Mathematics Solutions for Class 8 Maths Chapter 15 - Compound Interest

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Page No 90:

Question 1:

Find the amount and the compound interest.

No.	Principal (₹)	Rate (p.c.p.a.)	Duration (Years)
1	2000	5	2
2	5000	8	3
3	4000	7.5	2

Answer:

(1) Here, P = ₹ 2000; R = 5 % ; N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} = 2000 {(1 + \frac{5}{100})}^{2} = 2000 {(\frac{105}{100})}^{2} = 2000 {(\frac{21}{20})}^{2} = 2205 Rupees ∴ Compound Interest after 2 years, I = Amount - Principal = 2205 - 2000 = 205 Rupees$
Hence, Amount = ₹ 2205 and Compound interest = ₹ 205.

(2) Here, P = ₹ 5000; R = 8 % ; N = 3 years

$A = P {(1 + \frac{R}{100})}^{N} = 5000 {(1 + \frac{8}{100})}^{3} = 5000 {(\frac{108}{100})}^{3} = 5000 {(\frac{27}{25})}^{3} = 6298.56 Rupees ∴ Compound Interest after 3 years, I = Amount - Principal = 6298.56 - 5000 = 1298.56 Rupees$
Hence, Amount = ₹ 6298.56 and Compound interest = ₹ 1298.56

(3) Here, P = ₹ 4000; R = 7.5 % ; N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} = 4000 {(1 + \frac{7.5}{100})}^{2} = 4000 {(1 + \frac{75}{1000})}^{2} = 4000 {(\frac{1075}{1000})}^{2} = 4000 {(\frac{43}{40})}^{2} = 4622.50 Rupees ∴ Compound Interest after 2 years, I = Amount - Principal = 4622.50 - 4000 = 622.50 Rupees$
Hence, Amount = ₹ 4622.50 and Compound interest = ₹ 622.50

Page No 90:

Question 2:

Sameerrao has taken a loan of ₹12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?

Answer:

Here, P = ₹ 12500; R = 12 % ; N = 3 years

$A = P {(1 + \frac{R}{100})}^{N} = 12500 {(1 + \frac{12}{100})}^{3} = 12500 {(1 + \frac{3}{25})}^{3} = 12500 {(\frac{28}{25})}^{3} = 17561.60 Rupees$

Hence, he should pay an amount of ₹ 17561.60 to clear his loan.

Page No 90:

Question 3:

To start a business Shalaka has taken a loan of ₹8000 at a rate of 10 $\frac{1}{2}$ p.c.p.a. After two years how much compound interest will she have to pay?

Answer:

Here, P = ₹ 8000; R = 10 $\frac{1}{2}$ % ; N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} = 8000 {(1 + \frac{10 \frac{1}{2}}{100})}^{2} = 8000 {(1 + \frac{21}{200})}^{2} = 8000 {(\frac{221}{200})}^{2} = 9768.20 Rupees ∴ Compound Interest after 2 years, I = Amount - Principal = 9768.20 - 8000 = 1768.20 Rupees$

Hence, she will have to pay a Compound interest of ₹ 1768.20 after 2 years.

Page No 93:

Question 1:

On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.

Answer:

Here, P = Number of workers initially = 320
A = Number of workers after 2 years
R = Rate of increase of number of workers per year = 25 %
N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} = 320 {(1 + \frac{25}{100})}^{2} = 320 {(1 + \frac{1}{4})}^{2} = 320 {(\frac{5}{4})}^{2} = 500$

Hence, the number of workers after 2 years is 500.

Page No 93:

Question 2:

A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.

Answer:

Here, P = Number of sheeps initially = 200
A = Number of sheeps after 3 years
R = Rate of increase of number of sheeps per year = 8 %
N = 3 years

$A = P {(1 + \frac{R}{100})}^{N} = 200 {(1 + \frac{8}{100})}^{3} = 200 {(1 + \frac{2}{25})}^{3} = 200 {(\frac{27}{25})}^{3} = 251.94 = 252 (approx)$

Hence, the number of sheeps after 3 years is 252.

Page No 93:

Question 3:

In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.

Answer:

Here, P = Number of trees initially = 40,000
A = Number of trees after 3 years
R = Rate of increase of number of trees per year = 5 %
N = 3 years

$A = P {(1 + \frac{R}{100})}^{N} = 40000 {(1 + \frac{5}{100})}^{3} = 40000 {(1 + \frac{1}{20})}^{3} = 40000 {(\frac{21}{20})}^{3} = 46305$

Hence, the expected number of trees after 3 years is 46,305.

Page No 93:

Question 4:

The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year find the depreciation in price of the machine after two years.

