Simple Interest Questions and Answers

Q: How much interest can a person get on Rs. 8200 at 17.5% p.a. simple interest for a period of two years and six months?

\[ \text{SI} = \frac{P \times R \times T}{100} \]Two years and six months is 2.5 years.\[ \text{SI} = \frac{8200 \times 17.5 \times 2.5}{100} \]\[ = 82 \times 17.5 \times 2.5 = 82 \times 43.75 = \text{Rs. } 3587.50 \]The six months must be converted to 0.5 of a year, not left as a separate term.

Q: In what time a sum of money double itself at 3% per annum simple interest?

\[ \text{SI} = \frac{P \times R \times T}{100} \]Doubling means the interest equals the principal.\[ P = \frac{P \times 3 \times T}{100} \quad\Longrightarrow\quad 100 = 3T \]\[ T = \frac{100}{3} = 33\tfrac{1}{3} \text{ years} \]The rule is simply \( T = \tfrac{100}{R} \) for doubling under simple interest.

Q: At what rate per cent per annum will the simple interest on a sum of money be \( \dfrac{2}{5} \) of the sum in 10 years?

\[ \text{SI} = \frac{P \times R \times T}{100} \]\[ \frac{2}{5}P = \frac{P \times R \times 10}{100} \]The \( P \) cancels:\[ \frac{2}{5} = \frac{10R}{100} = \frac{R}{10} \quad\Longrightarrow\quad R = 4\% \]Checking: 4% for 10 years gives 40% of the sum, which is \( \tfrac{2}{5} \).

Q: If the annual rate of simple interest increases from 10% to 12 1/2 %, a man's yearly income increases by Rs. 1250. His principal in Rs. is?

\[ \text{SI} = \frac{P \times R \times T}{100} \]The extra income comes from the \( 12\tfrac{1}{2} - 10 = 2.5 \) percentage points, over one year.\[ \frac{P \times 2.5 \times 1}{100} = 1250 \]\[ 0.025P = 1250 \quad\Longrightarrow\quad P = \text{Rs. } 50{,}000 \]Checking: Rs. 5000 a year at 10% against Rs. 6250 at 12.5%.

Q: 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?

\[ \text{SI} = \frac{P \times R \times T}{100} \]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.

Q: The simple interest on a sum of money is 4/9 of the principal and the number of years is equal to the rate percent. Find the rate and the time?

\[ \text{SI} = \frac{P \times R \times T}{100} \]The time equals the rate, so write both as \( R \).\[ \frac{4}{9}P = \frac{P \times R \times R}{100} \]The \( P \) cancels:\[ \frac{4}{9} = \frac{R^{2}}{100} \quad\Longrightarrow\quad R^{2} = \frac{400}{9} \]\[ R = \frac{20}{3} = 6\tfrac{2}{3} \]So the rate is \( 6\tfrac{2}{3}\% \) and the time \( 6\tfrac{2}{3} \) years.

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61.

How much interest can a person get on Rs. 8200 at 17.5% p.a. simple interest for a period of two years and six months?

A.Rs. 3770
B.Rs. 2850.50
C.Rs. 3587.50
D.Rs. 3750

Answer: C

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Two years and six months is 2.5 years.

\[ \text{SI} = \frac{8200 \times 17.5 \times 2.5}{100} \]

\[ = 82 \times 17.5 \times 2.5 = 82 \times 43.75 = \text{Rs. } 3587.50 \]

The six months must be converted to 0.5 of a year, not left as a separate term.

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62.

The simple interest on a sum of money will be Rs.600 after 10 years. If the principal is trebled after 5 years what will be the total interest at the end of the tenth year?

A.Rs.800
B.Rs.900
C.Rs.1200
D.Rs.1500

Answer: C

Rs. 600 over ten years means Rs. 60 a year on the original principal.

First five years at the original principal:

\[ 5 \times 60 = \text{Rs. } 300 \]

Next five years with the principal trebled, so three times the yearly interest:

\[ 5 \times 180 = \text{Rs. } 900 \]

\[ \text{total} = 300 + 900 = \text{Rs. } 1200 \]

Trebling the whole Rs. 600 would give Rs. 1800; only the second half of the period is affected.

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63.

Peter invested an amount of Rs. 12,000 at the rate of 10% p.a simple interest and another amount at the rate of 20% p.a. simple interest. The total interest earned at the end of one year on the total amount invested became 14% p.a. Find the total amount invested?

