Sum of money at compound interest amounts of thrice itself in 3

years. In how many years

will it take 9 times itself.

• Let sum be P
  Amount after 3 yrs = 3P
  P(1+R/100)^3 = 3P
  (1+R/100)^3 = 3 ---- (i)
  Now, let P(1+R/100)^n = 9P
  (1+R/100)^n = 9 = 3^2
  But, from (i) 3 = (1+R/100)^3
  So, (1+R/100)^n = {(1+R/100)^3}^2 = (1+R/100)^6
  Therefore, n = 6
  sum will become 9 times of itself in 6 years.
  
  
• 11 years agoHelpfull: Yes(12) No(1)
• Ans: 6years
  
  3P=P(1+(R/100)^3)
  =>(1+(R/100)^3)=3;
  9P=P(1+(R/100)^n);
  =>((1+(R/100)^n))=9;
  3^(n/3)=3^2;
  =>n=3*2=6;
• 11 years agoHelpfull: Yes(2) No(0)