1. A certain some of money and Rs. 2420 in 2 years and Rs. 2662 in 3 years at same rate of compound interest, compounded annually. The rate of interest per annum is =
a) 6%
b) 8%
c) 9%
d) 10%
Explanation:
$$\eqalign{ & {\text{Amount after three years}} {\text{ = Rs. 2662}} \cr & {\text{Amount after two years}} {\text{ = Rs. 2420}} \cr & {\text{Net interest earned in the }}{{\text{3}}^{{\text{rd}}}}{\text{ year}} \cr & {\text{ = }}\,{\text{2662}} - {\text{2420}} \cr & {\text{ = Rs}}{\text{. 242}} \cr & {\text{Rate of interest (r)}} \cr & {\text{ = }}\frac{{242}}{{2420}} \times {\text{100 = 10% }} \cr} $$
(2nd year's amount is principal for 3rd year)
2. Kamal took Rs. 6800 as a loan which along with interest is to be repaid in two equal annual installment. If the rate of interest is $$12\frac{1}{2}$$ % compounded annually, then the value of each installment is =
a) Rs. 8100
b) Rs. 4150
c) Rs. 4050
d) Rs. 4000
Explanation:
$$\eqalign{ & {\text{Rate of interest}} \cr & {\text{r}} = {\text{12}}\frac{1}{2}\% = \frac{1}{8} \cr} $$
| Year | Principal | Installment | |
| I | 8×9 → | 9×9 | ......(i) |
| II | 64 → | 81 | ......(ii) |
Total principal = 72 + 64 = 136 units
136 units → 6800
1 units → 50
81 units → 4050
Each installment = Rs. 4050
3. A man invests Rs 4000 for 3 years at compound interest. After one year the money amounts to Rs. 4320. What will be the amount (to the nearest rupee) due at the end of 3 years ?
a) Rs. 4939
b) Rs. 5039
c) Rs. 5789
d) Rs. 6129
Explanation:
$$\eqalign{ & {\text{Le the rate be R }}\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{4000}}\left( {1 + \frac{{{\text{R }}}}{{100}}} \right) = 4320 \cr & 1 + \frac{{{\text{R }}}}{{100}} = \frac{{4320}}{{4000}} = \frac{{108}}{{100}} \cr & \frac{{{\text{R }}}}{{100}} = \frac{8}{{100}} \cr & {\text{R }} = 8 \cr & {\text{Amount after 3 yeras}} \cr & {\text{ = Rs}}{\text{. }}\left[ {4000 + {{\left( {1 + \frac{8}{{100}}} \right)}^3}} \right] \cr & {\text{ = Rs}}{\text{. }}\left( {4000 \times \frac{{27}}{{25}} \times \frac{{27}}{{25}} \times \frac{{27}}{{25}}} \right) \cr & {\text{ = Rs}}{\text{. }}\left( {\frac{{629856}}{{125}}} \right) \cr & {\text{ = Rs}}{\text{. }}5038.848 \approx 5039 \cr} $$
4. A sum of Rs. 13360 was borrowed at $${\text{8}}\frac{3}{4}$$ % per annum compound interest and paid back in two years in two equal annual installments. What was the amount of each installment ?
a) Rs. 5769
b) Rs. 7569
c) Rs. 7009
d) Rs. 7500
Explanation:
$$\eqalign{ & {\text{Rate of interest (r)}} \cr & {\text{ = 8}}\frac{3}{4}\% = \frac{7}{{80}} = \frac{{87 \to {\text{ Installment}}}}{{80 \to {\text{Principal}}}} \cr} $$
| ⇒ | I | 80×87 | → | 87×87 | ......(i) |
| ⇒ | II | 6400 | → | 7569 | ......(ii) |
Total principal = 6960 + 6400 = 13360
13360 units = Rs. 13360
1 units = Rs. 1
7569 units = Rs. 7569
Each installment = Rs. 7569
5. An amount of Rs. 10000 becomes Rs. 14641 in 2 years if the interest is compounded half yearly. What is the rate of compound interest p.c.p.a. ?
a) 10%
b) 12%
c) 16%
d) 20%
Explanation:
$$\eqalign{ & {\text{Let the rate be R% p}}{\text{.a}}{\text{. }} \cr & {\text{10000}}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^4} = 14641 \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{200}}} \right)^4} = \frac{{14641}}{{10000}} = {\left( {\frac{{11}}{{10}}} \right)^4} \cr & 1 + \frac{{\text{R}}}{{200}} = \frac{{11}}{{10}} \cr & \frac{{\text{R}}}{{200}} = \frac{1}{{10}} \cr & {\text{R}} = {\text{20% }} \cr} $$
6. What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?
a) Rs. 9000.30
b) Rs. 9720
c) Rs. 10123.20
d) Rs. 10483.20
Explanation:
$$\eqalign{ & {\text{Amount}} = Rs.\,\left[ {25000 \times {{\left( {1 + \frac{{12}}{{100}}} \right)}^3}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,\left( {25000 \times \frac{{28}}{{25}} \times \frac{{28}}{{25}} \times \frac{{28}}{{25}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,35123.20 \cr & {\text{C}}{\text{.I}}{\text{.}} = Rs.\left( {35123.20 - 25000} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,10123.20 \cr} $$
7. At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?
a) 6%
b) 6.5%
c) 7%
d) 7.5%
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{rate}}\,{\text{be}}\,R\% \,p.a. \cr & 1200 \times {\left( {1 + \frac{R}{{100}}} \right)^2} = 1348.32 \cr & \Rightarrow {\left( {1 + \frac{R}{{100}}} \right)^2} = \frac{{134832}}{{120000}} = \frac{{11236}}{{10000}} \cr & {\left( {1 + \frac{R}{{100}}} \right)^2} = {\left( {\frac{{106}}{{100}}} \right)^2} \cr & 1 + \frac{R}{{100}} = \frac{{106}}{{100}} \cr & R = 6\% \cr} $$
8. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is
a) 3
b) 4
c) 5
d) 6
Explanation:
$$\eqalign{ & P{\left( {1 + \frac{{20}}{{100}}} \right)^n} > 2P\,\,\, \Rightarrow \,\,\,{\left( {\frac{6}{5}} \right)^n} > 2 \cr & \left( {\frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}} \right) > 2 \cr & n = 4\,{\text{years}} \cr} $$
9. Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?
a) Rs. 8600
b) Rs. 8620
c) Rs. 8820
d) None of these
Explanation:
$$\eqalign{ & {\text{Amount}} = Rs.\left[ {8000 \times {{\left( {1 + \frac{5}{{100}}} \right)}^2}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,\left( {8000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,8820 \cr} $$
10. 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%
Explanation:
$$\eqalign{ & {\text{Amount}}\,{\text{of}}\,{\text{Rs}}{\text{.}}\,{\text{100}}\,{\text{for}}\,{\text{1}}\,{\text{year}}\,{\text{when}}\, {\text{compounded}}\,{\text{half - yearly}} \cr & = Rs.\,\left[ {100 \times {{\left( {1 + \frac{3}{{100}}} \right)}^2}} \right] \cr & = Rs.\,106.09 \cr & {\text{Effective}}\,{\text{rate}} = \left( {106.09 - 100} \right)\% \cr & = 6.09\% \cr} $$