NCERT Solutions For Class 8 Maths Chapter 11 Direct And Inverse Proportions

NCERT Class 8 Maths Chapter 11 Direct And Inverse Proportions Introduction

Question 1. Mohan prepares tea for himself and his sister. He uses 300 mL of water, 2 spoons of sugar, 1 spoon of tea leaves and, 50 mL of milk. How much quantity of each item will he need, if he has to make tea for five persons?

Solution:

Quantity of water required for 2 persons = 300 mL

Quantity of water required for 1 person = \(\frac{300}{2} \mathrm{~mL}\)

Quantity of water required for 5 persons \(\frac{300}{2} \times 5 \mathrm{~mL}\)

Quantity of sugar required for 2 persons = 2 spoons

Quantity of sugar required for 1 person \(\frac{2}{2} \text { spoons }\)

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Quantity of sugar required for 5 persons \(\frac{2}{2} \times 5=5 \text { spoons }\)

Quantity of tea leaves required for 2 persons = 1 spoon

Quantity of tea leaves required for 1 person \(\frac{1}{2} \text { spoons }\)

Quantity of tea leaves required for 5 persons \(=\frac{1}{2} \times 5=2 \frac{1}{2} \text { spoons }\)

Quantity of milk required for 2 persons = 50 mL

Quantity of milk required for 1 person = \(\frac{50}{2} \mathrm{~mL}\)

Quantity of milk required for 5 persons = \(\frac{50}{2} \times 5 \mathrm{~mL}=125 \mathrm{~mL}\)

Thus, Mohan will require 750 mL of water, 5 spoons of sugar, \(2 \frac{1}{2}\) Spoons of tea leaves, and 125 mL of milk for 5 persons.

Question 2. If two students take 20 minutes to arrange chairs for an assembly, then how much time would five students have to do the same job?

Solution:

Time taken by 2 students = 20 minutes

Time taken by 1 student = 20 x 2 minutes

Time taken by 5 students = \(\frac{20 \times 2}{5} \text { minutes }=8 \text { minutes }\)

We come across many such situations in our day-to-day life, where we need to see variation in one quantity bringing in variation in the other quantity. Quantities so related are called variables and the relation is called variation.

For example :

1. If the number of articles purchased increases, the total cost also increases.
2. The more money deposited in a bank, more is the interest earned.
3. As the speed of a vehicle increases, the time taken to cover the same distance decreases.
4. For a given job, the more the number of workers less will be the less time taken to complete the work.

Question 3. Write five more such situations where a change in one quantity leads to  change in another quantity.

Solution:

Five more such situations where a change in one quantity leads to a change in another quantity are as follows :

1. If the amount of the commodity purchased increases, the total cost also increases.
2. If the amount of loan taken from a bank increases, then the interest to be paid also increases.
3. At a particular temperature, if the pressure of gas increases, then the volume decreases.
4. If the number of pipes to fill a tank increases, the time taken to fill the tank decreases.
5. If the number of guests increases, the number of days for which a given amount of food will last decreases.

Question 4. How do we find out the quantity of each item needed by Mohan? or, the time five students take to complete the job?

Solution:

We need to study some concepts of variation as follows :

1. Direct variation
2. Inverse variation

Two quantities may be linked in two ways :

Both increase or decrease together proportionally.

If one increases, the other decreases proportionally, and vice-versa.

The first way is named a direct variation whereas the second way is named as an inverse variation.

1. Direct Proportion

If two quantities are related in such a way that an increase in one leads to a corresponding proportional increase in the other, then such a variation is called direct variation.

Thus, two numbers x and y are said to vary in direct proportion if

\(\frac{x}{y}\)= k or x = ky where k is a constant.

In this case, if ÿ1,y2 are the values ofy corresponding to the values x1, x2 of x respectively, then

⇒  \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Question 5. Let the consumption of petrol be x liters and the corresponding distance traveled by km. Now, complete, the following table:

We find that as the value increases, the value also increases in such a way that the ratio \(\frac{x}{y}\) does not change; it remains constant (say It). In this case, it is……(check it !)
Solution:

In this case, it is \(\frac{1}{15}\)

Question 6. Think of a few more examples for direct proportion.

Solution:

1. A few more examples of direct proportion are as follows :
2. Length of the cloth purchased and its total cost.
3. Number of months and total salary.
4. Number of hours of production and the amount of the commodity produced.

Take a clock and fix its minute hand, at 12.

Record the angle turned through by the minute hand from its original position and the time that has passed, in the following table

What do you observe about T and A ? Do they increase together? Is \(\frac{T}{A}\) same every time ?

Is the angle turned through by the minute hand directly proportional to the time that has passed? Yes; From the above table, you can also see

T₁ : T₂ = A₁ : A₂, because

T₁ : T₂ = 15 : 30 = 1:2

A₁ : A₂ = 90 : 180 = 1:2

Check if T₂ : T₃ = A₂ : A₃ and T₃ : T₄ = A₃ : A₄

You can repeat this activity by choosing your time interval.

We observe about T and A that they increase together and \(\frac{T}{A}\)is the same every time. A Yes, the angle turned by the minute hand is directly proportional to the time that has passed.

On checking, we find that

T₂ : T₃ = A₂ : A₃ = 2:3

T₃ : T₄ = A₃ : A₄ = 3:4

Ask your friend to fill in the following table and find the ratio of his age to the corresponding age of his mother:

What do you observe?

Do F and M increase (or decrease) together? Is — the same every time? No.

You can repeat this activity with other friends and write down your observations.

Solution:

We observe that F and M increase (or decrease) together. But — is not the same every
time.

Note: It is not. necessary that the variables increasing (or decreasing) together are always in direct proportion.

