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# M26-08

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Math Expert
Joined: 02 Sep 2009
Posts: 51102

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16 Sep 2014, 00:24
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Difficulty:

95% (hard)

Question Stats:

48% (02:14) correct 52% (02:40) wrong based on 184 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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Math Expert
Joined: 02 Sep 2009
Posts: 51102

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16 Sep 2014, 00:24
4
6
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, $$\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10$$.

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11 Oct 2016, 22:24
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, $$\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10$$.

Will sqrt(10^2) not yield +/- 10 ?
Math Expert
Joined: 02 Sep 2009
Posts: 51102

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11 Oct 2016, 23:00
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, $$\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10$$.

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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Joined: 03 May 2014
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09 Sep 2017, 00:58
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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Joined: 02 Jan 2018
Posts: 27
Location: United Kingdom
GMAT 1: 740 Q50 V41
GPA: 3.7

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07 Jul 2018, 05:23
The square root of 125 isn't 5? i don't follow this step - shouldn't it be 5 root 5?
Math Expert
Joined: 02 Sep 2009
Posts: 51102

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07 Jul 2018, 06:32
londonconsultant wrote:
The square root of 125 isn't 5? i don't follow this step - shouldn't it be 5 root 5?

Where do we have the square root of 125? We have 5*(a+b)=125 --> a + b = 25...
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Joined: 20 Jan 2016
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Schools: HBS '18
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26 Jul 2018, 08:00
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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Migatte no Gokui

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Joined: 26 Jun 2017
Posts: 376
Location: Russian Federation
Concentration: General Management, Strategy
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26 Jul 2018, 12:39
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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01 Aug 2018, 14:14
This is a great question. Is this considered +700 question type?
Intern
Joined: 12 Jan 2017
Posts: 35

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20 Aug 2018, 10:16
this one is a monster!
Re: M26-08 &nbs [#permalink] 20 Aug 2018, 10:16
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# M26-08

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