Compound Interest Quiz Set 004 | Maths Aptitude Quiz Questions and Mock Test on Compound Interest Set 4 | mock tests for Bank PO, SSB/SSC Exams, SO Clerical Exam, GATE Exam, LIC AO, GIC AO, etc.

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Question 1

The interest earned by an amount of Rs. 30000 @4% compounded annually is Rs. 2448. What is the period in years?

2 years.

3 years.

1 year.

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.

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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?

Rs. 64575.

Rs. 69400.

Rs. 69200.

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.

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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?

Rs. 624.

Rs. 724.

Rs. 524.

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.

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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?

Rs. 6180.

Rs. 6280.

Rs. 6080.

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.

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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?

Rs. 113905.

Rs. 114005.

Rs. 113805.

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.

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