For any numbers a and b, min(a, b) and max(a, b)

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For any numbers a and b, min(a, b) and max(a, b) represent the minimum and the maximum of a and b, respectively. If c > a, is a < b < c?

(1) min(max(a, b), c) = max(min(b, c), a)

(2) max(max(a, b), c) - min(min(b, c), a) > c - a

What's the best way to determine whether statement 1 is sufficient?

OA B

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by Jay@ManhattanReview » Tue Jan 23, 2018 11:42 pm
lheiannie07 wrote:For any numbers a and b, min(a, b) and max(a, b) represent the minimum and the maximum of a and b, respectively. If c > a, is a < b < c?

(1) min(max(a, b), c) = max(min(b, c), a)

(2) max(max(a, b), c) - min(min(b, c), a) > c - a

What's the best way to determine whether statement 1 is sufficient?

OA B
We are given that c > a.

We need to determine whether a < b < c.

(1) min(max(a, b), c) = max(min(b, c), a).

Case 1: Say a < b < c

min(max(a, b), c) = max(min(b, c), a) => min(b, c) = max(b, a) => b = b. The answer is Yes.

Case 2: Say a < c < b

min(max(a, b), c) = max(min(b, c), a) => min(b, c) = max(c, a) => c = c. The answer is No. No unique answer.

Insufficient.

(2) max(max(a, b), c) - min(min(b, c), a) > c - a

Case 1: Say a < b < c

max(max(a, b), c) - min(min(b, c), a) > c - a => max(b, c) - min(b, a) > c - a => c - a = c - a. This is an invalid case. Thus, it is not true that a < b < c.

Thus, the only case left is a < c < b. The answer is No. A unique answer. Sufficient.

Let's test this for the sake of clarity.

Case 2: Say a < c < b

max(max(a, b), c) - min(min(b, c), a) > c - a => max(b, c) - min(c, a) > c - a => b - a > c - a => b > c. This is a valid case. Thus, it is not true that a < b < c.

A third case is also possible. Let's test this for the sake of clarity.

Case 3: Say b < a < c

max(max(a, b), c) - min(min(b, c), a) > c - a => max(a, c) - min(b, a) > c - a => c - b > c - a => b < c. This is a valid case. Thus, it is not true that a < b < c. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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