Last visit was: 06 Sep 2026, 13:38 It is currently 06 Sep 2026, 13:38
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
   1   2   3   4 
805+ (Hard)|   Word Problems|                                 
User avatar
HarshavardhanR
User avatar
Joined: 16 Mar 2023
Last visit: 05 Sep 2026
Posts: 649
Own Kudos:
758
 [1]
Given Kudos: 99
Status:Independent GMAT Tutor
Affiliations: Ex - Director, Subject Matter Expertise at e-GMAT
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 649
Kudos: 758
 [1]
Last Sunday a certain store sold copies of Newspaper A for $1.00 each 30 Nov 2024, 23:09
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post


An algebraic way to approach this question (above).

Harsha
Show Spoiler
Attachment:
GMAT-Club-Forum-rc843mzw.png
GMAT-Club-Forum-rc843mzw.png [ 202.33 KiB | Viewed 456 times ]
User avatar
BelisariusTirto
User avatar
Joined: 03 Oct 2024
Last visit: 15 Apr 2025
Posts: 33
Own Kudos:
5
Given Kudos: 64
Posts: 33
Kudos: 5
Last Sunday a certain store sold copies of Newspaper A for $1.00 each 30 Dec 2024, 10:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Apologies for bumping such an old thread, I was wondering where the 1 in 1+1.25 "goes?" as in the original solution it becomes:

Reduce by \(a\) and simplify:


\(r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}\);


I assumed P goes to the numerator by multiplying it out of the denominator (100-p)/p then it becomes 100p after being multiplied by 100. Forming 100p/1+125-1.25p?



Bunuel
Marcoson
Quote:

\(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) --> reduce by \(a\) and simplify --> \(r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}\)

Hi Bunel, I don't get the reduction. How do you get rid of the \(a+\) in the denominator?

I only get this:
\(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) --> reduces to --> \(r=\frac{100 p}{a+1,25*(100-p)}\)

\(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\);

Factor out a from the denominator: \(r=\frac{a}{a(1 + 1.25*\frac{(100-p)}{p})}*100\).

Reduce it: \(r=\frac{1}{1 + 1.25*\frac{(100-p)}{p}}*100\).

Hope it's clear.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,166
Own Kudos:
839,565
Given Kudos: 111,337
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,166
Kudos: 839,565
Last Sunday a certain store sold copies of Newspaper A for $1.00 each 30 Dec 2024, 10:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BelisariusTirto
Apologies for bumping such an old thread, I was wondering where the 1 in 1+1.25 "goes?" as in the original solution it becomes:

Reduce by \(a\) and simplify:


\(r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}\);


I assumed P goes to the numerator by multiplying it out of the denominator (100-p)/p then it becomes 100p after being multiplied by 100. Forming 100p/1+125-1.25p?

Show more


\(p + 125 - 1.25p =\)

\(= 125 + (p -1.25p)=\)

\(= 125 -0.25p\)
User avatar
BelisariusTirto
User avatar
Joined: 03 Oct 2024
Last visit: 15 Apr 2025
Posts: 33
Own Kudos:
5
Given Kudos: 64
Posts: 33
Kudos: 5
Last Sunday a certain store sold copies of Newspaper A for $1.00 each 30 Dec 2024, 11:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ah, maybe I should have phrased it better. I was wondering how the 1 becomes p.
Bunuel
BelisariusTirto
Apologies for bumping such an old thread, I was wondering where the 1 in 1+1.25 "goes?" as in the original solution it becomes:

Reduce by \(a\) and simplify:


\(r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}\);


I assumed P goes to the numerator by multiplying it out of the denominator (100-p)/p then it becomes 100p after being multiplied by 100. Forming 100p/1+125-1.25p?



\(p + 125 - 1.25p =\)

\(= 125 + (p -1.25p)=\)

\(= 125 -0.25p\)
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,166
Own Kudos:
839,565
Given Kudos: 111,337
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,166
Kudos: 839,565
Last Sunday a certain store sold copies of Newspaper A for $1.00 each 30 Dec 2024, 11:10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BelisariusTirto
Ah, maybe I should have phrased it better. I was wondering how the 1 becomes p.
Show more

No idea what you mean...
User avatar
DanTheGMATMan
User avatar
Joined: 02 Oct 2015
Last visit: 05 Sep 2026
Posts: 385
Own Kudos:
278
Given Kudos: 10
Expert
Expert reply
Posts: 385
Kudos: 278
Last Sunday a certain store sold copies of Newspaper A for $1.00 each 21 Oct 2025, 07:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
nihilveritatis
User avatar
Joined: 28 Oct 2025
Last visit: 03 Sep 2026
Posts: 8
Given Kudos: 34
Posts: 8
Kudos: 0
Last Sunday a certain store sold copies of Newspaper A for $1.00 each 25 Jan 2026, 06:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Solved using algebra without substituting -

eq 1 - (A+B)*p/100 = A .....(given)
=> (A+B)*p = 100A
=> PA +PB = 100A
PB = A(100 - P)
B/A = (100-P)/ P ....(1)
eq 2 - (A+ 5b/4)*r/100 = A
Manipulate to get
=> B/A = 4(100 - r)/5r ...(2)
Equate the two equations to isolate r in terms of p
   1   2   3   4 
Moderator:
Bunuel
Math Expert
113166 posts