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Std 8 Mathematics · Chapter 14

14. Compound Interest

Solutions to Practice Sets 14.1 and 14.2 — amount, compound interest, growth and depreciation, with every calculation shown. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

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Practice Set 14.1

1Find the amount and the compound interest.

Formula: A = P × (1 + R100)N and compound interest I = A − P.

(1) P = ₹ 2000, R = 5, N = 2

A = 2000 × (1 + 5100)2 = 2000 × (105100)2 = 2000 × 2120 × 2120
= 5 × 21 × 21 = 2205
I = 2205 − 2000 = 205

Answer: Amount ₹ 2205, compound interest ₹ 205

(2) P = ₹ 5000, R = 8, N = 3

A = 5000 × (108100)3 = 5000 × (2725)3
= 5000 × 1968315625 = 19683 × 825 = 6298.56
I = 6298.56 − 5000 = 1298.56

Answer: Amount ₹ 6298.56, compound interest ₹ 1298.56

(3) P = ₹ 4000, R = 7.5, N = 2

A = 4000 × (1 + 7.5100)2 = 4000 × (10751000)2 = 4000 × (4340)2
= 4000 × 18491600 = 1849 × 52 = 4622.50
I = 4622.50 − 4000 = 622.50

Answer: Amount ₹ 4622.50, compound interest ₹ 622.50

2Sameerrao has taken a loan of ₹ 12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually, how many rupees should he pay to clear his loan?

P = ₹ 12500, R = 12, N = 3
A = 12500 × (1 + 12100)3 = 12500 × (2825)3
= 12500 × 2195215625 = 0.8 × 21952 = 17561.60

Answer: He should pay ₹ 17561.60

3To start a business Shalaka has taken a loan of ₹ 8000 at a rate of 1012 p.c.p.a. After two years how much compound interest will she have to pay?

P = ₹ 8000, R = 1012 = 10.5, N = 2
A = 8000 × (1 + 10.5100)2 = 8000 × (11051000)2 = 8000 × (221200)2
= 8000 × 4884140000 = 488415 = 9768.20
Compound interest = 9768.20 − 8000 = 1768.20

Answer: ₹ 1768.20

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Practice Set 14.2

The same formula works for any quantity that grows (or falls) by a fixed percentage every year. For a decrease (depreciation), R is taken as negative: A = P × (1 − R/100)N.

1On the construction work of a flyover bridge there were 320 workers initially. The number of workers was increased by 25% every year. Find the number of workers after 2 years.

P = 320, R = 25, N = 2
A = 320 × (1 + 25100)2 = 320 × (54)2 = 320 × 2516 = 500

Answer: 500 workers

2A shepherd has 200 sheep with him. Find the number of sheep with him after 3 years if the increase in the number of sheep is 8% every year.

P = 200, R = 8, N = 3
A = 200 × (108100)3 = 200 × (2725)3 = 200 × 1968315625 = 8 × 19683625 = 251.94…
The number of sheep must be a whole number, so it is about 252.

Answer: About 252 sheep

Note on the textbook answer: the key gives 242, but 200 × 1.08³ = 251.94, which is about 252. The figure 242 would come from 10% growth for 2 years (200 × 1.1²), not 8% for 3 years. Please verify with your printed edition.

3In 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 of 5% per year.

P = 40000, R = 5, N = 3
A = 40000 × (105100)3 = 40000 × (2120)3 = 40000 × 92618000 = 5 × 9261 = 46305

Answer: 46,305 trees

4The 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.

P = 250000, R = 10 (depreciation), N = 2
A = 250000 × (1 − 10100)2 = 250000 × (910)2 = 250000 × 81100 = 202500
Depreciation = 250000 − 202500 = 47500

Answer: Depreciation ₹ 47,500 (price after 2 years ₹ 2,02,500)

5Find 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.

A = ₹ 4036.80, R = 16, N = 2. First find P.
4036.80 = P × (116100)2 = P × (2925)2 = P × 841625
P = 4036.80 × 625841 = 4.8 × 625 = 3000
Compound interest = 4036.80 − 3000 = 1036.80

Answer: ₹ 1036.80

6A loan of ₹ 15000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a., find the amount to settle the loan after 3 years.

P = 15000, R = 12, N = 3
A = 15000 × (2825)3 = 15000 × 2195215625 = 0.96 × 21952 = 21073.92

Answer: ₹ 21073.92

7A principal amounts to ₹ 13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.

A = 13924, R = 18, N = 2
13924 = P × (118100)2 = P × 1392410000
P = 13924 × 1000013924 = 10000

Answer: Principal ₹ 10,000

8The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.

P = 16000, A = 17640, N = 2
17640 = 16000 × (1 + R100)2
(1 + R100)2 = 1764016000 = 441400 = (2120)2
1 + R100 = 2120 ∴ R100 = 120 ∴ R = 5

Answer: The rate of increase is 5 p.c.p.a.

9In how many years will ₹ 700 amount to ₹ 847 at a compound interest rate of 10 p.c.p.a.?

P = 700, A = 847, R = 10
847 = 700 × (1110)N
(1110)N = 847700 = 121100 = (1110)2 ∴ N = 2

Answer: In 2 years

10Find the difference between simple interest and compound interest on ₹ 20000 at 8 p.c.p.a.

The duration is not printed in the question; the textbook answer (₹ 128) is for 2 years, so N = 2.
Simple interest = PNR100 = 20000 × 2 × 8100 = 3200
Amount by compound interest = 20000 × (108100)2 = 20000 × 1.1664 = 23328
Compound interest = 23328 − 20000 = 3328
Difference = 3328 − 3200 = 128

Answer: ₹ 128

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