Last visit was: 24 Sep 2026, 23:41 It is currently 24 Sep 2026, 23:41
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
Back to Forum
Create Topic
805+ (Hard)|   Algebra|   Word Problems|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Sep 2026
Posts: 113,907
Own Kudos:
843,460
 [47]
Given Kudos: 111,815
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,907
Kudos: 843,460
 [47]
A shoe store sells a new model of running shoe for $32. At that price 14 Sep 2022, 07:18
3
Kudos
Add Kudos
44
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

57% (02:53) correct 43% (03:02) wrong based on 407 sessions

History

Date
Time
Result
Not Attempted Yet
A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes ?

A. 1440
B. 1600
C. 2460
D. 3920
E. 4000


Check twin question HERE.

Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
GMAT Club Tests Check Our Products Try Free Tests
­
ShowHide Answer
Official Answer
A
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Sep 2026
Posts: 113,907
Own Kudos:
843,460
 [3]
Given Kudos: 111,815
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,907
Kudos: 843,460
 [3]
A shoe store sells a new model of running shoe for $32. At that price 01 Apr 2024, 08:45
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Official Solution:

A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes?

A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000

Assume that to achieve the maximum difference between the current and estimated revenue, the price should be reduced by $2 \(x\) times. Then the price of a pair of shoes would be \(32 - 2x\) dollars, and a total of \(80 + 20x\) pairs of shoes would be sold, generating revenue equal to \((32 - 2x)(80 + 20x)\) dollars.

We want to maximize \((32 - 2x)(80 + 20x)\). Let's work with the expression:


\((32 - 2x)(80 + 20x)=\)

\(=-40(x - 16)(4 + x)=\)

\(=-40(x^2 - 12x - 64)\)

Complete the square for \(x^2 - 12 x - 64\):


\(=-40(x^2 - 12x + 36 -36 - 64)\)

\(=-40[(x - 6)^2 - 100)]\)

\(=-40(x - 6)^2 + 4,000\)

Since the value of \(-40(x - 6)^2\) is 0 or negative, the maximum value of \(-40(x - 6)^2 + 4,000\) is \(4,000\), that is when \(-40(x - 6)^2\) is 0. Therefore, the maximum amount by which the store can increase its weekly revenue from the shoes is \(4,000 - 32*80 = 1,440\) dollars.

Answer: A­
General Discussion
User avatar
ankitgoswami
User avatar
Joined: 20 Dec 2019
Last visit: 22 Jun 2026
Posts: 87
Own Kudos:
147
 [6]
Given Kudos: 74
Posts: 87
Kudos: 147
 [6]
A shoe store sells a new model of running shoe for $32. At that price 14 Sep 2022, 07:40
5
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes ?

A. 1440
B. 1600
C. 2460
D. 3920
E. 4000


Check twin question HERE.

 


Enjoy this brand new question we just created for the GMAT Club Tests.

To get 1,600 more questions and to learn more visit: user reviews | learn more

 

Show more


Present weekly revenue = 80 x $32 = $2560

So according to estimation, Sales will increase by 20 pairs for every $2 decrease in price
So the possibilities are
100 x $30 = $3000
120 x $28 = $3360
140 x $26 = ....
.
..
.
.
.
200 x $20 = $4000 (Hereafter it starts decreasing)

So the maximum amount by which the store can increase its weekly revenue = $4000 - $2560 = $1440

Ans is A
User avatar
SaquibHGMATWhiz
User avatar
GMATWhiz Representative
Joined: 23 May 2022
Last visit: 12 Jun 2024
Posts: 623
Own Kudos:
815
 [4]
Given Kudos: 6
Location: India
GMAT 1: 760 Q51 V40
Expert
Expert reply
GMAT 1: 760 Q51 V40
Posts: 623
Kudos: 815
 [4]
A shoe store sells a new model of running shoe for $32. At that price 15 Sep 2022, 01:14
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes ?

A. 1440
B. 1600
C. 2460
D. 3920
E. 4000
Show more
Solution:

  • When selling the shoe at $32, the store sells 80 pairs
  • Reducing the price by $2, increases the sale by 20
  • Let us assume that this reduction os $2 is done n times
  • So, the revenue, in that case, will be \((80+20n)(32-2n)\)
  • We need to maximise \((80+20n)(32-2n)\)
    \(⇒2560-160n+640n-40n^2\)
    \(⇒-40n^2+480n+2560\)
    \(⇒-40(n^2-12n-64)\)
    \(⇒-40(n^2-12n+6^2-6^2-64)\)
    \(⇒-40[(n-6)^2-36-64]\)
    \(⇒-40[(n-6)^2-100]\)
    \(⇒-40(n-6)^2+4000\)
  • We can infer that the maximum value of \(-40(n-6)^2+4000\)is 4000 i.e., when \(-40(n-6)^2=0\)
  • Increase in the revenue \(=4000-2560=$1440\)


Hence the right answer is Option A
User avatar
gmatophobia
User avatar
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 19 Sep 2026
Posts: 3,174
Own Kudos:
12,432
 [3]
Given Kudos: 1,860
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
Kudos: 12,432
 [3]
A shoe store sells a new model of running shoe for $32. At that price 15 Sep 2022, 22:37
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes ?

