Golab Roofing is considering the purchase of a crane that would cost $69,846, would have a useful life of 6 years, and would have no salvage value. The use of the crane would result in labor savings of $21,000 per year. The internal rate of return on the investment in the crane is closest to (Ignore income taxes.): Refer to Exhibit 12B-1 and Exhibit 12B-2, to determine the appropriate discount factor(s) using the tables provided.

20%

Explanation:

Purchase of Crane cost = $69,846

Salvage value = 0

Labor savings = $21,000 per year

Let us assume Internal rate of return = X %. At IRR, Present value of Inflows = Present value of outflows.

69,846 = 21,000/1.0X + 21,000/ 1.0X^2 + 21,000/1.0X^3 + 21,000/1.0X^4 + 21,000/1.0X^5 + 21,000/1.0X^6.

69,846 = 21,000 [1/1.0X + 1/1.0X^2 + 1/1.0X^3 + 1/1.0X^4 + 1/1.0X^5 + 1/1.0X^6]

69,846/21,000 = [1/1.0X + 1/1.0X^2 + 1/1.0X^3 + 1/1.0X^4 + 1/1.0X^5 + 1/1.0X^6]

69,846 / 21,000 = 3.326

From the given table (as attached below), period (6 of 20%) Indicates 3.326. So, the Internal rate of return = 20%.

answer may be d i think