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): \(100\)