For example :

Height and age of an individual

Height and weight of an individual

Height of mango tree and the number of mango fruits growing on its branches.

Question 7. Observe the following tables and find if x and y are directly proportional:

Solution:

We have \(\frac{20}{40}=\frac{1}{2}\)

⇒ \(\frac{17}{34}=\frac{1}{2}\)

⇒ \(\frac{14}{28}=\frac{1}{2}\)

⇒ \(\frac{11}{22}=\frac{1}{2}\)

⇒ \(\frac{8}{16}=\frac{1}{2}\)

⇒ \(\frac{5}{10}=\frac{1}{2}\)

⇒ \(\frac{2}{4}=\frac{1}{2}\)

We find that each ratio is the same.

Hence, x and y are directly proportional.

We have \(\frac{6}{4}=\frac{3}{2}\)

⇒ \(\frac{10}{8}=\frac{5}{4}\)

⇒ \(\frac{14}{12}=\frac{7}{6}\)

⇒ \(\frac{18}{16}=\frac{9}{8}\)

⇒ \(\frac{22}{20}=\frac{11}{10}\)

⇒ \(\frac{26}{24}=\frac{13}{12}\)

⇒ \(\frac{30}{28}=\frac{15}{14}\)

We find that all the ratios are not the Same.

Hence, x and y are not directly proportional.

⇒ \(\frac{5}{15}=\frac{1}{3}\)

⇒ \(\frac{8}{24}=\frac{1}{3}\)

⇒ \(\frac{12}{36}=\frac{1}{3}\)

⇒ \(\frac{15}{60}=\frac{1}{4}\)

⇒ \(\frac{18}{72}=\frac{1}{4}\)

⇒ \(\frac{20}{100}=\frac{1}{5}\)

We find that all the ratios are not the same.

Hence, x and y are not directly proportional.

Question 8. Principal = f1,000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time.

Solution:

1 year \(\frac{\mathrm{P} \times r \times t}{100}\)

⇒ \(₹ \frac{1000 \times 8 \times 1}{100}\)

₹ 80

We have \(\frac{80}{1}=80\)

2 years \(\frac{\mathrm{P} \times r \times t}{100}\)

⇒ \(₹ \frac{1000 \times 8 \times 2}{100}\)

₹ 160

\(\frac{160}{2}=80,\)

3 years \(\frac{\mathrm{P} \times r \times t}{100}\)

⇒\(₹ \frac{1000 \times 8 \times 3}{100}\)

₹ 240

⇒ \(\frac{240}{3}=80\)

We find that each ratio is the same.

Hence, simple interest changes in direct proportion with a period.

Compound Interest (in ?) for different periods

1 year \(\mathrm{P}\left(1+\frac{r}{100}\right)^t-\mathrm{P}\)

⇒ \(1000\left(1+\frac{8}{100}\right)-1000\)

⇒ \(1080-1000=₹ 80\)

We have \(\frac{80}{1}=80\)

2 years \(\mathrm{P}\left(1+\frac{r}{100}\right)^t-\mathrm{P}\)

v\(1000\left(1+\frac{8}{100}\right)^2-1000\)

₹ 166.40

⇒ \(\frac{166.40}{2}=83.20\)

3 years \(\mathrm{P}\left(1+\frac{r}{100}\right)^t-\mathrm{P}\)

⇒ \(1000\left(1+\frac{8}{100}\right)^3-1000\)

₹ 259.712

⇒ \(\frac{259.712}{3}=86.5706\)

We find that all the ratios are not the same.

Hence, compound interest does not change in direct proportion to the period.

Question 9. If we fix the period and the rate of interest, simple interest changes proportionally with the principal. Would there be a similar relationship for compound interest? Why?

Solution:

Case 1

⇒ \(\text { S.I. }=\frac{\mathrm{P} r t}{100}\)

⇒ \(\frac{\text { S.I. }}{\mathrm{P}}=\frac{r t}{100}\)

r and t are fixed

⇒ \(\frac{r t}{100} \text { is constant }\)

⇒ \(\frac{\mathrm{S} . \mathrm{I}}{\mathrm{P}}=\text { constant }\)

S.I. changes proportionally with principal

Case 2

⇒ \(\text { C.I. }=\mathrm{P}\left\{\left(1+\frac{r}{100}\right)^t-1\right\}\)

⇒ \(\frac{\mathrm{C} . \mathrm{I}}{\mathrm{P}}=\left(1+\frac{r}{100}\right)^t-1\)

r and tare fixed

⇒ \(\left(1+\frac{r}{100}\right)^t-1 \text { is constant }\)

⇒ \(\frac{\mathrm{C} . \mathrm{I}}{\mathrm{P}}=\text { constant }\)

C.I. changes proportionally with principal

There is a similar relationship for compound interest

Question 10. Take a map of your State. Note the scale used there. Using a ruler, measure the “map distance’’ between any two cities. Calculate the actual distance between them.

Solution:

Let the scale be 1 cm = 100 km

Let the distance between any two cities on the map be 5 cm.

Then, the actual distance between these two cities = 100 x 5 km = 500 km

Direct And Inverse Proportions Exercise 11.1

Question 1. Following are the car parking charges near a railway station up to

1. 4 hours ₹ 60
2. 8 hours ₹ 100
3. 12 hours ₹ 140
4. 24 hours ₹ 180

Check if the parking charges are in direct proportion to the parking time.

