A sum becomes 2.89 times of itself in 2 years when invested at compound interest (compounded annually). What is the annual rate of interest?

Correct option is D

Given:
The sum becomes 2.89 times itself in 2 years when invested at compound interest (compounded annually)
Formula Used:
A = $$P \left(1 + \frac{r}{100}\right)^n$$​​
Where:
A is the amount after n years.
P is the principal amount (initial sum).
r is the rate of interest per annum.
n is the number of years.

Solution:
Since the sum becomes 2.89 times itself, we can write:
A = $$2.89 \times P$$​​
2.89 P = $$P \left(1 + \frac{r}{100}\right)^2$$​​

2.89 = $$\left(1 + \frac{r}{100}\right)^2$$​​

​$$\sqrt{2.89} = 1 + \frac{r}{100} \\ \ \\1.7 = 1 + \frac{r}{100}$$​​

1.7 - 1 =  0.7 = \frac{r}{100}" data-font-size="1em" style="font-size: 16px">r100=>0.7=r100\frac{r}{100} => 0.7 = \frac{r}{100}100r​=>0.7=100r​​​
r = $$0.7 \times 100 = 70$$​​
Thus, the correct option is (d) 70 percent

Alternate Method:

P = 2.89 P → in 2 years 

100 P = 289 P

10 P = 17 P  → in 1 year

Rate % = $$\frac{7}{10} \times 100\%$$ = 70%​

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