# Compound Interest Quiz Set 014

### Question 1

An interest rate of 14% compounded half-annually is offered by a bank. An account holder deposits Rs. 60000 in the bank under this scheme. After six months he again deposits Rs 60000. What is the total amount that he will get after 1 year?

A

Rs. 132894.

B

Rs. 146476.

C

Rs. 146276.

D

Rs. 146576.

Soln.
Ans: a

Let P, A, r and n have their usual meanings. For the first deposit n = 2, and for the second deposit n = 1. So total amount is P × \$((1 + r/100)^2 + (1 + r/100))\$ = \$P/10000\$ × \$((100 + r)^2 + 100(100 + r))\$ = \$P/10000 × (100 + r)\$ × \$(100 + r + 100)\$ which equals \${P × (100 + r) × (200 + r)}/10000.\$ Putting r = 7 and P = 60000 and cancelling 10000, we get 6 × 107 × 207 = Rs. 132894. Please note that the rate of interest will be 1/2 because the compounding is half yearly.

### Question 2

An interest rate of 14% compounded half-annually is offered by a bank. An account holder deposits Rs. 30000 in the bank under this scheme. After six months he again deposits Rs 30000. What is the total amount that he will get after 1 year?

A

Rs. 66447.

B

Rs. 46476.

C

Rs. 76276.

D

Rs. 65577.

Soln.
Ans: a

Let P, A, r and n have their usual meanings. For the first deposit n = 2, and for the second deposit n = 1. So total amount is P × \$((1 + r/100)^2 + (1 + r/100))\$ = \$P/10000\$ × \$((100 + r)^2 + 100(100 + r))\$ = \$P/10000 × (100 + r)\$ × \$(100 + r + 100)\$ which equals \${P × (100 + r) × (200 + r)}/10000.\$ Putting r = 7 and P = 30000 and cancelling 10000, we get 3 × 107 × 207 = Rs. 66447. Please note that the rate of interest will be 1/2 because the compounding is half yearly.

### Question 3

The amount of Rs. 7000000 earns an interest of Rs. 1103375 @5% 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 7000000 + 1103375. So 8103375 = 7000000 × \$(105/100)^n\$. So \$8103375/7000000\$ = \$(105/100)^n\$, which can be put in the form \$(105/100)^3\$ = \$(105/100)^3\$, so n = 3 years.

### Question 4

The difference between compound interest(annual compounding) and simple interest for a period of 2 years is Rs. 8. What is the rate p.a. if principal is Rs. 20000?

A

2%.

B

4%.

C

3%.

D

5%.

Soln.
Ans: a

If d is the difference, r is the rate and P is the principal, then the shortcut formula for the difference between compound and simple interest over a period of 2 years is d = P × \$(r/100)^2\$. So rate = 100 × \$√{d/P}\$ = 100 × \$√{8/20000}\$ = 2%.

### Question 5

The interest earned by an amount of Rs. 90000 @2% compounded annually is Rs. 3636. 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 90000 + 3636. So 93636 = 90000 × \$(102/100)^n\$. So \$93636/90000\$ = \$(102/100)^n\$, which can be put in the form \$(102/100)^2\$ = \$(102/100)^n\$, so n = 2 years. 