# Compound Interest Quiz Set 004

### Question 1

The interest earned by an amount of Rs. 30000 @4% compounded annually is Rs. 2448. 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 30000 + 2448. So 32448 = 30000 × \$(104/100)^n\$. So \$32448/30000\$ = \$(104/100)^n\$, which can be put in the form \$(104/100)^2\$ = \$(104/100)^n\$, so n = 2 years.

### Question 2

An interest rate of 10% 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. 64575.

B

Rs. 69400.

C

Rs. 69200.

D

Rs. 69500.

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 = 5 and P = 30000 and cancelling 10000, we get 3 × 105 × 205 = Rs. 64575. Please note that the rate of interest will be 1/2 because the compounding is half yearly.

### Question 3

The compound amount after 2 years on a principal of Rs. P is same as that on a principal of Rs. (1224 - P) after 3 years, then what is P if the rate of interest is 4% p.a. compounded yearly?

A

Rs. 624.

B

Rs. 724.

C

Rs. 524.

D

Rs. 824.

Soln.
Ans: a

We have P × \$(1 + 4/100)^2\$ = (1224 - P) × \$(1 + 4/100)^3\$. Cancelling, we get P = (1224 - P) × (1 + 4/100). Simplifying, P = \${1224 × (100 + 4)}/(200 + 4)\$, which gives P = Rs. 624.

### Question 4

An amount P is invested for 2 years @6% p.a. The simple interest is Rs. 6000. What would be the compound interest on the same amount, at the same rate and for the same time, compounded annually?

A

Rs. 6180.

B

Rs. 6280.

C

Rs. 6080.

D

Rs. 6380.

Soln.
Ans: a

Let SI, P, r, t have usual meanings. Then, for 2 years, SI = (P × r × 2)/100. So P = \$(50 × SI)/r\$. The compound interest for 2 years by shortcut formula is \${P × r × (200 + r)}/10000\$. Putting P here, it becomes, \${{(50 × SI)/r} × r × (200 + r)}/10000\$ = \${SI × (r + 200)}/200\$ = \${6000 × (6 + 200)}/200\$ = Rs. 6180.

### Question 5

A bank offers an interest rate of 9% compounded annually. Initially I deposit Rs. 50000 in the bank under this scheme. After 1 year I again deposit Rs 50000. What is the total amount that I will get after 2 years?

A

Rs. 113905.

B

Rs. 114005.

C

Rs. 113805.

D

Rs. 114105.

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 = 9 and P = 50000 and cancelling 10000, we get 5 × 109 × 209 = Rs. 113905. 