Question

If a certain sum, invested under compound interest amounts to thrice as much at the end of the eighth year as it would at the end of the third year, then, the amount at the end of the 55^th years will be how many times that at the end of the 40^th year?
1. 8
2. 64
3. 27
4. 16
5. None of these

Answer 1

Correct Answer - Option 3 : 27

Formula used   :

Amount = P(1 + r/100)t

CALCULATION   :

Suppose the principal and the rate of interest per annum be denoted by p and r, respectively.

According to given condition –

⇒ A₈ = 3 × A₃

⇒ \({\text{P}}{\left( {1 + \frac{{\text{r}}}{{100}}} \right)^8} = 3{\text{P}} \times {\left( {1 + \frac{{\text{r}}}{{100}}} \right)^3}\)

⇒ \({\left( {1 + \frac{{\text{r}}}{{100}}} \right)^5} = 3\)

We need to find the value of \(\;\frac{{{A_{55}}}}{{{A_{40}}}}\)  :

⇒ \(\frac{{{A_{55}}}}{{{A_{40}}}} = \frac{{p{{\left( {1 + \frac{r}{{100}}} \right)}^{55}}}}{{p{{\left( {1 + \frac{r}{{100}}} \right)}^{40}}}}\)

⇒ \(\frac{{{A_{55}}}}{{{A_{40}}}} = {\left( {1 + \frac{r}{{100}}} \right)^{15}}\)

⇒ \(\frac{{{A_{55}}}}{{{A_{40}}}} = {\left( {1 + \frac{r}{{100}}} \right)^{\left( {5 \times 3} \right)}}\)

⇒ \(\frac{{{A_{55}}}}{{{A_{40}}}} = {\left( 3 \right)^3}\)

⇒ \(\frac{{{A_{55}}}}{{{A_{40}}}} = 27\)

∴ The amount at the end of the 55^th years will be 27 times that at the end of the 40^th year.