Solution:

We have

⇒ \(\frac{60}{4}=\frac{60 \div 4}{4 \div 4}=\frac{15}{1}\) HCF (60, 4) = 4

⇒ \(\frac{100}{8}=\frac{100 \div 4}{8 \div 4}=\frac{25}{2}\) HCF (100, 8) = 4

⇒ \(\frac{140}{12}=\frac{140 \div 4}{12 \div 4}=\frac{35}{3}\) HCF (140, 12) = 4

⇒ \(\frac{180}{24}=\frac{180 \div 12}{24 \div 12}=\frac{15}{2}\) HCF (180, 24) = 12

Since all the ratios are not the same, therefore, the parking charges are not in direct proportion to the parking time.

Question 2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of the base. In the following table, find the parts of the base that need to be added

Solution:

Let the number of parts of red pigments be x and the number of parts of the base is y.

As the number of parts of red pigments increases, several parts of the base also increase in the same ratio. So it is a case of direct proportion.

We make use of the relation of the type \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Here

x₁= 1

y₁ = 8 and x₂ = 4

Therefore \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \text { gives }\)

⇒ \(\frac{1}{8}=\frac{4}{y_2}\)

y₂ = 8 x 4

y₂ = 32

Hence, 32 parts of the base are needed to be added to 4 parts of red pigments.

Here,

x₁ = 1

y₄ = 8 and x3 = 7

Therefore \(\frac{x_1}{y_1}=\frac{x_3}{y_3}\) gives

⇒ \(\frac{1}{8}=\frac{7}{y_3}\)

y₃ = 8 x 7

y₃ = 56

Hence, 56 parts of the base are needed to be added to 7 parts of red pigments.

Here

x₁= 1

y₁= 8

x₄ = 12

Therefore,\(\frac{x_1}{y_1}=\frac{x_4}{y_4}\) gives

⇒ \(\frac{1}{8}=\frac{12}{y_4}\)

y4= 12 x 8

y₄ = 96

Hence, 96 parts of the base are needed to be added to 12 parts of red pigments.

Here,

x₁= 1

y₁ = 8

x₅ = 20

Therefore \(\frac{x_1}{y_1}=\frac{x_5}{y_5}\)gives

⇒ \(\frac{1}{8}=\frac{20}{y_5}\)

y₅ = 8 x 20

y₅ = 160

Hence, 160 parts of the base are needed to be added to 20 parts of red pigments.

Question 3. In Question 2 above, if part of the red pigment requires 75 mL of the base, how much red pigment should we mix with 1800 mL of the base?

Solution:

Let the number of parts of red pigments be and the amount of base be y mL.

As the number of parts of red pigment increases, the amount of base also increases in the same ratio. So it is a case of direct proportion.

We make use of the relation of the type.

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Here x1=1

y₂ = 75 and ,y₂ = 1800

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\) give

⇒ \(\frac{1}{75}=\frac{x_2}{1800}\)

⇒ \(\frac{1}{75} =\frac{x_2}{1800}\)

⇒ \(75 x_2 =1800 \)

⇒ \(x_2 =\frac{1800}{75}=\frac{600}{25}\)

x₂ = 24

Hence, 24 parts of the red pigment should be mixed with 1800 mL of base.

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow x_1 y_2=x_2 y_1\)

⇒ \(\frac{x_1}{x_2}=\frac{y_1}{y_2}\)

Question 4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Solution:

Let the machine fill x bottles in five hours. We put the given information in the form of a table as shown below :

The more the number of hours, the more the y2 number of bottles would be filled. So, the number of bottles filled and the number of hours are directly proportional to each other.

⇒ \(\frac{x_1}{x_2}=\frac{y_1}{y_2}\)

⇒ \(\frac{840}{x_2}=\frac{6}{5}\)

6×2 = 840 x 5

⇒ \(x_2=\frac{840 \times 5}{6}=700\)

Hence, 700 bottles will be filled in five hours.

Question 5. A photograph of bacteria enlarged 50,000 limes allies a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what, would be its enlarged length?

Solution:

Actual length of the bacteria \(\frac{5}{50000} \mathrm{~cm}=\frac{1}{10000} \mathrm{~cm}=10^{-4} \mathrm{~cm}\)

Let the enlarged length be cm. We put the given information in the form of a table as shown below :

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

⇒ \(\frac{50000}{5}=\frac{20000}{y_2}\)

50000 y₂ = 5 x 20000

⇒ \(y_2=\frac{5 \times 20000}{50000}=\frac{5 \times 2}{5}\)

y₂ = 2

Hence, its enlarged length would be 2 cm.

Question 6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m. high. If the length of the ship is 28 m, how long is the model ship?

Solution:

Let the length of the model ship be x₂ cm.

We form a table as shown below :

The more the length of the ship, the more would be the length of its mast. Hence, this is a case of direct proportion. That is,

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

⇒ \(\frac{28}{12}=\frac{x_2}{9}\)

12 x₂ = 28 x 9

⇒ \(x_2=\frac{28 \times 9}{12}=\frac{28 \times 3}{4}\)

x₂ = 21

Hence, the length of the model ship is 21 m

Question 7. Suppose 2 kg of sugar contains 9 Y.106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg ofsugar ?

Solution:

Suppose the amount of sugar is x kg and the number of crystals is y. We put the given information in the form of a table as shown below :

As the amount of sugar increases, the number of crystals also increases in the same ratio. So it is an ease of direct proportion. We make use of the relation of the type \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Here, x₁ = 2

y₁ = 9 x 106

x₂= 5

Therefore \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\) gives

⇒ \(\frac{2}{9 \times 10^6}=\frac{5}{y_2}\)

2y₂= 5 x 9 x 106

⇒ \(y_2=\frac{5 \times 9 \times 10^6}{2}\)

y₂ = 22.5 x 106

y₂ = 2.25 x 107

[Standard form]

Hence, there are 2.25 x 107 crystals of sugar in 5 kg of sugar

Here, .x₁ = 2

y₁ =9 X 106

x₃ = 1.2

Therefore \(\frac{x_1}{y_1}=\frac{x_3}{y_3}\) gives

⇒ \(\frac{2}{9 \times 10^6}=\frac{1.2}{y_3}\)

2y₃ = 1.2 x 9 x 106

2y₃= 10.8 x 106

⇒ \(y_3=\frac{10.8 \times 10^6}{2}\)

y₃ = 5.4 x 106

Hence, there are 5.4 x 106 crystals of sugar in 1.2 kg of sugar.

