[Math Revolution GMAT math practice question]
Is x^2>xy?
1) x>y
2) y>0
Is x^2>xy?
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- Max@Math Revolution
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$${x^2}\,\,\mathop > \limits^? \,\,\,xy\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\left( {x - y} \right)\,\,\mathop > \limits^? \,\,\,0$$Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is x^2>xy?
1) x>y
2) y>0
$$\left( 1 \right)\,\,x > y\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,y > 0\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,x > y > 0\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,x > 0 \hfill \cr
\,x - y > 0 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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x² > xy is true only if x is NONZERO, in which case we can safely divide by x², since the square of a nonzero value must be positive:Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is x^2>xy?
1) x>y
2) y>0
x²/x² > (xy)/x²
1 > y/x
Question stem, rephrased:
Is y/x < 1?
Statement 1: x > y
If x=2 and y=1, then y/x =1/2. so the answer to the rephrased question stem is YES.
If x=-1 and y=-2, then y/x = 2, so the answer to the rephrased question stem is NO.
INSUFFICIENT.
Statement 2:
No information about x.
INSUFFICIENT.
Statements combined:
Since x > y > 0, x is POSITIVE, enabling us to safely simplify x > y by dividing both sides by x:
x/x > y/x
1 > y/x
y/x < 1
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.
The correct answer is C.
Last edited by GMATGuruNY on Tue Nov 06, 2018 2:52 am, edited 2 times in total.
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The reasoning that comes pre-statements (shown in red) is not correct.GMATGuruNY wrote:Since x² > xy implies that x is NONZERO, we can safely divide by x², which must be equal to a POSITIVE VALUE:Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is x^2>xy?
1) x>y
2) y>0
x²/x² > (xy)/x²
1 > y/x
Question stem, rephrased:
y/x < 1?
Statement 1: x > y
If x=2 and y=1, then y/x =1/2. so the answer to the rephrased question stem is YES.
If x=-1 and y=-2, then y/x = 2, so the answer to the rephrased question stem is NO.
INSUFFICIENT.
Statement 2:
No information about x.
INSUFFICIENT.
Statements combined:
Since x > y > 0, x is POSITIVE, enabling us to safely simplify x > y by dividing both sides by x:
x/x > y/x
1 > y/x
y/x < 1
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.
The correct answer is C.
In other words, the original question stem and the question stem rephrased (proposed) are not equivalent.
I explain: x CAN be zero (only in the original question stem) and, being so, the question asked is answered in the negative (for any real value of y that satisfies the corresponding statement).
More explicitly:
(1) (x,y) = (0,-1) is a possible particular case for which the answer to the question asked (in the original question stem) is NO.
(2) (x,y) = (0,1) is a possible particular case for which the answer to the question asked (in the original question stem) is NO.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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For the purposes of a Y/N DS problem, the rephrase is the equivalent of the original question stem.fskilnik@GMATH wrote:The reasoning that comes pre-statements (shown in red) is not correct.GMATGuruNY wrote:Since x² > xy implies that x is NONZERO, we can safely divide by x², which must be equal to a POSITIVE VALUE:Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is x^2>xy?
1) x>y
2) y>0
x²/x² > (xy)/x²
1 > y/x
Question stem, rephrased:
y/x < 1?
In other words, the original question stem and the question stem rephrased (proposed) are not equivalent.
Original question stem: Is x² > xy?
The answer will be YES if y/x is less than 1..
The answer will be NO if y/x is NOT less than 1.
Thus, the original question stem can be rephrased as follows:
Is y/x < 1?
If x=0, the answer is NO, since y/x is undefined and thus not less than 1.
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- Max@Math Revolution
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=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
x^2>xy
=> x^2-xy > 0
=> x(x-y) > 0
=> x>0, x>y or x<0, x<y
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since we have x > y and x > 0 from x > y > 0 which are a combined inequality of both conditions, both conditions together are sufficient.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1) x > y
If x = 2, y = 1, then we have x^2>xy and the answer is 'yes'.
If x = -1, y = -2, then we have x^2<xy and the answer is 'no'.
Condition 2) y > 0
If x = 2, y = 1, then we have x^2>xy and the answer is 'yes'.
If x = 1, y = 2, then we have x^2<xy and the answer is 'no'.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
x^2>xy
=> x^2-xy > 0
=> x(x-y) > 0
=> x>0, x>y or x<0, x<y
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since we have x > y and x > 0 from x > y > 0 which are a combined inequality of both conditions, both conditions together are sufficient.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1) x > y
If x = 2, y = 1, then we have x^2>xy and the answer is 'yes'.
If x = -1, y = -2, then we have x^2<xy and the answer is 'no'.
Condition 2) y > 0
If x = 2, y = 1, then we have x^2>xy and the answer is 'yes'.
If x = 1, y = 2, then we have x^2<xy and the answer is 'no'.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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