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 0710:00 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 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0410:00 AM EDT
-11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.- 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 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 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
17 Jul 2019, 04:47
Kudos
Bookmarks
A
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:
63% (02:36) correct
37%
(02:15)
wrong
based on 277
sessions
History
Date
Time
Result
Not Attempted Yet
Product of three distinct positive integers p, q and r is 120. Is p + q > r?
(1) 4p + 5r = 50
(2) p < r
(1) 4p + 5r = 50
(2) p < r
18 Jul 2019, 03:45
Kudos
Bookmarks
This is a great question on equations. It tests whether you are able to break the stereotype of “Number of equations should be equal to number of variables” and solve it using logic.
For you to be able to do it though, you need to extract every bit of data that you possibly can from the statements and the question. In this question, we are lucky that the question mentions explicitly that the integers p,q and r are positive integers. And distinct too. This is a BIG clue.
Also, p*q*r = 120. For example, we could take p = 10, q = 2 and r = 6; or p = 2, q = 6 and r = 10. Depending on the relative values of p, q and r, we need to figure out if p+q > r.
From statement I, 4p + 5r = 50. Looking at the equation, if you thought that this equation cannot be solved because there are more variables than equations, you cannot be more wrong.
In 4p + 5r = 50, you need to understand that 4p will always be even, irrespective of the value of p. 50 is also even. Therefore, 5r HAS to be even since only EVEN + EVEN = EVEN.
5r can be even only when r is even.
So, the only values we need to try for r are the multiples of 120 which are even numbers and which satisfy the equation. This means, r = 2 or 4 or 6 or 8 or 10. The other even factors of 120 do not satisfy the equation given.
For r = 2, p = 10. Therefore, q = 6. In this case, p + q > r.
For r = 4, p = \(\frac{30}{4}\) which is not an integer. Ruled out.
For r = 6, p = 5. Therefore, q = 4. In this case also, p + q > r.
For r = 8, p = \(\frac{10}{4}\), not an integer. Ruled out.
For r = 10, p = 0, which is not a positive integer. Ruled out.
Although we have two cases, in both the cases, we are able to say p+q is always more than r. Statement I alone is sufficient. Possible answer options are A or D. Answer options B, C and E can be ruled out.
From statement II, p<r.
p = 3 and r = 4 is one case. Here, q = 10 and so p+q>r.
p = 3 and r = 10 is another case. Here, q = 4 and so p+q<r.
Statement II alone is insufficient. Answer option D can be eliminated.
The correct answer option is A.
When there is ONE linear equation with two variables, do not conclude straight away that it cannot be solved. Depending on what additional conditions are given, you may just be able to solve it, even uniquely sometimes.
So, the next time you are faced with an equation like 4p + 5r = 50, look for additional constraints.
Hope this helps!
For you to be able to do it though, you need to extract every bit of data that you possibly can from the statements and the question. In this question, we are lucky that the question mentions explicitly that the integers p,q and r are positive integers. And distinct too. This is a BIG clue.
Also, p*q*r = 120. For example, we could take p = 10, q = 2 and r = 6; or p = 2, q = 6 and r = 10. Depending on the relative values of p, q and r, we need to figure out if p+q > r.
From statement I, 4p + 5r = 50. Looking at the equation, if you thought that this equation cannot be solved because there are more variables than equations, you cannot be more wrong.
In 4p + 5r = 50, you need to understand that 4p will always be even, irrespective of the value of p. 50 is also even. Therefore, 5r HAS to be even since only EVEN + EVEN = EVEN.
5r can be even only when r is even.
So, the only values we need to try for r are the multiples of 120 which are even numbers and which satisfy the equation. This means, r = 2 or 4 or 6 or 8 or 10. The other even factors of 120 do not satisfy the equation given.
For r = 2, p = 10. Therefore, q = 6. In this case, p + q > r.
For r = 4, p = \(\frac{30}{4}\) which is not an integer. Ruled out.
For r = 6, p = 5. Therefore, q = 4. In this case also, p + q > r.
For r = 8, p = \(\frac{10}{4}\), not an integer. Ruled out.
For r = 10, p = 0, which is not a positive integer. Ruled out.
Although we have two cases, in both the cases, we are able to say p+q is always more than r. Statement I alone is sufficient. Possible answer options are A or D. Answer options B, C and E can be ruled out.
From statement II, p<r.
p = 3 and r = 4 is one case. Here, q = 10 and so p+q>r.
p = 3 and r = 10 is another case. Here, q = 4 and so p+q<r.
Statement II alone is insufficient. Answer option D can be eliminated.
The correct answer option is A.
When there is ONE linear equation with two variables, do not conclude straight away that it cannot be solved. Depending on what additional conditions are given, you may just be able to solve it, even uniquely sometimes.
So, the next time you are faced with an equation like 4p + 5r = 50, look for additional constraints.
Hope this helps!
18 Jul 2019, 11:51
Kudos
Bookmarks
Dillesh4096
The source JAMBOREE says, "Keep it simple".
So why not
Given- positive integers, pqr=120
We have an equation from st 1
possible set of (p,r)
(5,6)(10,2)
In each case p+q>r
A