Question 8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?

Solution:

Let. the distance covered in the map be x cm. We put the given information in the form of a table as shown below :

The more the actual distance covered on the road, the more the distance on the map. So, it is a case of direct proportion

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{18}{1}=\frac{72}{y_2}\)

⇒ \(y_2=\frac{72}{18}=4\)

Hence, the distance covered on the map would be 4 cm.

Question 9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long.

Solution:

Let the height of the vertical pole be * m and the length of the shadow be y m. We put the given information in the form of a table as shown below :

As the height of the vertical pole increases, the length of the shadow also increases in the same ratio. So it is a case of direct proportion.

We make use of the relation of the type \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \text {. }\)

Here xl = 5 m 60 cm = 5.60 m

yi= 3 m 20 cm = 3.20 m

x2 = 10 m 50 cm = 10.50 m

Therefore \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\) gives

⇒ \(\frac{5.6}{3.2}=\frac{10.5}{y_2}\)

5.6y2 = 3.2 x 10.5

⇒ \(y_{2 .}=\frac{3.2 \times 10.5}{5.6}=\frac{32 \times 105}{560}\)

⇒ \(\frac{4 \times 105}{70}=\frac{4 \times 15}{10}\)

y2= 6

Hence, the length of the shadow is 6 m.

(ii) Here xx = 5 m 60 cm = 560 cm

y1 = 3 m 20 cm = 320 cm

y3 = 5 m = 500 cm

Therefore \( \frac{x_1}{y_1} =\frac{x_3}{y_3} \text { gives } \frac{560}{320}=\frac{x_3}{500}\)

⇒ \(320 x_3 =560 \times 500\)

⇒ \(x_3 =\frac{560 \times 500}{320}=\frac{560 \times 50}{32}\)

⇒ \(\frac{70}{4} \times 50=\frac{3500}{4}\)

x3 = 875

Hence, the height of the pole is 8 m 75 cm.

Question 10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Solution:

Two quantities x and y which vary £ in direct proportion have the relation

⇒ \(x=k y \text { or } \frac{x}{y}=k\)

⇒ \(k =\frac{\text { Number of km it can travel }}{\text { time in hours }}\)

⇒ \(\frac{14}{\left(\frac{25}{60}\right)}=\frac{14 \times 60}{25}\)

⇒ \(\frac{14 \times 12}{5}=\frac{168}{5}\)

Now, x is the distance traveled in 5 hours

Using the relation x = ky, obtain

⇒ \(x=\frac{168}{5} \times 5\)

x = 168

Hence, it can travel 168 km in 5 hours.

Question 11. On a squared paper, draw five squares of different sides. Write the following information in a tabular form.

Find whether the length of a side is in direct proportion to the:

1. The perimeter of the square.
2. The area of the square.

Solution:

Since each \(\frac{L}{\mathrm{P}}\) is the same, so we find that the length of a side is in direct proportion to the perimeter of the square.

Since all \(\frac{L}{\mathrm{a}}\) are not the same, therefore, we find that the length of a side is not in direct proportion to the area of the square.

Question 12. The following ingredients are required to make halwa for 5 persons: SujiRawa = 250 g, Sugar = 300 g, Ghee = 200 g, Water = 500 ml. Using the concept of proportion, estimate the changes in the quantity of ingredients, to prepare halwa for your class.

Solution:

Suppose that there are 25 (= 5 x 5) students in the class. Then, this is a case of direct variation. Hence, we shall require five times the quantity of ingredients.

Question 3. Choose a scale and make a map of your classroom., showing windows, doors blackboard, etc. (An example is given here),

Solution:

Please make yourself.

Question 13. Take a few problems discussed so far under ‘direct variation’. Do you think that they can be solved by the ‘unitary method’?

Solution:

Yes, they can be solved by the unitary method

∴  75 mL of the base is required for

= 1 part of red pigment

∴  1 mL of the base is required for

\(\frac{1}{75}\) part of red pigment

∴  1800 mL of the base is required for

⇒ \(=\frac{1}{75} \times 1800=\frac{120}{5}\)

= 24 parts of red pigment

Number of bottles filled in 6 hours = 840

Number of bottles filled in 1 hour \(\)

⇒ \(\frac{840}{6}=140\)

∴  Number of bottles filled in 5 hours = 140 x 5 = 700

∴  Distance travelled in 25 minutes = 14 km

∴  Distance travelled in 1 minute\(=\frac{14}{25} \mathrm{~km}\)

∴  Distance travelled in 5 hours(= 5 x 60 minutes or 300 minutes)

⇒ \(\frac{14}{25} \times 300 \mathrm{~km}\)

14 x 12 = 168 km

Inverse Proportion

If two quantities are related in such a way that an increase in one quantity leads to a corresponding proportional decrease in the other and vice-versa, then such a variation is called inverse proportion.

Thus, two quantities x and y are said to vary in inverse proportion if xy = k where k is a constant of proportionality

In this case, if. y1, y2 are the values of y corresponding to the values xv x2
f x respectively,

⇒ \(x_1 y_1=x_2 y_2 \quad \text { or } \quad \frac{x_1}{x_2}=\frac{y_2}{y_1}\)

Question 14. Think of more such examples of pairs of quantities that vary in inverse proportion.

