## Compound Interest Questions and Answers Part-5

1. In what time will Rs 64000 amounts to Rs 68921 at 5% per annum interest being compounded half yearly ?
a) $$1\frac{1}{2}$$ years
b) 2 years
c) 3 years
d) $$2\frac{1}{2}$$ years

Explanation:
\eqalign{ & {\text{Amount}} = {\text{ }}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr & 68921 = 64000{\left( {1 + \frac{5}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr & \frac{{68921}}{{64000}} = {\left( {1 + \frac{1}{{40}}} \right)^{2 \times {\text{t}}}} \cr & {\left( {\frac{{41}}{{40}}} \right)^3} = {\left( {\frac{{41}}{{40}}} \right)^{2 \times {\text{t}}}} \cr & 2{\text{t = 3}} \cr & {\text{t = }}\frac{3}{2} \cr & {\text{t = 1}}\frac{1}{2}{\text{ years}} \cr}

2. When principal = Rs. S, rate of interest = 2r % p.a., then a person will get after 3 years at compound interest = ?
a) $${\text{Rs}}{\text{. }}\frac{{6{\text{Sr}}}}{{100}}$$
b) $${\text{Rs}}{\text{. S}}{\left( {1 + \frac{{\text{r}}}{{50}}} \right)^3}$$
c) $${\text{Rs}}{\text{. S}}{\left( {1 + \frac{{\text{r}}}{{100}}} \right)^3}$$
d) $${\text{Rs}}{\text{. 3S}}{\left( {1 + \frac{{\text{r}}}{{100}}} \right)^3}$$

Explanation:
\eqalign{ & {\text{Principal = Rs S}} \cr & {\text{Rate }}\% {\text{ = 2r}}\,\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{Time = 3 years}} \cr & {\text{A = P}}{\left( {1 + \frac{{\text{r}}}{{100}}} \right)^T} \cr & {\text{A = S}}{\left( {1 + \frac{{{\text{2r}}}}{{100}}} \right)^3} \cr & {\text{A = S}}{\left( {1 + \frac{{\text{r}}}{{50}}} \right)^3} \cr}

3. At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years ?
a) 6.5%
b) 4.5%
c) 6%
d) 7.5%

Explanation:
\eqalign{ & {\text{A = P }}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} \cr & 1348.32 = 1200{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \frac{{134832}}{{120000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \frac{{2809}}{{2500}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & {\left( {\frac{{53}}{{50}}} \right)^2} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \frac{{53}}{{50}} = 1 + \frac{{\text{R}}}{{100}} \cr & {\text{R}} = {\text{ 6% }} \cr}

4. On what sum of money will the difference between simple interest and compound interest for 2 years at 5% per annum be equal to Rs. 63 ?
a) Rs. 24600
b) Rs. 24800
c) Rs. 25200
d) Rs. 25500

Explanation:
\eqalign{ & {\text{Rate of interest = 5}}\% {\text{ per annum}} \cr & {\text{Time = 2 year}} \cr & P\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right] - \frac{{P \times r \times t}}{{100}}{\text{ = 63}} \cr & P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{P \times 5 \times 2}}{{100}}{\text{ = 63}} \cr & P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr & P\left[ {{{\left( {\frac{{105}}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr & P\left( {\frac{{11025 - 10000}}{{10000}}} \right) - \frac{{10P}}{{100}} = 63 \cr & \frac{{1025P}}{{10000}} - \frac{{10P}}{{100}} = 63 \cr & \frac{{1025P - 1000P}}{{10000}} = 63 \cr & 25P = Rs.630000 \cr & P = \frac{{630000}}{{25}} \cr & P = Rs. 25200 \cr}

5. The sum for 2 years given a compound interest of Rs. 3225 at 15% rate. Then the sum is =
a) Rs. 10000
b) Rs. 20000
c) Rs. 15000
d) Rs. 32250

