SSC BOARD PAPERS IMPORTANT TOPICS COVERED FOR BOARD EXAM 2024

# Compound Interest

##### Class 8th Mathematics (new) MHB Solution

###### Practice Set 14.1
Question 1.

Find the amount and the compound interest.

(a) Principal = 2000/-, Rate = 5% (p.c.p.a), Duration (n) = 2 years

2

A = 2000 (1+0.05)2

A = 2000 (1.05)2

A = 2000 (1.1025)

∴A= 2205/-

∴ C.I = A – P

∴ C.I = 2205 – 2000

C.I. = 205/-

Amount is 2205/- and Compound interest is 205/- .

b. Principal = 5000/-, Rate = 8% (p.c.p.a), Duration (n) = 3 years

3

A = 5000 (1 + 0.08)3

A = 5000 (1.08)3

A = 5000 (1.259712)

∴A= 6298.56/-

∵ C.I. = A - P

∴ C.I. = 6298.56 - 5000

C.I. = 1298.56/-

Amount is 6298.56/- and Compound interest is 1298.56/- .

c. Principal = 4000/-, Rate = 7.5% (p.c.p.a), Duration (n) = 2 years

A = 4000 (1 + 0.075)2

A = 4000 (1.075)2

A = 4000 (1.155625)

∴A= 4622.5/-

∵ C.I. = A - P

∴ C.I. = 4622.5 - 4000

C.I. = 622.5/-

Amount is 4622.5/- and Compound interest is 622.5/- .

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?

Principal = 12500/-, Rate = 12% (p.c.p.a), Duration (n) = 3 years

3

A = 12500 (1+ 0.12)3

A = 12500 (1.12)3

A = 12500 (1.404928)

A = 17561.60/-

Sameerao has to pay an amount of 17561.60/- .

Question 3.

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

Principal = 8000/-, Rate = 10.5% (p.c.p.a), Duration (n) = 2 years

2

A = 8000 (1+0.105)2

A = 8000 (1.105)2

A = 8000 (1.221025)

∴ A = 9768.2/-

∵ C.I. = A - P

∴ C.I. = 9768.2 - 8000

C.I. = 1768.2/-

∴ Shalaka has to pay a compound interest of 1768.2/- .

###### Practice Set 14.2
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.

Present number of workers = 320 workers, Rate (increase) = 25% (p.c.p.a), Duration (n) = 2 years

2

A = 320 (1+0.25)2

A = 320 (1.25)2

A = 320 (1.5625)

∴A= 500/-

∴ The number of workers after 2 years will be 500.

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.

Present number of sheeps (P) = 200 sheeps, Rate = 8% (p.c.p.a), Duration (n) = 3 years

3

A = 200 (1+0.08)3

A = 200 (1.08)3

A = 200 (1.259712)

∴A= 251.9424

A = 252 sheeps (Rounded off)

∴ The number of sheeps after 3 years is 252.

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.

Present Trees (P) = 40000 trees, Rate = 5% (p.c.p.a), Duration (n) = 3 years

3

A = 40000 (1+0.05)3

A = 40000 (1.05)3

A = 40000 (1.157625)

∴A= 46305/-

∴ The expected number of trees after 3 years will be 46305.

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.

Principal = 250000/-, Rate (decrement) = 10% (p.c.p.a), Duration (n) = 2 years

2

2

A = 250000 (1-0.1)2

A = 250000 (0.9)2

A = 250000 (0.81)

∴A= 202500/-

∵ C.I. = A - P

∴ Depreciation in Price (C.I.) = 202500 – 250000

Depreciation in Price (C.I.) = -47500/-

(-) sign denotes the depreciation in amount.

∴ Depreciation in Price of the machine after 2 years will be 47500/- .

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.

Amount= 4036.80/-, Rate = 16% (p.c.p.a), Duration (n) = 2 years

2

4036.80 = P (1+0.16)2

4036.80 = P (1.16)2

4036.80 = P (1.3456)

∴ P = 3000/-

∵ C.I. = A - P

∴ C.I. = 4036.80 - 3000

C.I. = 1036.80/-

Compound interest is 1036.80/- .

Question 6.

A 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.

Principal = 15000/-, Rate = 12% (p.c.p.a), Duration (n) = 3 years

3

A = 15000 (1+0.12)3

A = 15000 (1.12)3

A = 15000 (1.404928)

∴A= 21073.92/-

Amount to settle the loan after 3 years is 21073.92/- .

Question 7.

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

Amount= 13924/-, Rate = 18% (p.c.p.a), Duration (n) = 2 years

2

13924 = P (1+0.18)2

13924 = P (1.18)2

13924 = P (1.3924)

∴ A = 10000/-

∴ The principal is 10000/- .

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.

Present Population (P) = 16000/-, Rate = R% (p.c.p.a), Duration (n) = 2 years

Population after 2 years (A) =17640/-

2

∴R= 5%

∴ The population of that suburb will increase at the rate of 5% .

Question 9.

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

Principal = 700/-, Rate = 10% (p.c.p.a), Duration (n) = n years Amount = 847/-

1.21 =

1.21 =

∴ n = 2 years

∴ The number of years required to gain an amount of 847/- from a principal of 700/- is 2 .

Question 10.

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

Principal = 20000/-, Rate = 8% (p.c.p.a), Duration (n) = n years

For the first year, compound interest and simple interest will be same, so it will vary from second year, therefore assuming the duration as 2 years in the same case.

A = 20000 (1+0.08)2

A = 20000 (1.08)2

A = 20000 (1.1664)

∴ A = 23328/-

∵ C.I. = A - P

C.I. = 23328 – 20000

C.I. = 3328/-

S.I. = 3200/-

∴ Difference = C.I. – S.I.

Difference = 3328 – 3200

Difference = 128 /-

∴ The difference between simple interest and compound interest is 128/- .