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A manufacturer can save X dollars per unit in production

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A manufacturer can save X dollars per unit in production

by VJesus12 » Sat Feb 24, 2018 3:01 am
A manufacturer can save x dollars per unit in production costs by overproducing in certain seasons. If storage costs for the excess are y dollars per unit per day ( x > y), which of the following expresses the maximum number of days that n excess units can be stored before the storage costs exceed the savings on the excess units?

A) x−y
B) (x−y)n
C) x/y
D) xn/y
E) x/yn

The OA is the option C .

What are the equations that I should use here? Experts, could you show me how to solve this PS question? Please.
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by Jay@ManhattanReview » Fri Mar 09, 2018 1:16 am
VJesus12 wrote:A manufacturer can save x dollars per unit in production costs by overproducing in certain seasons. If storage costs for the excess are y dollars per unit per day ( x > y), which of the following expresses the maximum number of days that n excess units can be stored before the storage costs exceed the savings on the excess units?

A) x−y
B) (x−y)n
C) x/y
D) xn/y
E) x/yn

The OA is the option C .

What are the equations that I should use here? Experts, could you show me how to solve this PS question? Please.
Say the number of days that n excess units can be stored = p

Total saving out of overproduction of n excess units = xn
Total storage cost for p days of n excess units = ypn

Since the total saving must be greater than or equal to the total storage cost, we have xn ≥ ypn

Cancelling, n, we get x ≥ yp

Thus, p ≤ x/y. The maximum value of p (the number of days that n excess units can be stored) = x/y.

The correct answer: C

Hope this helps!

-Jay
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by swerve » Fri Mar 09, 2018 10:08 am
Hi VJesus12,

You can solve this question as follow,

Total Storage Costs for excess units when kept for m days = m * n * y
Total Savings = nX (X dollars for 1 unit, hence for n units n*X).

Question asks when will Total Storage costs > Total Savings
Therfore m*n*y >= nX

Hence m >= nX/ny = X/y. Hence C is the correct answer.

Regards!
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by Jeff@TargetTestPrep » Mon Mar 12, 2018 4:07 pm
VJesus12 wrote:A manufacturer can save x dollars per unit in production costs by overproducing in certain seasons. If storage costs for the excess are y dollars per unit per day ( x > y), which of the following expresses the maximum number of days that n excess units can be stored before the storage costs exceed the savings on the excess units?

A) x−y
B) (x−y)n
C) x/y
D) xn/y
E) x/yn
Let d denote the number of days that n excess units are stored.

Since each unit costs y dollars per day, n units will cost nyd dollars for storage of d days. And, since each unit saves x dollars , then the savings for n units is nx.. The break-even for this scenario, when the storage cost is equal to the savings, is:

nyd - nx = 0

We see, too, that if (nyd - nx) > 0, then the company will lose money (i.e., the cost of storage is greater than the savings). Let's study this inequality:

nyd - nx > 0

nyd > nx

dy > x

d > x/y

or

x/y < d

As soon as the ratio x/y is less than d, the storage costs are greater than the savings. Therefore, the (maximum) number of days the n units can be stored before the storage costs exceed the savings is x/y.

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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