Simple Interest - Problems and Answers with Detailed Explanation

Published on Monday, July 20, 2015
Today I am going to share practice questions for Simple Interest. Must download read SI and CI concepts here.

  1. On what sum of money lent out at 9% per annum simple interest for 6 years does the simple interest amount to Rs. 810?(RBI'82)
  2. A sum of Rs. 800 amounts to Rs. 920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to how much ?
  3. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.
  4. The simple interest on a sum of money is 4/9 of the principal. Find the rate percent and time, if both are numerically equal.(S.S.C. 2000)
  5. At what rate of interest per annum will a sum double itself in 8 years? ( Bank PO'86)
  6. At what rate percent per annum will a sum of money double in 16 years? ( R.R.B. 2003)
  7. A sum when reckoned at simple interest 21/2% per annum amounts to Rs. 630 after 2 years. Find the sum.
  8. A sum of Rs. 2 is lent to be paid back in 3 equal monthly installments of Re. 1 each. Find the rate per cent .
  9. What annual installment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest 
  10. What annual installment will discharge a debt of Rs 4,200 due in 5 years at 10% simple interest ? (AAO '82)
  11. A sum of Rs. 3,800 is lent out in two parts in such a way that the interest on one part at 8% for 5 years is equal to that on another part at 1/2% for 15 years. Find the sum lent out at 8%.
  12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate.(L.I.C. A.A.O 2003)
  13. If rs. 85 amounts to Rs. 95 in 3 years, what Rs. 102 will amount to in 5 years at the same rate percent?

# Solution 

1. 1500
r×t = S.I/Principal×100 ⇒9×6 = 810/P×100 ⇒ 81000/54 ⇒ 1500

2. 992
S.I = Amount - Principal ⇒S.I = 920 - 800 = 120⇒  r×t = S.I/Principal×100 ⇒ r×3 = 120/800 ×100
r = 5% ⇒ Now r+3% ⇒ 5+3 = 8% ⇒  r×t = S.I/Principal×100 ⇒ 8×3 = S.I/800 ×100 ⇒192
⇒ 800+192
⇒ 992

3. 6000
⇒ [P×(R+2)×3/100] = [P×(R+2)×3/100] = 360
⇒ 3PR + 6P - 3PR = 36000 ⇒ 6P = 36000 ⇒ P = 6000

4. R = 62/3% T= 6 yrs 8 months
⇒  r×t = S.I/Principal×100 ⇒ R×R = 4x/9/x×100 ⇒ R2 = 400x/9
⇒ R = 20/3

5. R= 12 1/2%
S.I=A-P ⇒ Let P = x ⇒SI = 2x - x ⇒ S.I = x
⇒  r×t = S.I/Principal×100 ⇒ r×8= x/x×100 ⇒ R = 100/8

6. R = 61/4 % ( Same As Q5.)

7. P = 600
A = P + S.I
⇒ A = P + [ 1+ RT/100  ]
⇒ 630 = P [ 1+ 2×5/100×2] ⇒630 = P × 105/108 ⇒ P = 600

8. 400%
Borrowed ( debt) amount = M, n = no. of installments , amount of equal installments = x ,
rate of percent = r (quarterly = r/4 % , monthly = r/12 % ).
M = nx [ 1 + r(n-1)/ 200 ]
2 = 3×1 [ 1 + r(3-1)/200×12 ]
⇒x =400 %

9. Rs.325 (Same As Q 8&10)

10. 700 per year
M = nx [ 1 + r(n-1)/ 200 ]
Here, M = Rs. 4200, n = 5 , y = 1 , r  = 10% , x = annual installment = ?
4200 = 5x [ 1 + 10×4 /200 ]
x = Rs. 700 per year.

11. P = Rs. 600
Let the sum lent out at 8% = P1 ; sum lent out at 1/2 % = P2 [since (SI)1 = (SI)2 ]
P1 × 8×5 =P2 × 1/2 ×15 ⇒ P1/P2 = 15/2*8*5 = 3/16
P1 = Sum lent out at 8% = 3/(3+16) ⇒ 3800 = Rs. 600

12. 900

13. A = 122

    About me

    ramandeep singh

    My name is Ramandeep Singh. I authored the Quantitative Aptitude Made Easy book. I have been providing online courses and free study material for RBI Grade B, NABARD Grade A, SEBI Grade A and Specialist Officer exams since 2013.

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