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05 Jan 2022, 10:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
63% (01:24) correct
37%
(01:17)
wrong
based on 146
sessions
History
Date
Time
Result
Not Attempted Yet
Does a^b lie between 0 and 1, exclusive?
(1) |a| < 1
(2) b is a positive odd integer
(1) |a| < 1
(2) b is a positive odd integer
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
05 Jan 2022, 12:49
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Strategy: Since point 1) deals with "a", start here.
1) |a| < 1 means that "a" must be a fraction (either pos or neg - due to the absolute value bars). This validates that a is a fraction possibly between 0 and 1 but it also could be positioned between -1 and 0, depending on the value of b. INSUFFICIENT.
2) b is a positive odd integer. This does not tell us anything about "a". However, it does tell us that "a's" sign will remain constant (whether neg or pos). INSUFFICIENT.
1/2) We know that "a" is a fraction, either positive or negative and that "b" is an odd integer, which keeps its sign. Therefore, "a^b" could either be a positive fraction between 0 and 1 or a negative fraction between -1 and 0. INSUFFICIENT.
ANSWER = E.
1) |a| < 1 means that "a" must be a fraction (either pos or neg - due to the absolute value bars). This validates that a is a fraction possibly between 0 and 1 but it also could be positioned between -1 and 0, depending on the value of b. INSUFFICIENT.
2) b is a positive odd integer. This does not tell us anything about "a". However, it does tell us that "a's" sign will remain constant (whether neg or pos). INSUFFICIENT.
1/2) We know that "a" is a fraction, either positive or negative and that "b" is an odd integer, which keeps its sign. Therefore, "a^b" could either be a positive fraction between 0 and 1 or a negative fraction between -1 and 0. INSUFFICIENT.
ANSWER = E.
05 Jan 2022, 22:13
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(1) |a| < 1
|a| < 1 implies that
-1 < a < 1
For the sake for simplicity, lets assume b as an integer.
When b is positive and even
0 < \(a^b\) < 1
When b is positive and odd
0 < \(a^b\) < 1; when a is > 0
-1 < \(a^b\) < 0; when a is < 0
2) b is a positive odd integer
Doesn't provide any information on the value of a.
0 < \(a^b\) < +infinity; when a is > 1
-infinity < \(a^b\) < -1; when a is < 1
Combining we get -
0 < \(a^b\) < 1; when a is > 0
-1 < \(a^b\) < 0; when a is < 0
Hence E
|a| < 1 implies that
-1 < a < 1
For the sake for simplicity, lets assume b as an integer.
When b is positive and even
0 < \(a^b\) < 1
When b is positive and odd
0 < \(a^b\) < 1; when a is > 0
-1 < \(a^b\) < 0; when a is < 0
2) b is a positive odd integer
Doesn't provide any information on the value of a.
0 < \(a^b\) < +infinity; when a is > 1
-infinity < \(a^b\) < -1; when a is < 1
Combining we get -
0 < \(a^b\) < 1; when a is > 0
-1 < \(a^b\) < 0; when a is < 0
Hence E
