Simple Interest Questions and Answers

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: A money lender finds that due to a fall in the annual rate of interest from 8% to 7 3/4 % his yearly income diminishes by Rs. 61.50, his capital is?

\[ \text{SI} = \frac{P \times R \times T}{100} \]The fall is \( 8 - 7\tfrac{3}{4} = 0.25 \) percentage points, over one year.\[ \frac{P \times 0.25 \times 1}{100} = 61.50 \]\[ 0.0025P = 61.50 \quad\Longrightarrow\quad P = \text{Rs. } 24{,}600 \]A quarter of a percentage point is a small slice, so it takes a large capital to produce a Rs. 61.50 drop.

Q: 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?

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

Important Instructions

131.

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.

Discuss in Forum

132.

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%.

Discuss in Forum

133.

A money lender finds that due to a fall in the annual rate of interest from 8% to 7 3/4 % his yearly income diminishes by Rs. 61.50, his capital is?

A.Rs. 22,400
B.Rs. 23,800
C.Rs. 24,600
D.Rs. 26,000

Answer: C

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

The fall is \( 8 - 7\tfrac{3}{4} = 0.25 \) percentage points, over one year.

\[ \frac{P \times 0.25 \times 1}{100} = 61.50 \]

\[ 0.0025P = 61.50 \quad\Longrightarrow\quad P = \text{Rs. } 24{,}600 \]

A quarter of a percentage point is a small slice, so it takes a large capital to produce a Rs. 61.50 drop.

Discuss in Forum

134.

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?

A.Rs. 1050
B.Rs. 2050
C.Rs. 3000
D.None of these

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.

Discuss in Forum

Simple Interest Questions and Answers

Important Instructions

Topics

Other Subjects

About Us

Simple Interest Questions and Answers

Important Instructions

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications