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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 - An aggregate of cash adds up to Rs 5200 in 5 years and to Rs 5680 in 7 years at basic premium. The rate of interest per annum is:
Answer : D
Explanation
S.I for 2 years = Rs. (5680-5200) = Rs. 480 S.I for 1 year = Rs. (480/2) = 240 S.I for 5 year = (240*5) = 1200 Rs. Principle = Rs. (5200-1200) = 4000. Rate = (1200*100/4000*5) %p.a. = 6% p.a.
Q 2 - A sure whole is contributed on basic interest, if it trebles in 10 years, what is the rate of interest?
Answer : B
Explanation
Let the sum be Rs. x then, S.I = Rs. (3x-x) = rs. 2x ∴ x*R/100*10= 2x = R = 20% p.a.
Q 3 - Mr. A loans 40% of the whole at 15% p.a., half of rest aggregate at 10% p.a. also, the rest at 18% p.a. Simple interest. What might be the rate of interest per annum figured in general total?
Answer : D
Explanation
Let the whole sum be Rs 100. Sum at 15% p.a. =Rs 40, Rest=Rs 60. Sum at 10% p.a. =Rs 30, sum at 18% p.a. =Rs 30. S.I. on Rs 100 for 1 years = (40*15/100*1) + (30*10/100*1) + (30*18/100*1) = Rs (6+3+ 5.4) =Rs 14.4. Required rate = 14.4%p.a.
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 - With a given of rate of basic interest, the proportion of standard and measure of a sure timeframe is 4:5 , following 3 years , with the same rate of interest , the proportion of the rule and sum gets to be 5:7 , the rate of interest per annum is :
Answer : B
Explanation
After t years, let P =Rs. 4x and amount = Rs. 5x.
P + S.I for t years = Rs. 5x
P: {p+S.I for (t+3) years} = 5:7= 1: 7/5 = 4x: (7/5*4x) = 4x: 28x/5
∴ P+S.I. For (t+3) years = Rs. 28x/5
On subtracting (i) from (ii), we get:
S.I for 3 years = Rs. (28x/5 -5x) = Rs. 3x/5
S.I on Rs. 4x for 3 years = 3x/5
∴ Rate = {(100*3x/5)/ (4x*3)} % p.a. = 5% p.a.
Q 6 - Rs 24000 is lent out in three parts. The first part is lent out at 4% per annum for 5 years, the second at 5% for 3 years and the third at 3% for 4 years. The total interest earned on each part is equal. Find the sum in each part.
Answer : A
Explanation
P1 : P2 : P3 = 1/r1t1 : 1/r2t2 : 1/r3t3 = 1/(4*5) : 1/(5*3) : 1/(3*4) P1 : P2 : P3 = 3 : 4 : 5 Hence P1 = 24000*3/3 + 4 + 5 = Rs 6000 P2 = 24000*4/3 + 4 + 5 = Rs 8000 P3 = 24000*5/3 + 4 + 5 = Rs 10000
Q 7 - A sum of money becomes 7⁄6 of itself in 3 years at a certain rate of simple interest. The rate per annum is?
Answer : B
Explanation
Let sum be z. Then, Amount = 7z⁄6 S.I. = (7z⁄6 - 6) = z⁄6 years = 3 years. therefore Rate = 100 x z⁄z x 6 x 3 = 50⁄9 = 55⁄9
Q 8 - David invested certain amount in three different schemes P,Q, and R with the rate of interest 10% p.a., 12% p.a., and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme R was 150% of the amount invested in scheme P and 240% of the amount invested in scheme Q, what was the amount invested in scheme Q?
Answer : C
Explanation
Let a, b ,c be the amounts invested in schemes P, Q, R respectively. Then, a x 10 x 1⁄100 + b x 12 x 1⁄100 + c x 15 x 1⁄100 = 3200 10a + 12b + 15c = 320000.....(i) c = 240% of b = 12b⁄5 ...(ii) and c = 150% of a = 3a⁄2 a = 2c⁄3 = (2⁄3 x 12⁄5)b = 8b⁄5 .....(iii) from (i), (ii), (iii), we have. 16b + 12b + 36b = 320000 64b = 320000 b = 5000 Sum invested in Scheme Q = Rs. 5000.
Q 9 - Divide Rs. 2379 into 3 parts so that their amounts after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is?
Answer : A
Explanation
Let the parts be a,b and [2379 - (a + b)] a + (a x 25⁄100) = b + (b x 35⁄100) = c + (c x 45⁄100) = 11a⁄10 = 23b⁄20 = 6c⁄5 = k a = 10k⁄11
b = 20k⁄23
c = 5k⁄6 But, a + b + c = 2379 10k⁄11 + 20k⁄23 + 5k⁄6 = 2379 1380k + 1320k + 1265k = 2379 x 11 x 23 x 6 k = 2379 x 11 x 23 x 6⁄3965 = 3 x 11 x 23 x 6⁄5 a = 828