Solution:

Some more such examples of pairs of quantities that vary in inverse proportion are as follows :

Number of machines to produce a given number of articles and the number of days required for production.

Number of periods a day in a school and the length of the period.

Number of workers to build a wall and The number of hours to build the wall

Question 15. Take a squared paper and arrange 48 counters on it in different numbers of rows shown below

What do you observe? As R increases, C decreases.

1. Is R1: R₂ = C₂: C₁?
2. Is R₃:R₄ = C₄:C₃ ?
3. Are R and C inversely proportional to each other?

Try this activity with 36 counters.

Solution:

So \(\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{2}{3}\)

⇒ \( \frac{\mathrm{C}_2}{\mathrm{C}_1} =\frac{16}{24}=\frac{2}{3} \)

⇒ \( \frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{\mathrm{C}_2}{\mathrm{C}_1} \)

⇒ \(\mathrm{R}_1: \mathrm{R}_2 =\mathrm{C}_2: \mathrm{C}_1\)

R1: R₂ = C₂: C₁

⇒ \(\frac{\mathrm{R}_3}{\mathrm{R}_4}=\frac{4}{6}=\frac{2}{3}\)

⇒ \(\frac{\mathrm{C}_4}{\mathrm{C}_3}=\frac{8}{12}=\frac{2}{3}\)

⇒ \(\frac{\mathrm{R}_3}{\mathrm{R}_4}=\frac{\mathrm{C}_4}{\mathrm{C}_3}\)

R₃:R₄ = C₄:C₃

Yes! R and C are inversely proportional to each other.

Let us arrange 36 counters in different numbers of rows as shown below :

Let us try this activity with 36 counters as follows :

We observe that as R increases, C decreases.

R₁ : R₂ = 2:3

C₂ : C4= 12 : 18 = 2 : 3

R₁:R₂ = C₂ : C₁

R₃ : R₄ = 4:6 = 2:3

C₄ : C₃ = 6:9 = 2:3

R₃:R₄= c₄-c₃

Yes! R and C are inversely proportional to each other.

Question 16. Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.

Solution:

x₁y₁ = 50 x 5 = 250

x₂y₂ = 40 x6 = 240

So, x₁y₁ ≠ x₂y₂

Hence, x and y are not in inverse proportion

x₁y₁= 100 x 60 = 6000

x₂y₂ = 200 x 30 = 6000

x₃y₃ = 300 x 20 = 6000

x₄ÿ₄ = 400 x 15 = 6000

So , x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄

Hence, x and y are in inverse proportion.

x₁y₁= 90 x 10 = 900

x₂y₂ = 60 x 15 = 900

x₃y₃ = 45 x 20 = 900

x₄ÿ₄ = 30 x 25 = 750

As x₁y₁ = x₂y₂ = x₃y₃ ≠ x₄y₄

So, x and y are not in inverse proportion.

Note: When two quantities x and y are in direct proportion (or vary directly) they are also written as x y.

When two quantities x andy are in inverse proportion (or vary inversely) they are also written as x «c \(\frac{1}{y}\)

Direct And Inverse Proportions Exercise 11.2

Question 1. Which of the following are in inverse proportion 1

1. The number of workers on a job and the time to complete the job.
2. The time taken for a journey and the distance traveled at a uniform speed,
3. Area cultivated land and the crop harvested,
4. The time taken for a fixed journey and the speed of the vehicle.
5. The population of a country and the area, of land per person.

Solution:

The number of workers on a job and the time to complete the job are in inverse
proportion, since as the number of workers increases, then the time to complete the job decreases proportionally.

The time taken for a journey and the distance traveled at a uniform speed are not in inverse proportion, since for a longer distance, more time will be required.

The area of cultivated land and the crop harvested is not in inverse proportion, since for more area of cultivated land, more crops will be needed to be harvested.

The time taken for a fixed journey and the speed of the vehicle are in inverse proportion, since the speed of the vehicle, proportionally less would be the time to cover a fixed journey.

The population of the country and the area of land per person are in inverse proportion,
since for more population of the country, the area of land per person would be proportionally less.

Question 2. In a Television game show, the prize money of f1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners.

Solution:

Let the prize (in T) for each winner bey3, y4, y5, y6, and y1 corresponding to the number of winners 4, 5, 8, 10, and 20 respectively. Then,

1 X 1,00,000 = 4 X y3 \(y_3=\frac{1,00,000}{4}=25,000\)

1 x 1,00,000 = 5 x y4 \(y_3=\frac{1,00,000}{5}=20,000\)

1 x 1,00,000 = 8 x y5 \(y_3=\frac{1,00,000}{8}=12,000\)

1 x 1,00,000 = 10 x y6 \(y_3=\frac{1,00,000}{10}=10,000\)

1 x 1,00,000 = 20 x y7 \(y_3=\frac{1,00,000}{20}=5,000\)

As the number of winners increases, amount of prize decreases. So the prize money given to an individual winner is inversely proportional to the number of winners.

Question 3. Rehrnan is making any paira wheel of using spokes in such a way that the angles between any pair of consecutive equal. Help equal him spokes by completing the following table:

1. Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
2. Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
3. How many spokes would be needed, if the angle between a pair of-consecutive spokes is 40°?

Solution:

Let the angles (in degree) between a pair of consecutive spokes be y₃, y₄, and y₅ respectively. Then,

⇒ \(y_3=\frac{360^{\circ}}{8}=45^{\circ}\)

⇒ \(y_4=\frac{360^{\circ}}{10}=36^{\circ}\)

⇒ \(y_5=\frac{360^{\circ}}{12}=30^{\circ}\)

Yes; The number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion.

|4 x 90° = 6 x 60° = 8 x 45° = 10 x 36° = 12 x 30°

Let the angle (in degrees) between a pair of consecutive spokes on a wheel with 15 spokes be y₆.

