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Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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29 May 2015, 05:54
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Is \(x > y\)? (1) \(a*x^4 + a*y < 0\) (2) \(a*x^3 > a*y^3\)
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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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01 Jun 2015, 04:19
Bunuel wrote: Is x > y?
(1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3 OFFICIAL SOLUTION:Is x > y?(1) a*x^4 + a*y < 0. Factor out a: a(x^4 + y) < 0. Since x^4 + y = nonnegative + nonnegative = nonnegative, then for a*nonnegative to be negative, a must be negative. We know nothing about x and y. Not sufficient. (2) a*x^3 > a*y^3. If a is positive, then when reducing by it, we'd get x^3 > y^3, or which is the same x > y BUT if a is negative, then when reducing by negative value and flipping the sign, we'd get x^3 < y^3, or which is the same x < y. Not sufficient. (1)+(2) Since from (1) a < 0, then from (2) x < y. Sufficient, Answer: C.
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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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29 May 2015, 06:21
Hello all My attempt:Statement 1:\(a*x^4 + a*y < 0\) \(a*(x^4 + y) < 0\) Either \(a\) is \(ve\) or the \(expression\) in the brackets is \(ve\). But the \(expression\) in the bracket will always be \(+ve\) as \(x^4\) is \(+ve\) irrespective of the sign of \(x\) and \(y\) is always \(+ve\). The only conclusion from this statement is that \(a\) is \(ve\). Therefore standalone statement 1 can't answer the question. Statement 2:\(a*x^3 > a*y^3\) \(a*(x^3  y^3) > 0\) From this statement either \(a\) and the \(expression\) in the bracket are both \(+ve\) or both \(ve\) so we cannot form any relationship between \(x\) and \(y\). Therefore standalone statement 2 can't answer the question. Combining the statement 1 & 2:If \(a\) is \(ve\) then as per the second statement \((x^3  y^3)\) is also \(ve\) \(x^3  y^3 <0\) \(x^3 < y^3\) \(x < y\) Therefore we can answer by combining the two statements. Hence I will go with option \(C\)
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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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29 May 2015, 07:10
Ans B
Explanation :
(1) a*x^4 + a*y < 0
x^4+ y < 0 x^4 < y
Assuming x=2, y =1 > 16<1 (not possible) x=1,y=2 > 1<2 (not possible) x=1,y=2 > 1<2 (not possible) x=2,y=1 > 16<1 (not possible)
or
Another way to look at this is
x^4<y
on the left side of the equation the value will always be +ve on the right side the value will always be ve Hence, this inequality will always be false no matter what we do.
Therefore this statement is not sufficient
(2) a*x^3 > a*y^3
x^3>y^3 (eliminating a on both sides)
The above mentioned equation is possible only when x>y
Therefore 2nd statement is sufficient.



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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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29 May 2015, 08:36
Quote: Is x > y?
(1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3 My attempt: 1.) a*x^4 + a*y < 0 ==> a(x^4+y)<0 ==> We know, x^4+y will always be positive. ==>So, a < 0. Still we have nothing about x and y. So, Insufficient. (2) a*x^3 > a*y^3 ==> The result we get from this inequality depends on the value of a. if a>0, x^3>y^3 ==> x>y. But if a<0, y^3>x^3 ==> x<y. So, Insufficient. Now, combining both 1 and 2, we get, a < 0, and so x < y. So x is not greater than y. Hence (C).



