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Compound Interest - MCQ 1 - GMAT MCQ


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20 Questions MCQ Test Quantitative for GMAT - Compound Interest - MCQ 1

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Compound Interest - MCQ 1 - Question 1

The compound interest on a certain sum for 2 years is Rs. 786 and S.I. is Rs. 750. If the sum is invested such that the S.I. is Rs. 1296 and the number of years is equal to the rate per cent per annum, Find the rate of interest?

Detailed Solution for Compound Interest - MCQ 1 - Question 1

CI for 2 years = Rs. 786
SI for 2 years = Rs. 750
36/360 * 100 = 10%
P for first year = 3600
P*x*x/100 = 1296
x = 6%

Compound Interest - MCQ 1 - Question 2

Hari took an educational loan from a nationalized bank for his 2 years course of MBA. He took the loan of Rs.5 lakh such that he would be charged at 7% p.a. at CI during his course and at 9% CI after the completion of the course. He returned half of the amount which he had to be paid on the completion of his studies and remaining after 2 years. What is the total amount returned by Hari?

Detailed Solution for Compound Interest - MCQ 1 - Question 2

5,00,000 * (1.07)² = 572450
Returned amount = 286225
After two years = 286225 * (1.09)² = 340063
Total amount = 286225 + 340063 = 626288

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Compound Interest - MCQ 1 - Question 3

Rs.20,000 was invested by Mahesh in a FD @ 10% pa at CI. However every year he has to pay 20% tax on the CI. How much money does Mahesh have after 3 years?

Detailed Solution for Compound Interest - MCQ 1 - Question 3

(20000*(1.08)³)=25194

Compound Interest - MCQ 1 - Question 4

Leela takes a loan of Rs. 8400 at 10% p.a. compounded annually which is to be repaid in two equal annual installments. One at the end of one year and the other at the end of the second year. The value of each installment is?

Detailed Solution for Compound Interest - MCQ 1 - Question 4

8400 = x*(210/121) ⇒ 4840

Compound Interest - MCQ 1 - Question 5

A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.723 more, if the interest was payable half yearly than if it was payable annually. The sum is ____

Detailed Solution for Compound Interest - MCQ 1 - Question 5

sum – Rs.x
C.I. compounded half yearly = (4641/10000)x
C.I. compounded annually = (11/25)x
(4641/10000)x – (11/25)x = 723
x = 30000

Compound Interest - MCQ 1 - Question 6

A sum of Rs.7140 is to be divided between Anita and Bala who are respectively 18 and 19 yr old, in such a way that if their shares will be invested at 4% per annum at compound interest, they will receive equal amounts on attaining the age of 21 year. The present share of Anita is

Detailed Solution for Compound Interest - MCQ 1 - Question 6

Amount got by Anita after 3 yr = Amount got by Bala after 2 yr
x*(26/25)³ = (7140 – x)*(26/25)
26/25 = 7140 – x / x
x = 3500

Compound Interest - MCQ 1 - Question 7

Suresh borrows Rs.6375 to be paid back with compound interest at the rate of 4 % pa by the end of 2 year in two equal yearly installments. How much will each installment will be?

Detailed Solution for Compound Interest - MCQ 1 - Question 7

25x/26 + 625/676x = 6375
x = (6375 * 676)/1275 = 3380

Compound Interest - MCQ 1 - Question 8

A sum of Rs. 8400 was taken as loan. This is to be paid in two equal annual installments. If the rate of interest be 20% compounded annually, then the value of each installment is

Detailed Solution for Compound Interest - MCQ 1 - Question 8

Let value of each installment be X.
X/(1 + 20/100) + X/(1 + 20/100)² = 8400
⇒ X(5/6 + 25/36) = 8400
⇒ X(56/36) = 8400
X = 5400

Compound Interest - MCQ 1 - Question 9

During the first year the population of a village is increased by 5% and the second year it is diminished by 5%. At the end of the second year its population was 31500. What was the population at the beginning of the first year?

Detailed Solution for Compound Interest - MCQ 1 - Question 9

x * 105/100 * 95/100 = 31500
x = 31500 * 100/105 * 100/95
D = 31578

Compound Interest - MCQ 1 - Question 10

If Rs. 7200 amounts to Rs.10368 at compound interest in a certain time , then Rs. 7200 amounts to what in half of the time?