The smaller the number of spikes, the more the angle between a pair of consecutive spokes.

So, this is a case of inverse proportion.

Hence, 4 x 90° = 15 x y

⇒ \(y_6=\frac{4 \times 90^{\circ}}{15}\) x1y1 = x_2y2

y₆ = 24°

Hence, the angle between a pair of consecutive spokes on a wheel with 15 spokes is 24°.

Let x spokes be needed.

The lesser the number of spokes, the more will be the angle between a pair of consecutive spokes, bo, this is a case of inverse proportion.

Hence, 4 x 90° = x x 40°

⇒ \(x=\frac{4 \times 90^{\circ}}{40}\) x1y1 = x₂y₂

X = 9

Hence, 9 spokes would be needed, if the angle between a pair of consecutive spokes is 40″,

Question 4. If a box of sweets is divided among 21 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?

Solution:

Suppose that each would get y2 sweets

Thus, we have the following table :

The fewer the number of children, the more the number of sweets each would get. So, this is a case of inverse proportion.

Hence 24 x 5 = 20 x y₂

⇒ \(y_2=\frac{24 \times 5}{20}\)

y₂ = 6

Hence, each would get 6 sweets, if the number of children is reduced by 4.

Question 5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:

Suppose that the food would last for two days. We have the following table :

We note that the more the number of animals, the lesser will be the number of days for which the food will last. Therefore, this is a case of inverse proportion.

So, 20 x 6 = 30 x y₂

⇒ \(y_2=\frac{20 \times 6}{30}\)

y₂ = 4

Hence, the food would last for 4 days, if there were 10 more animals in his cattle.

Question 6. A contractor estimates that 3 persons could rewire Dasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?

Solution:

Suppose that they take y₂ days to complete the job. We have the following table :

The more the number of persons, the lesser will be the number of days required to complete the job. So, this is a case of inverse proportion.

Hence, 3 x 4 =4 x y₂

⇒ \(y_2=\frac{3 \times 4}{4}\)

y₂ = 3

Hence, they would take 3 days to complete the job.

Question 7. A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Solution:

Suppose that y₂ boxes would be filled. We have the following table:

The smaller the number of bottles, the more the number of boxes required to be filled. So, this is a case of inverse proportion.

Hence, 12 x 25 = 20 x y₂

⇒ \(y_2=\frac{12 \times 25}{20}=3 \times 5\)

y₂= 15

Hence, 15 boxes would be filled if the same batch is packed using 20 bottles in each box.

Question 8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?

Solution:

Suppose that x2 machines would be required. We have the following table :

The smaller the number of machines, the more will be the number of days to produce the same number of articles.

So, this is a case of inverse proportion.

Hence, 42 x 63 = x₂ x 54

⇒ \(x_2=\frac{42 \times 63}{54}=\frac{21 \times 7}{3}\)

x₂= 49

Hence, 49 machines would be required to produce the same number of articles in 54 days.

Question 9. A car takes 2 hours to reach a destination by traveling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?

Solution:

Let it take two hours. We have the following table :

Lesser the speed, the more the number of hours to reach the destination. So, this is a case of inverse proportion.

Hence, 60 x 2 = 80 x y₂

⇒ \(y_2=\frac{60 \times 2}{80}=\frac{3 \times 2}{4}\)

⇒ \(y_2=\frac{3}{2}=1 \frac{1}{2}\)

Thus, \(1 \frac{1}{2}\) hours would be taken when the car travels at the speed of 80 km/h.

Question 10. Two people could fit new windows in a house in 3 days.

1. One of the people fell ill before the work started. How long would the job take now?
2. How many people would be needed to fit the windows in one day?

Solution:

Let the job would take y2 days.

We have the following table :

The more the number of people, the lesser the number of days to do the job. So, the number of persons and number of days vary in inverse proportion.

So, 2 x 3= 1 x y₂

Thus, the job would now take 6 days when one of the persons fell ill before the work started.

Let y3 persons be needed.

We have the following table :

Clearly, the more the number of people, the lesser the number of days to do the job. So, the number of persons and number of days vary in inverse proportion.

So, 3 x 2 = 1 x y₃

y₃= 6

Thus, 6 people would be needed to fit the windows in one day.

Question 11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?

Solution:

Let each period be two minutes long. We have the following table :

Wo note that the more the number of periods, the lesser would be the length of each period. Therefore, this is a case of inverse proportion.

So, 8 x 45 = 9 x y₂

⇒ \(y_2=\frac{8 \times 45}{9}\)

y₂ = 40

Hence, each period would be 40 minutes long if the school had 9 periods a day.

Question 12. Take a sheet of paper. Fold it as shown in the figure. Count the number of parts and the area of a part in each case.

Tabulate your observations and discuss them with your friends. Is it a case of inverse proportion? Why?

Solution:

We see that

⇒ \(1 \times 1=2 \times \frac{1}{2}=4 \times \frac{1}{4}=8 \times \frac{1}{8}=16 \times \frac{1}{16}=1 \text { (= constant) }\)

So yes it is a case of inverse proportion because the greater the number of parts lesser the area of each part.

Question 13. Take a few containers of different sizes with circular bases, Fill the same amount of water in each container. note the diameter of each container and the respective height at which the water levels stand. Tabulate your observations. Is it a case of inverse proportion?

