Question

A sum of money amounts to Rs. 6690 after 3 years and Rs.10,035 after 6 years on compound interest. The sum is :

• A. Rs. 4460
• B. Rs. 5460
• C. Rs. 6640
• D. Rs. 3420

Hint:

When dealing with amounts at different time periods under compound interest, dividing the equations can help eliminate the principal and solve for the rate or a power of \( (1+r) \).

Answer 1

The Correct Option is A

Step 1: Use the formula for the amount under compound interest.
Amount (A) = \( P(1 + r)^n \), where P is the principal, r is the rate of interest per annum, and n is the number of years.
Let the principal be \( P \) and the rate of interest be \( r \).
According to the problem:
$$P(1 + r)^3 = 6690 \quad \cdots (1)$$
$$P(1 + r)^6 = 10035 \quad \cdots (2)$$
Step 2: Divide equation (2) by equation (1).
$$\frac{P(1 + r)^6}{P(1 + r)^3} = \frac{10035}{6690}$$
$$(1 + r)^{6-3} = (1 + r)^3 = \frac{10035}{6690}$$
Step 3: Simplify the fraction \( \frac{10035}{6690} \).
Divide both numerator and denominator by 5: \( \frac{2007}{1338} \)
Divide both numerator and denominator by 3: \( \frac{669}{446} \)
So, \( (1 + r)^3 = \frac{669}{446} = 1.5 \) (approximately)
Step 4: Substitute the value of \( (1 + r)^3 \) back into equation (1). $$P(1.5) = 6690$$ Step 5: Solve for the principal \( P \). $$P = \frac{6690}{1.5} = \frac{6690}{\frac{3}{2}} = 6690 \times \frac{2}{3}$$ $$P = 2230 \times 2 = 4460$$ The sum (principal) is \( \rupee 4460 \).