Question

a sum of money triples itself at compound rate of interest in 10 years . in how many years will it become nine times ​

Answer 1

Given :-

• A sum of Money Triples Itself at CI in 10years.

To Find :-

• In how many Years it wi be 9 Times ?

Solution :-

Let ,

→ Principal = P

→ Rate % = R

→ Time = 10 years .

→ Amount = 3P

So,

→ P [ 1 + (R/100) ]^10 = 3P

→ [1 + (R/100) ]^10 = 3 ------------------- Equation (1).

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Now, Let it becomes 9 times in T years.

So,

→ P [ 1 + (R/100) ]^T = 9P

→ [ 1 + (R/100) ]^T = 9

→ [ 1 + (R/100) ]^T = (3)²

Now, Putting value of 3 from Equation (1) Here , we get,

→ [ 1 + (R/100) ]^T = { [ 1 + (R/100) ]^10 }²

→ [ 1 + (R/100) ]^T = [ 1 + (R/100) ]^20

Now, we know that, if a^b = a^c Than, b = c,

So,

→ T = 20 Years (Ans).

Hence, The Sum of Money will be 9 Times in 20 years.

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Answer 2

Solution :

=> Assume Principal = P

=> Rate % = R

=> Time = 10 years .

=> So, Amount = 3P

Than, using CI formula

=> P [ 1 + (R/100) ]^10 = 3P

=> [1 + (R/100) ]^10 = 3 ------------------- Equation (1).

Now, Let it becomes 9 times in T years.

So,

=> P [ 1 + (R/100) ]^T = 9P

=> [ 1 + (R/100) ]^T = 9

=> [ 1 + (R/100) ]^T = (3)²

Now, Putting value of 3 from Equation (1) Here , we get,

=> [ 1 + (R/100) ]^T = { [ 1 + (R/100) ]^10 }²

=> [ 1 + (R/100) ]^T = [ 1 + (R/100) ]^20

=> T = 20 Years .