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

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Math Expert
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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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24 Aug 2018, 00:53
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Difficulty:

35% (medium)

Question Stats:

72% (01:16) correct 28% (02:22) wrong based on 32 sessions

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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

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Joined: 18 Jul 2018
Posts: 193
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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Updated on: 24 Aug 2018, 01:03
11+w+x+y = 70. w+x+y = 59

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

Originally posted by Afc0892 on 24 Aug 2018, 01:01.
Last edited by Afc0892 on 24 Aug 2018, 01:03, edited 1 time in total.
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Posts: 860
WE: Supply Chain Management (Energy and Utilities)
If the average (arithmetic mean) of 3, 8, w, x, and y is 14, then the  [#permalink]

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24 Aug 2018, 01:03
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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24 Aug 2018, 01:06
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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24 Aug 2018, 02:00
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

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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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24 Aug 2018, 11:10
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, 11:10
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