A sum of money doubles itself at compound interest in 15 yrs. In how many years it will become eight times?

• 2p=p(1+ r/100)^15 => 2=(1 +r/100)^15
  
  8p=p(1+ r/100)^t => 8=2^3=(1 +r/100)^t
  => [(1 +r/100)^15]^3=(1 +r/100)^t
  => t=45 years
  
• 10 years agoHelpfull: Yes(25) No(0)
• CI=p(1+R/100)^15=2p
  = (1+R/100)^15 --->(1)
  
  if it is 8P, n=?
  P(1+R/100)^n=8P
  (1+R/100)^n=8
  (1+R/100)^n=2^3
  
  sub 2=(1+R/100)^15
  
  (1+R/100)^n=((1+R/100)^15)^3
  
  (1+R/100)^n=(1+R/100)^45
  
  n=45
  
  So after 45 years it ll become 8 times
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• Ans:45 years
  initially -> x
  aftr 15 yrs -> 2x
  aftr 30 yrs -> 4x
  aftr 45 yrs -> 8x
• 10 years agoHelpfull: Yes(3) No(0)
• we know that
  A=p(1+r/100)^n
  here given is : A=2p
  there for
  2p=p(1+r/100)^15
  now cancel p from both side
  2=(1+r/100)^15
  next step
  2^4=(1+r/100)^15*4
  ans is 60 years
• 10 years agoHelpfull: Yes(0) No(1)
• sorry actually it make 8 times not 16
  we know that
  A=p(1+r/100)^n
  here given is : A=2p
  there for
  2p=p(1+r/100)^15
  now cancel p from both side
  2=(1+r/100)^15
  next step
  2^3=(1+r/100)^15*3
  ans is 45 years
  
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• X = 45 years
  
  2p = (1 + r/100)^15
  8p = (1 + r/100)^x
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• 45 HMMMMMM.....
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• let sum is p
  after 15 yrs money doubles i.e 2p
  after 30 yrs money again doubles i.e 4p
  after 45 yrs again it doubles i.e 8p
  so the money 8 times after 45 yrs
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• Let the initial money be "x" then after 15 years it will be "2x"
  then after more 15 years it will be "4x" then after 15 years it will be "8x"
  Hence 3 times 15=45
• 10 years agoHelpfull: Yes(0) No(0)