Question

A sum of money triples itself in 15 years 6 months. In how many years would it double itself?

Answer 1

Please see the attachment

Answer 2

Answer:

$\frac{31}{4} years$

Step-by-step explanation:

Let the principal be x

We are given that A sum of money triples itself

So, Amount = 3x

SI = 3x-x = 2x

Time = 15 years 6 months = $15 + \frac{6}{12} = \frac{31}{2}$

$SI = \frac{P \times R \times T}{100}$

$2x= \frac{x \times R \times \frac{31}{2}}{100}$

$\frac{2x \times 100}{x \times\frac{31}{2}}=R$

$\frac{400}{31}=R$

Now we are supposed to find  In how many years would it double itself?

Amount = 2x

SI = 2x-x = x

So, $SI = \frac{P \times R \times T}{100}$

$x = \frac{x \times\frac{400}{31} \times T}{100}$

$\frac{x \times 100}{{x \times\frac{400}{31}}= T$

$\frac{31}{4}= T$

Hence it would double itself in $\frac{31}{4} years$