Detailed Solution for Compound Interest - MCQ 1 - Question 10

Let rate = R% and time = n year
Then, 10368 =7200(1+R/100)n
⇒ (1+R/100)n = 10368/7200 = 1.44
∴ (1 + R/100)n/2 = √1.44 = 1.2
∴ Required amount for n/2 yr
= 7200(1+ R/100)n/2
= 7200 x 1.2 = Rs. 8640

Compound Interest - MCQ 1 - Question 11

A part of 70000 is lent out at 10% annum. The rest of the amount is lent out at 5% per annum after one year. The ratio of interest after 3 years from the time when first amount was lent out is 1:2. Find the second part that was lent out at 5%.

Detailed Solution for Compound Interest - MCQ 1 - Question 11

10*3*x/5*2*y = 1/2
x/y = 1/6
6/7*70000 = 60000

Compound Interest - MCQ 1 - Question 12

There is 50% increase in an amount in 5 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

Detailed Solution for Compound Interest - MCQ 1 - Question 12

In S.I,
Let P=100, I=50, T=5 yrs
R = 50*100/100*5 = 10%
In C.I, P = 12000, T=3 yrs, R= 10%
C.I = [12000*(1 + 10/100)^3 – 1 ] C.I = 3972.

Compound Interest - MCQ 1 - Question 13

Karthik lends a certain amount to Vignesh on simple interest for two years at 20%. Vignesh gives this entire amount to Kamal on compound interest for two years at the same rate annually. Find the percentage earning of Vignesh at the end of two years on the entire amount.

Detailed Solution for Compound Interest - MCQ 1 - Question 13

SI=20*2=40%
CI=20+20+(400/100)=44%
Diff = 44-40=4%

Compound Interest - MCQ 1 - Question 14

A man borrows 3000 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 fourth Year to clear all his debt?

Detailed Solution for Compound Interest - MCQ 1 - Question 14

After one year amount = 3000 *110/100 = 3300
He pays 1000 back, so remaining = 3300-1000 = 2300
After two year amount = 2300 *110/100 = 2530
He pays 1000 back, so remaining = 2530-1000 = 1530
After three year amount = 1530*110/100 = 1683
He pays 1000 back, so remaining = 1683-1000= 683
After fouth year = 683 * 110/100 = 751.3

Compound Interest - MCQ 1 - Question 15

Rahul saves an amount of 800 every year and then lent that amount at an interest of 10 percent compounded annually. Find the amount after 3 years.

Detailed Solution for Compound Interest - MCQ 1 - Question 15

800*(11/10)³=1064.8
800*(11/10)²=968
800*(11/10)=880
Total amount =2912.8

Compound Interest - MCQ 1 - Question 16

Find the compound interest at the rate of 8% for 3 years on that principal which in 3 years at therate of 10% per annum gives 300 as simple interest.

Detailed Solution for Compound Interest - MCQ 1 - Question 16

SI =300
Per yr = 100
Rate = 10%
C.I = 1000*(108/100)³ -1000
C.I = 259.712

Compound Interest - MCQ 1 - Question 17

The difference between the total simple interest and the total compound interest compounded annually at the same rate of interest on a sum of money at the end of two years is Rs. 450. What is definitely the rate of interest per cent per annum?

Detailed Solution for Compound Interest - MCQ 1 - Question 17

Difference = Pr /(100)2
= (450×100×100)/(P×r2)
P is not given

Compound Interest - MCQ 1 - Question 18

The CI on Rs.6000 for 3 years at 8% for first year, 7% for second year, 6% for the third year will be

Detailed Solution for Compound Interest - MCQ 1 - Question 18

A = 6000*108/100*107/100*106/100
= 6000*1.08*1.07*1.06
= 7349.616 = 7350
CI = 7350-6000 = 1350

Compound Interest - MCQ 1 - Question 19

Venkat and Vidhya have to clear their respective loans by paying 2 equal annual instalments of Rs.30000 each. Venkat pays at 10% pa of SI and Vidhyapays at 10% CI pa. What is the difference in their payments ?

Detailed Solution for Compound Interest - MCQ 1 - Question 19

D =[(30,000 *110/100*110/100) – 30,000] – 30,000 *10*2/100
=[36300-30000]- 6000
=6300 – 6000
D = 300

Compound Interest - MCQ 1 - Question 20

The difference between interest received by Vivek and Vimal is Rs.405 on Rs.4500 for 3 years. What is the difference in rate of interest ?

Detailed Solution for Compound Interest - MCQ 1 - Question 20

4500*3/100(R1-R2) = 405
R1-R2 = 405*100/13500 = 3%

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