Is ∣x∣<1?
1) x = 1/(3+y^2)
2) y =-2
The OA is the option A.
Why is sufficient the first statement? We don't know the value of y. <i class="em em-neutral_face"></i>
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Is ∣x∣<1 ?
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GMATGuruNY
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Statement 1:M7MBA wrote:Is ∣x∣<1?
1) x = 1/(3+y^2)
2) y =-2
Since the square of a value must be NONNEGATIVE, y² ≥ 0.
If y²=0, then x = 1/3.
If y²>0, then x = 1/(more than 3), implying that x < 1/3.
Since x≤1/3, the answer to the question stem is YES.
SUFFICIENT.
Statement 2:
No information about x.
INSUFFICIENT.
The correct answer is A.
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Hello M7MBA.
I will solve it as follows:
1) x = 1/(3+y^2)
We have the following equations: $$y^2\ge0\ \ and\ \ \ \ 0<1<3\ \ \Rightarrow\ \ \ 0<\ 1<3+y^2\ \ \ \Rightarrow\ \ 0<\frac{1}{3+y^2}<1$$ $$Since\ x=\frac{1}{3+y^2}\ \ \ \Rightarrow\ \ \ \ 0 < x < 1\ \ \ \Leftrightarrow\ \ \ \left|x\right| < 1.$$ Therefore, this statement is SUFFICIENT.
2) y =-2
This statement doesn't tell us anything about x. Therefore, is NOT SUFFICIENT.
The correct answer is the option A. <i class="em em---1"></i><i class="em em-sunglasses"></i>
I will solve it as follows:
1) x = 1/(3+y^2)
We have the following equations: $$y^2\ge0\ \ and\ \ \ \ 0<1<3\ \ \Rightarrow\ \ \ 0<\ 1<3+y^2\ \ \ \Rightarrow\ \ 0<\frac{1}{3+y^2}<1$$ $$Since\ x=\frac{1}{3+y^2}\ \ \ \Rightarrow\ \ \ \ 0 < x < 1\ \ \ \Leftrightarrow\ \ \ \left|x\right| < 1.$$ Therefore, this statement is SUFFICIENT.
2) y =-2
This statement doesn't tell us anything about x. Therefore, is NOT SUFFICIENT.
The correct answer is the option A. <i class="em em---1"></i><i class="em em-sunglasses"></i>
