GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A manufacturer can save x dollars per unit in production cos
[#permalink]
Updated on: 17 Sep 2022, 12:56
Updated on: 17 Sep 2022, 12:56
4
1
27
Bookmarks
00:00
Question Stats:
50% (02:21) correct
49% (02:16) wrong based on 198 sessions
Hide Show timer Statistics
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) \(\frac{x}{yn}\)
A) \(x - y\)
B) \((x - y)n\)
C) \(\frac{x}{y}\)
D) \(\frac{xn}{y}\)
E) \(\frac{x}{yn}\)
GRE Instructor
Joined: 19 Jan 2020
Status:Entrepreneur | GMAT, GRE, CAT, SAT, ACT coach & mentor | Founder @CUBIX | Edu-consulting | Content creator
Posts: 117
Given Kudos: 0
GPA: 3.72
Re: A manufacturer can save x dollars per unit in production cos
[#permalink]
03 Feb 2020, 09:46
03 Feb 2020, 09:46
9
1
5
Bookmarks
Asif1234 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 ) ( y x > , 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) n(x-y) (C) x/y (D) xn/y (E) x/yn
(A) x − y (B) n(x-y) (C) x/y (D) xn/y (E) x/yn
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 \((y > x)\), 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?
Amount saved per unit = \($x\)
=> Amount saved for \(n\) units = \($(nx)\)
Extra cost per unit per day = \($y\)
=> Extra cost for \(n\) units per day = \($(ny)\)
Let the required number of days in storage be \(d\)
=> Extra cost for \(n\) units in \(d\) days = \($(nyd)\)
Thus, for storage costs to remain less than or be equal to the savings on the excess units, we have:
\(nyd ≤ nx\)
\(=> d ≤ x/y\)
Thus, maximum number of days = \(x/y\)
Answer C
General Discussion
Re: A manufacturer can save x dollars per unit in production cos
[#permalink]
28 Aug 2022, 18:01
28 Aug 2022, 18:01
1
The problem states y>x. If the storage cost per unit per day (y) exceeds the savings per unit (x), then I don't understand how a unit could be stored even one day without losing money. Shouldn't the answer just be 0?
Example: x=5, y=6 (since y>x).
I produce a widget for $10. But if I produce extra, I can do it for $5, thus saving $5.
It will cost me $6 per day to store the widget. This already exceeds the $5 in savings. (If I store the widget for 1 day, it costs me $5 to produce it plus $6 to store it = $11 now instead of $10 if I had just not overproduced it.)
Example: x=5, y=6 (since y>x).
I produce a widget for $10. But if I produce extra, I can do it for $5, thus saving $5.
It will cost me $6 per day to store the widget. This already exceeds the $5 in savings. (If I store the widget for 1 day, it costs me $5 to produce it plus $6 to store it = $11 now instead of $10 if I had just not overproduced it.)
