# Aptitude - Simple Interest Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to **Simple Interest**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

Q 1 - A man puts cash in three unique plans for a long time, 10 years and 12years at 10%, 12% and 15% simple premium, separately. At the completion of every plan he gets the same interest. The proportion of his speculations is:

### Answer : D

### Explanation

Let the required ratio be x: 1: y. Then, S.I on Rs. x for 6 years at 10 % p.a. = S.I on Re 1 for 10 years at 12%p.a. X*10/100*6 = 1*12/100*10 => x= 120/60= 2 S.I. on Rs.1 for 10 years at 12% p.a. = S.I. on Rs. y for 12 years at 15% p.a. ∴ (1*12/100*10) = (y* 15/100*12) => y = 120/180= 2/3 Required ratio = 2:1:2/3 = 6:3:2

Q 2 - The total of cash that will give Re 1 as basic premium for each at the rate of 5% for every annum is:

### Answer : C

### Explanation

Annual interest = Rs. 365, Rate = 5% p.a. Sum = (100*S.I/R*T) = Rs. (100*365/5*1) = Rs. 7300.

Q 3 - x, y, z are three entireties of cash such that y is the simple interest on x and z is the basic on y for the same time and rate. Which of the accompanying is right?

### Answer : D

### Explanation

y= S.I. On Rs. x = Rs. (x*R*T/100); z=S.I on Rs. y = Rs. (y*r*t/100) ∴ y/z = x/y => y^{2}= zx.

Q 4 - A man loses Rs 55.50 yearly when the yearly rate of interest tumbles from 11.5% to 10%. His capital is:

### Answer : A

### Explanation

Let the capital be Rs. x. then, (x*23/2*1/100*1) ?(x*10*1/100*1) = 55.50 => 23x/200 ?x/10 =111/2 => 23x-20x =11100 => 3x= 11100 => x = 3700 Hence, the capital is Rs. 3700.

Q 5 - A sum of money doubles itself in 20 years. What is the rate of interest?

### Answer : B

### Explanation

Here, I = P. Given T = 20 years. R = ? We know, I = PTR/100 Or, R = I*100/PT = P*100/P*20 = 5%

Q 6 - Vishwas borrowed a total amount of Rs 30,000 part of it at 12 % per annum and remaining at 10% per annum. If at the end of two years he paid in all Rs 36,480 to settle the loan amount, what was the amount borrowed at 12 % per annum?

### Answer : D

### Explanation

Let the sum borrowed at 12% per annum be Rs x. So, sum borrowed at 10% per annum = Rs (30000 - x) Simple Interest = Rs 36480 ? Rs 30000 = Rs 6480 According to the question, x*2*12/100 + (30000 - x)*2*10/100 = 6480 or, 24x + 600000 - 20x = 648000 or, 4x = 48000 or, x = 12000

Q 7 - Simple interest on a certain amounts is ^{9}⁄_{16} of the principal. If the numbers representing the rate of interest in percent and time in years be equal, then time, for which the principal is lent out, is?

### Answer : C

### Explanation

Let sum be z. Then,S.I =^{9z}⁄_{16}Let rate = R% and Time = R years therefore^{z x R x R}⁄_{100}R^{2}^{900}⁄_{16}R =^{30}⁄_{4}= 7^{1}⁄_{2}years.

Q 8 - If the annual ROI increases from 10% to 12^{1}⁄_{2}%, a man's yearly income increases by Rs. 1250. his principal (in Rs) is?

### Answer : B

### Explanation

Let the sum be z. Then, (z x^{25}⁄_{2}x^{1}⁄_{100}) - (^{z x 10 x 1}⁄_{100}) = 1250 = 25z - 20z = 250000

= 5z = 250000 z = 50000

Q 9 - A sure entirety of cash adds up to rs. 756 in 2 years and to Rs. 873 in 7/2 years. Discover the aggregate and the rate of hobby.

### Answer : A

### Explanation

Sum in 7/2 years = Rs. 873, sum in 2 years= Rs. 756 S.I for 3/2 year = Rs. (873-756) = Rs. 117 S.I for a long time = Rs. (117*2/3*2) = rs. 156. Standard = (sum in 2 year) - (s.I for 2 year) = Rs. (756-156)= Rs. 600 Presently P = 600 Rs. T= 2 years. What's more, S.I = 156 Rs. ∴ R = (100*S.I)/P*T = (100*156)/ (600*2) = 13% P.a. Henceforth, aggregate = Rs. 600 and Rate = 13% P.a.