# Compound Interest - MCQ 2

## 20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Compound Interest - MCQ 2

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Attempt Compound Interest - MCQ 2 | 20 questions in 40 minutes | Mock test for Quant preparation | Free important questions MCQ to study Quantitative Aptitude for Competitive Examinations for Quant Exam | Download free PDF with solutions
QUESTION: 1

### A sum of rupees 3903 is divided between P and Q such that the share of P at the end of 8 years is equal to the share of Q after 10 years. Find the share of P if rate of interest is 4% compounded annually.

Solution:

P*(1 + 4/100)8 = (3903 – P)*(1 + 4/100)10

QUESTION: 2

### A man borrows 2000 rupees at 10% compound interest. At the end every year he pays rupees 1000 back. How much amount should he pay at the end of the third Year to clear all his debt?

Solution:

After one year amount = 2000*110/100 = 2200
He pays 1000 back, so remaining = 2200 – 1000 = 1200
After second year = 1200*110/100 = 1320
He pays 1000 back, so remaining = 1320 – 1000 = 320
After third year = 320*110/100 = 352

QUESTION: 3

### A sum of rupees 3200 is compounded annually at the rate of 10 paisa per rupee per annum. Find the compound interest payable after 2 years.

Solution:

Rate of interest is 10 paisa per rupee per annum. So for 100 rupees it is 1000 paise i.e. 10 percent
Now, CI = 3200(1+10/100)2 – 3200 = 672

QUESTION: 4

What sum of money will amount to rupees 1124.76 in 3 years, if the rate of interest is 5% for the first year, 4% for the second year and 3% for the third year?

Solution:

1124.76 = p*(105/100)*(104/100)*(103/100)

QUESTION: 5

Riya saves an amount of 500 every year and then lent that amount at an interest of 10 percent compounded annually. Find the amount after 3 years.

Solution:

Total amount = 500*(1+10/100)3 + 500*(1+10/100)2 + 500*(1+10/100) = 1820.5

QUESTION: 6

A sum of 3000 becomes 3600 in 3 years at 15 percent per annum. What will be the sum at the same rate after 9 years?

Solution:

3600 = 3000*(1+15/100)3 (1+15/100)3 = 6/5
Amount = 3000*[(1+15/100)3]3 Amount = 3000*(6/5)3 = 5184

QUESTION: 7

On a certain sum of money, after 2 years the simple interest and compound interest obtained are Rs 400 and Rs 600 respectively. What is the sum of money invested?

Solution:

400 = P*(R/100)*2
600 = P*(1+R/100)2 – P
Solve both equations to get P

QUESTION: 8

A sum of money becomes Rs 35,280 after 2 years and Rs 37,044 after 3 years when lent on compound interest. Find the principal amount.

Solution:

37044 = p*(1 +r/100)3
35280 = p*(1 + r/100)2
Divide both equations to get the value of r and then substitute in any equation to get P

QUESTION: 9

A sum of money is lent for 2 years at 10% p.a. compound interest. It yields Rs 8.81 more when compounded semi-annually than compounded annually. What is the sum lent?

Solution:

8.81 = p*(1+5/100)4 – p*(1+10/100)2

QUESTION: 10

A sum of rupees 4420 is to be divided between raj and parth in such a way that after 5 years and 7 years respectively the amount they get is equal. The rate of interest is 10 percent. Find the share of raj and parth

Solution:

Let the share of raj and parth be R and P
R*(1+10/100)5 = (4420 – R)*(1+10/100)7
We get R = 2420, so P = 2000

QUESTION: 11

The difference between compound interest compounded every 6 months and simple interest after 2 years is 248.10. The rate of interest is 10 percent. Find the sum

Solution:

Explanation : P*(1+5/100)4 – P – P*(10/100)*2 = 248.10
P = 16000

QUESTION: 12

A person earns an interest of 240 on investing certain amount at Simple interest for 2 years at 5 percent amount. If the rate of interest is compounded annually then how much more interest will be gain by the person at same rate of interest and on the same sum.

Solution:

240 = P*(5/100)*2, P = 2400
CI = 2400(1+5/100)2 – 2400 = 246
So, 246 – 240 = 6

QUESTION: 13

Find the least number of years in which the sum put at 25% rate of interest will be more than doubled.

Solution:

Amount > = P*(1+25/100)n
Amount = p*(5/4)n
For n = 4, (625/256) which is greater than 2.

QUESTION: 14

A sum of rupees 4420 is to be divided between rakesh and prakash in such a way that after 5 years and 7 years respectively the amount they get is equal. The rate of interest is 10 percent. Find the share of rakesh and prakash

Solution:

Let the share of rakesh and prakash be R and P
R*(1+10/100)5 = (4420 – R)*(1+10/100)7 We get R = 2420, so P = 2000

QUESTION: 15

The simple interest on a certain sum of money for 4 years at 15 percent per annum is 600. Find the compound interest in the same sum at 10 percent interest for 2 years

Solution:

600 = p*4*(15/100), P = 1000
CI = 1000(1+10/100)2 – 1000 = 210

QUESTION: 16

Find the effective annual rate of 10 percent per annum compounded halfyearly

Solution:

Take principal as 100 and then calculate,
A = 100*(1+5/100)2
A = 110.25
So effective rate is 10.25

QUESTION: 17

A sum of rupees 3200 is compounded annually at the rate of 25 paise per rupee per annum. Find the compound interest payable after 2 years.

Solution:

Rate of interest is 25 paise per rupee per annum.
So for 100 rupees it is 2500 paise i.e. 25 percent
Now, CI = 3200(1+25/100)2 – 3200 = 1800

QUESTION: 18

A sum of 3000 becomes 3600 in 3 years at 15 percent per annum. What will be the sum at the same rate after 9 years.

Solution:

3600 = 3000*(1+15/100)3
(1+15/100)3 = 6/5
Amount = 3000*[(1+15/100)3]3
Amount = 3000*(6/5)3 = 5184

QUESTION: 19

Priya saves an amount of 500 every year and then lent that amount at an interest of 10 percent compounded annually. Find the amount after 3 years.

Solution:

Total amount = 500*(1+10/100)3 + 500*(1+10/100)2 + 500*(1+10/100)
= 1820.5

QUESTION: 20

A man borrows 10000 rupees at 20 % compound interest for 3 years. If every year he pays 2000 rupees as repayment. How much amount is still left to be paid by the man?

Solution:

Amount to be paid at the end of three years = 10000*(1+20/100)3 = 17280
Amount paid as instalment by the man = 2000*(1+20/100)2 + 2000*(1+20/100) + 2000 = 7280
So remaining amount = 10000

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