# Simple Interest Quiz Set 020

### Question 1

A sum of Rs. 300 is lent in two parts. One at 7% p.a. and one at 2% p.a. What is the amount lent at 7% if the total simple interest at the end of 5 years is Rs. 155?

A

Rs. 500.

B

Rs. 600.

C

Rs. 400.

D

Rs. 700.

Soln.
Ans: a

Let the amount x be lent at r1% and remaining (P - x) at r2%, and let the sum of interests be I. Then, I = \${x × r_1 × 5}/100\$ + \${(P - x) × r_2 × 5}/100\$, which simplifies to 100I = 5 × (\$x × (r_1 - r_2) + P × r_2\$). Putting r1 = 7, r2 = 2, P = 300, I = 155, we get x = Rs. 500.

### Question 2

The simple interest on a certain principal sum @9% for a period of 7 years is Rs. 3087. What is the sum?

A

Rs. 4900.

B

Rs. 5000.

C

Rs. 4800.

D

Rs. 5100.

Soln.
Ans: a

P = \$(I × 100)/(R × T)\$. Solving, we get P = \$(3087 × 100)/(9 × 7)\$ = Rs. 4900.

### Question 3

An investor puts an amount of Rs. 2300 in a simple interest scheme. If the rate of interest is 4% per month, how long does he have to wait for getting an amount of Rs. 2944?

A

\${7/12}\$ year.

B

\${2/3}\$ year.

C

\${3/4}\$ year.

D

\${5/6}\$ year.

Soln.
Ans: a

The interest is I = 2944 - 2300 = 644. So T = \$(I × 100)/(R × P)\$. Solving, we get T = \$(644 × 100)/(4 × 2300)\$ = 7 months.

### Question 4

A certain amount is split into two parts. The first part is invested at 8% p.a. and the second at 5% p.a. What is the total amount if the total simple interest at the end of 3 years is Rs. 357, and if the amount invested at 8% is Rs. 1800?

A

Rs. 1300.

B

Rs. 1400.

C

Rs. 1200.

D

Rs. 1500.

Soln.
Ans: a

Let the amount x be invested at r1% and remaining (P - x) at r2%, and let the sum of interests be I. Then, I = \${x × r_1 × 3}/100\$ + \${(P - x) × r_2 × 3}/100\$, which simplifies to 100I = 3 × (\$x × (r_1 - r_2) + P × r_2\$). Putting r1 = 8, r2 = 5, x = 1800, I = 357, we get P = Rs. 1300.

### Question 5

The interest on a certain principal sum is 4/81 times the sum. What is R, the rate of interest, if the time is R years?

A

\$2{2/9}\$%.

B

\$2{1/3}\$%.

C

\$2{4/9}\$%.

D

\$2{1/9}\$%.

Soln.
Ans: a

I = P × (4/81), so we can write P × (4/81) = P × (R/100) × R. Cancelling P and solving for R, we get, R = \$√{100 × 4/81}\$ = \$2{2/9}\$%.