# Simple Interest Quiz Set 008

### Question 1

A sum of Rs. 500 is lent in two parts. One at 9% p.a. and one at 4% p.a. What is the amount lent at 9% if the total simple interest at the end of 4 years is Rs. 120?

A

Rs. 200.

B

Rs. 300.

C

Rs. 250.

D

Rs. 400.

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 × 4}/100\$ + \${(P - x) × r_2 × 4}/100\$, which simplifies to 100I = 4 × (\$x × (r_1 - r_2) + P × r_2\$). Putting r1 = 9, r2 = 4, P = 500, I = 120, we get x = Rs. 200.

### Question 2

The interest on Rs. 16000 @9% for a certain number of days starting from Jan 1, 2001 is Rs. 288. How many days?

A

73 days.

B

74 days.

C

72 days.

D

75 days.

Soln.
Ans: a

We have r = 9%, P = 16000, I = 288, so t = \${100 × 288}/{16000 × 9}\$. We get t = 1/5. The given year is not a leap year. So it has 365 days. Since 365/5 = 73, the number of days is 73.

### Question 3

A sum of Rs. 34300 is divided into three parts such that simple interest on these parts at 10% p.a. after 5, 11 and 18 years, respectively, is same. What is the amount of the smallest part?

A

Rs. 5500.

B

Rs. 5600.

C

Rs. 5400.

D

Rs. 5700.

Soln.
Ans: a

We should use the shortcut technique here. If r1, t1, r2, t2 and r3, t3 be the rates and times for three parts with same interest amount, then the three parts must be in the ratio \$1/{r_1 t_1} : 1/{r_2 t_2} : 1/{r_3 t_3}\$. In our case r1 = r2 = r3 = 10, which cancels, so the ratio is \$1/t_1 : 1/t_2 : 1/t_3\$. The product of denominators is 5 × 11 × 18 = 990. Thus, the three parts are in the ratio \$198 : 90 : 55\$. The parts are: 34300 × \$55/{198 + 90 + 55}\$, 34300 × \$90/{198 + 90 + 55}\$ and 34300 × \$198/{198 + 90 + 55}\$, which are 19800, 9000 and 5500. The smaller is Rs. 5500.

### Question 4

An investor puts an amount of Rs. 3100 in a simple interest scheme. If the rate of interest is 9%, how long does he have to wait for getting an amount of Rs. 4216?

A

4 years.

B

5 years.

C

3 years.

D

6 years.

Soln.
Ans: a

The interest is I = 4216 - 3100 = 1116. So T = \$(I × 100)/(R × P)\$. Solving, we get T = \$(1116 × 100)/(9 × 3100)\$ = 4 years.

### Question 5

A sum of Rs. 1540 is divided into two parts such that simple interest on these parts at 10% p.a. after 2 and 9 years, respectively, is same. What is the amount of the smaller part?

A

Rs. 280.

B

Rs. 380.

C

Rs. 180.

D

Rs. 480.

Soln.
Ans: a

We should use the shortcut technique here. If r1, t1, r2, t2 be the rates and times for two parts with same interest amount, then the two parts must be in the ratio \$1/{r_1 t_1} : 1/{r_2 t_2}\$. In our case r1 = r2 = 10, which cancels, so the ratio is \$1/t_1 : 1/t_2\$. Thus, the two parts are in the ratio \$t_2 : t_1\$. The parts are: 1540 × \$9/{2 + 9}\$, and 1540 × \$2/{2 + 9}\$, which are 1260 and 280. The smaller is Rs. 280.