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If xy=y, |x|+|y|=?

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If xy=y, |x|+|y|=?

by Max@Math Revolution » Fri Oct 05, 2018 12:18 am

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[Math Revolution GMAT math practice question]

If xy=y, |x|+|y|=?

1) x=-1
2) y=0
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Source: — Data Sufficiency |
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If xy=y, |x|+|y|=?

by fskilnik@GMATH » Fri Oct 05, 2018 7:21 am

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Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If xy=y, |x|+|y|=?

1) x=-1
2) y=0
$$xy = y\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,y\left( {x - 1} \right) = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( * \right)\,\,\,\,\,\left\{ \matrix{
\,\,y = 0 \hfill \cr
\,\,{\rm{OR}} \hfill \cr
\,\,x = 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,$$
$$? = \left| x \right| + \left| y \right|$$
$$\left( 1 \right)\,\,\,x = - 1\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,y = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = \left| { - 1} \right| + \left| 0 \right|\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\,\,\,\,\,\,$$
$$\left( 2 \right)\,\,\,y = 0\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,0\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{1}}\,\, \hfill \cr} \right.\,$$

The correct answer is therefore (A).

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Sun Oct 07, 2018 5:21 pm

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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.

Modifying the original condition:
The equality xy=y is equivalent to y = 0 or x = 1 as shown below:

xy=y
=> xy-y=0
=> y(x-1)=0
=> y = 0 or x = 1

Since we have 2 variables (x and y) and 1 equation (xy=y), D is most likely to be the answer.


Condition 1)
Since x = -1 from condition 1) and y = 0 or x = 1 from the original condition, y = 0.
Thus, |x| + |y| = |-1| + |0| = 1.
Condition 1) is sufficient.

Condition 2)
If x = 1 and y = 0, then |x|+|y| = 1.
If x = 2 and y = 0, then |x|+|y| = 2.
Since we don't have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

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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