# Quants Questions : Compound Interest Set 2

Hello 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. What annual payment will discharge a debt of 1025 due in 2 years at the rate of 5% compound interest?
A)550
B)551.25
C)560
D)560.35
E)None of these
Explanation : 2. In what time will Rs. 64,000 amount to Rs.68921 at 5% per annum interest being compounded half yearly?
A)3 years
B)2 years
C)2(1/2) years
D)1(1/2) years
E)None of these
Explanation :
R = 2.5%, A = 68921 , P = 64000 and t= 2n
A/P = [1+(R/100)]^2n
68921/64000 = [1+(2.5/100)]^2n
(41/40)^3 = (41/40)^2n
2n = 3
n=3/2 =1(1/2) years

3. If the difference between the  CI and SI on a sum of money at 5% per annum  for 2years is Rs.16.Find the Simple Interest ?
A)180
B)460
C)520
D)640
E)None of these
Explanation :
16 =P*( 5/100)^2
P = 6400
SI =(6400*5*2)/100 =640

4. The difference between the SI and CI on Rs.5000  at 10%per annum for 2 year is
A)24
B)35
C)50
D)56
E)None of these
Explanation :
d =  p (r/100)^2
= 5000 (10/100)^2
d = 50

5. The difference  SI and CI  on Rs.1000 for 1 year at 10%per annum reckoned Half yearly is
A)2
B)2.5
C)3
D)2.4
E)None of these
Explanation :
SI = (1000/100 =100
CI = [1000(1+(5/100))^2] – 1000 = 102.5
D = 102.5 – 100 = 2.5

6. Compound interest on a certain sum of money at 20% per annum for 2 years is Rs.5984.What is the SI on the same money at 9% per annum for 6 years ?
A)7320
B)7233
C)7433
D)7344
E)None of these
Explanation :
5984 = p [(1+(20/100)^2) -1]
P = (5984*25)/11 = 13600
SI = (13600*9*6)/100 = 7344

7. The difference between  CI and SI on an amount Rs. 15000 for 2 year is Rs.96.What is the rate of interest per annum ?
A)12
B)10
C)8
D)7
E)None of these
Explanation : 8. The effective annual rate of interest corresponding to the nominal rate of 4% per annum payable half yearly  is
A)4
B)4.4%
C)4.04%
D)4.2%
E)None of these
Explanation :
Let p = 100
CI = [100+(1+[2/100]^2)]
= 100 × (102/100)× (102/100)
= 104.04
Effective rate = 104.04 – 100 = 4.04%

9. Rohit  borrowed Rs. 1200 at  12% PA .He repaid Rs. 500 at the end of 1 year. What is the amount required to pay at the end of 2nd  year to discharge his loan which was calculated in CI
A)945.28
B)1106.00
C)1107.55
D)1100.65
E)None of these
Explanation :
CI at the end of 1st year = 1200 *( 1+(12 /100)) = 1344
CI= 1344-1200 = 144
500 paid then remaining amount = 1344 – 500 = 844
At the end of  2nd year
844*[(1+(12/100)] =945.28

10. A sum of money invested at CI  to Rs.800 in 3 years and to Rs.840 in  4 years.Find rate of interest PA ?
A)6%
B)5%
C)4%
D)2%
E)None of these