# PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions InText Questions and Answers.

## PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These : [Textbook Page No. 204]

1. Observe the following tables and find if x and y are directly proportional.

Question (i)

 X 20 17 14 11 8 5 2 y 40 34 28 22 16 10 4

Solution:
\begin{aligned} &\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} \end{aligned}
The value of $$\frac{\text {x}}{\text {y}}$$ is same for different values of x and y. So these values x and y are directly proportional.

Question (ii)

 X 6 10 14 18 22 26 30 y 4 8 12 16 20 24 28

Solution:
\begin{aligned} &\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} \end{aligned}
The values of $$\frac{\text {x}}{\text {y}}$$ are different for different values of x and y respectively. So these values of x and y are not directly proportional.

Question (iii)

 X 5 8 12 15 18 20 y 15 24 36 60 72 100

Solution:
The values of $$\frac{\text {x}}{\text {y}}$$ are different for different values of x and y respectively.
So these values of x and y are not directly proportional.

2. Principal = ₹ 1000, Rate = 8 % per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.

 Time Period 1 year 2 years 3 years Simple Interest (in ₹) Compound Interest (in ₹)

Solution:
Simple interest : SI = $$\frac{P \times R \times T}{100}$$
For calculation:
P = ₹ 1000, R = 8 %, T = …………….

 Time (T) → 1 year: T = 1 2 years : T = 2 3 years : T = 3 Simple interest SI = $$\frac{P \times R \times T}{100}$$ ₹ $$\frac{1000 \times 8 \times 1}{100}$$ = ₹ 80 ₹ $$\frac{1000 \times 8 \times 2}{100}$$ = ₹ 160 ₹ $$\frac{1000 \times 8 \times 3}{100}$$ = ₹ 240 $$\frac{\text { SI }}{\text { T }}$$ $$\frac {80}{1}$$ = 80 $$\frac {160}{2}$$ = 80 $$\frac {240}{3}$$ = 80

Here, the ratio of simple interest with time period is same for every year.
Hence, simple interest changes in direct proportion with time period.
Compound interest:
For calculation:
P = ₹ 1000, R = 8 %, T = ……………

 Time → 1 year : n = 1 A = P(1 + $$\frac {R}{100}$$)n CI = A – P A = 1000(1 + $$\frac {8}{100}$$)1 = 1000 × $$\frac {108}{100}$$ = 1080 ∴ CI = 1080 – 1000 = ₹ 80 $$\frac{\text { CI }}{\text { T }}$$ $$\frac {80}{1}$$
 Time → 2 years : n = 2 A = P(1 + $$\frac {R}{100}$$)n CI = A – P A = 1000(1 + $$\frac {8}{100}$$)2 = 1000 × $$\frac {108}{100}$$ × $$\frac {108}{100}$$ = ₹ 1166.40 ∴ CI = ₹ 1166.40 – 1000 = ₹ 166.40 $$\frac{\text { CI }}{\text { T }}$$ $$\frac {166.40}{2}$$
 Time → 3 years : n = 3 A = P(1 + $$\frac {R}{100}$$)n CI = A – P A = 1000(1 + $$\frac {8}{100}$$)3 = 1000 × $$\frac {108}{100}$$ × $$\frac {108}{100}$$ × $$\frac {108}{100}$$ = ₹ 1259.712 ∴ CI = ₹ 1259.712 – ₹ 1000 = ₹ 259.712 $$\frac{\text { CI }}{\text { T }}$$ $$\frac {259.712}{3}$$

Here, the ratio of CI and T is not same.
Thus, compound interest is not proportional with time period.

Think, Discuss and Write: [Textbook Page No. 204]

1. If we fix time period and the rate of interest, simple interest changes proportionally with principal. Would there be a similar relationship for compound interest? Why?
Solution:
Time period (T) and rate of interest (R) are fixed, then
Simple interest = $$\frac {PRT}{100}$$ = P × Constant
So simple interest depends on principal. Simple interest changes proportionally with principal.
Now, compound interest = P(1 + $$\frac {R}{100}$$)T – P
i.e., A – P
= P [(1 + $$\frac {R}{100}$$T – 1]
= P × Constant
So compound interest depends on principal.
If principal increases or decreases, then compound interest will also increases or decreases.
Thus, compound interest changes with principal.

Think, Discuss and Write : [Textbook Page No. 209]

1. Take a few problems discussed so far under ‘direct variation’. Do you think that they can be solved by ‘unitary method’?
Solution:
Yes, each problem can be solved by unitary method.
e.g. Question 4 of Exercise: 13.1
Number of bottles filled in 6 hours = 840
∴ The number of bottles filled in 1 hour = $$\frac {840}{6}$$ = 140
The number of bottles filled in 5 hours = 140 × 5 = 700
Thus, 700 bottles will it fill in five hours.

Try These : [Textbook Page No. 211]

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

Question (i)

 X 50 40 30 20 y 5 6 7 8

Solution:
x1 = 50 and y1 = 5
∴ x1y1 = 50 × 5
∴ x1y1 = 250

x2 = 40 and y2 = 6
∴ x2y2 = 40 × 6
∴ x2y2 = 240

x3 = 30 and y3 = 7
∴ x3y3 = 30 × 7
∴ x3y3 = 210

x4 = 20 and y4 = 8
∴ x4y4 = 20 × 8
∴ x4y4 = 160
Now 250 ≠ 240 ≠ 210 ≠ 160
∴ x1y1 ≠ x2y2 ≠ x3y3 ≠ x4y4
∴ x and y are not in inverse proportion.

Question (ii)

 X 100 200 300 400 y 60 30 20 15

Solution:
x1 = 100 and y1 = 60
∴ x1y1 = 100 × 60
∴ x1y1 = 6000

x2 = 200 and y2 = 30
∴ x2y2 = 200 × 30
∴ x2y2 = 6000

x3 = 300 and y3 = 20
∴ x3y3 = 300 × 20
∴ x3y3 = 6000

x4 = 400 and y4 = 15
∴ x4y4 = 400 × 15
∴ x4y4 = 6000
Now x1y1 = x2y2 = x3y3 = x4y4
∴ x and y are in inverse proportion.

Question (iii)

 X 90 60 45 30 20 5 y 10 15 20 25 30 35

Solution:
x1 = 90 and y1 = 10
∴ x1y1 = 90 × 10
∴ x1y1 = 900

x2 = 60 and y2 = 15
∴ x2y2 = 60 × 15
∴ x2y2 = 900

x3 = 45 and y3 = 20
∴ x3y3 = 45 × 20
∴ x3y3 = 900

x4 = 30 and y4 = 25
∴ x4y4 = 30 × 25
∴ x4y4 = 750

x5 = 20 and y5 = 30
∴ x5y5 = 20 × 30
∴ x5y5 = 600

x6 = 30 and y6 = 35
∴ x6y6 = 5 × 35
∴ x6y6 = 175

Now x1y1 = x2y2 = x3y3 ≠ x4y4 ≠ x5y5 ≠ x6y6
∴ x and y are in inverse proportion.