GATE 2018 EC GA Q.6: solution and answer

GATE 2018 · ECGA Q.6 · 2 marks · MCQ

Leila aspires to buy a car worth Rs. 10,00,000 after 5 years. What is the minimum amount in Rupees that she should deposit now in a bank which offers 10% annual rate of interest, if the interest was compounded annually?

(A)
5,00,000
(B)
6,21,000
(C)
6,66,667
(D)
7,50,000

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Answer and solution

ANSWER

(B) 6,21,000. The present value is $10, 00, 000/1. 1^{5} = 10, 00, 000/1.61051 \approx 6, 20, 921$ , and 6,21,000 is the smallest option that covers it.

Official answer: BSource: Official answer key, GATE 2018 EC (IIT Guwahati)

With interest compounded annually, the amount after $n$ years is $A = P (1 + \frac{r}{100})^{n}$ . Here $A = 10, 00, 000$ , $r = 10$ and $n = 5$ , and we need the principal $P$ .
Growth factor over 5 years, by squaring and multiplying: $1. 1^{2} = 1.21$ , $1. 1^{4} = 1.2 1^{2} = 1.4641$ , so $1. 1^{5} = 1.4641 \times 1.1 = 1.61051$ .
Divide to get the present value: $P = \frac{10 , 00 , 000}{1.61051} \approx 6, 20, 921$
So she must deposit at least about Rs. 6,20,921. Of the options, 6,21,000 is the smallest one that is at least this much.
Check: $6, 21, 000 \times 1.61051 \approx 10, 00, 127 \geq 10, 00, 000$ , so it is enough. Option (A) is not: $5, 00, 000 \times 1.61051 \approx 8, 05, 255 < 10, 00, 000$ . Options (C) and (D) are enough too, but both are well above the minimum.
Answer: (B) 6,21,000.

KEY IDEA

This is a present-value problem. Undo compound growth by dividing the target amount by $(1 + \frac{r}{100})^{n}$ . Work out $1. 1^{5}$ as $(1. 1^{2})^{2} \times 1.1 = 1.61051$ instead of estimating it.

WATCH OUT FOR

Using simple interest gives $P (1 + 5 \times 0.1) = 1.5 P$ , so $P \approx 6, 66, 667$ , which is option (C). Another mistake is to take off 10% a year from the target, or to divide by $1.5$ instead of $1.61051$ . Under compound interest, (C) is enough money but it is not the minimum.

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GATE 2018 EC GA Q.6: solution and answer

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