Question 135

# A sum of money was invested for 14 years was in Scheme A which offers simple interest at a rate of 8% p.a. The amount received from Scheme A after 14 years was then invested for two years in Scheme B which offers compound interest (compounded annually) at a rate of 10% p.a. If the interest received from Scheme B was Rs. 6,678, what was the sum invested in Scheme A?

Solution

Let the sum invested in scheme A = $$Rs. 100x$$

Time = 14 years and rate = 8% under simple interest

=> $$S.I. = \frac{P \times R \times T}{100}$$

= $$\frac{100x \times 8 \times 14}{100} = 112x$$

=> Amount invested in Scheme B = $$100x + 112x = Rs. 212x$$

Time = 2 years and rate = 10% under compound interest.

$$C.I. = P [(1 + \frac{R}{100})^T - 1]$$

=> $$6678 = 212x [(1 + \frac{10}{100})^2 - 1]$$

=> $$6678 = 212x [(\frac{11}{10})^2 - 1] = 212x (\frac{21}{100})$$

=> $$0.21x = \frac{6678}{212} = 31.5$$

=> $$x = \frac{31.5}{0.21} = 150$$

$$\therefore$$ Sum invested in scheme A = $$100 \times 150 = Rs. 15,000$$