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

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

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New post 23 Aug 2018, 23:53
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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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New post 24 Aug 2018, 00:00
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.

A is the answer.
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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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New post 24 Aug 2018, 00:01
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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New post 24 Aug 2018, 01:30
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.

The best answer is 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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New post 24 Aug 2018, 11:13
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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New post 25 Aug 2018, 02:49

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.

Answer: A

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Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera   [#permalink] 25 Aug 2018, 02:49
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