Question

A certain sum of money becomes 625/256 times of itself in 1 year. Find the rate of interest per annum if interest is compounded quarterly?

• A. 5
• B. 25
• C. 50
• D. 100

Answer 1

The Correct Option is D

The formula for compound interest is given by: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ Where: - $A$ is the amount after $t$ years - $P$ is the principal amount (initial sum of money) - $r$ is the rate of interest per annum - $n$ is the number of times interest is compounded per year - $t$ is the time in years In this case, we are given that the amount becomes $\frac{625}{256}$ times the initial amount ($A = \frac{625}{256}P$) in 1 year, and the interest is compounded quarterly ($n = 4$). Substituting the given values, we have: $$\frac{625}{256}P = P \left(1 + \frac{r}{4}\right)^{4 \cdot 1}$$ Simplifying: $$\frac{625}{256} = \left(1 + \frac{r}{4}\right)^4$$ Taking the fourth root of both sides: $$\sqrt[4]{\frac{625}{256}} = 1 + \frac{r}{4}$$ Now, solving for $r$: $$\frac{5}{4} = 1 + \frac{r}{4}$$ Subtracting 1 from both sides: $$\frac{1}{4} = \frac{r}{4}$$ Multiplying both sides by 4: $$1 = r$$ So, the rate of interest per annum is 100%. Therefore, the answer is 100%, which corresponds to the option: 100.

Answer 2

Let, Principal P = x
A = $(\frac {625}{256})x$
T = 1 Year = 4 Quarter
Rate of interest = r%
p.a. = $\frac r4$ % p.quarter

Amount $A = P(1 + \frac {r}{100})^T$

$(\frac {625}{256})x$ $= x(1+ \frac {r}{400})^4$

$(\frac 54)^4 = (1+ \frac {r}{400})^4$

$\frac 54 = 1 + \frac {r}{400}$

$\frac 54 - 1 = \frac {r}{100}$

$\frac {1}{4} = \frac {r}{400}$

$4r= 400$
$r = 100 \%$ p.a.

So, the correct option is (D): $100$