BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Absolute value + Geometry

Expert replies
Source: — Data Sufficiency |

by 4GMAT_Mumbai » Tue Apr 20, 2010 3:08 am
Very interesting sum. Thanks Apoorva.

Statment 1:

If x < 0 and y < 0; then the LHS of the expression will be 0. Taking simple examples works. let x = -2 and y = -3.
If x < 0 and y > 0; then the LHS of the expression will be 0. let x = -2 and y = 3.
If x > 0 and y < 0; then the LHS of the expression will be 0. let x = 2 and y = -3.
[spoiler]Only IF x > 0 and y > 0; will the LHS be > 0. Hence, (x,y) lies in Quadrant 1.[/spoiler]

Statement 2:

-y < |y| implies that y is a +ve number.

If -x < a negative number; then x has to be a +ve number. [spoiler]Thus, (x,y) lies in Q1. [/spoiler]

Hence D

Hope this helps.
Join the discussion

by ajith » Tue Apr 20, 2010 4:18 am
apoorva.srivastva wrote:In which quadrant of the coordinate plane does the point (x, y) lie?

(1) |xy| + x|y| + |x|y + xy > 0
(2) -x < -y < |y|

please explain st 1.

OA is
D
1) there are 4 cases a) x and y positive b) x and y negative c) x positive and y negative d) x negative and y positive

In each of the cases a,b,c and d (x,y) will lie on a different quadrant

in the case a

1) is satisfied

In the case b) c) and d)

1) is not satisfied since the sum is ZERO

so 1) is good enough to conclude that the point is in the first quadrant

2) it is apparent that y is +ve and since -x< -y => x>y x also should be +ve
Sufficient too

D
Always borrow money from a pessimist, he doesn't expect to be paid back.
Join the discussion

by sk818020 » Tue Apr 20, 2010 8:59 am
The way I thought about (1) is:

You can rewrite |xy| + x|y| + |x|y + xy as:

( lxl + x ) ( lyl + y)

Which tells you:

If x is negative, then lxl + x = 0. If x positive, lxl + x = 2x.

and

If y is negative, then lyl + y = 0. If y is postive, lyl + y = 2y.

Thus, if ( lxl + x ) ( lyl + y) > 0, then neither y nor x can be negative because if they were ( lxl + x ) ( lyl + y) = 0.

Further, if x nor y is negative, then (x,y) must be in quadrant I (or top right quadrant) because they are both positive.

Hope that helps.
Join the discussion