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2^{-8}<x<2^{-7}
[#permalink]
13 Aug 2026, 13:19
13 Aug 2026, 13:19
Expert Reply
00:00
Question Stats:
71% (01:33) correct
28% (00:50) wrong based on 7 sessions
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\(2^{-8}<x<2^{-7} \)
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{1}{16x}\) |
\(8\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2928
Given Kudos: 1257
GPA: 3.39
2^{-8}<x<2^{-7}
[#permalink]
18 Aug 2026, 09:46
18 Aug 2026, 09:46
1
Expert Reply
Explanation
We know:
\(2^{-8}<x<2^{-7}\)
So:
\(\frac{1}{256}<x<\frac{1}{128}\)
We need to compare:
Quantity A=\(\frac{1}{16x}\)
Since x is between \(\frac{1}{256}\) and \(\frac{1}{128}\), multiply by \(\frac{1}{16}\):
1/[16(1/256)]>1/16x>1/[16(1/128)]
Thus: \(16>\frac{1}{16x}>8\)
Therefore: Quantity A is greater
Answer: A
We know:
\(2^{-8}<x<2^{-7}\)
So:
\(\frac{1}{256}<x<\frac{1}{128}\)
We need to compare:
Quantity A=\(\frac{1}{16x}\)
Since x is between \(\frac{1}{256}\) and \(\frac{1}{128}\), multiply by \(\frac{1}{16}\):
1/[16(1/256)]>1/16x>1/[16(1/128)]
Thus: \(16>\frac{1}{16x}>8\)
Therefore: Quantity A is greater
Answer: A
gmatclubot
2^{-8}<x<2^{-7} [#permalink]
18 Aug 2026, 09:46
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