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# M26-08

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Math Expert
Joined: 02 Sep 2009
Posts: 42280

Kudos [?]: 132904 [0], given: 12391

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16 Sep 2014, 01:24
Expert's post
5
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

35% (00:44) correct 65% (03:00) wrong based on 17 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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 132904 [0], given: 12391

Math Expert
Joined: 02 Sep 2009
Posts: 42280

Kudos [?]: 132904 [0], given: 12391

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16 Sep 2014, 01:24
Expert's post
1
This post was
BOOKMARKED
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, $$\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10$$.

_________________

Kudos [?]: 132904 [0], given: 12391

Manager
Joined: 17 Sep 2014
Posts: 60

Kudos [?]: 14 [0], given: 5

Location: India
GMAT 1: 710 Q49 V38
GPA: 3.65

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

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

Kudos [?]: 14 [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 42280

Kudos [?]: 132904 [0], given: 12391

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

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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Kudos [?]: 132904 [0], given: 12391

Intern
Joined: 03 May 2014
Posts: 14

Kudos [?]: [0], given: 3

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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

Kudos [?]: [0], given: 3

Re: M26-08   [#permalink] 09 Sep 2017, 01:58
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# M26-08

Moderators: Bunuel, chetan2u

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