GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2018, 02:44

### 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

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Joined: 29 Nov 2012
Posts: 775
A manufacturer can save X dollars per unit in production cos  [#permalink]

### Show Tags

16 Sep 2013, 03:26
2
7
00:00

Difficulty:

65% (hard)

Question Stats:

60% (02:10) correct 40% (02:08) wrong based on 165 sessions

### HideShow 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) $$\frac{x}{yn}$$

Please Explain

_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Math Expert
Joined: 02 Sep 2009
Posts: 49954
Re: A manufacturer can save X dollars per unit in production cos  [#permalink]

### Show Tags

16 Sep 2013, 03:37
2
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.
_________________
Manager
Joined: 29 Aug 2013
Posts: 74
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Re: A manufacturer can save X dollars per unit in production cos  [#permalink]

### Show Tags

16 Sep 2013, 04:03
2
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)
Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: A manufacturer can save X dollars per unit in production cos  [#permalink]

### Show Tags

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
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: A manufacturer can save X dollars per unit in production cos &nbs [#permalink] 12 Mar 2018, 16:37
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.