If a, b, and c are nonzero integers and z = b^c>, is a^z negative?
(1) abc is an odd positive number.
(2) | b + c | < | b | + | c |
Isn't Statement 1 sufficient?
OA C
If a, b, and c are nonzero integers
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Statement 1: abc is an odd positive numberlheiannie07 wrote:If a, b, and c are nonzero integers and z = b^c, is a^z negative?
(1) abc is an odd positive number.
(2) | b + c | < | b | + | c |
Case 1: a=1, b=1, and c=1, with the result that abc = (1)(1)(1) = 1 and that z = b^c = 1¹ = 1
In this case, a^z = 1¹ = 1, so the answer to the question stem is NO.
Case 2: a=-1, b=-1, and c=1, with the result that abc = (-1)(-1)(1) = 1 and that z = b^c = (-1)¹ = -1
In this case, a^z = (-1)¯¹ = -1, so the answer to the question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.
Statement 2: | b + c | < | b | + | c |
No information about a.
INSUFFICIENT.
Since an absolute value cannot be negative, both sides of the inequality in Statement 2 are NONNEGATIVE.
Thus, we can safely square both sides:
(| b + c |)² < (| b | + | c |)²
b² + c² + 2bc < b² + c² + 2|b||c|
bc < |b||c|.
The resulting inequality holds true only if b and c have DIFFERENT SIGNS.
Thus, bc < 0.
Statements combined:
Since bc < 0 and abc = odd positive number, we get two cases:
Case 1: a = negative odd integer, b = negative odd integer, c = positive odd integer
Case 2: a = negative odd integer, b = positive odd integer, c= negative odd integer
Since in each case b and c are both odd, we get:
z = b^c = (odd integer)^(odd integer) = NOT EVEN.
Thus:
a^z = (negative)^(non-even exponent) = negative.
SUFFICIENT.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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