ARITHMETICINTEREST & INSTALLMENTSMATHS

QUESTIONS ON COMPOUND INTEREST

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QUESTIONS ON COMPOUND INTEREST

QUERY 1

A sum of Rs 1000 after 3 years at compound interest becomes a certain amount that is equal to the amount that is the result of a depreciation from Rs 1728. Find the difference between the rates of CI and depreciation, CI being given 10%. p.a.

A) 8.33
B) 2.00
C) 1.67
D) 2.75

MAHA GUPTA
The concept of DEPRECIATION is just the same as of COMPOUND INTEREST. The only difference is that in compound interest the amount increases whereas in depreciation it decreases. Hence we take 1 – r/100 instead of 1 + r/100 in depreciation.

Now according to the question
1728(1 – r/100)³ = 1000(1 + 10/100)³; where ‘r’ is the rate of depreciation

=> [(100 – r)/100]³ = (11³)/1728
=> (100 – r)/100 = 11/12
=> 1200 – 12r = 1100
=> r = 100/12 = 8.33 (approx)

So the needed difference of rates = 10 – 8.33 = 1.67 (approx) [option ‘C’]

QUERY 2

A sum of money amounts to Rs 9990 after 3 years and to Rs 14985 after 6 years on compound interest. Find the sum.

A) 6600
B) 6660
C) 6000
D) 7500

MAHA GUPTA
In 3 years i.e. (6 -3) Rs 9990 becomes Rs 14985; means 14985/9990 = 3/2 of itself

Means Rs 9990 also is 3/2 of the initial sum because it also took 3 years to be such

Hence the required sum = 9990*(2/3) = 6660 (option ‘B’)

QUERY 3

An amount of money at compound interest grows up to Rs 3840 in 4 years and up to Rs 3936 in 5 years. Find the rate of interest.

A) 2.5%
B) 2%
C) 3.5%
D) 2.05%

MAHA GUPTA
The amount of money becomes 3840 in 4 years and 3936 in 5 years means 3840 becomes 3936 in 1 year at simple interest.

Thus interest on 3840 for 1 year at simple interest = 3936 – 3840 = 96

Therefore rate of interest = (Interest*100)/(Principal*Time) = (96*100)/(3840*1) = 9600/3840 = 2.5 (option ‘A’)

QUERY 4

A sum of money becomes 4 times in 6 years at compounded interest, in how much time it will become 8 times?

A) 8 years
B) 9 years
C) 10 years
D) 12 years

MAHA GUPTA
Let the number of years in which the sum will become 8 times = n

Formula:
4n/6 = 8
=> 22n/6 = 23
=> 2n/6 = 3
=> n = 9 (option ‘B’)

Hence, the number of years in which the sum will become 8 times = 9 years (option ‘B’)

HOW IS THE FORMULA GOT
Letting the sum be Re 1
Thus, amount in the first case = 1*4 = Rs 4
And amount in the second case = 1*8 = Rs 8

Therefore,
4 = 1*(1 + r/100)6 => 41/6 = 1 + r/100                                          —- (i)

8 = 1*(1 + r/100)n => 81/n = 1 + r/100                                         —- (ii)

From (i) and (ii) 
41/6
  = 81/n
= 4n/6
 = 8

QUERY 5

A bank gives compound interest on deposits at the rate of 5% for the first year, 6% for the second year and 10% for the third year; find the effective rate of interest for one year. Also find the initial deposit if the sum after three years is Rs 12243.

A) 21.57%,         Rs 11000
B) 21.22%,         Rs 13000
C) 22%,              Rs 15000
D) 22.43%,        Rs 10000

MAHA GUPTA
First find here the one time interest which is done by the following formula. Remember it applies for two terms at a time only
a + b + (a*b)/100

Now taking a = 5 and b = 6, the rate of interest = 5 + 6 + (5*6)/100 = 11.3

Now by taking a = 11.3 and b = 10, the rate of interest = 11.3 + 10 + (11.3 + 10)/100 = 22.43

So, we can say that the effective rate of interest is 22.43% (option ‘D’)

Therefore, the sum 100 + 22.43 = 122.43

But actual sum is Rs 12243

Hence the initial deposit = (100/122.43)*12243 = 10000 (option ‘D’)

NOTE: After getting this rate of interest it’ll be easier if you do the some by answer options given.

QUERY 6

The salary of an employee of a company increases every month by 4%. If his salary in August was Rs 6300, what would be his approximate salary in the month of October in the same year?

A) Rs 6600
B) Rs 6814
C) Rs 6875
D) Rs 7000

MAHA GUPTA
This is a case as we see in compound interest. But this is just a 2 month affair, we need not apply the formula meant for compound interest here as it’s a little time consuming. So better if it’s done applying simple approach. Look how:

Salary in the month of September = 6300 + 4% of 6300 = 6300 + 252 = 6552

Salary in October: 6552 + 4% of 6552 = 6552 + 262 .08 = 6814.08 = 6814 (approx) [option ‘B’]

QUERY 7

The least number of complete years in which a sum of money put out at 20% compound interest will be more than double is?

A) 3 years
B) 4 years
C) 5 years
D) 6 years

MAHA GUPTA
According to the question
P(1 + 20100)n ˃ 2P
= (6/5)n ˃ 2

We see only n = 4 can satisfy the given condition

So, n = 4 years (option ‘B’)

QUERY 8

The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is?

A) 6.06%
B) 6.07%
C) 6.08%
D) 6.09%

MAHA GUPTA
As earlier the interest was payable half-yearly, the rate will be 6/2 = 3% and terms = 2

Now letting the sum Rs 100, the effective rate of interest if it’s payable annually =
100(1 + 3100)² – 100
= 6.09 (option ‘D’)

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QUESTIONS ON SIMPLE INTEREST (PART-I)

Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)