If difference between compound interest and simple interest on a sum at 10% P.a. for 2 years is Rs.51 then sum is

a) Rs.5000
b) Rs.5100
c) Rs.5800
d) Rs.6000

• p(r/100)^2=C.I-S.I
  P(10/100)^2=51
  5100 ANS
  
• 8 years agoHelpfull: Yes(2) No(0)
• FORMULA: CI------------ p[ ( (1+r/100)^t)-p ]
  SI-------------ptr/100
  
  given CI-SI=51
  p[ (1+10/100)^2)-p] - p*2*10/100 = 51
  p [ (11/10*11/10) -1- (1/5) ] = 51
  p (0.01) = 51
  p= 51/0.01
  p= 5100.
• 8 years agoHelpfull: Yes(1) No(0)
• Hello everyone
  from ans.we can solve this problem.
  take 5100 as sum
  5100*10/100=510 which is interest
  and interest is for 2 years is=510+510=1020
  now,for compound interst ,we have to add the interest of first year with principle to calculate the c.i for 2nd year
  so 5100+510=5610
  5610*10/100=561
  so compound interest =510+561=1071
  then.the diff between C.I and S.I is=1071-1020=51
  it means our approach to solve this is right
  
• 8 years agoHelpfull: Yes(0) No(0)