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# If the average (arithmetic mean) of x, y, and 20 is 11, then the avera

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Joined: 02 Sep 2009
Posts: 52138
If the average (arithmetic mean) of x, y, and 20 is 11, then the avera  [#permalink]

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23 Aug 2018, 22:53
00:00

Difficulty:

5% (low)

Question Stats:

100% (01:05) correct 0% (00:00) wrong based on 37 sessions

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If the average (arithmetic mean) of x, y, and 20 is 11, then the average of 2x + 3, 2y - 4, and 8 is

A. 11
B. 12
C. 13
D. 14
E. 15

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Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera  [#permalink]

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23 Aug 2018, 23:00
Given $$\frac{{X+Y+20}}{3}$$ = 11.
X+Y+20 = 33.
X+Y = 13.

Now the average of,
$$\frac{{2X+3+2Y-4+8}}{3}$$ = $$\frac{{2(X+Y)+7}}{3}$$ = $$\frac{{26+7}}{3}$$ = 11.

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Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera  [#permalink]

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23 Aug 2018, 23:01
Bunuel wrote:
If the average (arithmetic mean) of x, y, and 20 is 11, then the average of 2x + 3, 2y - 4, and 8 is

A. 11
B. 12
C. 13
D. 14
E. 15

Given x+y+20=11*3=33
Or, x+y=13

We have 2x+3+2y-4+8=2(x+y)+7=2*13+7=33

So average=33/3=11

Ans .(A)

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Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera  [#permalink]

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24 Aug 2018, 00:30
Bunuel wrote:
If the average (arithmetic mean) of x, y, and 20 is 11, then the average of 2x + 3, 2y - 4, and 8 is

A. 11
B. 12
C. 13
D. 14
E. 15

$$\frac{x + y + 20}{3} = 11$$

x + y + 20 = 33

x + y = 13.

Now we have to find out the average of the following:

$$\frac{2x + 3 + 2y- 4 + 8}{3}$$

=$$\frac{2(x + y) - 4 + 11}{3}$$

= $$\frac{2(13) + 7}{3}$$

= 11.

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Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera  [#permalink]

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24 Aug 2018, 10:13
Bunuel wrote:
If the average (arithmetic mean) of x, y, and 20 is 11, then the average of 2x + 3, 2y - 4, and 8 is

A. 11
B. 12
C. 13
D. 14
E. 15

$$x + y + 20 = 33$$

Or, $$x + y = 13$$

Or, $$2x + 2y = 26$$

So, $$(2x + 3 )+ (2y - 4) + 8 = ( 2x + 2y ) + ( 3 - 4 ) + 8= 26 - 1 + 8= 33$$

Hence, Average of 2x + 3, 2y - 4, and 8 is $$\frac{33}{3} = 11$$, Answer must be (A) 11
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Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera  [#permalink]

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25 Aug 2018, 01:49

Solution

Given:
• Average (arithmetic mean) of x, y, and 20 is 11

To find:
• The average of 2x + 3, 2y - 4, and 8

Approach and Working:
As the average of x, y, and 20 is 11, we can say
• 20 + x + y = 11 x 3 = 33
Or, x + y = 33 – 20 = 13

Hence, the average of (2x + 3), (2y – 4), and 8 is:
• $$\frac{1}{3}$$ (2x + 3 + 2y – 4 + 8)
= $$\frac{1}{3}$$ [2 (x + y) + 7]
= $$\frac{1}{3}$$ (2 x 13 + 7)
= $$\frac{1}{3}$$ x 33 = 11

Hence, the correct answer is option A.

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Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera &nbs [#permalink] 25 Aug 2018, 01:49
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