Is x y? (1) x y^3 (2) x/y 1

Question

Is x > y?

(1) x > y^3

(2) x/y > 1

IanStewart · Answer

When x and y are positive, we can easily get a 'yes' answer to the question (if x = 4 and y = 1, say). But x and y can be negative, and then we can get a 'no' answer to the question. If we have x = -4, and y = -2, then x/y > 1, and x > y^3, but x is not greater than y. So the answer is E.

kunalbean · Answer

Hi! Why is B wrong? If X/Y > 1 it should mean that x > y ?If x/y > 0 then we could have said that there can be values between 0 and 1?

IanStewart · Answer

x/y > 1 does not mean x > y. I gather you're multiplying by y on both sides of the inequality. But you can't do that unless you know whether y is positive or negative. If y is positive, then we can multiply by y and we do not need to flip the inequality. Then it is true that x > y. But if y is negative, then when you multiply by y on both sides, you need to flip the inequality. So if x/y > 1 is true, and y < 0 is true, then x < y is true. You can confirm that by testing any negative numbers at all -- let x = -3 and y = -1, for example. Then x/y > 1 is true, and x < y is true.

kunalbean · Answer

Yes. didnt think of 1 as a possibility .

firas92 · Answer

(1) x > y^3

If x=-1, y=-2

x>y and x>y^3

If x=-3, y=-2

x<y and x>y^3

1 is insufficient

(2) x/y > 1

If x=-2 and y=-1

x<y and x/y>1

If x=2 and y=1

x>y and x/y>1

2 is insufficient

(1)+(2)

If x=-3, y=-2

x<y and x>y^3 and x/y>1

If x=2, y=1

x>y and x>y^3 and x/y>1

(1)+(2) is insufficient

Answer is (E)

Posted from my mobile device

hiranmay · Answer

Is x > y?

(1) x > y^3 --> insuff: if x = 10 & y =2, then x >y^3 & x> y: yes. But if x = 0.1 & y =0.2, then x >y^3 & x< y: no

(2) x/y > 1--> insuff: x/y > 1 => xy/y^2>1 => xy > y^2=> xy-y^2>0=>y(x-y)> 0, so y> 0 & x-y>0 or y<0 & x-y <0 => 0<y<x i.e if y> 0: yes, but x<y<0 i.e if y< 0: no

combining (1) & (2) also, we can say if y> 0 also, we can say from case-1, that it's insuff.
Answer: E

crackGMAT760 · Answer

Bunuel   VeritasKarishma , could you please help me in how to approach questions like these? I get confused in choosing values to test the options.

Thank you.

KarishmaB · Answer

Use a combination of approaches, not just number plugging. Use some logic and some number plugging. Also, keep in mind some number properties - how x, x^2, sqrt(x), x^3 are related in diff regions of the number line. Is x > y? (1) x > y^3 This tells us that x is greater than y^3. Is it necessary that y^3 must be greater than y? Well it will be in case y > 1 but not otherwise. So not sufficient to say whether x is greater than y since we don't have any range of values for y. (2) x/y > 1 Easy to see that if we multiply both sides by y, we get x > y but only if y is positive. If y is negative, this relation will reverse to x < y. So again not sufficient to say whether x > y since we don't have any range of values for y. So we see that if say y = 10, x will be more than 1000. Answer = Yes. But if y is say -10, x could be -11. Answer = No. Answer (E) Also check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -question/ Note, ?x = sqrt(x) in the post. There is some formatting issue.

sa119 · Answer

Hi, you can see this recording of a session conducted by crackverbal on acing inequalities. I really found it helpful. Link: https://youtu.be/j-jGsD2dweI Posted from my mobile device

bumpbot · Answer

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Is x > y? (1) x > y^3 (2) x/y > 1

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