A manufacturer can save X dollars per unit in production cos

Question

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) $\frac{x}{y}$
D) $\frac{xn}{y}$
E) $\frac{x}{yn}$

Please Explain

A)  x - y 
B)  (x - y)n 
C)  \frac{x}{y} 
D)  \frac{xn}{y} 
E)  \frac{x}{yn}

Please Explain

Bunuel · Accepted Answer

You save x dollars per unit and have additional cost of y dollars per unit per day. So, the maximum number of days that 1 excess unit can be stored before the storage costs exceed the savings on the excess units is x/y days (for example if you save $10 per unit and have additional cost of $5 per unit per day, then you can store for 10/5=2 days).

Answer: C.

Hope it's clear.

shameekv · Answer

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 ----- (c)

JeffTargetTestPrep · Answer

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

dave13 · Answer

Bunuel  but if i save 10$ per unit and produce only one unit and the cost of storage for two days of ONE unit will be 10$, and this contradicts the statement that X>Y cause in this case saving is 10$ and storage costs for two days is10 $

Bunuel · Answer

What's your question? Anyway, the storage costs for the excess are y dollars per unit  per day . So, y there is cost PER DAY.

GMATGuruNY · Answer

To guarantee that the correct answer yields an integer value for the number of days, the prompt should include the following condition in blue:

Let n=2 excess units.
Let x=$100 production savings per unit, implying that the total production savings for 2 units = 2*100 = $200.
Let y=$10 daily storage cost per unit, implying that the daily storage cost for 2 units = 2*10 = $20.

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? 
Since the storage costs may not exceed the $200 in production savings, the maximum total storage cost = $200.
At a rate of $20 per day, the greatest number of storage days that can be purchased for $200  = \frac{200}{20} = 10 .

The correct answer must yield 10 when x=100, y=10 and n=2.
Only C works:
 \frac{x}{y} = \frac{100}{10} = 10

C

arushi118 · Answer

Is it possible to comprehend that the storage costs are not per unit and are just based on the total number of days we store? As it has not been said so explicitly.

A manufacturer can save X dollars per unit in production cos

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