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x=y^3 and y>1
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02 Jun 2020, 02:06
02 Jun 2020, 02:06
Expert Reply
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Question Stats:
32% (01:39) correct
67% (01:21) wrong based on 194 sessions
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\(x=y^3\) and \(y>1\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(x^y\) |
\(y^x\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: x=y^3 and y>1
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02 Jun 2020, 03:22
02 Jun 2020, 03:22
5
Expert Reply
This question picking numbers does not work very well
Better to use a general approach
We do know that \(x=y^3\)
Substitute in the two quantities
we do have
QA \(y^{{y{^3}}\)
QB \(y^{y+y+y}\)
We do have the same base so is the same to write
\(y^3>3y\)
\(y^3-3y>0\)
\(y(y^2-3)>0\)
Now
\(y>0\) which is false because we do know that \(y>1
\)
So we have to consider only \(y^2>3\)
Take the square root we do have that
\(y>1.7\)
We do know that \(y>1\)
However, we do have also \(y>1.7\)
Therefore y can be 1.1 or 2. Below 1.7 or greater than 1.7
The answer is D
Better to use a general approach
We do know that \(x=y^3\)
Substitute in the two quantities
we do have
QA \(y^{{y{^3}}\)
QB \(y^{y+y+y}\)
We do have the same base so is the same to write
\(y^3>3y\)
\(y^3-3y>0\)
\(y(y^2-3)>0\)
Now
\(y>0\) which is false because we do know that \(y>1
\)
So we have to consider only \(y^2>3\)
Take the square root we do have that
\(y>1.7\)
We do know that \(y>1\)
However, we do have also \(y>1.7\)
Therefore y can be 1.1 or 2. Below 1.7 or greater than 1.7
The answer is D
General Discussion
Re: x=y^3 and y>1
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02 Jun 2020, 02:07
02 Jun 2020, 02:07
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
