A sum of Rs. 12500 amounts to Rs. 15500 in 4 years at simple interest. What is the rate of interest?
Answer: D
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ \text{SI} = 15500 - 12500 = \text{Rs. } 3000 \]
\[ 3000 = \frac{12500 \times R \times 4}{100} = 500R \]
\[ R = 6\% \]
The interest is 24% of the principal over four years, which is 6% a year.
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Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
The number of years equals the rate, so write both as \( R \).
\[ 432 = \frac{1200 \times R \times R}{100} = 12R^{2} \]
\[ R^{2} = 36 \quad\Longrightarrow\quad R = 6 \]
So the rate is 6% and the period 6 years. The data are complete, so "cannot be determined" is a trap; the negative root is discarded as a rate.
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A man took loan from a bank at the rate of 12% p.a. S.I. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ 5400 = \frac{P \times 12 \times 3}{100} = 0.36P \]
\[ P = \frac{5400}{0.36} = \text{Rs. } 15{,}000 \]
Three years at 12% comes to 36% of the principal, so the principal is a little under three times the interest.
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What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?
Answer: C
Present worth is the sum that would grow to Rs. 132 over the two years.
\[ \text{PW}\left(1 + \frac{5 \times 2}{100}\right) = 132 \]
\[ 1.1 \times \text{PW} = 132 \quad\Longrightarrow\quad \text{PW} = \text{Rs. } 120 \]
Taking 10% off Rs. 132 would give Rs. 118.80 - the second option, and wrong, because the interest accrues on the present worth, not on the future amount.
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A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9% p.a. in 5 years. What is the sum?
Answer: D
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ 4016.25 = \frac{P \times 9 \times 5}{100} = 0.45P \]
\[ P = \frac{4016.25}{0.45} = \text{Rs. } 8925 \]
Five years at 9% comes to 45% of the sum, so the principal is a little over twice the interest.
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Rs. 800 becomes Rs. 956 in 3 years at a rate of S.I. If the rate of interest is increased by 4%, what amount will Rs. 800 become in 3 years?
Answer: C
S.I. = (956 - 800) = Rs. 156.
Rate = (100 * 156) / (800 * 3) = 6 1/2 %
Now rate = (6 1/2 + 4) = 10 1/2 %
New S.I. = (800 * 21/2 * 3/100) = Rs. 252
New amount = (800 + 252) = Rs. 1052.
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A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
Answer: D
The extra interest from two more percentage points over seven years would be
\[ \frac{P \times 2 \times 7}{100} = 0.14P \]
which needs the principal. All that is known is
\[ \frac{P \times R \times 7}{100} = 1750 \quad\Longrightarrow\quad PR = 25{,}000 \]
one equation in two unknowns. A sum of Rs. 5000 at 5% and a sum of Rs. 2500 at 10% both fit, yet they would give Rs. 700 and Rs. 350 of extra interest respectively.
So the answer cannot be determined.
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What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
With the sum and the rate the same, simple interest is proportional to time.
\[ \frac{\text{SI}_{6}}{\text{SI}_{9}} = \frac{6}{9} = \frac{2}{3} \]
So the ratio is \( 2 : 3 \). Neither the sum nor the rate is needed - both cancel - so "data inadequate" is a trap.
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Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160, how much money did he borrow?
Answer: A
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Beyond eight years leaves \( 11 - 8 = 3 \) years at 13%, so the three slabs are:
\[ P\left(\frac{6 \times 3 + 9 \times 5 + 13 \times 3}{100}\right) = 8160 \]
\[ P\left(\frac{18 + 45 + 39}{100}\right) = 1.02P = 8160 \]
\[ P = \text{Rs. } 8000 \]
The final slab covers only the last three years, not all eleven.
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An automobile financier claims to be lending money at S.I., but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes?
Answer: B
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 - 100) = 10.25%.
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