A. 1440
B. 1600
C. 2460
D. 3920
E. 4000

 


Enjoy this brand new question we just created for the GMAT Club Tests.

To get 1,600 more questions and to learn more visit: user reviews | learn more

 

Show more

We are given that for every $2 decrease in price, 20 additional pairs of shoes are sold.

Current Revenue = 32 * 80 = $2560

Let's assume that we need to decrease the price by 2x to obtain maximum revenue.

At that price, the number of pair of shoes that will be sold is (100 + 20x)

Revenue = (100 + 2x) (32-2x)

We see that this is a quadratic equation, and a downward parabola. The maxima will lie at \(\frac{-b}{2a}\)

Upon solving x = 6

Revenue at x = 6

= (32 - 12)(80 + 120) = 4000

Difference = 4000 - 2560 = $1440

Option A

Detailed working is shown in the attached image.
Attachments

Screenshot 2022-09-16 110103.png
Screenshot 2022-09-16 110103.png [ 226.28 KiB | Viewed 10764 times ]

User avatar
emupton
User avatar
Joined: 11 Feb 2024
Last visit: 07 Jun 2024
Posts: 1
Own Kudos:
2
 [2]
Given Kudos: 8
Posts: 1
Kudos: 2
 [2]
A shoe store sells a new model of running shoe for $32. At that price 10 May 2024, 08:00
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Another way to solve:
Once you get to the quadratic equation (x^2−12x−64) 
Factor to: (x-16)(x+4)
Solve for x's: x = 16, x = -4
As is parabola, maximum point is between the two x-values, so (16+(-4))/2 = 6

Then continue as other solutions:
new revenue = (80+20(6))*(32-2(6)) = 4000
old revenue = 32*80 = 2560
4000 - 2560 = 1440
 ­
User avatar
sayan640
User avatar
Joined: 29 Oct 2015
Last visit: 18 Sep 2026
Posts: 1,040
Own Kudos:
924
Given Kudos: 793
GMAT 1: 570 Q42 V28
Products:
My Reviews
GMAT 1: 570 Q42 V28
Posts: 1,040
Kudos: 924
A shoe store sells a new model of running shoe for $32. At that price 15 Sep 2024, 03:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The store will sell 20 additional pairs of shoes each week for every $2 reduction in the price.
Lets say , this happens for 'c' times.

New revenue will be = 32*80 * (1 + c/4) * (1 - c/16)

Because increase in shoes = 20/80 = 1/4
Decrease in price = 2/32 = 1/16

Hence for maximising , we need to maximise = (1 + c/4) * (1 - c/16)
IF we differentiate , we get the value of c = 6

At c= 6 , the revenue will be maximum.
The maximum revenue will be = 32.80 . ( 1 + 6/4) * ( 1-6/16) = 4000

Initial revenue = 32 * 80 = 2560
Hence , the maximum by which we can increase = 4000 - 2560 = 1440

ans A.
Kindly check KarishmaB GMATinsight
User avatar
Sujithz001
User avatar
Joined: 09 Jun 2024
Last visit: 06 Feb 2026
Posts: 99
Own Kudos:
47
Given Kudos: 75
Posts: 99
Kudos: 47
A shoe store sells a new model of running shoe for $32. At that price 06 Oct 2024, 07:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey Bunuel / KarishmaB / gmatophobia

Might be a beginner question - I'm facing difficulty understanding the solution from this step,





[ltr]
1. Why are we manipulating the quadratic equation, rather the solving it upfront?

2. Would we not reach the soln. by solving it directly? If yes, why?


Thanks a lot in advance![/ltr]



Bunuel
Official Solution:

A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes?

A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000


Assume that to achieve the maximum difference between the current and estimated revenue, the price should be reduced by $2 \(x\) times. Then the price of a pair of shoes would be \(32 - 2x\) dollars, and a total of \(80 + 20x\) pairs of shoes would be sold, generating revenue equal to \((32 - 2x)(80 + 20x)\) dollars.

