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.