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  • Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

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If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the

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If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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New post 23 Aug 2018, 23:53
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E

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If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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New post Updated on: 24 Aug 2018, 00:03
11+w+x+y = 70. w+x+y = 59

\(\frac{(w+2+x-3+y+8)}{3}\) = \(\frac{(59+7)}{3}\) = 22.

E is the answer.
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Originally posted by Afc0892 on 24 Aug 2018, 00:01.
Last edited by Afc0892 on 24 Aug 2018, 00:03, edited 1 time in total.
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If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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New post 24 Aug 2018, 00:03
Bunuel wrote:
If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the average of w + 2, x - 3, and y + 8 is

A. 11
B. 13.2
C. 17
D. 19.4
E. 22


Given, 3+8+w+x+y=14*5=70
Or, w+x+y=70-11=59

Average of w + 2, x - 3, and y + 8 is:-
\(\frac{w+2+x-3+y+8}{3}=\frac{w+x+y+7}{3}=\frac{59+7}{3}=\frac{66}{3}=22\)

Ans. (E)
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If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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New post 24 Aug 2018, 00:06
Bunuel wrote:
If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the average of w + 2, x - 3, and y + 8 is

A. 11
B. 13.2
C. 17
D. 19.4
E. 22


Since, the arithmetic mean of 3, 8, w, x, and y is 14
\(\frac{w + x + y + 3 + 8}{5}= 14\) -> \(w + x + y = 14*5 - 11 = 70 -11 = 59\)

Average of w + 2, x - 3, and y + 8 can be calcluated as follows: \(\frac{59 + 10 - 3}{3} = \frac{66}{3} = 22\)

Therefore, the average w + 2, x - 3, and y + 8 is 22(Option E)
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If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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New post 24 Aug 2018, 01:00
Bunuel wrote:
If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the average of w + 2, x - 3, and y + 8 is

A. 11
B. 13.2
C. 17
D. 19.4
E. 22



Given ,

\(\frac{3 + 8 + x + y + w}{5}\) = 14

w + x + y = 70 - 11

w + x + y = 59.

Question:

\(\frac{(w + 2) + ( x - 3) + ( y + 8)}{3}\)

= \(\frac{w + x + y + 7}{3}\)

= \(\frac{59 + 7}{3}\)

=\(\frac{66}{3}\)

= 22

The best answer is E.
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Re: If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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New post 24 Aug 2018, 10:10
Bunuel wrote:
If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the average of w + 2, x - 3, and y + 8 is

A. 11
B. 13.2
C. 17
D. 19.4
E. 22

\(3 + 8 + w + x + y = 70\)

Or, \(w + x + y = 59\)

So, the average of w + 2, x - 3, and y + 8 is

\(\frac{( w + x + y ) + ( 2 - 3 + 8 )}{3} = \frac{59 + 7}{3} = \frac{63}{3} = 22\) , Answer must be (E)
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Re: If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the &nbs [#permalink] 24 Aug 2018, 10:10
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