We want to maximize \((32 - 2x)(80 + 20x)\). Let's work with the expression:

\((32 - 2x)(80 + 20x)=\)

\(=-40(x - 16)(4 + x)=\)

\(=-40(x^2 - 12x - 64)\)
Complete the square for \(x^2 - 12 x - 64\):

\(=-40(x^2 - 12x + 36 -36 - 64)\)

\(=-40[(x - 6)^2 - 100)]\)

\(=-40(x - 6)^2 + 4,000\)
Since the value of \(-40(x - 6)^2\) is 0 or negative, the maximum value of \(-40(x - 6)^2 + 4,000\) is \(4,000\), that is when \(-40(x - 6)^2\) is 0. Therefore, the maximum amount by which the store can increase its weekly revenue from the shoes is \(4,000 - 32*80 = 1,440\) dollars.


Answer: A­
Show more
Show Spoiler
Attachment:
GMAT-Club-Forum-v9g7ur1j.png
GMAT-Club-Forum-v9g7ur1j.png [ 16.16 KiB | Viewed 3372 times ]
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Sep 2026
Posts: 113,907
Own Kudos:
843,460
Given Kudos: 111,815
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,907
Kudos: 843,460
A shoe store sells a new model of running shoe for $32. At that price 06 Oct 2024, 09:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sujithz001
Hey Bunuel / KarishmaB / gmatophobia

Might be a beginner question - I'm facing difficulty understanding the solution from this step,





[ltr]
1. Why are we manipulating the quadratic equation, rather the solving it upfront?

2. Would we not reach the soln. by solving it directly? If yes, why?


Thanks a lot in advance![/ltr]



Bunuel
Official Solution:

A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes?

A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000


Assume that to achieve the maximum difference between the current and estimated revenue, the price should be reduced by $2 \(x\) times. Then the price of a pair of shoes would be \(32 - 2x\) dollars, and a total of \(80 + 20x\) pairs of shoes would be sold, generating revenue equal to \((32 - 2x)(80 + 20x)\) dollars.

We want to maximize \((32 - 2x)(80 + 20x)\). Let's work with the expression:

\((32 - 2x)(80 + 20x)=\)

\(=-40(x - 16)(4 + x)=\)

\(=-40(x^2 - 12x - 64)\)
Complete the square for \(x^2 - 12 x - 64\):

\(=-40(x^2 - 12x + 36 -36 - 64)\)

\(=-40[(x - 6)^2 - 100)]\)

\(=-40(x - 6)^2 + 4,000\)
Since the value of \(-40(x - 6)^2\) is 0 or negative, the maximum value of \(-40(x - 6)^2 + 4,000\) is \(4,000\), that is when \(-40(x - 6)^2\) is 0. Therefore, the maximum amount by which the store can increase its weekly revenue from the shoes is \(4,000 - 32*80 = 1,440\) dollars.


Answer: A­
Show more

To solve, we need an equation, like \(-40(x^2 - 12x - 64) = 0\), which could be solved for x to get specific values: x = -4 or x = 16. However, in this case, we are dealing with an expression, \(-40(x^2 - 12x - 64)\), for which we need to find the maximum value.

To do this, we manipulated the expression to break it into a sum of two parts: a non-positive term \(-40(x - 6)^2\) and 4,000. This setup helps us see that the maximum value of the expression occurs when \(-40(x - 6)^2\) is 0, meaning the overall maximum is 0 + 4,000 = 4,000.
Show Spoiler
Attachment:
GMAT-Club-Forum-thynizpe.png
GMAT-Club-Forum-thynizpe.png [ 16.16 KiB | Viewed 3326 times ]
User avatar
Globethrotter
User avatar
Joined: 12 Sep 2015
Last visit: 14 Nov 2025
Posts: 26
Own Kudos:
18
Given Kudos: 56
Posts: 26
Kudos: 18
A shoe store sells a new model of running shoe for $32. At that price 25 Jul 2025, 22:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post

A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes ?

Bunuel A faster way to get to the answer is with a little bit different approach without actually solving any equation.
Just write down the possible pairs as shown below and observe the point at which the pattern starts to reverse. That is the inflection point and thus the point of maximum value.

See below, the green value is the inflection point. Observe the values start inversing at that point(for eg. 22*180 becomes 18*220)
3280
30100
28120
26140
24160
22180
20200
18220
16240
14260
12280
10300

So the max value is 20*200 = 4000, Store currently selling at 32*80 = 2560.
Max amount by which store can increase its revenue = 4000 - 2560 = 1440.

Answer is A
Moderator:
Bunuel
Math Expert
113907 posts