In which quadrant of the coordinate plane does the point (x,y) lie?

OFFICIAL SOLUTION:

In which quadrant of the coordinate plane does the point (x,y) lie?

(1) xy = 7. This implies that x and y are either both negative or both positive, thus (x,y) lies either in III quadrant or in I quadrant. Not sufficient.

(2) 5x + 7y < -1/2. Both x and y can be negative or one of them can be negative and another positive. Not sufficient.
Notice that from this statement it's NOT possible both x and y to be positive: 5x + 7y = positive + positive = positive, not negative number -1/2.

(1)+(2) Since (2) rules out possibility of both x and y being positive, then from (1) we are left with negative x and y: (x,y) lies either in III quadrant. Sufficient.

Answer: C.
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Stmt 1: xy = 7

Since the product of x and y coordinates is +7, x and y must be both positive or both negative.
So point (x, y) will be either in the first quadrant(where x and y are both positive)or third quadrant (where x and y both are negative)
Not suff

Stmt 2: 5x + 7y < -1/2
Either x or y or both has to be negative to satisfy the above equation which means point (x,y) can be in any quadrant other than quadrant I.
Three cases can happen
x is negative y is positive. (Eg: x =-5 and y =1) In this case (x,y) will be in quadrant II
x is negative y is negative. (Eg: x =-1 and y = -1) In this case (x,y) will be in quadrant III
x is positive y is negative. (Eg: x = 1 and y = -1) In this case (x,y) will be in quadrant IV.

Not suff

Combining stmt 1 and 2:
From 1: Point (x, y) will be either in the first quadrant(where x and y are both positive)or third quadrant (where x and y both are negative)
From 2: When x and y are both positive 5x + 7y is always positive (So it does not satisfy 5x + 7y < -1/2) So (x,y) can never be in quadrant I.
It has to be in quadrant III.
Suff


Answer C

1) xy = 7, x and y can both be positive or negative, i.e. they can both be in 1st quadrant or 3rd quadrant ---- Insufficient

2) 5x + 7y < -1/2 => 10x + 14y < -1, this can be true if both x and y are negative or only y is negative ---- Insufficient

Combining 1 & 2, x and y both must have to be negative to satisfy xy = 7 and 10x + 14y < -1 => They have to be in third quadrant ---- Sufficient


Hence answer is C

Hi all

My attempt:

xy=7 is a hyperbola
and
5x + 7y = -1/2 is a line. For the inequality 5x + 7y < -1/2 the concerned area is given as per the attached diagram.

Statement 1: the quadrant could be I or III. Not solvable.
Statement 2: the quadrant could be II or III or IV. Not solvable.

But combining the two statements the solution lies on the hyperbola curve in the third quadrant only. Therefore solvable. Thus answer is C.

p.s. please note that when we substitute y=7/x in the equation of the line we get a quadratic equation whose discriminant is negative and thus no real roots exist meaning that the line does not intersect the hyperbola.

edited because of a silly mistake

(1) xy = 7
x=1
y=7
Quadrant I

x=-1
y=-7
Quadrant III

Insufficient

(2) 5x + 7y < -1/2
This graph covers Quadrants II, III, IV
Insufficient

Combined,
when x=-7 and y=-1 , 5x + 7y < -1/2 is also true. This lies in Quadrant III. other points work as well (-2, -7/2)
Sufficient

Answer: C

Ans: C

Solution:
1) xy=7 means there are two possibilities (x, y) = (+,+) OR (-,-), Hence it can be either first or third Q. [Not Sufficient]
2) 5x + 7y < -1/2 line which satisfy this equation contains points from 2nd, 3rd and 4th Q. [Not Sufficient]

Combining both: Common ans is Third Q. [Sufficient]
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1. xy=7 ; x & both can be negative or positive. and as per graph its an hyperbola lying in 1st and 3rd quadrant; so insufficient
2. 5x+7y< -1/2 ; the value/range of this line lies in 2nd, 3rd & 4th quadrant, so not sufficient

1+2 only 3rd quadrant the common range/area will be. so (x,y) will be in 3rd quadrant. sufficient. Hence answer is C

Thanks,

Bunuel what do we say to the point lying on any on the axes? if point lies on any of the axes then we say that point doesnt lie on any quadrant right?

You won't need this for the GMAT but points on the axes do not belong to any quadrant, they lie on the quadrant boundaries.
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