If |x| = |2y| , what is the value of x - 2y?
(1) x + 2y = 6
(2) xy>0
Modulus
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Dear Deepthi Subbu,Deepthi Subbu wrote:If |x| = |2y| , what is the value of x - 2y?
(1) x + 2y = 6
(2) xy>0
I'm happy to help with this.
From the prompt information, we have no idea whether both x & y are positive, both negative, or one positive and one negative.
Statement #1: x + 2y = 6
If x & y are both positive, then x = 2y, so x + 2y = x + x = 6 ----> x = 2y = 3, and x - 2y = 0
If one is positive and one is negative, then x = - 2y, and then x + 2y would have to have a sum of zero, and it would not be possible to equal 6.
If both are negative, the sum x + 2y would be negative, and it would not be possible to equal 6.
Thus, this statement requires that x & y are both positive, and therefore we can calculate the numerical value of x - 2y. This statement, alone and by itself is sufficient.
Statement #2: (xy) > 0
This tells us, x & y are either both positive or both negative. We know x = 2y, which means x - 2y = 0. Voila! This statement, alone and by itself is sufficient
Answer = D
Does this make sense?
Mike
Last edited by Mike@Magoosh on Fri Aug 23, 2013 12:34 pm, edited 1 time in total.
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Statement 1: x + 2y = 6Deepthi Subbu wrote:If |x| = |2y| , what is the value of x - 2y?
(1) x + 2y = 6
(2) xy>0
Thus, x = 6-2y.
Substituting x = 6-2y into |x| = |2y|, we get:
|6-2y| = |2y|.
Case 1: Signs unchanged
6-2y = 2y
6 = 4y
y = 3/2.
Thus, x = 6-2y = 6 - 2 * (3/2) = 3.
In this case, x-2y = 3 - 2 * (3/2) = 0.
Case 2: Signs of one side changed
-6+2y = 2y
-6=0.
Not possible.
Since only Case 1 is possible, x-2y=0.
SUFFICIENT.
Statement 2: xy>0
In other words, x and y have the SAME SIGN.
We must also satisfy the constraint that |x| = |2y|.
If y=1, then x=2.
In this case, x-2y = 2 - (2*1) = 0.
If y=-1, then x=-2.
In this case, x-2y = -2 - (2 * -1) = 0.
I'm almost convinced: x-2y = 0.
One more case to be safe:
If y=100, then x=200.
In this case, x-2y = 200 - (2*100) = 0.
Since x-2y = 0 in every case, SUFFICIENT.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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