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605-655 (Medium)|   Algebra|      
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rahulp11
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If x > 0 and y > 0, which of the following is equal to 21 Feb 2022, 04:10
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

62% (01:33) correct 38% (01:48) wrong based on 216 sessions

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If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann
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E
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chetan2u
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If x > 0 and y > 0, which of the following is equal to 21 Feb 2022, 06:35
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rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann
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I. Algebraic

Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)


\((\frac{1}{\sqrt{x}+ \sqrt{x+y}})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)



II. Substitute some value for x and y.
Take x as 9 and y as 16..

\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)

(A)\(\frac{1}{y}=\frac{1}{16}\)….No
(B)\(\sqrt{2x + y}\)…..Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)…No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)…..Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)…Yes


E
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If x > 0 and y > 0, which of the following is equal to 03 Jun 2023, 13:29
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Choose numbers that are easily manipulated. In this case x= 1 and y =8.

the original expression in the question leads to 1/4.

Test the options.

Only option E, gives the same value as in the question. Therefore, option E is the answer.
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If x > 0 and y > 0, which of the following is equal to 05 Jun 2023, 23:25
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rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann

I. Algebraic

Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)


\((\frac{1}{\sqrt{x}+ \sqrt{x+y)=(\frac{\sqrt{x+y}-\sqrt{x{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)



II. Substitute some value for x and y.
Take x as 9 and y as 16..

\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)

(A)\(\frac{1}{y}=\frac{1}{16}\)….No
(B)\(\sqrt{2x + y}\)…..Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)…No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)…..Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)…Yes


E
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In the first solution you used the (a^2-b^2) formula for conjugate. However, I not understanding one thing as I took the Conjugate of √x-√x+y.
Shouldn’t we take a= √ x and b= √ x=y?
Can you please explain this part?

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If x > 0 and y > 0, which of the following is equal to 17 Jan 2025, 18:48
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Method isn't full proof. If x=4 y=5 then you'd mistakenly get answer A?

1/2+3 = 1/5

chetan2u
rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann

I. Algebraic

Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)


\((\frac{1}{\sqrt{x}+ \sqrt{x+y}})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)



II. Substitute some value for x and y.
Take x as 9 and y as 16..

\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)

(A)\(\frac{1}{y}=\frac{1}{16}\)....No
(B)\(\sqrt{2x + y}\).....Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)...No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\).....Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)...Yes


E
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If x > 0 and y > 0, which of the following is equal to 18 Jan 2025, 03:31
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I would recommend the algebraic approach. However, if you go with the second approach, for the numbers you pick you will have to test each option - for x=4 y=5 you will find option A and E both hold true, in that case you will have to check both with different numbers and select the one that holds true for any set of numbers.
Mgerman42
Method isn't full proof. If x=4 y=5 then you'd mistakenly get answer A?

1/2+3 = 1/5

chetan2u
rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann

I. Algebraic

Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)


\((\frac{1}{\sqrt{x}+ \sqrt{x+y}})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)



II. Substitute some value for x and y.
Take x as 9 and y as 16..

\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)

(A)\(\frac{1}{y}=\frac{1}{16}\)....No
(B)\(\sqrt{2x + y}\).....Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)...No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\).....Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)...Yes


E
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