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.

this is brilliant. is that how you usually tackle x<y or x>y questions? by looking at whether x-y is + or -?

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.

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:

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:

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.

sorry to say but i could not understand any of the methods, please help me experts.

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.

Thanks for the graphs Mitch! That works.