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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
12 Jan 2022, 09:15
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:
32% (02:52) correct
68%
(03:01)
wrong
based on 85
sessions
History
Date
Time
Result
Not Attempted Yet
How many ordered pairs of positive integers (x,y) satisfy both of the inequalities \(2x+y<10\) and \(x-y>-2\)?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
12 Jan 2022, 09:53
Kudos
Bookmarks
Bunuel
We have the inequalities y < -2x + 10 and y < x + 2. I'd draw a picture here; the inequality y < -2x + 10 just describes all the points below the line y = -2x + 10, and the inequality y < x + 2 is all the points below the line y = x + 2. If you draw both lines, remembering x and y are positive, we want all the points with integer coordinates contained in a quadrilateral with vertices at (0, 0), (0, 2), (5, 0) and another point in the first quadrant we don't need to solve for. From the picture, you can see which of the two lines produces the relevant boundary for different values of x.
When x = 1, because y < x + 2, substituting for x we know y < 3, so we have two points, (1, 1) and (1, 2)
When x = 2, again because y < x + 2, we know y < 4, and we have three points
When x = 3, it's now the other line that matters: y < -2x + 10, so y < 4, and we have three points
When x = 4, y < -2x + 10, and y < 2, so we have only the one point (4, 1)
When x = 5 or x > 5, we have no points, because the inequality y < -2x + 10 tells us y is negative
Counting our points, we have nine of them.
20 Jan 2025, 23:46
Kudos
Bookmarks
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
