What is the difference between the C.I. on Rs. 5000 for 1 1/2 years at 4% per annum compounded yearly and half-yearly?
Answer: A
C.I. when interest is compounded yearly
= [5000 * (1 + 4/100) * (1 + (1/2 * 4)/100]
= 5000 * 26/25 * 51/50 = Rs. 5304
C.I. when interest is compounded half-yearly
= [5000 * (1 + 2/100)²]
= (5000 * 51/50 * 51/50 * 51/50) = Rs. 5306.04 Difference = (5306.04 - 5304) = Rs. 2.04
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The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
Answer: C
Let the rate of interest be R% p.a.
4400{[1 + R/100]² - 1} = 11193.60
[1 + R/100]² = (44000 + 11193.60)/44000
[1 + R/100]² = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]² = 784/625 = (28/25)2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
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The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?
Answer: A
Over two years the gap between compound and simple interest is
\[ \text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^{2} \]
\[ 96 = 15000 \times \frac{R^{2}}{10000} = 1.5R^{2} \]
\[ R^{2} = 64 \quad\Longrightarrow\quad R = 8\% \]
The rate is fully determined, so "cannot be determined" is a trap: for a two-year period the difference, the principal and the rate are locked together by that single relation.
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Ramesh gets double the amount in 9 years when invested at compound interest. In how many years will the amount become four times itself?
Answer: D
Compound growth is multiplicative, so doubling twice is the same as quadrupling.
\[ \left(1 + \frac{R}{100}\right)^{9} = 2 \]
\[ \left(1 + \frac{R}{100}\right)^{18} = \left[\left(1 + \frac{R}{100}\right)^{9}\right]^{2} = 2^{2} = 4 \]
So it takes 18 years. The rate never needs to be found.
This multiplicative behaviour is what separates compound from simple interest: under simple interest, doubling in 9 years means quadrupling in 27, since there the growth is additive.
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A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is:
Answer: B
\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]
Half-yearly compounding at 5% per annum means 2.5% per half-year. The January deposit earns for two half-years and the July deposit for one.
\[ A = 1600(1.025)^{2} + 1600(1.025) \]
\[ = 1600(1.050625) + 1640 = 1681 + 1640 = \text{Rs. } 3321 \]
\[ \text{interest} = 3321 - 3200 = \text{Rs. } 121 \]
Each deposit must be tracked from its own date; treating the full Rs. 3200 as present all year would give Rs. 162.
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Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.64000?
Answer: A
An increase of \( \tfrac{1}{8} \) each year multiplies the value by \( \tfrac{9}{8} \).
\[ A = 64000 \times \left(\frac{9}{8}\right)^{2} = 64000 \times \frac{81}{64} \]
\[ = 1000 \times 81 = \text{Rs. } 81{,}000 \]
The 64 in the principal cancels the \( 8^{2} \) in the denominator exactly, which is why the answer is so clean. Adding \( \tfrac{1}{8} \) of the original twice would give only Rs. 80,000.
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At the end of 2 years difference between simple and compound interest is Rs. 2166. If the principal is Rs. 60000, then what is the rate of interest?
Answer: C
\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]
For two years the difference is
\[ \text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^{2} \]
\[ 2166 = 60000 \times \frac{R^{2}}{10000} = 6R^{2} \]
\[ R^{2} = 361 \quad\Longrightarrow\quad R = 19\% \]
The rate appears squared, so a perfect square is the signal that the arithmetic is on track - here \( 19^{2} = 361 \).
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A sum of money place at compound interest doubles itself in 4 years. In how many years will it amount to eight times itself?
Answer: C
\[ \left(1 + \frac{R}{100}\right)^{4} = 2 \]
Eight is \( 2^{3} \), so three doubling periods are needed:
\[ \left(1 + \frac{R}{100}\right)^{12} = 2^{3} = 8 \]
\[ 3 \times 4 = 12 \text{ years} \]
Multiplying the four years by eight would give 32 - the mistake of treating compound growth as if it were proportional to time.
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A sum of money is put out at compound interest for 2 years at 20%. It would fetch Rs.482 more if the interest were payable half-yearly, then it were pay able yearly. Find the sum.
Answer: D
\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]
Compare the two schemes over the same two years.
Yearly: \( P(1.2)^{2} = 1.44P \)
Half-yearly - 10% per half-year over four periods:
\[ P(1.1)^{4} = 1.4641P \]
\[ 1.4641P - 1.44P = 0.0241P = 482 \]
\[ P = \frac{482}{0.0241} = \text{Rs. } 20{,}000 \]
More frequent rests always earn a little more on the same nominal rate; here the gain is 2.41% of the principal.
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Ravi invested certain amount for two rates of simple interests at 6% p.a. and 7% p.a. What is the ratio of Ravi's investments if the interests from those investments are equal?
Answer: D
Let the two investments be \( P_{1} \) at 6% and \( P_{2} \) at 7%, over the same period \( n \). Equal interest means
\[ \frac{P_{1} \times 6 \times n}{100} = \frac{P_{2} \times 7 \times n}{100} \]
\[ 6P_{1} = 7P_{2} \quad\Longrightarrow\quad \frac{P_{1}}{P_{2}} = \frac{7}{6} \]
The investments are in the inverse ratio of the rates: the lower rate needs the larger sum to yield the same return. The time cancels, so it need not be known.
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