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If 0.005<k<0.02
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08 Aug 2026, 08:16
08 Aug 2026, 08:16
Expert Reply
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Question Stats:
36% (01:46) correct
63% (02:42) wrong based on 11 sessions
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If \(0.005<k<0.02 \)
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 |
\(\frac{0.02}{1-0.02 }\) |
\(\frac{k}{1-3k}\) |
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
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Re: If 0.005<k<0.02
[#permalink]
13 Aug 2026, 13:16
13 Aug 2026, 13:16
1
Expert Reply
Explanation
We need compare: \(A=\frac{0.02}{1-0.02}=\frac{0.02}{0.98}≈0.0204\)
For Quantity B:
\(B=\frac{k}{1-3k}\)
Since \(0.005<k<0.02,\) B increases as \(k\) increases. Its maximum occurs as k approaches 0.02:
\(B<\frac{0.02}{1-3(0.02)}=\frac{0.02}{0.94}≈0.0213\)
But B can be less than or greater than A.
For example:
\(k=0.01: B≈0.0103<A\)
\(k=0.019: B≈0.0201<A\)
\(k=0.0195: B≈0.02075>A\)
So the relationship cannot be determined.
Answer: D
We need compare: \(A=\frac{0.02}{1-0.02}=\frac{0.02}{0.98}≈0.0204\)
For Quantity B:
\(B=\frac{k}{1-3k}\)
Since \(0.005<k<0.02,\) B increases as \(k\) increases. Its maximum occurs as k approaches 0.02:
\(B<\frac{0.02}{1-3(0.02)}=\frac{0.02}{0.94}≈0.0213\)
But B can be less than or greater than A.
For example:
\(k=0.01: B≈0.0103<A\)
\(k=0.019: B≈0.0201<A\)
\(k=0.0195: B≈0.02075>A\)
So the relationship cannot be determined.
Answer: D
gmatclubot
Re: If 0.005<k<0.02 [#permalink]
13 Aug 2026, 13:16
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