emmak wrote:

A manufacturer makes umbrellas at the cost of c dollars per umbrella, and sells them at the retail price of r dollars each. How many umbrellas can it afford to sell at the below-cost rate of b dollars per umbrella and still make a 100% profit on its lot of x total umbrellas?

A. \(\frac{b(2c-r)}{(x-r)}\)

B. \(\frac{2x(c-r)}{(b-r)}\)

C. \(\frac{x(2c-r)}{(b-r)}\)

D. \(\frac{2b(c-r)}{(x-r)}\)

E. \(\frac{2(xc-r)}{(x-r)}\)

This is a tough one to use the INPUT-OUTPUT approach, but here is goes:

Let c = $2 (it cost $2 to make each umbrella)

Let x = 10 (we make 10 umbrellas)

Let r = $5 (the retail price is $5 per umbrella)

Let b = $0 (the below-cost sale price is $0 per umbrella)

So, the manufacturer made 10 umbrellas at the cost of $2 per umbrella. So the total cost =

$20We need a 100% profit. So, we must earn

$40 in revenue. In other words, we must sell 8 umbrellas at $5 each.

This means we can "sell"

2 umbrellas at the below-cost sale price of $0 each.

At this point, we must plug x = 2, x = 10, r = 5 and b = 10 into each expression and see which one yields an OUTPUT of

2A. \(\frac{b(2c-r)}{(x-r)}\) =

0 ELIMINATE A

B. \(\frac{2x(c-r)}{(b-r)}\) =

12 ELIMINATE B

C. \(\frac{x(2c-r)}{(b-r)}\) =

2 KEEP C

D. \(\frac{2b(c-r)}{(x-r)}\) =

0 ELIMINATE D

E. \(\frac{2(xc-r)}{(x-r)}\) =

6 ELIMINATE E

Answer : C

Cheers,

Brent

_________________

Brent Hanneson – Founder of gmatprepnow.com

Brent also tutors students for the GMAT