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M26-08

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M26-08  [#permalink]

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New post 16 Sep 2014, 01:24
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48% (02:14) correct 52% (02:40) wrong based on 179 sessions

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Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?

A. 5
B. 10
C. 15
D. 20
E. cannot be determined

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Re M26-08  [#permalink]

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New post 16 Sep 2014, 01:24
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Official Solution:

Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?

A. 5
B. 10
C. 15
D. 20
E. cannot be determined


Rearranging both expressions we'll get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have that \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\) then \((a-b)(a+b)=5*(a+b)=125\), which means that \(a+b=25\).

Thus we get two equations with two unknowns: \(a+b=25\) and \(a-b=5\). Solving for \(a\) gives \(a=15=\sqrt{5x}\) so, \(x=45\). Solving for \(b\) gives \(b=10=\sqrt{z-3y}\).

Finally, \(\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).


Answer: B
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Re: M26-08  [#permalink]

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New post 11 Oct 2016, 23:24
Bunuel wrote:
Official Solution:

Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?

A. 5
B. 10
C. 15
D. 20
E. cannot be determined


Rearranging both expressions we'll get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have that \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\) then \((a-b)(a+b)=5*(a+b)=125\), which means that \(a+b=25\).

Thus we get two equations with two unknowns: \(a+b=25\) and \(a-b=5\). Solving for \(a\) gives \(a=15=\sqrt{5x}\) so, \(x=45\). Solving for \(b\) gives \(b=10=\sqrt{z-3y}\).

Finally, \(\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).


Answer: B


Will sqrt(10^2) not yield +/- 10 ?
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Re: M26-08  [#permalink]

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New post 12 Oct 2016, 00:00
avdgmat4777 wrote:
Bunuel wrote:
Official Solution:

Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?

A. 5
B. 10
C. 15
D. 20
E. cannot be determined


Rearranging both expressions we'll get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have that \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\) then \((a-b)(a+b)=5*(a+b)=125\), which means that \(a+b=25\).

Thus we get two equations with two unknowns: \(a+b=25\) and \(a-b=5\). Solving for \(a\) gives \(a=15=\sqrt{5x}\) so, \(x=45\). Solving for \(b\) gives \(b=10=\sqrt{z-3y}\).

Finally, \(\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).


Answer: B


Will sqrt(10^2) not yield +/- 10 ?


No.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{16}=4\), NOT +4 or -4. In contrast, the equation \(x^2=16\) has TWO solutions, +4 and -4. Even roots have only a positive value on the GMAT.

Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
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Re: M26-08  [#permalink]

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New post 09 Sep 2017, 01:58
Is 10^2=100 not the same as x^2=100. In which case both would have two solutions.

Alternatively the sqrt of 100 only has one solution, 10
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Re: M26-08  [#permalink]

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New post 07 Jul 2018, 06:23
The square root of 125 isn't 5? i don't follow this step - shouldn't it be 5 root 5?
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Re: M26-08  [#permalink]

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New post 07 Jul 2018, 07:32
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Re: M26-08  [#permalink]

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New post 26 Jul 2018, 09:00
The equations we are given are:

Equation 1 - 5x - 125 = Z - 3y
Equation 2 -Sqroot(Z-3y) = Sqroot(5x) - 5

Squaring Equation 2 we get:
Equation 3: z - 3y = 5x - 10sqroot(5x) + 25

Substituting equation 1 into 3 we get:

5x - 125 = 5x - 10sqroot(5x) + 25 ---------> x = 45

We are asked to determine the value

Sqroot((45(Z-3y)/x))

X = 45 so it cancels out with the 45 in numerator.

Therefore only Sqroot(Z-3y) remains.

From Equation 2 we know that Sqroot(Z-3y) = Sqroot(5x) - 5

Since x = 45 ---> Sqroot(5* 45 ) - 5 = 10
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M26-08  [#permalink]

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New post 26 Jul 2018, 13:39
Bunuel wrote:
Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?

A. 5
B. 10
C. 15
D. 20
E. cannot be determined


I like Bunuel's method better, but here what I did.


(1) \(\sqrt{\frac{45(z-3y)}{x}}\) = \(\sqrt{\frac{45}{x}}\)*(\(\sqrt{5x}-5\))

Now.

\(\sqrt{5x}-5-\sqrt{z-3y}=0\). Square both parts:
(2) z-3y = 5x - 10\(\sqrt{5x}\) + 25

Then.

Given \(5x=125-3y+z\), substitute 5x into (2) --->
----> z - 3y= 125 - 3y + z - 10\(\sqrt{5x} + 25\) ---> 150-10\(\sqrt{5x}\) = 0 ---> x = 45.
Substitute in in (1)
\(\sqrt{\frac{45}{x}}\)*(\(\sqrt{5x}-5\)) = 10

Sorry about that - have to learn how to put these symbols right
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Re: M26-08  [#permalink]

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New post 01 Aug 2018, 15:14
This is a great question. Is this considered +700 question type?
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Re: M26-08  [#permalink]

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New post 20 Aug 2018, 11:16
this one is a monster!
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Re: M26-08 &nbs [#permalink] 20 Aug 2018, 11:16
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