# Compound Interest Quiz Set 008

### Question 1

The compound interest on a certain sum of money at 5% for a period of 2 years is Rs. 3075. What is the SI on this sum if the rate is halved, and time doubled?

A

Rs. 3000.

B

Rs. 3100.

C

Rs. 2900.

D

Rs. 3200.

Soln.
Ans: a

The shortcut formula is CI = Pr(r + 200)/10000. Putting CI = 3075, r = 5, we get 3075 = \${P × 5 × (5 + 200)}/10000\$. We can solve it to get P = Rs. 30000. The SI = P × (2 × t) × (r / 200) = P × t × (r / 100). Putting t = 2, r = 5 and P = 30000, we get SI = Rs. 3000.

### Question 2

The compound amount after 2 years on a principal of Rs. P is same as that on a principal of Rs. (404 - P) after 3 years, then what is P if the rate of interest is 2% p.a. compounded yearly?

A

Rs. 204.

B

Rs. 304.

C

Rs. 104.

D

Rs. 404.

Soln.
Ans: a

We have P × \$(1 + 2/100)^2\$ = (404 - P) × \$(1 + 2/100)^3\$. Cancelling, we get P = (404 - P) × (1 + 2/100). Simplifying, P = \${404 × (100 + 2)}/(200 + 2)\$, which gives P = Rs. 204.

### Question 3

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 4

The amount of Rs. 2000000 earns an interest of Rs. 185454 @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 2000000 + 185454. So 2185454 = 2000000 × \$(103/100)^n\$. So \$2185454/2000000\$ = \$(103/100)^n\$, which can be put in the form \$(103/100)^3\$ = \$(103/100)^3\$, so n = 3 years.

### Question 5

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.

Updated on 2017-05-17.

Posted by Parveen(Hoven),
Aptitude Trainer