Simple Interest Quiz Set 007 | Maths Aptitude Quiz Questions and Mock Test on Simple Interest Set 7 | mock tests for Bank PO, SSB/SSC Exams, SO Clerical Exam, GATE Exam, LIC AO, GIC AO, etc.

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Question 1

The interest on a certain principal sum is 4/9 times the sum. What is R, the rate of interest, if the time is R years?

$6{2/3}$%.

7%.

$7{1/3}$%.

$6{1/3}$%.

Soln.
Ans: a

I = P × (4/9), so we can write P × (4/9) = P × (R/100) × R. Cancelling P and solving for R, we get, R = $√{100 × 4/9}$ = $6{2/3}$%.

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Question 2

A sum of Rs. 1540 is divided into two parts such that simple interest on these parts at 10% p.a. after 6 and 5 years, respectively, is same. What is the amount of the smaller part?

Rs. 700.

Rs. 800.

Rs. 600.

Rs. 900.

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 × $5/{6 + 5}$, and 1540 × $6/{6 + 5}$, which are 700 and 840. The smaller is Rs. 700.

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Question 3

A sum of Rs. 1300 is split into two parts. The first part is invested at 12% p.a. and the second at 4% p.a. What is the amount invested at 12% if the total simple interest at the end of 4 years is Rs. 368?

Rs. 500.

Rs. 600.

Rs. 400.

Rs. 700.

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 = ${x × r_1 × 4}/100$ + ${(P - x) × r_2 × 4}/100$, which simplifies to 100I = 4 × ($x × (r_1 - r_2) + P × r_2$). Putting r₁ = 12, r₂ = 4, P = 1300, I = 368, we get x = Rs. 500.

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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?

4 years.

5 years.

3 years.

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.

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Question 5

A sum of Rs. 1100 is lent in two parts. One at 13% p.a. and one at 6% p.a. What is the amount lent at 13% if the total simple interest at the end of 5 years is Rs. 715?

Rs. 1100.

Rs. 1200.

Rs. 1000.

Rs. 1300.

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 = 13, r₂ = 6, P = 1100, I = 715, we get x = Rs. 1100.

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