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A manufacturer can save X dollars per unit in production cos

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A manufacturer can save X dollars per unit in production cos  [#permalink]

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New post 16 Sep 2013, 03:26
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Question Stats:

60% (02:10) correct 40% (02:08) wrong based on 173 sessions

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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
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Re: A manufacturer can save X dollars per unit in production cos  [#permalink]

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New post 16 Sep 2013, 03:37
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fozzzy 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) \(\frac{x}{y}\)
D) \(\frac{xn}{y}\)
E) \(\frac{x}{yn}\)

Please Explain


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.
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Re: A manufacturer can save X dollars per unit in production cos  [#permalink]

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New post 16 Sep 2013, 04:03
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fozzzy 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) \(\frac{x}{y}\)
D) \(\frac{xn}{y}\)
E) \(\frac{x}{yn}\)

Please Explain


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)
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Re: A manufacturer can save X dollars per unit in production cos  [#permalink]

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New post 12 Mar 2018, 16:37
fozzzy 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) \(\frac{x}{y}\)
D) \(\frac{xn}{y}\)
E) \(\frac{x}{yn}\)


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
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Re: A manufacturer can save X dollars per unit in production cos   [#permalink] 12 Mar 2018, 16:37
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