# Quants Questions : Compound Interest Set 1

[dropcap]H[/dropcap]ello Aspirants. Welcome to Online Maths in AffairsCloud.com. Here we are creating question sample in compound Interest, which is common for all the IBPS,SBI exam and other competitive exams. We have included Some questions that are repeatedly asked in exams !!
1.The Simple interest on a certain sum for 2 years at 10% per annum is Rs. 90. The corresponding compound interest is:
A.97
B.90
C.94.50
D.100
E.None of these
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###### Answer & Explanation

Answer –  C (94.50)

Explanation:

Sum =       100 x90 = Rs. 450
2 x 10
C.I. = Rs.450 x (1+10)^ 2 – 450 = Rs.94.50
100

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2.Manoj borrowed a certain sum from Anuj at a certain rate of simple interest for 2 years. He lent this sum to Rakesh at the same rate of interest compounded annually for the same period. At the end of two years, he received Rs. 4200 as compound interest but paid Rs. 4000 only as simple interest. Find the rate of interest?
A.10%
B.20%C.25%
D.30%
E.None of these
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###### Answer & Explanation

Answer – A (10%)

Explanation –
Suppose the sum borrowed = Rs X
Rate of interest = R% Time = 2 years
4000=[X x R x 2 ]/100 [simple interesr formula] RX=200,000………….(1)
compound interest= P(1+r/n)^nt
X[1+ (R/100)]^2= X+4200
after solving and using value of RX
we get R=10%

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3.If the difference between the simple interest and compound interests on some principal amount at 20% for 3 years is Rs. 48, then the principal amount is
A.636
B.650
C.375
D.400
E.None of these
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###### Answer & Explanation
Answer – C (375)
Explanation – For three years
Sum = Difference x (100)^3
r^2(300 + r)
= 48 x (100)^3
20^2 (300 + 20)
= Rs. 375

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4.What sum invested for 2 years at 12% compounded annually will grow to Rs. 4390.40…..
A.4000
B.3875
C.3800
D.3500
E.None of these
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###### Answer & Explanation
Answer – D (3500)
Explanation –
P(1+ R)^T = Amount
100

P (1+12)^2= 4390.40
100

P (112)^2 = 4390.40
100
P = 43904000= 3500.
112 x 112

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5.A sum of money is borrowed and paid back in two annual installments of Rs. 882 each allowing 5% compound interest. The sum borrowed was:
A.1620
B.1640
C.1680
D.1700
E.None of these
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###### Answer & Explanation
Answer – B (1640)
Explanation –
=
882 + 882
(1+ 5 ) ( 1+ 5 )^2
100       100
= 882 x 20 + 882 x 400
21              441
= Rs. 1640

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6.Divide Rs. 3903 between A and B, so that A’s Share at the end of 7 years may equal to B’s share at the end of 9 years, compound interest being at 4 percent.
A.2018 and 1885
B.2028 and 1875
C.2008 and 1895
D.2038 and 1865
E.None of these
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###### Answer & Explanation
Answer – B (2028 and 1875)
Explanation –
We have (A’s present share) (1+ 4 )^ 7= (B’s present share) (1+ 4 )^9
100                                          100
A’s present share =(1+ 4)^2
B’s present share       100
=( 26 )^2=676
25        625
Dividing Rs. 3903 in the ratio of 676:625
A’s present share = 676 of Rs. 3903
(676+625)
= Rs. 2028
B’s present share = Rs. 3903 – Rs. 2028
=Rs. 1875

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7.If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 27 times in:
A.8 years
B.12 years
C.24 years
D.36 years
E.None of these
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###### Answer & Explanation
Answer – B (12 years)
Explanation –
P(1 + R )^4 = 3P
100
(1 + R)^4 = 3 …(i)
100

Let P
(1 + R ) ^n = 27P
100
(1 + R )n = 27 = (3)^3 = ( 1 + R )^(4×3)
100 100 [using (i)]

(1 + R )^n ( 1 +R )^12
100 100

n = 12
Required time = 12 years.

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