If x 0, and |x| |y|, which of the following must be true?

Question

If x < 0, y > 0, and |x| > |y|, which of the following must be true? A. x > y B. y^2 > x^2 C. x^3 > y^2 D. –x < y E. x < –y

gmat6nplus1 · Accepted Answer

say x=-6 and y=3; say x=-0,5 and y=0,3. Plug them into the answer choices and the only one that must be true is E.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION: Correct Answer: (E) Let’s go through each answer choice: (A) can never be true, since no negative is greater than a positive. (B) doesn’t have to be true – consider what would happen if x = -2 and y = 1. (C) can never be true, as x^3 must be negative, and y^2 must be positive. (D) can never be true, since if x < 0, -x is the same thing as |x|, and |x| > y. (E) can be manipulated by multiplying both sides by -1, which gives us –x > y. Remember that x < 0, so –x = |x|, and y is positive, so |y| = y. Thus –x > y is the same statement as |x| > |y|, and (E) must be true.

sytabish · Answer

x<0, y>0, and |x| > |y|

The above statements mean that the magnitude of x is greater than the magnitude of y but x is negative.
So, if we make both of them -ve by changing the sign of y to -ve i.e. -y , then -y, which has a magnitude less than x, will be greater than x.
Hence (E).

U can also solve this problem fairly quickly if u plot x and y on a number line.
x will be at a greater distance from 0 on the left hand side than y on the right hand side.

Thanks!

chetan2u · Answer

the three inequalities tell us ..
 x is -ive and y is +ive..
 x is farther than y from 0 on a line..
so ans E

Atul11 · Answer

First, I will put the variables in a tabular format. Then try to plug in values to the variables which satisfy all the given conditions . Match the answer choices . e.g. 
X Y !X! !Y! -X -Y
-4 3 4 3 4 -3

From the answer choices, you can see that only option E matches. i.e. X<-Y

EMPOWERgmatRichC · Answer

Hi All,

This question involves Number Properties and can be solved by TESTing VALUES.

We're given three facts to work with:
1) X < 0
2) Y > 0
3) |X| > |Y|

We're asked "which of the following MUST be true", which is the same as asking "which of the following is ALWAYS TRUE no matter how many different examples we can come up with?" We can TEST VALUES to prove which of the answers is sometimes NOT true (and you'll likely find that the right pair of values can be used to eliminate MORE than one answer)....

Answer A: X > Y
IF...X = -3, Y = 2, then this answer is NOT true. Eliminate A.

Answer B: Y^2 > X^2
If...X = -3, Y = 2, then this answer is NOT true. Eliminate B.

Answer C: X^3 > Y^2
IF...X = -3, Y = 2, then this answer is NOT true. Eliminate C.

Answer D: -X < Y
If...X = -3, Y = 2, then this answer is NOT true. Eliminate D.

Since we've eliminated 4 answers, there only answer that's left must be the correct one.

Final Answer:

Show Spoiler

E

GMAT assassins aren't born, they're made,
Rich

dumbodingo · Answer

Used negation to get to E. Since x must be negative, A, B, C, and D are out.
But is the sign correct in option E? I understand that -(-x) > y is the same as |x| > |y|, but E says x < -y.

chetan2u · Answer

Hi.. Whenever you multiply both sides of INEQUALITY by '-', you change the sign of INEQUALITY. That is -x>y......-*-x.<-y....x<-y Reason is the properties of number when NEGATIVE is OPPOSITE of when POSITIVE. Larger the numeric value, larger the number 8>4 While when negative, larger the numeric value, smaller the number -8<-4 Example of -x>y -2>-7.....-*-2<-*-7....2<7

generis · Answer

Translate: x < 0 --> x is negative y > 0 --> y is positive |x| > |y| --> the "positive" value of x, -(-x), is greater than y. Magnitude of x > magnitude of y Or, x is more negative than y is positive: on the number line, x's distance from 0, to the left, is farther than y's distance from 0, to the right. Pick numbers and check answers. x = -3, y = 2 Plug in A. x > y: FALSE. -3 is not > 2 B. y^2 > x^2: FALSE. 4 is not > than 9 C. x^3 > y^2: FALSE. -27 is not greater than 4 D. –x < y. FALSE. -(-3) = 3. 3 is not less than 2 E. x < –y: TRUE. -3 is less than -2 Answer E Assess signs x = -3, y = +2 A. x > y: (-) > (+)?? NO B. y^2 > x^2. Integers raised to even powers = positive. Small (+) > Bigger (+)?? NO C. x^3 > y^2: (-) raised to odd power = negative. (-) > (+)?? NO D. –x < y. For (x), the negative, or opposite (which is one thing the "-" sign means), of a negative is positive. So bigger positive < smaller positive?? NO E. x < –y. For y, the negative of a positive is negative. More negative < less negative? YES. <-- --- ------- --> More to the left of zero = smaller Answer E

dave13 · Answer

Guten Tag Generis

Option D is not correct because if if x= -3 and y= 2 -3 < 2 then by multiplying by -1 i get 3 > - 2 and since y>0, hence it cant be true. Am i thinking correctly ?

thank you !

kaladin123 · Answer

I drew a number line.
 https://i.ibb.co/PD5KpCL/nl.jpg

x < -y

Hence E.

playthegame · Answer

I love graphs so I did a graph method below. You can also elimiate answer choices as you know that x is negative so -x > y that is the same as E. For graphical method let's say red line is graph of |x| and light blue line is graph of |y| in the second quadrant. You can see that when you draw the lines in all the quadrants you can get a sense of the two lines. So when x is less than 0 we need to look at the second and third quadrant hence you deduce that x is less than - y. I would not apply this method here as there is a simpler way out. The graphing helps for some other problems and helps to solve faster as you can do lot's with a simple graph. https://gmatclub.com/forum/download/file.php?id=81499 GMAT-Club-Forum-9e7v1l6q.png

RakshitaR · Answer

if it would have been

x^2 > y^2

then it would be the correct option?

Bunuel · Answer

Yes, because |x| > |y| when squared directly gives x^2 > y^2.

FATT · Answer

Please correct me if im wrong but we are given x<0 ; y>0 and |x| > |y| so plugging in what we know we can say that -x > y which become x <-y so option E

If x < 0, y > 0, and |x| > |y|, which of the following must be true?

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