mg- 6 g

Quantity of medicine in 5 strips or 1 box = (6 x 5) g = 30 g

Quantity of medicinein 32 boxes = (30 x 32)g = 960g

The total weight in grams of medicine in 32 such boxes = 960g