>> Aptitude >> Compound interest

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Contents:

- Aptitude
- Approximation
- Average
- Boat and Stream
- Compound interest
- Discount
- Linear Equations
- Mensuration
- Mixture and Allegation
- Number series
- Number System
- Partnership
- Percentage
- Permutation and combination
- Pipes and Cisterns
- Probability
- Problem on ages
- Profit and Loss
- Ratio and Proportions
- Simple and compound interest
- Time and Distance
- Time and work
- Trains
- Unitary Method
- Word problems
- Work and Wages

1). A sum amounts to Rs. 1352 in 2 yr at 4% compound interest. The sum is
Using the formula, A = \( \Large P \left( 1+ \frac{R}{100}\right)^{n} \) | ||||

2). What will be the present worth of RS. 169 due in 2 yr at 4% pa compound interest?
Given, R = 4%, n = 2 yr and A =RS.169 P = ? | ||||

3). What amount will be received on a sum 0f Rs. 1750 in 2 1/2 Yr, if the interest is compounded at the rate of 8% pa?
Given, P = RS. 1750,R = 8%, n = 2 and \( \Large \frac{a}{b} \) = \( \Large \frac{1}{2} \) | ||||

4). What amount will be received on a sum of Rs. 15000 in 1 1/4 yr at 12% pa, if interest is compounded quarterly?
Given, P = RS. 15000, R = 12% | ||||

5). The compound interest on a sum of RS. 4000 becomes RS. 630.50 in 9 months. Find the rate of interest, if interest is compounded quarterly.
Given, P = RS. 4000, n = 9 months = \( \Large \frac{3}{4}yr \) and CI= RS. 630. Amount = P + CI = 4000 + 630.50 = rs. 4630.50 According to the formula, Amount = \( \Large P \left(1+ \frac{R}{100\times 4}\right)^{4n} \) => 4630.50=\( \Large 4000 \left(1+ \frac{R}{400}\right)^{4\times 3/4} \) => 4630.50=\( \Large 4000 \left(\frac{400+R}{400}\right)^{3} \) => \( \Large \frac{4630.50}{4000} \)=\( \Large \left(\frac{400+R}{400}\right)^{3} \) => \( \Large \frac{9261}{8000} \)=\( \Large \left(\frac{400+R}{400}\right)^{3} \) => \( \Large \left(\frac{21}{20}\right)^{3} \)=\( \Large \left(\frac{400+R}{400}\right)^{3} \) => \( \Large \frac{400+R}{400} \)=\( \Large \frac{21}{20} \) => 400+R=\( \Large 21\times 20 \)=420 R=420-400=20% | ||||

6). The population of a city increases at the rate of 5% pa. If the present population of the city is 185220, then what was its population 3 yr ago?
Given, P=185220,R=5% (increases) and n=3yr. | ||||

7). The population of a country is 10 crore and it is the possibility that the population will become 13.31 crore in 3 yr. What will be the the annual rate per cent on this growth?
Given, P = 10 crore and population after 3 yr = 13.31 crore | ||||

8). A sum of RS. 400 amounts to RS. 441 in 2 yr. What will be its amount, if the rate of interest is increased by 5%?
According to the given conditign 441=\( \Large 400 \left(1+ \frac{R}{100}\right)^{2} \) => \( \Large \frac{441}{400} \)= \( \Large \left(1+ \frac{R}{100}\right)^{2} \) => \( \Large \left(\frac{21}{20}\right)^{2} \)=\( \Large \left(1+ \frac{R}{100}\right)^{2} \) => \( \Large \frac{21}{20} \)=\( \Large 1+ \frac{R}{100} \) => \( \Large \frac{21}{20}-1=\frac{R}{100} \) => \( \Large \frac{R}{100}=\frac{1}{20} \) R=5% New rate = 5+5=10% Amount=\( \Large 400 \left(1+ \frac{10}{100}\right)^{2} \) =\( \Large 400\times 11/10\times 11/10 \)=RS.484 | ||||

9). A sum, at the compound rate of interest, becomes 2 1/2 times in 6 yr. The same sum becomes what times in 18 yr?
If sum is x, then x becomes \( \Large \frac{5}{2}x \) in 6 yr. \( \Large \frac{5}{2}x \) becomes \( \Large \frac{25}{4}x \) in 12 yr. \( \Large \frac{25}{4}x \) becomes \( \Large \frac{125}{8}x \) in 18 yr. The sum becomes \( \Large \frac{125}{8}x \) times 1n 18 yr. | ||||

10). What is the difference between compound interest and simple interest for 2 yr on the sum of RS. 1250 at 4% pa?
Given, P = RS. 1250, n = 2yr and R = 4% |