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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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29 May 2015, 10:47
benjaminmishra wrote: Ans B
Explanation :
(1) a*x^4 + a*y < 0
x^4+ y < 0 x^4 < y
Assuming x=2, y =1 > 16<1 (not possible) x=1,y=2 > 1<2 (not possible) x=1,y=2 > 1<2 (not possible) x=2,y=1 > 16<1 (not possible)
or
Another way to look at this is
x^4<y
on the left side of the equation the value will always be +ve on the right side the value will always be ve Hence, this inequality will always be false no matter what we do.
Therefore this statement is not sufficient
(2) a*x^3 > a*y^3
x^3>y^3 (eliminating a on both sides)
The above mentioned equation is possible only when x>y
Therefore 2nd statement is sufficient. Hi benjaminmishra, Just a minor suggestion on your analysis. You have eliminated a from both sides of the inequality without taking into consideration its sign. You may want to have a relook at this step of your solution. Hope it helps Regards Harsh
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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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29 May 2015, 12:19
EgmatQuantExpert wrote: benjaminmishra wrote: Ans B
Explanation :
(1) a*x^4 + a*y < 0
x^4+ y < 0 x^4 < y
Assuming x=2, y =1 > 16<1 (not possible) x=1,y=2 > 1<2 (not possible) x=1,y=2 > 1<2 (not possible) x=2,y=1 > 16<1 (not possible)
or
Another way to look at this is
x^4<y
on the left side of the equation the value will always be +ve on the right side the value will always be ve Hence, this inequality will always be false no matter what we do.
Therefore this statement is not sufficient
(2) a*x^3 > a*y^3
x^3>y^3 (eliminating a on both sides)
The above mentioned equation is possible only when x>y
Therefore 2nd statement is sufficient. Hi benjaminmishra Just a minor suggestion on your analysis. You have eliminated a from both sides of the inequality without taking into consideration its sign. You may want to have a relook at this step of your solution. Hope it helps Regards Harsh Hi Harsh, Thanks for the suggestion ... taking a second look at Jackal's explanation makes a little more sense. Guess I was in a hurry to arrive at an answer



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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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30 May 2015, 12:46
If x^3 > y^3, is it necessary that x > y? The question does not say anything about x, y being integers.



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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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30 May 2015, 14:07
My option Answer C. The question looked simple, but it is very important to quantify all the variable a, x and y. From Option A; you can only say a= negative. we cannot determine X or Y. From Option B. Three primary solutions 1. if A>0; Then X> Y If 2. If A<0 and X<0 Then X<Y If A>0 and X>0 Then Y can be + or . Option B difficult to say. Both options together, you can say A=Ve, then X>Y. I used a full A4 paper to solve this question. God bless me on the real GMAT. Bunuel wrote: Is x > y?
(1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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30 May 2015, 19:30
Tina1785 wrote: If x^3 > y^3, is it necessary that x > y? The question does not say anything about x, y being integers. Hello Tina1785This is a nice query and I think all of us should be clear in our minds about this. I am pasting one handwritten note it might be helpful. Let us look at two broad cases \(Case I\): \(x^3<y^3\) \(Case II\): \(x^2<y^2\) These cases if you notice are the ones which confuse us. So let's look at it closer. Also let us assume only non zero real numbers between \(1\) and \(+1\). For numbers not in this range the solution also follows the same approach. \(Case I\): \(x^3<y^3\) In this case no matter what the variable \(x\) or \(y\) is. They will never cross the \(0\) line because as you guessed it right the sign is the reason. Look at the left side of the attached picture. If \(x<y\) then no matter what \(x^3<y^3\) will hold true. \(Case II\): \(x^2<y^2\) This is the case which creates confusion in our minds and rightly so. The sign changes when we square the variable and moreover the combined change in absolute value and sign may change the relationship between \(x\) and \(y\) in any way. Look at the right side of the attached diagram. In the first you see a clean inverse relationship, in the second same relationship but in the last two you will see that the absolute value matters. Conclusion:Don't be worried when the \(variables\) are raised to an \(odd\) \(power\) but be very careful when they are raised to \(even\) \(power\). p.s. the bottom left diagram has 0.009 it is supposed to be 0.027
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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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31 May 2015, 06:52
Thanks Jackal for taking the time to post the reply. It was helpful.



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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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Re: Is x > y? (1) a*x^4 + a*y < 0 (2) a*x^3 > a*y^3
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