Solution:

Let the same amount of water in each container be V. Then

⇒ \(V=\pi\left(\frac{d_1}{2}\right)^2 h_1=\pi\left(\frac{d_2}{2}\right)^2 h_2=\pi\left(\frac{d_3}{2}\right)^2 h_3\)

d₁²h₁ = d²2h₂ = d_3²h₃

d₁h₁= d₂h₂ = d₃h3

So, it is not a case of inverse proportion.

It does not imply that

Direct And Inverse Proportions Multiple-Choice Question And Solutions

Question 1. 10 meters of cloth cost? 1000. What will 4 meters cost?

1. ₹ 400
2. ₹ 800
3. ₹ 200
4. ₹ 100

Solution: 1. ₹ 400

⇒ \(\frac{10}{1000}=\frac{4}{?} \quad \Rightarrow ?=400\)

Question 2. 15 books weigh 6 kg. What will 6 books weigh?

1. 1.2 kg
2. 2.4 kg
3. 3.8 kg
4. 3 kg

Solution: 2. 2.4 kg

⇒ \(\frac{15}{6}=\frac{6}{?} \quad \Rightarrow ?=2.4\)

Question 3. Does a horse eat 18 kg of corn in 12 days? How much does he eat in 9 days?

1. 11.5 kg
2. 12.5 kg
3. 13.5 kg
4. 14.5 kg

Solution: 3. 13.5 kg

⇒ \(\frac{12}{18}=\frac{9}{?} \quad \Rightarrow ?=\frac{18 \times 9}{12}\)

= 13.5

Question 4. 8 g of sandalwood cost ₹ 40. What will 10 g cost?

1. ₹ 30
2. ₹ 36
3. ₹ 48
4. ₹ 50

Solution: 4. ₹ 50

⇒ \(\frac{8}{40}=\frac{10}{?} \quad \Rightarrow ?=50\)

Question 5. 20 trucks can hold 150 metric tonnes. How much will 12 trucks hold?

1. 80 metric tonnes
2. 90 metric tonnes
3. 60 metric tonnes
4. 40 metric tonnes.

Solution: 2. 90 metric tonnes

⇒ \(\frac{20}{150}=\frac{12}{?} \quad \Rightarrow ?=90 .\)

Question 6. 120 copies of a book cost ₹ 600. What will 400 copies cost?

1. ₹ 1000
2. ₹ 2000
3. ₹ 3000
4. ₹ 2400

Solution: 2. ₹ 2000

⇒  \(\frac{120}{600}=\frac{400}{?} \Rightarrow ?=2000 .\)

Question 7. The rent of 7 hectares is ₹ 875. What is the rent of 16 hectares?

1. ₹ 2000
2. ₹ 1500
3. ₹ 1600
4. ₹ 1200

Solution: 1. ₹ 2000

⇒ \(\frac{7}{875}=\frac{16}{?} \quad \Rightarrow ?=\frac{875 \times 16}{7}\)

= 2000

Question 8. A boy runs 1 km in 10 minutes. How long will it take to run 600 m?

1. 2 minutes
2. 3 minutes
3. 4 minutes
4. 6 minutes

Solution: 4. 6 minutes

⇒ \(\frac{1000}{10}=\frac{600}{?} \Rightarrow ?=6\)

Question 9. A shot travels 90 m in 1 second. How long will it take to go 225 m?

1. 2 seconds
2. 2.5.seconds
3. 4 seconds
4. 3.5 seconds

Solution: 2. 2.5.seconds

⇒ \(\frac{90}{1}=\frac{225}{?} \Rightarrow ?=2.5 .\)

Question 10. 3 knives cost ₹ 63. What will 17 knives cost?

1. ₹ 357
2. ₹ 375
3. ₹ 537
4. ₹ 573

Solution: 1. ₹ 357

⇒ \(\frac{3}{63}=\frac{17}{?} \quad \Rightarrow \quad ?=357\)

Question 11. 15 men can mow 40 hectares of land in 1 day. How much will 6 men mow in 1 day?

1. 16 hectares
2. 12 hectares
3. 20 hectares
4. 24 hectares

Solution: 1. 16 hectares

⇒ \(\frac{15}{40}=\frac{6}{?} \quad \Rightarrow ?=\frac{40 \times 6}{15}=16 .\)

Question 12. A man walks 20 km in 5 hours. How long would it take to walk 32 km?

1. 3 hours
2. 4 hours
3. 6 hours
4. 8 hours

Solution: 4. 8 hours

⇒ \(\frac{20}{5}=\frac{32}{?} \quad \Rightarrow ?=8\)

Question 13. What is the cost of 50 sticks at ₹ 24 per score?

1. ₹ 30
2. ₹ 40
3. ₹ 50
4. ₹ 60

Solution: 4. ₹ 60

⇒ \(\frac{20}{24}=\frac{50}{?} \quad \Rightarrow ?=60 \text {. }\)

Question 14. A train travels 60 km in 1 hour. How long will it take to go 150 km?

1. 2 hours
2. 3 hours
3. 2.5 hours
4. 4 hours.

Solution: 3. 2.5 hours

⇒ \(\frac{60}{1}=\frac{150}{?} \quad \Rightarrow ?=2.5\)

Question 15. If 3 quintals of coal cost ₹ 6000, what is the cost of 120 kg?

1. ₹ 1200
2. ₹ 2400
3. ₹ 3600
4. ₹ 4800

Solution: 2. ₹ 2400

⇒ \(\frac{300}{6000}=\frac{120}{?} \Rightarrow ?=2400\)

Question 16. If 20 cows eat as much as 15 oxen, how many cows will eat as much as 36 oxen?

1. 40
2. 44
3. 45
4. 48

Solution: 4. 48

⇒ \(\frac{15}{20}=\frac{36}{?} \quad \Rightarrow ?=\frac{20 \times 36}{15}\)= 48

Question 17. The fare for a journey of 40 km is ₹ 25. How much can be traveled for ₹ 40?

