Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 0410:00 AM PDT
-11:00 AM PDT
What does test anxiety really look like on the GMAT? For many test takers, it does not look like a panic attack. Instead, it shows up as brain fog, rereading, negative comparisons, mental noise....
Sep 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0401:30 AM EDT
-02:30 AM EDT
This Labor Day, turn your hard work into real score gains. From 615→715 in 15 days to 705 in 30 days, our students prove what's possible with personalized, expert guidance. Save up to 25% on 1-on-1 tutoring.
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 07
12:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT 
Sep 0712:30 AM EDT
-01:30 AM EDT
Tejas studied hard but kept repeating the same mistakes. After 3 attempts at 645, a focused 4-week strategy helped him identify and fix what was holding him back—leading to a 695 with just 12–15 hours of study per week.- Sep 07
10:00 AM PDT
-11:00 AM PDT
A complete walkthrough of GMAT Club’s free 12-week GMAT study plan: what you study each week, how progress and goals are tracked, how the error log works, and how your practice unlocks paid tools at no cost. - Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway - Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 1008:00 PM PDT
-09:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
09 Aug 2026, 22:38
Kudos
Bookmarks
E
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:
35% (02:09) correct
65%
(02:12)
wrong
based on 55
sessions
History
Date
Time
Result
Not Attempted Yet
For how many integers \(p\) is the square root of \(\frac{p}{50-p}\) an integer ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) More than 4
(A) 1
(B) 2
(C) 3
(D) 4
(E) More than 4
09 Aug 2026, 23:29
Kudos
Bookmarks
Trying out numbers is a good way to start this problem. The answers tell us that there are only a small number of solutions. In addition, approaching this problem algebraically can be confusing.
We can try out integers for p, and we'll see that p=0 gives one solution. We might have some insight, for example that p=25 also works, but we're unlikely to be confident in finding all the solutions with this approach as we wouldn't know how high to go.
A more efficient way is to consider what numbers the expression could be equal to. If the square root of the expression is an integer, then the expression itself must be equal to an integer squared. So set the expression equal to 1, then 4, then 9, then 16, then 25, then 49. Some of these will work and some won't. To make this efficient, remember that you're looking for solutions where p is an integer, so you don't need to fully solve or do all the arithmetic. By the time you find 5 solutions for p you can stop, knowing that the answer must be E.
To solve this more theoretically, you can pick a letter for an integer (e.g. k) and set the expression equal to k. Square both sides and rearrange to make p the subject, and then you can deduce solutions for k (and thus for p). However, this strategy is longer and more arduous than trying out examples.
We can try out integers for p, and we'll see that p=0 gives one solution. We might have some insight, for example that p=25 also works, but we're unlikely to be confident in finding all the solutions with this approach as we wouldn't know how high to go.
A more efficient way is to consider what numbers the expression could be equal to. If the square root of the expression is an integer, then the expression itself must be equal to an integer squared. So set the expression equal to 1, then 4, then 9, then 16, then 25, then 49. Some of these will work and some won't. To make this efficient, remember that you're looking for solutions where p is an integer, so you don't need to fully solve or do all the arithmetic. By the time you find 5 solutions for p you can stop, knowing that the answer must be E.
To solve this more theoretically, you can pick a letter for an integer (e.g. k) and set the expression equal to k. Square both sides and rearrange to make p the subject, and then you can deduce solutions for k (and thus for p). However, this strategy is longer and more arduous than trying out examples.
10 Aug 2026, 04:10
Kudos
Bookmarks
Quicker approach (I think!)
Given: Sqrt (p/(50-p)) = integer
=> p/(50-p) = square of an integer = k^2.
We could simply start listing squares of integers from 0 (0,1,4,9,16, 25, etc.), and for each value, quickly calculate p based on
[p/(50-p) = k^2]. We just need to get integer "p".
No need to check further - we got more than 4. Choice E, already.
Sidenote: Also, p cannot be more than 49. The very next integer value is 50, for which p/50-p would be not defined. And any further integer values of p would lead to a negative value within the sqrt (p/50-p) term. Again, this is not possible (k is an integer, not an imaginary number). So, that is another reason why we can confidently stop at 49.
That said, here is another way (longer, traditional solve).
Explanation:
- Given: Sqrt (p/(50-p)) = integer = k (say). Square both sides to get rid of the pesky sqrt and get p in terms of k.
- We get p = (50) * (k^2)/(k^2+1)
- What is interesting here -> k^2 and k^2+1 are consecutive numbers. Hence, they have no common factors.
- Hence, the only way p will be an integer is if the denominator is a factor of 50 (else we will always be stuck with a denominator term).
- This restricts the possible values of the denominator to factors of 50.
- 50 has 6 factors - 1,2,5,10,25,50. For all these values, we should an integer p. But we should also ensure that sqrt(p/(50-p)) is also an integer. The table in the image shows the quick calculations here. We can clearly see that yet again, we get more than 4 legit P values that satisfy the conditions. Thus, choice E.
---
Harsha
Given: Sqrt (p/(50-p)) = integer
=> p/(50-p) = square of an integer = k^2.
We could simply start listing squares of integers from 0 (0,1,4,9,16, 25, etc.), and for each value, quickly calculate p based on
[p/(50-p) = k^2]. We just need to get integer "p".
| k^2 | k | p | 50-p |
| 0 | 0 | 0 | 50 |
| 1 | 1 | 25 | 25 |
| 4 | 2 | 40 | 10 |
| 9 | 3 | 45 | 5 |
| 16 | 4 | not an integer | discard |
| 25 | 5 | not an integer | discard |
| 36 | 6 | not an integer | discard |
| 49 | 7 | 49 | 1 |
No need to check further - we got more than 4. Choice E, already.
Sidenote: Also, p cannot be more than 49. The very next integer value is 50, for which p/50-p would be not defined. And any further integer values of p would lead to a negative value within the sqrt (p/50-p) term. Again, this is not possible (k is an integer, not an imaginary number). So, that is another reason why we can confidently stop at 49.
That said, here is another way (longer, traditional solve).
Explanation:
- Given: Sqrt (p/(50-p)) = integer = k (say). Square both sides to get rid of the pesky sqrt and get p in terms of k.
- We get p = (50) * (k^2)/(k^2+1)
- What is interesting here -> k^2 and k^2+1 are consecutive numbers. Hence, they have no common factors.
- Hence, the only way p will be an integer is if the denominator is a factor of 50 (else we will always be stuck with a denominator term).
- This restricts the possible values of the denominator to factors of 50.
- 50 has 6 factors - 1,2,5,10,25,50. For all these values, we should an integer p. But we should also ensure that sqrt(p/(50-p)) is also an integer. The table in the image shows the quick calculations here. We can clearly see that yet again, we get more than 4 legit P values that satisfy the conditions. Thus, choice E.
---
Harsha
