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x<y<z , where x, y and z are the lengths of the edges of a cuboid
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22 Aug 2026, 00:59
22 Aug 2026, 00:59
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100% (02:12) wrong based on 4 sessions
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x<y<z , where x, y and z are the lengths of the edges of a cuboid.
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 |
Volume of the cuboid with edges \( (x+12), y,(z+7)\) |
Volume of the cuboid with edges \( (x+7), y,(z+12)\) |
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
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Re: x<y<z , where x, y and z are the lengths of the edges of a cuboid
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30 Aug 2026, 10:15
30 Aug 2026, 10:15
1
Expert Reply
Solution
Since \(y\) is common to both, we only need to compare:
\(A=(x+12)(z+7)\) and \(B=(x+7)(z+12)\)
\(A=xz+7x+12z+84, B=xz+12x+7z+84\)
\(A-B=(7x+12z)-(12x+7z), A-B=5z-5x=5(z-x)\)
Given:
\(x<y<z\)
we know \(z>x\), so:
\(5(z-x)>0\)
Therefore,
Quantity A is greater
Since \(y\) is common to both, we only need to compare:
\(A=(x+12)(z+7)\) and \(B=(x+7)(z+12)\)
\(A=xz+7x+12z+84, B=xz+12x+7z+84\)
\(A-B=(7x+12z)-(12x+7z), A-B=5z-5x=5(z-x)\)
Given:
\(x<y<z\)
we know \(z>x\), so:
\(5(z-x)>0\)
Therefore,
Quantity A is greater
gmatclubot
Re: x<y<z , where x, y and z are the lengths of the edges of a cuboid [#permalink]
30 Aug 2026, 10:15
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