It is currently 11 Sep 2026, 08:57
Home
Messages
Tests
Schools
Events
Close
GMAT 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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
avatar
fozzzy
avatar
Joined: 29 Nov 2012
Last visit: 17 May 2015
Posts: 573
Posts: 573
A manufacturer can save X dollars per unit in production cos
Post URL
16 Sep 2013, 02:26
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
Official Answer and Stats are available only to registered users. Register/Login.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

59% (02:18) correct 41% (01:59) wrong based on 305 sessions

History

Date
Time
Result
Not Attempted Yet
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
ShowHide Answer
Official Answer
Official Answer and Stats are available only to registered users. Register/Login.
3
Kudos
Add Kudos
11
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 11 Sep 2026
Posts: 113,452
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,452
A manufacturer can save X dollars per unit in production cos
Post URL
16 Sep 2013, 02:37
fozzzy
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
Show more

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.

_________________
New to the GMAT Club?
Signature Read More
4
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
shameekv
User avatar
Joined: 29 Aug 2013
Last visit: 11 Aug 2020
Posts: 50
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE:Programming (Computer Software)
GMAT 2: 540 Q44 V20
Posts: 50
A manufacturer can save X dollars per unit in production cos
Post URL
16 Sep 2013, 03:03
fozzzy
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
Show more

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)
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
JeffTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,973
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert
Expert reply
Posts: 2,973
A manufacturer can save X dollars per unit in production cos
Post URL
12 Mar 2018, 15:37
fozzzy
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) \(\frac{x}{yn}\)
Show more

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

_________________
See why Target Test Prep is the top rated GMAT course on GMAT Club. Read Our Reviews
Signature Read More
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
dave13
User avatar
Joined: 09 Mar 2016
Last visit: 09 Jul 2026
Posts: 1,081
Posts: 1,081
A manufacturer can save X dollars per unit in production cos
Post URL
28 Apr 2020, 04:04
Bunuel
fozzzy
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.
Show more

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 $
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 11 Sep 2026
Posts: 113,452
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,452
A manufacturer can save X dollars per unit in production cos
Post URL
28 Apr 2020, 04:12
dave13
Bunuel
fozzzy
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.

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 $
Show more

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.

_________________
New to the GMAT Club?
Signature Read More
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
GMATGuruNY
User avatar
Joined: 04 Aug 2010
Last visit: 04 Sep 2026
Posts: 1,348
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,348
A manufacturer can save X dollars per unit in production cos
Post URL
28 Apr 2020, 05:25
To guarantee that the correct answer yields an integer value for the number of days, the prompt should include the following condition in blue:

fozzzy
A manufacturer can save x dollars per unit in production costs by overproducing in certain seasons. Storage costs for the excess are y dollars per unit per day, where x/y is an integer. 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) \(\frac{x}{yn}\)
Show more

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:
\(\frac{x}{y} = \frac{100}{10} = 10\)

Show Spoiler
C

_________________
GMAT and GRE Tutor for over 20 years
Recent success stories for my students:admissions into Booth, Kellogg, HBS, Wharton, Tuck, Fuqua, Emory and others.
Available for live sessions in NYC and remotely all over the world
For more information, please email me at GMATGuruNY at gmail
Signature Read More
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
arushi118
User avatar
Joined: 21 Jul 2024
Last visit: 11 Sep 2026
Posts: 280
Location: India
Concentration: Leadership, General Management
Schools: INSEAD Aug '26 LBS '26 Kellogg '26
GPA: 8.2/10
Products:
My Reviews
Schools: INSEAD Aug '26 LBS '26 Kellogg '26
Posts: 280
A manufacturer can save X dollars per unit in production cos
Post URL
01 Mar 2026, 02:18
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.
Bunuel


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.
Show more
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NEW TOPIC
POST REPLY
Downloads
My Bookmarks
Reviews
Moderator:
Bunuel
Math Expert
113452 posts