Karl invested Rs.4000 at 10% interest compounded annually. The interest he will receive after 2 years is:

• int=amount-pri
  amount=p(1+r/100)^n
  =4000*(1+10/100)^2
  =4000*(110/100)*(110/100)
  =4840
  int=4840-4000=840
• 12 years agoHelpfull: Yes(30) No(0)
• Amount= P(1+10/100)^2
  A=4000(110/100*110/100)
  A=4840
  CI=A-P=4840-4000=840 Rs
  
• 12 years agoHelpfull: Yes(8) No(0)
• int=p{(1+r/100)^n-1}
  =4000*{(1+10/100)^2-1}
  =4000*[(110/100)*(110/100)}-1]
  =840
  int=840
• 12 years agoHelpfull: Yes(4) No(0)
• 800
  here p=4000,t=2,r=10 find i
  we know i=ptr/100
  i=4000*2*10/100
  i=800
• 12 years agoHelpfull: Yes(2) No(21)
• ci=p{(1+r/100)^t-1}
  ==>4000{(1+10/100}^2-1}
  ==>840
• 12 years agoHelpfull: Yes(1) No(0)
• ci=p{(1+r/100)^t-1}
  ==>4000{(1+10/100}^2-1}
  ==>840
• 12 years agoHelpfull: Yes(1) No(0)
• please mail me to: bhanuvasamsetti.v@gmail.com
  
• 12 years agoHelpfull: Yes(0) No(6)
• We know that interest for C.I and S.I for the first year is always same so
  For first year Interest will 10% of 4000=400
  For second year interest will be 400+10% of 400=440
  Total interest will be 400+440=840
• 8 years agoHelpfull: Yes(0) No(0)