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If f(x)=x+√x and z=y^2, is f(z)=y^2+y?

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If f(x)=x+√x and z=y^2, is f(z)=y^2+y? [#permalink]

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New post 20 Dec 2017, 00:59
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[GMAT math practice question]

If \(f(x)=x+√x\) and \(z=y^2\), is \(f(z)=y^2+y\)?

1) \(z=4\)
2) \(y>0\)
[Reveal] Spoiler: OA

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Re: If f(x)=x+√x and z=y^2, is f(z)=y^2+y? [#permalink]

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New post 20 Dec 2017, 04:14
f(z)=z+z^1/2
z=y^2
z^1/2=|y|
substituting we get
case 1: y>0
f(z)=y^2+y
case 2: y<0
f(z)=y^2-y

So reasoning boils down to reveal whether y is +\-ve.

(1)z=4
f(z)=z+z^1/2=4+2=6
|y|=z^1/2=|2|==+\- 2
case 1: y=2
f(z)=y^2+y^1/2
6=4+2=6 OK
case 2: y=-2
6=4-2=2 no solution
Insufficient
(2)
y>0 Sufficient

Answer B
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If f(x)=x+√x and z=y^2, is f(z)=y^2+y? [#permalink]

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New post 22 Dec 2017, 05:43
=>

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, and then recheck the question.

Modifying the question:
\(f(z)=y^2+y\)
\(⇔ z + \sqrt{z} = y^2 + y\)
\(⇔ y^2 + \sqrt{y^2} = y^2 + y\)
\(⇔ \sqrt{y^2} = y\)
\(⇔ |y| = y\)
\(⇔ y ≥0\)

Since condition 1) tells us nothing about y, and condition 2) tells us that y > 0, only condition 2) is sufficient.

Therefore, the answer is B.

Answer: B
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If f(x)=x+√x and z=y^2, is f(z)=y^2+y?   [#permalink] 22 Dec 2017, 05:43
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