Question

32A sum of money lent out at CI. at a certain
rate per annum becomes three times of itself
in 8 years. Find in how many years will the
money become twenty-seven times of itself at
the same rate of interest pa.​

Answer 1

Given :-

• Sum of money becomes 3 times in 8 years.
• Rate is compounded Annually.

To Find :-

• in how many years it will be 27 Times. ?

Solution :-

First Lets Try to Solve with Basic Method .

Let ,

→ Principal = P

→ Rate % = R

→ Time = 8 years .

→ Amount = 3P

So,

→ P [ 1 + (R/100) ]^8 = 3P

→ [1 + (R/100) ]^8 = 3 ------------------- Equation (1).

________________

Now, Let it becomes 27 times in T years.

So,

→ P [ 1 + (R/100) ]^T = 27P

→ [ 1 + (R/100) ]^T = 27

→ [ 1 + (R/100) ]^T = (3)³

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

→ [ 1 + (R/100) ]^T = { [ 1 + (R/100) ]^8 }³

→ [ 1 + (R/100) ]^T = [ 1 + (R/100) ]^24

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

So,

→ T = 24 Years (Ans).

Hence, The Sum of Money will be 27 Times in 24 years.

______________________________

Answer 2

Given

In 8 years = rate become 3 times

To find

n Years= rate become 27 times

Solution

Let the sum of money be X and the rate of interest per annum r%

Amount=$\boxed{\boxed{\sf{P({1+\frac{r}{100}})^n}}}$

According To the question

In 8 years = the amount 3x

X$({1+\frac{r}{100}})^8$=3x

=>$({1+\frac{r}{100}})^8$=3-------(1)

After n years money become 27 times

X$({1+\frac{r}{100}})^n$=27x

=>$({1+\frac{r}{100}})^n$=3³

From equation (1) we write it as

{$({1+\frac{r}{100}})^8$}³=$({1+\frac{r}{100}})^n$

=>$({1+\frac{r}{100}})^{24}$=($({1+\frac{r}{100}})^n$

n = 24

$\boxed{\boxed{\sf{n=24 }}}$