# Compound Interest Quiz Set 012

### Question 1

What is the difference between the simple interest and compound interest at the rate of 12% for 1 year? The compounding is half-yearly, and the principal is Rs. 60000.

A

Rs. 216.

B

Rs. 316.

C

Rs. 116.

D

Rs. 416.

Soln.
Ans: a

The simple interest SI = (P × r)/100 = (60000 × 12)/100 = Rs. 7200. Compound interest will have half interest rate and n = 2. By shortcut formula, we have CI = \${P × R × (R + 200)}/10000\$ = \${60000 × 6 × (6 + 200)}/10000\$ = Rs. 7416. The difference = Rs. 216.

### Question 2

The amount of Rs. 2000000 earns an interest of Rs. 590058 @9% 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 2000000 + 590058. So 2590058 = 2000000 × \$(109/100)^n\$. So \$2590058/2000000\$ = \$(109/100)^n\$, which can be put in the form \$(109/100)^3\$ = \$(109/100)^3\$, so n = 3 years.

### Question 3

A bank offers an interest rate of 6% compounded annually. Initially I deposit Rs. 90000 in the bank under this scheme. After 1 year I again deposit Rs 90000. What is the total amount that I will get after 2 years?

A

Rs. 196524.

B

Rs. 196624.

C

Rs. 196424.

D

Rs. 196724.

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 = 6 and P = 90000 and cancelling 10000, we get 9 × 106 × 206 = Rs. 196524.

### Question 4

An amount P is invested for 2 years @5% p.a. The simple interest is Rs. 7000. What would be the compound interest on the same amount, at the same rate and for the same time, compounded annually?

A

Rs. 7175.

B

Rs. 7275.

C

Rs. 7075.

D

Rs. 7375.

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\$ = \${7000 × (5 + 200)}/200\$ = Rs. 7175.

### Question 5

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. 