Bunuel wrote:
Is rst > 1?
(1) rs > t > 1
(2) (rs)² + t² > rst
Target question: Is 1 < rstASIDE: I prefer to rearrange inequalities so that the right side is greater than the left side.
Statement 1: 1 < t < rsLet's use some
number sense.
This tells us that rs is greater than 1
IF t were to EQUAL 1, then rst would be greater than 1 (since rs is already greater than 1)
So, since t is GREATER THAN 1, we can be certain that rst is greater than 1
In other words,
1 < rstSince we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: rst < (rs)² + t² There are several values of r, s and t that satisfy statement 2. Here are two:
Case a: r = 2, s = 2 and t = 2. This satisfies the condition that rst < (rs)² + t². In this case
rst = (2)(2)(2) = 8. So, 1 < rst (i.e., rst is greater than 1)Case b: r = 1, s = 1 and t = 1. This satisfies the condition that rst < (rs)² + t². In this case
rst = (1)(1)(1) = 1. So, 1 = rst (i.e., rst is NOT greater than 1)Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Answer:
Cheers,
Brent
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