MAAJ wrote:In the xy-plane, region R consist of all the points (x, y) such that 2x + 3y ≤ 6. Is the point (r, s) in region R?
(1) 3r + 2s ≤ 6
(2) r ≤ 3 and s ≤ 2
Not satisfied with the answer from the OG, particularly on the second statement... it's easier to solve by picking numbers but can some1 try to solve it by graphing? I'm curious about the graphic of statement 2...
Correct Answer [spoiler](E)[/spoiler]
While this looks like a geometry problem, it can be treated as an algebra problem. If (r,s) is in region R, then it must satisfy the conditions of the inequality. Thus, the question can be rewritten as:
Is 2r + 3s ≤ 6?
Statement 1: 3r + 2s ≤ 6
r=0 and s=3 works, because 3(0) + 2(3) ≤ 6. Is 2(0) + 3(3) ≤ 6? No.
r=2 and s=0 works, because 3(2) + 2(0) ≤ 6. Is 2(2) + 3(0) ≤ 6? Yes.
Since the answer can be both no and yes, insufficient.
Statement 2: r ≤ 3 and s ≤ 2
r=2 and s=0 works, because 2 ≤ 3 and 0 ≤ 2. Is 2(2) + 3(0) ≤ 6? Yes.
r=3 and s=2 works, because 3 ≤ 3 and 2 ≤ 2. Is 2(3) + 3(2) ≤ 6? No.
Since the answer can be both yes and no, insufficient.
Statements 1 and 2 together:
We saw above that r=2 and s=0 satisfy both statements. Is 2(2) + 3(0) ≤ 6? Yes.
To change the answer from yes to no, we need to maximize r and s. The largest combination that satisfies both statements is
r=2/3 and s=2. Is 2(2/3) + 3(2) ≤ 6? No.
Since the answer can be both yes and no, insufficient.
The correct answer is
E.
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