Explanation: Interest for 2 years at the rate of 15%
\eqalign{ & {\text{ = 15 + 15 + }}\frac{{15 \times 15}}{{100}} = 32.25\% \cr & {\text{According to question,}} \cr & {\text{32}}{\text{.25}}\% {\text{ = 3225}} \cr & {\text{100}}\% {\text{ = }}\frac{{3225}}{{32.25}} \times 100 \cr & = 100 \times 100 = 10000 \cr}

6. The compound interest on Rs. 5000 for 3 years at 10% p.a. will amount to = ?
a) Rs. 1654
b) Rs. 1655
c) Rs. 1600
d) Rs. 1565

Explanation:
\eqalign{ & {\text{Principal = Rs. 5000}} \cr & {\text{Time = 3 years}} \cr & {\text{Rate = 10}}\% {\text{ = }}\frac{1}{{10}} \cr & {\text{Principal}}\,\,\,\,\,\,\,{\text{Amount}} \cr & \,\,\,\,\,\,\,\,\,\,{\text{10}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{11}} \cr & \,\,\,\,\,\,\,\,\,\,{\text{10}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{11}} \cr & \,\,\,\,\,\,\,\,\,\,{\text{10}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{11}} \cr & \underbrace {\overline {\,\,\,\,\,\,1000\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1331}}\,\,\,\,\,} }_{331{\text{ units}}} \cr & 1000{\text{ units = Rs 5000}} \cr & 1{\text{ units = Rs 5}} \cr & 331{\text{ units = 331}} \times {\text{5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = Rs. 1655}} \cr}

7. If the compound interest on a certain sum for two years at 12% per annum is Rs. 2544, the simple interest on it at the same rate for 2 years will be = ?
a) Rs. 2400
b) Rs. 2500
c) Rs. 2480
d) Rs. 2440

Explanation:
\eqalign{ & {\text{Rate = 12}}\% \cr & {\text{Time = 2 years}} \cr & {\text{Effective rate of CI for 2 years}} \cr & {\text{ = 12 + 12 + }}\frac{{12 \times 12}}{{100}} \cr & = 25.44\,\% \cr & {\text{Effective rate of SI for 2 years}} \cr & {\text{ = 12}} \times 2{\text{ = 24}}\,\% \cr & {\text{Required SI}} \cr & {\text{ = }}\frac{{2544}}{{25.44}} \times {\text{24}} = {\text{ Rs. 2400}} \cr}

8. A sum becomes Rs. 2916 in 2 years at 8% per annum compound interest. The simple interest at 9% per annum for 3 years on the same amount will be = ?
a) Rs. 600
b) Rs. 675
c) Rs. 650
d) Rs. 625

Explanation:
\eqalign{ & {\text{Amount = Rs}}{\text{. 2916}} \cr & {\text{Time = 2 years }} \cr & {\text{Rate = 8}}\% \cr & {\text{Effective rate }}\% {\text{ CI for 2 years}} \cr & {\text{ = 8 + 8 + }}\frac{{8 \times 8}}{{100}} = 16.64\% \cr & {\text{Required sum}} \cr & {\text{ = }}\frac{{2916}}{{\left( {100 + 16.64} \right)}} \times 100 = {\text{Rs}}{\text{. }}2500 \cr & {\text{Required simple interest}} \cr & {\text{ = }}\frac{{2500 \times 9 \times 3}}{{100}} \cr & = {\text{Rs}}{\text{. }}675 \cr}

9. A sum of money is compound interest became doubles itself in 15 years. It will become eight times of itself in =
a) 45 years
b) 48 years
c) 54 years
d) 60 years

\eqalign{ & P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{15}} = 2P \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{15}} = 2 \cr & {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 8P \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 8 = {2^3} = {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^{15}}} \right\}^3} \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{45}} \cr & \Rightarrow n = 45 \cr}
\eqalign{ & P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8P \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8 \cr & {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16P \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16 = {2^4} = {\left( {{2^3}} \right)^{\frac{4}{3}}} \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( 8 \right)^{\frac{4}{3}}} \cr & \Rightarrow {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3}} \right\}^{\frac{4}{3}}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^4} \cr & \Rightarrow n = 4 \cr}