The number x is the average (arithmetic mean) of the positiv

"the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b"

Why is it that the translation for this is not 1/Y = [1/a+b]/2 ?

Would someone be able to clarify this?

The reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b:

\(\frac{1}{y}= \frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{a+b}{2ab}\) --> \(y=\frac{2ab}{a+b}\)

Got it so next it would be:

[a + b]/ 2 - 2ab/[a+b] = > but how would this not become (a+b)^2 - 2ab/[2(a+b)]??

The number x is the average (arithmetic mean) of the positive numbers a and b: \(x=\frac{a+b}{2}\);

The reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b: \(\frac{1}{y}= \frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{a+b}{2ab}\) --> \(y=\frac{2ab}{a+b}\);

\(x-y=\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{(a+b)(a+b)}{2(a+b)}-\frac{2*2ab}{2(a+b)}=\frac{a^2+2ab+b^2-4ab}{2(a+b)}=\frac{a^2-2ab+b^2}{2(a+b)}=\frac{(a-b)^2}{2(a+b)}\)

Great Question.
Here is my solution to this one =>

Using =>

\(Mean = \frac{Sum}{#}\)



\(x=\frac{a+b}{2}\)

\(\frac{2}{y}={\frac{1}{a}+\frac{1}{b}}\)

\(y=\frac{2ab}{a+b}\)


\(x-y=\frac{(a+b)^2-4ab}{2(a+b)}\)
=> \(\frac{(a-b)^2}{2(a+b)}\)


Hence B

Hi All,

While this is a "thick" question, it can be solved by TESTing Values.

We're told that…
X = average of A and B
The reciprocal of Y = the average of the reciprocals of A and B

A = 2
B = 4

X = (2+4)/2 = 3

1/Y = (1/2 + 1/4)/2 = (3/4)/2 = 3/8
1/Y = 3/8
Y = 8/3

We're asked for the value of X - Y:
X - Y =
3 - 8/3 =
9/3 - 8/3 =
1/3

So we're looking for an answer that equals 1/3 when we use X = 2 and Y = 4

From the answer choices, we can quickly eliminate C, D and E (none of them have a denominator that is (or could become) a "3"; so we're down to the first 2 answers.

Answer A: (4 + 16)/12 = 20/12 = 5/3
Answer B: (-2)^2/12 = 4/12 = 1/3

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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