Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
Dec 1511:00 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.- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options. - Dec 21
11:00 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
05 Aug 2013, 00:56
Kudos
Bookmarks
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:
30% (02:28) correct
70%
(02:19)
wrong
based on 438
sessions
History
Date
Time
Result
Not Attempted Yet
ax + by=10
ay + bx=12
which one is greater here, x or y ?
(1) a > b
(2) a/b > 1
ay + bx=12
which one is greater here, x or y ?
(1) a > b
(2) a/b > 1
solution:
From equation (1), x/(10/a) + y/(10/b) = 1
Length of the Excerpt from X axis = 10/a
Length of the Excerpt from Y axis = 10/b
From statement (1) , a>b that’s why 10/a < 10/b
We know for this line, x = 10/a and y= 10/b (from the co-ordinate system)
That’s why Y>X .
For equation (2) we can have the same result using the same technique.
So statement (1) is sufficient.
From statement (2) when a is negative then b is also negative and a<b .
Or when a is positive then b is also positive and a>b .
Using this sort of data through the above technique will give us double answer.
That’s why it’s not sufficient at all.
So Answer is (A)
From equation (1), x/(10/a) + y/(10/b) = 1
Length of the Excerpt from X axis = 10/a
Length of the Excerpt from Y axis = 10/b
From statement (1) , a>b that’s why 10/a < 10/b
We know for this line, x = 10/a and y= 10/b (from the co-ordinate system)
That’s why Y>X .
For equation (2) we can have the same result using the same technique.
So statement (1) is sufficient.
From statement (2) when a is negative then b is also negative and a<b .
Or when a is positive then b is also positive and a>b .
Using this sort of data through the above technique will give us double answer.
That’s why it’s not sufficient at all.
So Answer is (A)
blueseas
Current Student
Joined: 14 Dec 2012
Last visit: 15 Jan 2019
Posts: 577
Own Kudos:
Given Kudos: 197
Location: India
Concentration: General Management, Operations
Schools: PGDM (GM) '15 (A)
GMAT 1: 700 Q50 V34
GPA: 3.6
Asifpirlo
ax + by=10
ay + bx=12
which one is greater here, x or y ?
(1) a > b
(2) a/b > 1
ay + bx=12
which one is greater here, x or y ?
(1) a > b
(2) a/b > 1
ax + by=10
bx + ay=12
subtracting these 2
(b-a)x + (a-b)y =2
or
(a-b)x - (a-b)y = -2
(a-b)(x-y) = -2
statement 1:
a>b and
(a-b)(x-y) = -2
(+)*(-)= -ve
therefore x-y<0
hence x<y
sufficient
statement 2:
a/b>1
(a-b)/b > 0
now a-b can be positive or negative
hence insufficient
hence A
28 Mar 2014, 06:11
Kudos
Bookmarks
After replying to a PM, I thought I have to make a few comments on this question.
This question is about comparing two numbers \(x\) and \(y\).
The best way to do this, is to look at the difference of the two:
if \(x - y = 0\), the two numbers are equal
if \(x - y > 0\), then \(x\) is greater than \(y\)
if \(x - y < 0\), then \(x\) is less than \(y\)
The "trap" here is the given system of the two equations. We don't need to solve it.
The two numbers are totally irrelevant, the only important thing is that 10 < 12.
Therefore, we can write that
\(ax + by < ay + bx\), or \(ax - ay - bx + by < 0\).
After factorisation we get
\(a(x - y) - b(x - y) = (x - y)(a - b) < 0\).
This means that \(x - y\) and \(a - b\) have opposite signs and that neither one can be 0.
From (1), \(a > b\) or \(a - b > 0\), so, necessarily \(x - y < 0\) and \(y\) is greater than \(x\).
Sufficient.
From (2), if \(b > 0\), then \(a > b\) (multiply the given inequality by \(b > 0\), direction preserved).
Then \(a - b > 0\), so \(x - y\) must be negative and \(x\) is smaller than \(y\).
If \(b < 0\), then \(a < b\) (multiply the given inequality by \(b < 0\), direction is reversed).
Then \(a - b < 0\), so \(x - y\) must be positive and \(x\) is greater than \(y\).
Not sufficient.
Hence, Answer A.
Remark:
When dealing with the inequality \(a/b > 1\), since \(b\) cannot be 0, we can multiply by \(b^2\), which is for sure positive and get
\(ab > b^2\)
This leads to \(ab - b^2\) > 0 or \(b(a - b) > 0\).
Now, we can easily see that if \(b > 0\), then \(a > b\).
When \(b < 0\), then \(a < b\).
Suggestion:
An inequality involving a fraction with unknown sign of the denominator can always be multiplied by the square of that
denominator, and the direction of the inequality is preserved.
This question is about comparing two numbers \(x\) and \(y\).
The best way to do this, is to look at the difference of the two:
if \(x - y = 0\), the two numbers are equal
if \(x - y > 0\), then \(x\) is greater than \(y\)
if \(x - y < 0\), then \(x\) is less than \(y\)
The "trap" here is the given system of the two equations. We don't need to solve it.
The two numbers are totally irrelevant, the only important thing is that 10 < 12.
Therefore, we can write that
\(ax + by < ay + bx\), or \(ax - ay - bx + by < 0\).
After factorisation we get
\(a(x - y) - b(x - y) = (x - y)(a - b) < 0\).
This means that \(x - y\) and \(a - b\) have opposite signs and that neither one can be 0.
From (1), \(a > b\) or \(a - b > 0\), so, necessarily \(x - y < 0\) and \(y\) is greater than \(x\).
Sufficient.
From (2), if \(b > 0\), then \(a > b\) (multiply the given inequality by \(b > 0\), direction preserved).
Then \(a - b > 0\), so \(x - y\) must be negative and \(x\) is smaller than \(y\).
If \(b < 0\), then \(a < b\) (multiply the given inequality by \(b < 0\), direction is reversed).
Then \(a - b < 0\), so \(x - y\) must be positive and \(x\) is greater than \(y\).
Not sufficient.
Hence, Answer A.
Remark:
When dealing with the inequality \(a/b > 1\), since \(b\) cannot be 0, we can multiply by \(b^2\), which is for sure positive and get
\(ab > b^2\)
This leads to \(ab - b^2\) > 0 or \(b(a - b) > 0\).
Now, we can easily see that if \(b > 0\), then \(a > b\).
When \(b < 0\), then \(a < b\).
Suggestion:
An inequality involving a fraction with unknown sign of the denominator can always be multiplied by the square of that
denominator, and the direction of the inequality is preserved.
