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Variables x and y are non-zero integers. If x = 2y3 /z what happens to

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BrendaMSL
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Variables x and y are non-zero integers. If x = 2y3 /z what happens to [#permalink] New post   08 Aug 2020, 13:53
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Variables x and y are non-zero integers. If x = 2y^3 /z what happens to z when x is halved and y is doubled?
(a) z is not changed
(b) z is halved
(c) z is doubled
(d) z is multiplied by 4
(e) z is multiplied by 16


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E
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CrackverbalGMAT
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Re: Variables x and y are non-zero integers. If x = 2y3 /z what happens to [#permalink] New post   09 Aug 2020, 06:56
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WBogey wrote:
Hi CrackVerbalGMAT,
I think there is a typo in your answer. Please let me know if I'm because I think this changes the answer as well...

Greetingss
CrackVerbalGMAT wrote:
Given that x = \(\frac{y^3}{z}\) --> Should be x = \(\frac{2y^3}{z}\)


Heloo WBogey. There was an error in terms of copying the question while solving it. I took it as \(y^3\) instead of \(2y^3\), and have edited the same.

There is no change in the answer.

y becomes 2y and not 4y, because the question says \(2y^3\) and not \((2y)^3\).

If the question was \((2y)^3\), y would double and become 4y, but not in this case above.

Hope this helps.
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Variables x and y are non-zero integers. If x = 2y3 /z what happens to [#permalink] New post   Updated on: 09 Aug 2020, 06:50
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Given that x = \(\frac{2y^3}{z}\)

z = \(\frac{2y^3}{x}\)

x becomes x/2 and y becomes 2y

Substituting these in the above equation, we get \(\frac{2*(2y)^3}{\frac{x}{2}}\) = \(\frac{2 * 8y^3 * 2}{x}\) = 16 *\(\frac{2y^3}{x}\) = 16z

or z is multiplied by 16

Option E

Arun Kumar

Originally posted by CrackverbalGMAT on 08 Aug 2020, 19:38.
Last edited by CrackverbalGMAT on 09 Aug 2020, 06:50, edited 1 time in total.
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Variables x and y are non-zero integers. If x = 2y3 /z what happens to [#permalink] New post   09 Aug 2020, 05:23
Hi CrackVerbalGMAT,
I think there is a typo in your answer. Please let me know if I'm because I think this changes the answer as well...

Greetingss
CrackVerbalGMAT wrote:
Given that x = \(\frac{y^3}{z}\) --> Should be x = \(\frac{2y^3}{z}\)

z = \(\frac{y^3}{x}\)

x becomes x/2 and y becomes 2y 4y

Substituting these in the above equation, we get \(\frac{(2y)^3}{\frac{x}{2}}\) = \(\frac{8y^3 * 2}{x}\) = \(\frac{16y^3}{x}\) = 16z

or z is multiplird by 16

Option E

Arun Kumar
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Re: Variables x and y are non-zero integers. If x = 2y3 /z what happens to [#permalink] New post   24 Aug 2021, 09:33
\(x = \frac{2y^3 }{z}\)

\(z = \frac{2y^3 }{x}\)

x is halved and y is doubled
New \(z = \frac{2(2y)^3 }{x/2} = 16*\frac{2y^3 }{x} = 16z\)

Hence, OA is (E).
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Re: Variables x and y are non-zero integers. If x = 2y3 /z what happens to [#permalink]
24 Aug 2021, 09:33
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