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negative numbers equations..

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negative numbers equations..

by topspin360 » Sat Aug 04, 2012 2:31 pm
How do we do this one efficiently? Would graphical representation work in 2 mins? Testing numbers just takes way too long....



If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
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Source: — Data Sufficiency |
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by eagleeye » Sat Aug 04, 2012 2:54 pm
topspin360 wrote:How do we do this one efficiently? Would graphical representation work in 2 mins? Testing numbers just takes way too long....

If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4
This is how I did it:
We are told that both x and y are negative numbers.
We need to find if x<y => if x-y < 0.

With this in mind, let's look at the statements:


(1) 3x+4<2y+3
3x+4<2y+3
We need x-y, let's do a bit of juggling.
We have 3x+4<2y+3
=> 3x-2y < 3-4
=> 3x-3y < -1-y
=> 3(x-y) < -1-y
=> x-y < 1/3(-1-y). Now, we know that y is negative. If y is = -1, x-y < 0, If y < -1, it may be positive or negative. we can't say. Insufficient.


(2) 2x−3<3y−4
Let's do the same as above:

2x−3<3y−4
2x-3y <-4+3
=> 3x-3y < x-1
=> x-y < 1/3(x-1)
Since x is negative, RHS is -ve. Hence x-y <0. Sufficient.

B is correct. :)
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by topspin360 » Sun Aug 05, 2012 9:04 am
this is brilliant. is that how you usually tackle x<y or x>y questions? by looking at whether x-y is + or -?
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by eagleeye » Sun Aug 05, 2012 11:25 am
topspin360 wrote:this is brilliant. is that how you usually tackle x<y or x>y questions? by looking at whether x-y is + or -?
Yes.
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by GMATGuruNY » Mon Aug 06, 2012 4:33 am
topspin360 wrote:
If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4
Graphing is an efficient way to solve here.

Question rephrased: In quadrant III, is the graph represented by the statement above y=x?
Treat each statement as a LINE.
Calculate the the y-intercept and the point where y=x.

Statement 1: 3x+4<2y+3
3x+1 = 2y
y = (3/2)x + 1/2.

The y-intercept = 1/2.

When y=x:
x = (3/2)x + 1/2
(-1/2)x = 1/2
x=-1, implying that y=-1.

y > (3/2)x + 1/2 is the region ABOVE y = (3/2)x + 1/2.
Here's the graph:
Image
In quadrant III, y > (3/2)x + 1/2 -- represented by the blue region -- is sometimes ABOVE y=x and sometimes BELOW y=x.
INSUFFICIENT.

Statement 2: 2x−3<3y−4
2x+1 = 3y
y = (2/3)x + 1/3.
The y-intercept = 1/3.

When y=x:
x = (2/3)x + 1/3
(1/3)x = 1/3
x=1, implying that y=1.

y > (2/3)x + 1/3 is the region ABOVE y = (2/3)x + 1/3.
Here's the graph:
Image
In quadrant III, y > (2/3)x + 1/3 -- represented by the blue region -- will always be ABOVE y=x.
SUFFICIENT.

The correct answer is B.
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by thevenus » Tue Aug 07, 2012 5:10 am
sorry to say but i could not understand any of the methods, please help me experts.
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by GMATGuruNY » Tue Aug 07, 2012 11:16 am
thevenus wrote:sorry to say but i could not understand any of the methods, please help me experts.
For many test-takers, the easiest and most efficient approach will be to plug in.
If x and y are negative numbers, is x<y?
To make it easier to choose numbers, rephrase each statement so that x is in terms of y.
Since this problem is restricted to NEGATIVE NUMBERS, be sure to test 3 cases: an extreme value less than -1, -1 itself, and a negative fraction.

Statement 1: 3x+4<2y+3
Thus, x < (2y-1)/3.
If y=-10, then x < -7.
It's possible that x = -8, in which case x>y.
It's possible that x= -11, in which case x<y.
INSUFFICIENT.

Statement 2: 2x-3<3y-4
Thus, x < (3y-1)/2.
If y=-10, then x < -31/2.
If y=-1, then x < -2.
If y=-1/3, then x < -1.
Since x<y in each case, SUFFICIENT.

The correct answer is B.
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by topspin360 » Fri Aug 10, 2012 6:45 pm
Thanks for the graphs Mitch! That works.
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