Answer:

Here, P = Cost price of the machine = 2,50,000
A = Cost price after 2 years
I = Depreciation in price after 2 years
R = Rate of depreciation = 10 %
N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} = 250000 {(1 + \frac{- 10}{100})}^{2} = 250000 {(1 - \frac{1}{10})}^{2} = 250000 {(\frac{9}{10})}^{2} = 202500$

Also,
I = P − A
= 250000 − 202500
= 47500

Hence, the depreciation in price of the machine after two years is Rs 47,500.

Page No 93:

Question 5:

Find the compound interest if the amount of a certain principal after two years is ₹4036.80 at the rate of 16 p.c.p.a.

Answer:

Here, P = Principal
A = ₹ 4036.80
I = Compound Interest
R = 16 %
N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} \Rightarrow 4036.80 = P {(1 + \frac{16}{100})}^{2} \Rightarrow 4036.80 = P {(1 + \frac{4}{25})}^{2} \Rightarrow 4036.80 = P {(\frac{29}{25})}^{2} \Rightarrow P = \frac{4036.80 \times 25 \times 25}{29 \times 29} \Rightarrow P = 3000$

Also,
I = A − P
= 4036.80 − 3000
= 1036.80

Hence, the compound interest is ₹ 1036.80

Page No 93:

Question 6:

A loan of ₹15000 was taken on compound interest. If the rate of compound interest s 12 p.c.p.a. find the amount to settle the loan after 3 years.

Answer:

Here, P = Principal = ₹ 15000
A = Amount
R = 12 %
N = 3 years

$A = P {(1 + \frac{R}{100})}^{N} = 15000 {(1 + \frac{12}{100})}^{3} = 15000 {(1 + \frac{3}{25})}^{3} = 15000 {(\frac{28}{25})}^{3} = 21073.92$

Hence, the amount is ₹ 21073.92

Page No 93:

Question 7:

A principal amounts to ₹13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.

Answer:

Here, P = Principal
A = ₹ 13924
R = 18 %
N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} \Rightarrow 13924 = P {(1 + \frac{18}{100})}^{2} \Rightarrow 13924 = P {(1 + \frac{9}{50})}^{2} \Rightarrow 13924 = P {(\frac{59}{50})}^{2} \Rightarrow P = \frac{13924 \times 50 \times 50}{59 \times 59} \Rightarrow P = 10000$

Hence, the principal is ₹ 10000.

Page No 93:

Question 8:

The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.

Answer:

Here, P = Population of a suburb = 16000
A = Population after two years =17640
R = R %
N = 2 years

$A = P {(1 + \frac{R}{100})}^{N} \Rightarrow 17640 = 16000 {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{17640}{16000} = {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{441}{400} = {(1 + \frac{R}{100})}^{2} \Rightarrow {(\frac{21}{20})}^{2} = {(1 + \frac{R}{100})}^{2} \Rightarrow 1 + \frac{R}{100} = \frac{21}{20} \Rightarrow \frac{R}{100} = \frac{21}{20} - 1 \Rightarrow \frac{R}{100} = \frac{1}{20} \Rightarrow R = 5$

Hence, the rate of increase in the population is 5 p.c.p.a.

Page No 93:

Question 9:

In how many years ₹700 will amount to ₹847 at a compound interest rate of 10 p.c.p.a.

Answer:

Here, P = Principal = ₹ 700
A = Amount = ₹ 847
R = 10 %
N = N years

$A = P {(1 + \frac{R}{100})}^{N} \Rightarrow 847 = 700 {(1 + \frac{10}{100})}^{N} \Rightarrow \frac{847}{700} = {(1 + \frac{1}{10})}^{N} \Rightarrow \frac{121}{100} = {(\frac{11}{10})}^{N} \Rightarrow {(\frac{11}{10})}^{2} = {(\frac{11}{10})}^{N} \Rightarrow N = 2$

Hence, the number of years is 2 years.

Page No 93:

Question 10:

Find the difference between simple interest and compound interest on ₹20000 at 8 p.c.p.a.

Answer:

Disclaimer: In the question "Time" is not given. So the question is solved taking time as 2 years, because simple interest and compound interest will be same for one year.

Here, P = Principal = ₹ 20000
R = 8 %
N = 2 years

$Simple interest = \frac{P \times R \times N}{100} = \frac{20000 \times 8 \times 2}{100} = 3200 Amount = P {(1 + \frac{R}{100})}^{N} = 20000 {(1 + \frac{8}{100})}^{2} = 20000 {(1 + \frac{2}{25})}^{2} = 20000 {(\frac{27}{25})}^{2} = 23328 Compound interest = 23328 - 20000 = 3328 Compound interest - Simple interest = 3328 - 3000 = 128$

Hence, the difference between simple interest and compound interest is ₹ 128.

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