A.Rs. 20,000
B.Rs. 22,000
C.Rs. 24,000
D.Rs. 25,000

Answer: A

Let the second amount be Rs. x. Then,

(12000 * 10 * 1)/100 + (x * 20 * 1)/100 = [(12000 + x) * 14 * 1] / 100

120000 + 20x = 16800 + 14x

x = 8000

Total investment = 12000 + 8000 = Rs. 20,000.

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64.

In what time a sum of money double itself at 3% per annum simple interest?

A.29 years
B.33 1/3 years
C.23 1/3 years
D.13 1/3 years

Answer: B

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Doubling means the interest equals the principal.

\[ P = \frac{P \times 3 \times T}{100} \quad\Longrightarrow\quad 100 = 3T \]

\[ T = \frac{100}{3} = 33\tfrac{1}{3} \text{ years} \]

The rule is simply \( T = \tfrac{100}{R} \) for doubling under simple interest.

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65.

At what rate per cent per annum will the simple interest on a sum of money be \( \dfrac{2}{5} \) of the sum in 10 years?

A.4%
B.5 2/3 %
C.6%
D.6 2/3 %

Answer: A

\[ \text{SI} = \frac{P \times R \times T}{100} \]

\[ \frac{2}{5}P = \frac{P \times R \times 10}{100} \]

The \( P \) cancels:

\[ \frac{2}{5} = \frac{10R}{100} = \frac{R}{10} \quad\Longrightarrow\quad R = 4\% \]

Checking: 4% for 10 years gives 40% of the sum, which is \( \tfrac{2}{5} \).

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66.

If the annual rate of simple interest increases from 10% to 12 1/2 %, a man's yearly income increases by Rs. 1250. His principal in Rs. is?

A.45000
B.50000
C.60000
D.65000

Answer: B

\[ \text{SI} = \frac{P \times R \times T}{100} \]

The extra income comes from the \( 12\tfrac{1}{2} - 10 = 2.5 \) percentage points, over one year.

\[ \frac{P \times 2.5 \times 1}{100} = 1250 \]

\[ 0.025P = 1250 \quad\Longrightarrow\quad P = \text{Rs. } 50{,}000 \]

Checking: Rs. 5000 a year at 10% against Rs. 6250 at 12.5%.

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67.

What time will be taken by an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

A.2 years
B.1 years
C.3 years
D.4 years

Answer: A

P= Rs 900

Rate = 4.5*

SI= Rs 81

=> SI= Time*P*rate

calculating for rate,

=> Time= SI/rate*P

= 81/4.5%*900

=> 2 years

Thus, it will take 2 years to yield Rs 81.

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68.

A certain sum earns simple interest of Rs. 800 in 2 years at a certain rate of interest. If the same sum earns compound interest of Rs. 840 in the same period of 2 years, what must be the rate of interest?

A.5% p.a.
B.7.5% p.a.
C.10% p.a.
D.12.5% p.a.

Answer: C

Simple interest of Rs. 800 over two years is Rs. 400 a year. Compound interest exceeds it by

\[ 840 - 800 = \text{Rs. } 40 \]

which is one year's interest on that first Rs. 400:

\[ 400 \times \frac{R}{100} = 40 \quad\Longrightarrow\quad R = 10\% \text{ p.a.} \]

The sum works out at Rs. 4000, and \( 4000(1.21 - 1) = 840 \) confirms it - but the principal is not needed to answer.

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69.

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?

A.3.6
B.6
C.18
D.Cannot be determined

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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70.

The simple interest on a sum of money is 4/9 of the principal and the number of years is equal to the rate percent. Find the rate and the time?

A.6 2/3 years; 6 2/3 %
B.5 1/3 years; 5 1/3 %
C.4 2/3 years; 4 2/3 %
D.None

Answer: A

\[ \text{SI} = \frac{P \times R \times T}{100} \]

The time equals the rate, so write both as \( R \).

\[ \frac{4}{9}P = \frac{P \times R \times R}{100} \]

The \( P \) cancels:

\[ \frac{4}{9} = \frac{R^{2}}{100} \quad\Longrightarrow\quad R^{2} = \frac{400}{9} \]

\[ R = \frac{20}{3} = 6\tfrac{2}{3} \]

So the rate is \( 6\tfrac{2}{3}\% \) and the time \( 6\tfrac{2}{3} \) years.

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