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If the average (arithmetic mean) of x, y, and 20 is 11, then the avera
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23 Aug 2018, 23:53
23 Aug 2018, 23:53
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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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A. 11
B. 12
C. 13
D. 14
E. 15
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Director
Joined: 18 Jul 2018
Posts: 925
Location: India
Concentration: Operations, General Management
Schools: IIMB EPGP'22 (M)
GMAT 1: 590 Q46 V25 
GMAT 2: 690 Q49 V34
WE:Engineering (Energy and Utilities)
Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera
[#permalink]
24 Aug 2018, 00:00
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.
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.
Director
Joined: 01 Oct 2017
Status:Learning stage
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WE:Supply Chain Management (Energy and Utilities)
Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera
[#permalink]
24 Aug 2018, 00:01
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
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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PKN
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VP
Joined: 31 Oct 2013
Posts: 1260
Concentration: Accounting, Finance
GPA: 3.68
WE:Analyst (Accounting)
Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera
[#permalink]
24 Aug 2018, 01:30
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
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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Location: India
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WE:Business Development (Commercial Banking)
Re: If the average (arithmetic mean) of x, y, and 20 is 11, then the avera
[#permalink]
24 Aug 2018, 11:13
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
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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Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
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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
25 Aug 2018, 02:49
Expert Reply
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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