Aptitude question and answer on Compound interest

In this section we are going to discuss about compound interest problems. Not just the overview of the topic, but also you are going to learn the important formulas on compound interest along with explanation which is easy to understand.

1
A sum of money placed at compount interest doubles itself in 5 years. It will amount to eight times itself in
A.    10 years
B.    12 years
C.    15 years
D.    20 years

Answer : C.  15 years

Explanation:
\(P \bigg(1+{R \over 100} \bigg )^5=2P\)
=> \(\bigg(1+{R \over 100} \bigg )^5=2\) ------------  (1)

let    \(P \bigg(1+{R \over 100} \bigg )^n=8P\)
\(\dot{..}\)  \( \bigg(1+{R \over 100} \bigg )^n=8\)
=>\( \bigg(1+{R \over 100} \bigg )^n =2^3\)
=>\( \bigg(1+{R \over 100} \bigg )^n =\Bigg\{ \bigg(1+{R \over 100}\bigg)^5 \Bigg \}^3\) --------------------using (1)
=>\( \bigg(1+{R \over 100} \bigg )^n = \bigg(1+{R \over 100}\bigg)^{15}\)
\(\dot{..}\) n= 15 years.
 

2
What annual payment will discharge a debt of Rs. 7620 due in 3 years at \(16{2 \over 3}\)% per annum compound interest ?
A.    Rs.2540
B.    Rs.3430
C.    Rs.3260
D.    Rs.3380

Answer : B.   Rs.3430

Explanation:
Let each instalment be Rs. x. Then

\({x \over \bigg( {1+ {50 \over 3 \times 100}}\bigg)}+\)\({x \over \bigg( {1+ {50 \over 3 \times 100}}\bigg)^2}+\)\({x \over \bigg( {1+ {50 \over 3 \times 100}}\bigg)^3}+\)=7620.

or               \({6x \over7}+{36x \over 49}+{216x \over 343}=7620\)

or               \(294x+252x+216x=7620 \times 343\)

or            \(x={7620 \times 343 \over 762}=3430\)

\(\dot{..}\)    Amount of each instalment = Rs. 3430 .

3
A sum of money becomes Rs. 13380 after 3 years and Rs. 20070 after 6 years on compound interest. The sum is
A.    Rs.8800
B.    Rs.8890
C.    Rs.8920
D.    Rs.9040

Answer : C.  Rs.8920

Explanation:
Let the sum be Rs. x. 
Then           \(x \bigg(1+{R \over 100}\bigg)^3=13380\)
and            \(x \bigg(1+{R \over 100}\bigg)^6=20070\)
On dividing we get 

\(\bigg( 1+{R\over100}\bigg)^3\)=\(\bigg( {20070 \over 13380 }\bigg)\)=\(\bigg( {3 \over 2 }\bigg)\)
\(\dot{..}\)  \(x{3 \over2}\)=13380
=>   \(x=\bigg(13380 \times {2 \over 3} \bigg)\)=8920
Hence, the sum is Rs. 8920

4
Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.
A.    611
B.    612
C.    613
D.    614

Answer : B.   612

Explanation:
Amount = Rs\( [7500 \times(1+({4\over 100})^2]\)

= Rs \(\bigg[7500 \times ({26\over25}) \times ({26\over25})\bigg]\)

= Rs. 8112.
      
Therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.
 

5
Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually
A.    3108
B.    3109
C.    3110
D.    3111

Answer : B.   3109

Explanation:
Time = 2  years  4  months = \(2({4\over12})\) years = \(2({1\over3})\) years.
                      
Amount = Rs  \([8000 \times (1+({15\over100}))^2 \times (1+({1\over3} \times{15\over100})]\)

=Rs. \(\bigg[8000 \times ({23\over20}) \times ({23\over20}) \times ({21\over20})\bigg]\)

= Rs. 11109.                            .

\(\dot{..}\) C.I. = Rs. (11109 - 8000) = Rs. 3109.
 

6
Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the interest being compounded half-yearly.
A.    Rs. 624.32
B.    Rs. 724.32
C.    Rs. 824.32
D.    Rs. 924.32

Answer : C.  Rs. 824.32

Explanation:
Principal = Rs. 10000; Rate = 2%  per half-year;  Time = 2 years = 4 half-years.

Amount =Rs \(\bigg [10000 \times (1+({2\over100}))4\bigg] \)

= Rs \((10000\times(51/50) \times (51/50) \times (51/50)\times (51/50))\)            

= Rs. 10824.32.

\(\dot{..}\)C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.