Is 3 x 3 ? (1) x^2 + y^2 = 9 (2) x^2 + y 9

[BTGmoderatorDC]

$$Is\ -3\le x\le3?$$

(1) \(x^2 + y^2 = 9\)
(2) \(x^2 + y\le9\)


OA A

Source: Official Guide

[Brent@GMATPrepNow]

$$Is\ -3\le x\le3?$$

(1) \(x^2 + y^2 = 9\)
(2) \(x^2 + y\le9\)

Target question: Is -3 ≤ x ≤ 3?

Statement 1: x² + y² = 9
First we need to recognize that x² and y² are greater than or equal to 0 for all possible values of x and y
Considering this property, let's find the minimum and maximum possible values of x
In order to maximize the value of x we must minimize the value of y²
The minimum value of y² occurs when y = 0.
When y = 0, we get: x² + 0² = 9, which is the same as x² = 9, which means x = 3 or x = -3
For all other values of y, x will be BETWEEN 3 and -3
So, the answer to the target question is YES, it IS the case that -3 ≤ x ≤ 3
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x² + y ≤ 9
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 0 and y = 0. In this case, the answer to the target question is YES, it IS the case that -3 ≤ x ≤ 3
Case b: x = 4 and y = -10. In this case, the answer to the target question is NO, it is NOT the case that -3 ≤ x ≤ 3
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

[swerve]

$$Is\ -3\le x\le3?$$

(1) \(x^2 + y^2 = 9\)
(2) \(x^2 + y\le9\)


OA A

Source: Official Guide

Statement 1:

\(x^2+y^2=9\quad\Rightarrow\quad x^2=9-y^2\)

As \(y^2 \geq 0 \quad\Rightarrow\quad 0 \leq x^2 \leq 9 \quad\Rightarrow\quad -3 \leq x \leq 3\) Sufficient \(\,\Large{\color{green}\checkmark}\)

Statement 2

\(x^2+y=9\quad\Rightarrow\quad x^2=9-y\)

Case 1, when \(9 \geq y \geq 0\)

\(0 \leq x^2 \leq 9 \quad \Rightarrow \quad -3 \leq x \leq 3\)

Case 2, when \(y < 0\)

\(x^2 > 9 \quad \Rightarrow \quad x > 3\) or \(x < -3\) Not sufficient \(\,\Large{\color{red}\chi}\)

Therefore, __A__