Quants Questions : Compound Interest Set 3

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 is the difference between the CI and SI on Rs.6500 at the rate of 4 pcpa in 2 yrs ?
   A)9.4
   B)10.4
   C)15.4
   D)14.5
   
   Answer

   B)10.4
   Explanation :
   SI = 6500*4*2/100 = 520
   CI = 6500*(104/100)*(104/100) = 7030.4 = 530.4
   530.4 – 520 = 10.4
   
   
   
2. The CI on a certain sum at 10% pa for 2yrs is Rs.6548. What is SI on the sum of money at 7%pa for 4 yrs(approx) ?
   A)8700
   B)8731
   C)8370
   D)8470
   
   Answer

   B)8731
   Explanation :
   CI = p [ {1 + (10/100)}2 – 1]
   6548 = p [(110/100) ² – 1]
   6548 =p (121 – 100)/100
   P = 654800/21 =31,181
   SI = 31181*7*4/100 = 8731
   
   
   
3. A sum of money amount to Rs.900 in 3 yrs and Rs.1080 in 4yrs at CI. What is the rate of CI pa ?
   A)10%
   B)15%
   C)20%
   D)18%
   
   Answer

   C)20%
   Explanation :
   (1080-900/900)*100
   180*100/900 = 20%
   
   
   
4. Find the CI on Rs.4500 at 4% pa for 2 yrs compounded half yearly(approx) ?
   A)963
   B)369
   C)639
   D)360
   
   Answer

   C)639
   Explanation :
   CI = P [(1+(r/100)ⁿ – 1]
   = 4500[(1+2/100)⁴ – 1
   = 4500[ (102*102*102*102/100*100*100*100) – 1]
   = 4500[1.082 – 1] = 4500*0.082 = 369
   
   
   
5. In how many years will a sum of Rs.15625 at 8% pa compounded semi annually become Rs.17,576?
   A)2(1/3)yrs
   B)2yrs
   C)1(1/2)yrs
   D)1yr
   
   Answer

   C)1(1/2)yrs
   Explanation :
   15625*(1+(4/100))²ⁿ = 17576
   15625*(104/100) ²ⁿ = 17576
   (26/25) ²ⁿ = 17576/15625 = (26/27)³
   2n = 3
   N =3/2 = 1(1/2) yrs
   
   
   
6. If 40% increase in an amount in 4 years at SI. What will be the CI of Rs.10,000 after 2 yrs at the same rate ?
   A)1000
   B)1100
   C)2000
   D)2100
   
   Answer

   D)2100
   Explanation :
   P = 100, SI = 40,  T = 4
   R = 100*40 / 100*4 = 10%pa
   CI = 10,000(110/100)² = 10000*11*11/100 = 12100
   12100 – 10000 = 2100
   
   
   
7. A sum of money is borrowed and paid in 2 annual instalments of Rs.642 each allowing 4% CI. The sum borrowed was
   A)1120
   B)1211
   C)1112
   D)1121
   
   Answer

   B)1211
   Explanation :
   P = [642/(104/100)] + [642/(104/100)²)
   = 642*(25/26) + 642 (25/26)²
   = 617.3 + 593.6 = 1210.9 = 1211
   
   
   
8. The least number of complete years in which a sum of money put out at 10% CI will be more than  doubled is
   A)8yrs
   B)6yrs
   C)4yrs
   D)7yrs
   
   Answer

   A)8yrs
   Explanation :
   P(110/100)ⁿ  > 2P
   (11/10) ⁿ > 2P
   1.1*1.1*1.1*1.1*1.1*1.1*1.1*1.1 = 2.14 > 2
   N = 8
   
   
   
9. A sum of money doubles itself at CI in 10yrs. In how many yrs will it become 4 times ?
   A)10yrs
   B)20yrs
   C)15yrs
   D)22yrs
   
   Answer

   B)20yrs
   Explanation :
   P(1+(R/100))¹⁰ = 2P
   1+(R/100)¹⁰ = 2
   (1+(R/100))ⁿ = 4 = 2²
   2*10 = 20yrs
   
   
   
10. At what rate of CI pa will a sum of Rs.1000 becomes Rs.1040.4 in 2yrs ?
    A)2%
    B)1%
    C)1.5%
    D)3%
    
    Answer

    A)2%
    Explanation :
    1000*(1+(R/100)) ² = 1040.4
    (1+(R/100)) ² = 1040.4/1000
    (1+(R/100)) ² = 10404/10000 = (102/100) ²
    R = 2%