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

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Joined: 16 Feb 2012
Posts: 174
Concentration: Finance, Economics
The number x is the average (arithmetic mean) of the positiv  [#permalink]

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17 Aug 2013, 00:43
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55% (hard)

Question Stats:

69% (02:18) correct 31% (02:23) wrong based on 359 sessions

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The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x – y = ?

A. $$\frac{{a^2 + b^2}}{{2(a + b)}}$$

B. $$\frac{{(a - b)^2}}{{2(a + b)}}$$

C. $$\frac{{a + b}}{{2}}$$

D. $$\frac{{a - b}}{{2}}$$

E. 0

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Re: The number x is the average (arithmetic mean) of the positiv  [#permalink]

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17 Aug 2013, 00:51
1
Stiv wrote:
The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x – y = ?
A $$\frac{{a^2 + b^2}}{{2(a + b)}}$$

B$$\frac{{(a - b)^2}}{{2(a + b)}}$$

C $$\frac{{a + b}}{{2}}$$

D $$\frac{{a - b}}{{2}}$$

E 0

Assume a=1,b=3 Thus x = 2. Again, $$\frac{2}{y} = \frac{1}{1}+\frac{1}{3}$$

Thus,$$y = \frac{3}{2} \to x-y = 2-\frac{3}{2} = \frac{1}{2}$$. Only option B satisfies.
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Re: The number x is the average (arithmetic mean) of the positiv  [#permalink]

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17 Aug 2013, 11:14
Stiv wrote:
The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x – y = ?
A $$\frac{{a^2 + b^2}}{{2(a + b)}}$$

B$$\frac{{(a - b)^2}}{{2(a + b)}}$$

C $$\frac{{a + b}}{{2}}$$

D $$\frac{{a - b}}{{2}}$$

E 0

x-y= (a+b)/2 - 2ab/(a+b) = (a-b)^2/(2(a+b)
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Re: The number x is the average (arithmetic mean) of the positiv  [#permalink]

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19 Jun 2016, 12:46
"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?
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Joined: 02 Sep 2009
Posts: 51215
The number x is the average (arithmetic mean) of the positiv  [#permalink]

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19 Jun 2016, 12:50
GMAT01 wrote:
"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:

$$\frac{1}{y}= \frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{a+b}{2ab}$$ --> $$y=\frac{2ab}{a+b}$$
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The number x is the average (arithmetic mean) of the positiv  [#permalink]

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19 Jun 2016, 13:38
Bunuel wrote:
GMAT01 wrote:
"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)]??
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Joined: 02 Sep 2009
Posts: 51215
Re: The number x is the average (arithmetic mean) of the positiv  [#permalink]

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19 Jun 2016, 20:43
1
2
GMAT01 wrote:
Bunuel wrote:
GMAT01 wrote:
The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x – y = ?

A. $$\frac{{a^2 + b^2}}{{2(a + b)}}$$

B. $$\frac{{(a - b)^2}}{{2(a + b)}}$$

C. $$\frac{{a + b}}{{2}}$$

D. $$\frac{{a - b}}{{2}}$$

E. 0

"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: $$\frac{1}{y}= \frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{a+b}{2ab}$$ --> $$y=\frac{2ab}{a+b}$$;

$$x-y=\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{(a+b)(a+b)}{2(a+b)}-\frac{2*2ab}{2(a+b)}=\frac{a^2+2ab+b^2-4ab}{2(a+b)}=\frac{a^2-2ab+b^2}{2(a+b)}=\frac{(a-b)^2}{2(a+b)}$$
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Re: The number x is the average (arithmetic mean) of the positiv  [#permalink]

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17 Dec 2016, 19:41
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

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Re: The number x is the average (arithmetic mean) of the positiv  [#permalink]

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06 Mar 2018, 12:39
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

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Re: The number x is the average (arithmetic mean) of the positiv &nbs [#permalink] 06 Mar 2018, 12:39
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