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If xy^2 = 12 and xy = 4, then x =

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If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 01 May 2018, 05:23
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

85% (00:44) correct 15% (01:41) wrong based on 62 sessions

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Re: If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 01 May 2018, 05:46
Bunuel wrote:
If \(xy^2 = 12\) and \(xy = 4\), then x =


A. 1

B. 2

C. \(\sqrt{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{4}{3}\)


\(xy^2 = 12\) and \(xy = 4\)
now replace \(xy = 4\) in \(y*xy\) = 12
y = 3
Replacing y in \(xy\)= 12

X = \(\frac{4}{3}\)

Option E
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Re: If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 01 May 2018, 05:51
Bunuel wrote:
If \(xy^2 = 12\) and \(xy = 4\), then x =


A. 1

B. 2

C. \(\sqrt{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{4}{3}\)



Given : \(xy^2 = 12\) or \(xy*y = 12\) -----(I)

&

\(xy = 4\)-----(II)

Using I and II

\(4y = 12\)

\(y = 3\)

Therefore , \(x (3)^2 = 12\)

\(x = \frac{12}{9}\)

\(x = \frac{4}{3}\)

(E)
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If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 02 May 2018, 08:13
Bunuel wrote:
If \(xy^2 = 12\) and \(xy = 4\), then x =


A. 1

B. 2

C. \(\sqrt{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{4}{3}\)

\(xy^2 = 12\) and \(xy=4\)

\(x*y*y = 12\), i.e., \((x*y)*y=12\)
\(4*y=12\)
\(y=3\)

\(x=?\)
\(xy=4\), and \(y=3\)
\(x*3=4\)
\(x =\frac{4}{3}\)

Answer E
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Re: If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 02 May 2018, 10:27
Bunuel wrote:
If \(xy^2 = 12\) and \(xy = 4\), then x =


A. 1

B. 2

C. \(\sqrt{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{4}{3}\)


Let’s first establish that both x and y must be positive. Since xy^2 = 12, we see that x must be positive. And since xy = 4, we see that y must also be positive.

Dividing the first equation by the second, we have:

y = 3

Substituting y = 3 into the second equation, we have:

x(3) = 4

x = 4/3

Alternate Solution:

Let’s rewrite the first equation as:

x(xy) = 12

We are given that xy = 9, so we can substitute:

x(9) = 12

x = 12/9 = 4/3

Answer: E
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Re: If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 20 Aug 2018, 03:04
Bunuel wrote:
If \(xy^2 = 12\) and \(xy = 4\), then x =


A. 1

B. 2

C. \(\sqrt{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{4}{3}\)



1.divide:
\(xy^2 = 12\)
/\(xy = 4\)
y=3
2. substitute:
x=4/3 => E
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Re: If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 20 Aug 2018, 06:55
Bunuel wrote:
If \(xy^2 = 12\) and \(xy = 4\), then x =


A. 1

B. 2

C. \(\sqrt{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{4}{3}\)


\(y = \frac{xy^2}{xy}\)

So, \(y = \frac{12}{4}\)

Or, \(y = 3\)

We know that \(xy^2 = 12\)

Or, \(9x = 12\)

Or, \(x = \frac{4}{3}\), Hence Answer must be (E)
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Re: If xy^2 = 12 and xy = 4, then x =  [#permalink]

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New post 20 Aug 2018, 09:10
\(xy^2 = 12\)
\(xy*y = 12\)
\(4y = 12\)
\(y=3\)
so
\(3x=4\)
\(x=\frac{4}{3}\)
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Re: If xy^2 = 12 and xy = 4, then x = &nbs [#permalink] 20 Aug 2018, 09:10
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