Question

An engineer is planning to buy a truck that will costs $38,000 in 5 years from now. His plan is to pay immediately half of the truck cost. The remaining will be paid based on an end-of-year payments starting with a basic payment of $2,000 yearly, and increasing yearly by the same amount B. With an interest at 6% per year, the yearly increase B is nearest to:

		$1,700
		$2,400
		$1,000
		$400

Answer 1

The correct answer is (D) $400

In order to find B we have to find that value at which present worth of both payment equalize.

Present worth formula for a payment incur after year n is given by:

P = A/(1+r)ⁿ

n = time in years(here n = 5), r = interest rate = 0.06 and A = Amount after n years = 38000 and P = Present value

Hence P in first case = 32000/(1+ 0.06)⁵ = 23912.27

Using above formula Present Value in second case is given by:

P = 23912.27/2 + 2000/(1+0.06) + (2000 + B)/(1+0.06)² + (2000 + 2B)/(1+0.06)³ + (2000 + 3B)/(1+0.06)⁴ + (2000 + 4B)/(1+0.06)⁵

Here P also equals 23912.27

Hence we have :

23912.27 = 23912.27/2 + 2000/(1+0.06) + (2000 + B)/(1+0.06)² + (2000 + 2B)/(1+0.06)³ + (2000 + 3B)/(1+0.06)⁴ + (2000 + 4B)/(1+0.06)⁵

=> 23912.27 = 23912.27/2 + 2000/(1+0.06) + 2000/(1+0.06)² + 2000/(1+0.06)³ + 2000/(1+0.06)⁴ + 2000/(1+0.06)⁵ + B(1/(1+0.06)² + 2/(1+0.06)³ + 3/(1+0.06)⁴ + 4/(1+0.06)⁵)

=> 11956.135 = 8424.73 + 7.93B

=> B = 400(Closest answer)

Hence, the correct answer is (D) $400