# Compound Interest Quiz Set 003

### Question 1

What is the amount receivable on Rs. 3000000 after 9 months, invested at a rate of 24% compounded quarterly?

A

Rs. 3573048.

B

Rs. 3573148.

C

Rs. 3572948.

D

Rs. 3573248.

Soln.
Ans: a

In this case r = \$24/4\$% and n = 3 because compounding is quarterly, and in 9 months there are three quarters. So A = 3000000 × \$(1 + 6/100)^3\$, which equals 3 × 106 × 106 × 106, i.e., Rs. 3573048.

### Question 2

The interest earned by an amount of Rs. 40000 @5% compounded annually is Rs. 4100. What is the period in years?

A

2 years.

B

3 years.

C

1 year.

D

1/2 year.

Soln.
Ans: a

The amount is 40000 + 4100. So 44100 = 40000 × \$(105/100)^n\$. So \$44100/40000\$ = \$(105/100)^n\$, which can be put in the form \$(105/100)^2\$ = \$(105/100)^n\$, so n = 2 years.

### Question 3

The compound amount after 3 years on a principal of Rs. x is same as that on a principal of Rs. (621 - x) after 4 years, then what is x if the rate of interest is 7% p.a. compounded yearly?

A

Rs. 321.

B

Rs. 421.

C

Rs. 221.

D

Rs. 521.

Soln.
Ans: a

We have x × \$(1 + 7/100)^3\$ = (621 - x) × \$(1 + 7/100)^4\$. Cancelling, we get x = (621 - x) × (1 + 7/100). Simplifying, x = \${621 × (100 + 7)}/(200 + 7)\$, which gives x = Rs. 321.

### Question 4

What is the amount receivable on Rs. 2000000 after 9 months, invested at a rate of 28% compounded quarterly?

A

Rs. 2450086.

B

Rs. 2450186.

C

Rs. 2449986.

D

Rs. 2450286.

Soln.
Ans: a

In this case r = \$28/4\$% and n = 3 because compounding is quarterly, and in 9 months there are three quarters. So A = 2000000 × \$(1 + 7/100)^3\$, which equals 2 × 107 × 107 × 107, i.e., Rs. 2450086.

### Question 5

The amount of Rs. 3000000 earns an interest of Rs. 278181 @3% compounded annually. What is the investment period in years?

A

3 years.

B

2 years.

C

1 year.

D

1/2 year.

Soln.
Ans: a

The amount is 3000000 + 278181. So 3278181 = 3000000 × \$(103/100)^n\$. So \$3278181/3000000\$ = \$(103/100)^n\$, which can be put in the form \$(103/100)^3\$ = \$(103/100)^3\$, so n = 3 years.