[Anonymous]

Wiley Question –

The Bread Company is planning to purchase a new machine which it will depreciate on a straight-line basis over a 10-year period. A full year’s depreciation will be taken in the year acquisition. The machine is expected to produce cash flow from operations, net of income, of $3000 in each of the 10 years. The accounting (book value) rate of return is expected to be 10% on the initial increase in required investment. The cost of the new machine will be

A: 30,000

B: 12,000

C: 15,000

D: 13,500

Answer: 15,000. The accounting rate of return equals accounting net income over book value. The book value of the new machine would be its cost. The $3,000 cash flow net of income taxes does not reflect the 10% straight-line depreciation. The solutions approach is to set up a formula in which cost is equal to $3,000 minus depreciation (which is 10% of cost) divided by the 10% rate of return. The numerator is the expected increase in accounting income. The denominator is the capitalization rate, 10%. Solving the formula indicates that the cost of the machine is $15,000.

Cost = $3,000 – 0.10 (cost)0.10

0.10 cost = $3,000 – 0.10 cost

0.20 cost = $3,000

Cost = $15,000

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This doesn’t even make sense.

 

 

[CPA soon]

Bump

FAR - 71, 68, 74, (8/31/14) 78 ✔
REG - 67, 71, 71, (10/18/14) 78 ✔
BEC - (11/29/14) 86 ✔
AUD - 73, (4/4/15) 86 ✔

I can't believe this is over! 2 years and 3 months..

[Kimboroni]

It's an algebra problem based on this: “cost is equal to $3,000 minus depreciation (which is 10% of cost) divided by the 10% rate of return”

There's an operation symbol missing from the pasted solution, so that may be part of the confusion. And I think using the word “cost” as the variable is confusing, so let's try it again using C as the variable instead.

STEP 1

C = [3000 – (.10 x C)] / .10

STEP 2

.10 x C = 3000 – (.10 x C)

STEP 3

.20 x C = 3000

STEP 4

C = 15000

In step 1, you set up the equation. .10 is the denominator on the entire right side of the equation.

For step 2, you convert the division problem into a multiplication problem. It's the same as turning 2 = 6 / 3 into 2 * 3 = 6. They are two ways of representing the same relationship.

For step 3, you add .10 x C to both sides of the equation. You want to get all of the terms with C on the same side of the equation.

For step 4, you divide both sides by .20 to solve for C, which is the cost.

AUD 84 (1/9/14-Wiley books/TB + free materials)
FAR 83 (5/21/14-the above + NINJA 10 Pt Combo Lite)
REG 84 (7/9/14-Wiley books/TB + NINJA Audio/FC/Notes)
BEC 76 (10/5/14-Wiley books/TB + NINJA Audio/FC)

Disclaimer: My ninja avatar is not meant to imply that I have any affiliation with this site other than being a forum member. That's a pic of a T-shirt that my daughter gave me for my birthday. 🙂

[Author]

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