# Compound Interest

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

##### Class 8^{th} Mathematics (new) MHB Solution

**Practice Set 14.1**

- Find the amount and the compound interest.No.Principal (₹ )Rate (p.c.p.a.)Duration…
- Sameerrao has taken a loan of ₹ 12500 at a rate of 12 p.c.p.a. for 3 years. If the…
- To start a business Shalaka has taken a loan of ₹ 8000 at a rate of 10 {1}/{2}…

**Practice Set 14.2**

- On the construction work of a flyover bridge there were 320 workers initially. The…
- A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if…
- In a forest there are 40,000 trees. Find the expected number of trees after 3 years if…
- The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year…
- Find the compound interest if the amount of a certain principal after two years is₹…
- A loan of ₹ 15000 was taken on compound interest. If the rate of compound interest is…
- A principal amounts to ₹ 13924 in 2 years by compound interest at 18 p.c.p.a. Find the…
- The population of a suburb is 16000. Find the rate of increase in the population if the…
- In how many years ₹ 700 will amount to ₹ 847 at a compound interest rate of 10 p.c.p.a.…
- Find the difference between simple interest and compound interest on ₹ 20000 at 8…

###### Practice Set 14.1

**Question 1.**

Find the amount and the compound interest.

**Answer:**

(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?

**Answer:**

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?

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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/- .