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Bunuel
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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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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This is a great question. Is this considered +700 question type?
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this one is a monster!
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I think this is a high-quality question and I agree with explanation.
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love this official solution!
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Hello, I have arrived at correct solution by using second equation only. Am I missing something?

SqRoot (z-3y)= SqRoot(5x)-5

Substituting this in question

SqRoot(45/x) X (SqRoot(5x)-5)
SqRoot(225)-5
=10
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Hello, I have arrived at correct solution by using second equation only. Am I missing something?

SqRoot (z-3y)= SqRoot(5x)-5

Substituting this in question

SqRoot(45/x) X (SqRoot(5x)-5)
SqRoot(225)-5

=10

Yes. Check again the red part.

\(\sqrt{ \frac{45}{x} } *(\sqrt{5x}-5)\) does not equal to \(\sqrt{225}-5\)
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I think this is a high-quality question.
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how is the root of 125 be equal to 5
Bunuel
Official Solution:

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

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


By rearranging both expressions, we get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Let's denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have \(a^2-b^2=125\) 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 implies that \(a+b=25\).

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

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


Answer: B
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how is the root of 125 be equal to 5
Bunuel
Official Solution:

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

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


By rearranging both expressions, we get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Let's denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have \(a^2-b^2=125\) 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 implies that \(a+b=25\).

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

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


Answer: B

The solution does not say that the square root of 125 is 5.

It rearranges \(5x=125-3y+z\) to get \(5x-(z-3y)=125\) and rearranges \(\sqrt{5x}-5-\sqrt{z-3y}=0\) to get \(\sqrt{5x}-\sqrt{z-3y}=5\).
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What's the intuition for rearranging and assuming a and b values?
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What's the intuition for rearranging and assuming a and b values?
The two equations contain similar expressions: 5x and z - 3y in the first, and √(5x) and √(z - 3y) in the second. So we first rearrange both to get those expressions on one side:

Since the second equation has the square roots of the expressions in the first, we recognize that the root terms are square roots of those like terms. So we assign a = √(5x) and b = √(z - 3y) to connect the two equations using a^2 and b^2.
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