At what rate percent of simple interest will a sum of money double itself in 12 years?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Doubling means the interest equals the principal.
\[ P = \frac{P \times R \times 12}{100} \quad\Longrightarrow\quad 100 = 12R \]
\[ R = \frac{25}{3} = 8\tfrac{1}{3}\% \]
The rule for doubling under simple interest is \( R = \tfrac{100}{T} \).
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Kabir paid Rs. 9600 as interest on a loan he took 5 years ago at 16% rate of simple interest. What was the amount he took as loan?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ 9600 = \frac{P \times 16 \times 5}{100} = 0.8P \]
\[ P = \frac{9600}{0.8} = \text{Rs. } 12{,}000 \]
Over five years at 16% the interest comes to 80% of the loan, which is why the principal is only a little above the interest itself.
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How much time will take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ 81 = \frac{450 \times 4.5 \times T}{100} = 20.25\,T \]
\[ T = \frac{81}{20.25} = 4 \text{ years} \]
The interest is 18% of the principal, and at 4.5% a year that takes exactly four years.
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The price of a T.V. set worth Rs. 20000 is to be paid in 20 installments of Rs. 1000 each. If the rate of interest be 6% per annum, and the first installment be paid at the time of purchase, then the value of the last installment covering the interest as well will be?
Answer: D
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Carry everything forward to the date of the last installment, 19 years after purchase.
The debt grows for 19 years:
\[ 20000\left(1 + \frac{6 \times 19}{100}\right) = 20000 \times 2.14 = \text{Rs. } 42{,}800 \]
The first 19 installments of Rs. 1000 earn interest for 19, 18, …, 1 years:
\[ 1000\left[19 + \frac{6}{100}(19 + 18 + \cdots + 1)\right] = 1000\left[19 + 0.06 \times 190\right] \]
\[ = 1000(19 + 11.4) = \text{Rs. } 30{,}400 \]
\[ \text{last installment} = 42800 - 30400 = \text{Rs. } 12{,}400 \]
That figure is not among the choices, so the answer is none of these.
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Sam borrowed some money from his friend at simple interest of 6% per annum. He returned his friend Rs. 15600. After how much time did Sam return the money if he borrowed Rs. 12000?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ \text{SI} = 15600 - 12000 = \text{Rs. } 3600 \]
\[ 3600 = \frac{12000 \times 6 \times T}{100} = 720\,T \]
\[ T = 5 \text{ years} \]
The Rs. 15,600 is the amount, so the principal must be subtracted before the formula can be used.
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A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is?
Answer: D
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Both loans share one rate \( R \), so add the two interests.
\[ \frac{5000 \times R \times 2}{100} + \frac{3000 \times R \times 4}{100} = 2200 \]
\[ 100R + 120R = 2200 \quad\Longrightarrow\quad 220R = 2200 \]
\[ R = 10\% \]
Checking: Rs. 1000 from B and Rs. 1200 from C.
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A certain sum becomes Rs. 20720 in four years and 24080 in six years at simple interest. Find sum and rate of interest?
Answer: B
Let the interest for one year be x.
As amount = Principal + Interest, we have
P + 4x = 20720 --- (1) ; P + 6x = 24080 --- (2)
Solving the equations (1) and (2), we can get
P = Rs. 14000 and x = Rs. 1680
Interest for one year on Rs. 14000 is Rs. 1680
So, R = (100 * 1680)/(14000 * 1) = 12% p.a.
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A sum of Rs. 4000 amounts to Rs. 4600 in 5 years at a certain rate of simple interest. What would be the amount, if the rate of interest is increased by 3 %.
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Find the original rate first.
\[ \text{SI} = 4600 - 4000 = 600 \quad\Longrightarrow\quad 600 = \frac{4000 \times R \times 5}{100} = 200R \]
\[ R = 3\% \]
At \( 3 + 3 = 6\% \):
\[ \text{SI} = \frac{4000 \times 6 \times 5}{100} = 1200 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 5200 \]
The extra 3% over 5 years adds \( 4000 \times 0.15 = 600 \), doubling the interest.
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Mr. Karan borrowed a certain amount at 6% per annum simple interest for 9 years. After 9 years, he returned Rs. 8110/-. Find out the amount that he borrowed.
Answer: D
Let us assume Mr. Karan borrowed amount is Rs. A. (the principal)
By formula of simple interest, S.I. = PRT / 100
Where P = The principal, R = Rate of interest as a %, T = Time in years
S.I. = (P * 6 * 9)/100 = 54 P/100
Amount = Principal + S.I.
8110 = P + (54P/100)
8110 = (100P + 54P)/100
8110 = 154P/100
P = (8110 * 100)/154 = Rs. 5266.233
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If Re.1 amounts to Rs.9 over a period of 20 years. What is the rate of simple interest?
Answer: D
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Growing from Re. 1 to Rs. 9 means interest of Rs. 8.
\[ 8 = \frac{1 \times R \times 20}{100} = 0.2R \]
\[ R = 40\% \]
The amount is 9 times the principal, so the interest is 8 times it - using 9 here would give 45%.
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