1. 32 km
2. 64 km
3. 50 km
4. 60 km

Solution: 2. 64 km

⇒ \(\frac{25}{40}=\frac{40}{?} \Rightarrow ?=\frac{40 \times 40}{25}\)= 74

Question 18. Apala types 200 words in half an hour. How many words will she type in 12 minutes?

1. 80
2. 50
3. 100
4. 60

Solution: 1. 80

⇒ \(\frac{30}{200}=\frac{12}{?} \quad \Rightarrow ?=80 \text {. }\)

Question 19. A labourer is paid? 400 for 2 days of work. If he works for 5 days, how much will he get?

1. ₹ 1000
2. ₹ 800
3. ₹ 750
4. ₹ 900

Solution: 1. ₹ 1000

⇒ \(\frac{2}{400}=\frac{5}{?} \quad \Rightarrow ?=1000 .\)

Question 20. A machine in a soft drink factory fills 600 bottles in 5 hours. How many bottles will it fill in 2 hours?

1. 120
2. 180
3. 150
4. 240

Solution: 4. 240

⇒ \(\frac{5}{600}=\frac{2}{?} \quad \Rightarrow ?=240 .\)

Question 21. If 8 men can do a piece of work in 20 days, in how many days could 20 men do the same work?

1. 6 days
2. 8 days
3. 4 days
4. 10 days

Solution: 2. 8 days

⇒ \(8 \times 20=20 \times ? \quad \Rightarrow ?=8\)

⇒ \(5 \times 3=15,7 \times 3=21 \)

Question 22. If an amount of food last for 40 days for 120 men, how long will it last for 80 men at the same rate?

1. 50 days
2. 60 days
3. 80 days
4. 100 days

Solution: 2. 60 days

⇒ \( 40 \times 120=80 \times ? \quad \Rightarrow ?=60 \)

Question 23. If 18 women can reap a field in 7 days, in what time can 6 women reap the same field?

1. 15 days
2. 21 days
3. 30 days
4. 36 days

Solution: 2. 21 days

⇒ \(18 \times 7=6 \times ? \quad \Rightarrow ?=21 .\)

Question 24. 10 men can dig a trench in 15 days. How long will 3 men take?

1. 50 days
2. 60 days
3. 100 days
4. 75 days

Solution: 1. 50 days

⇒ \( 10 \times 15=3 \times ? \quad \Rightarrow ?=50\)

Question 25. 3 lambs finish eating turnips in 8 days. In how many days will 2 lambs finish them?

1. 6
2. 8
3. 10
4. 12

Solution: 4. 12

\(3 \times 8=2 \times ? \Rightarrow ?=12 .\)

Question 26. 6 pipes are required to fill a tank in 1 hour. How long will it take if only 5 pipes of the same type are used?

1. 75 minutes
2. 72 minutes
3. 80 minutes
4. 90 minutes.

Solution: 2. 72 minutes

⇒ \( 6 \times 60=5 \times? \)

⇒ \(?=72 \text { minutes. }\)

Question 27. 40 cows can graze a field in 16 days. How many cows will graze the same field in 10 days?

1. 60
2. 64
3. 80
4. 75

Solution: 2. 64

⇒ \(40 \times 16=? \times 10 \quad \Rightarrow ?=64 .\)

Question 28. If x = ky and when y = 4, x = 8 then k =

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

Solution: 2. 2

⇒ \(8=4 k \quad \Rightarrow k=2 .\)

Question 29. The constant of variation, if x «= y, from the following table, is

x     6     12     15    21

y     2      4       5      7

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

Solution: 3. 3

⇒ \(2 \times 3=6,4 \times 3=12, \\\)

Question 30. x and y vary inversely with each other. If x = 15 when y- 6, then the value of x when = 15 is

1. 2
2. 4
3. 5
4. 6

Solution: 4. 4

⇒ \( 15 \times 6=? \times 15 \quad \Rightarrow ?=6 \)

Direct And Inverse Proportions True-False

Write whether the following statements are True or False:

1. The distance traveled by a CNG vehicle and the amount of CNG consumed are directly proportional: True

2. If and q are in inverse proportion, then (p+l) and (q- 1) are also in inverse proportion: True

3. If Apala can finish a work in n days, then the amount of work done by her in one day is \(\frac{1}{y}\): True

4. If two quantities p and q vary inversely, then pq remains constant: True

5. If x and y are in direct proportion, then \(\frac{1}{x} \text { and } \frac{1}{y}\) are also in direct proportion: True

Direct And Inverse Proportions Fill in the Blanks

1. If xy = 1, then x and y vary ______ with each other: Inversely

2. The radius and the circumference of a circle vary ______ with each other.
sheets of the same: Directly

3. If the weight of 10 sheets of an A-4 size paper sheet is 30 g, then ______
the paper would weigh 90 g: 30

4. If the amount of work finished by Meenu in one day is \(\frac{1}{m}\) then the whole work will be finished by her in______days: m

5. If the speed remains constant, then the distance traveled is ______ proportional to the time: Directly

6. The area occupied by 10 postal stamps is 40 cm2. Find the area occupied by 100 such stamps: 400cm2

7. In a camp, there is enough flour for 50 persons for 40 days. How long will the food last, if 30 more people join the camp: 25days

8. 20 persons can reap a field in 15 days. How many more persons should be engaged to reap the same field in 10 days: 30

9. A car is traveling 80 km per hour. Find the distance traveled by car in 3 hours, if the speed remains constant: 240km

10. If x varies inversely as y and x = 20 when y = 30. Then, find y when x = 50: 12