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The average (arithmetic mean) of 14, 19 and 24 is 7 more than the aver

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The average (arithmetic mean) of 14, 19 and 24 is 7 more than the aver  [#permalink]

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New post 12 May 2017, 05:04
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A
B
C
D
E

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  5% (low)

Question Stats:

95% (01:22) correct 5% (00:39) wrong based on 42 sessions

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Re: The average (arithmetic mean) of 14, 19 and 24 is 7 more than the aver  [#permalink]

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New post 12 May 2017, 05:12
Bunuel wrote:
The average (arithmetic mean) of 14, 19 and 24 is 7 more than the average of 2, 12 and

A. 11
B. 19
C. 22
D. 33
E. 50


14+19+24 = 57
Average = 57/3 = 19
Reduce 7 from 19 = 12 is the average of 2,12,...
Since 2nd no is itself 12 & 2 is 10 less than 12 So 3rd no is 10 more than 12 i.e. 22

Answer C

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Re: The average (arithmetic mean) of 14, 19 and 24 is 7 more than the aver  [#permalink]

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New post 12 May 2017, 13:24
Bunuel wrote:
The average (arithmetic mean) of 14, 19 and 24 is 7 more than the average of 2, 12 and

A. 11
B. 19
C. 22
D. 33
E. 50


14, 19, 24 is an arithmetic sequence.

Median = mean (average) = 19

19 is 7 more than average of second set of numbers. 19 is 7 more than 12

Other set of numbers

Second set (2,12,x) of numbers' average is 12

Use either weighted averages (value of x in relation to 12 must offset exactly 2's relation to 12) or A=n*s

Weighted average: 12 - 2 = +10, so x must be 12 + 10

OR

A*n = 12*3 = 36 is sum of (2 + 12 + x)

14 + x = 36
x = 22 Answer C

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Re: The average (arithmetic mean) of 14, 19 and 24 is 7 more than the aver  [#permalink]

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New post 29 Sep 2018, 09:48
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Bunuel wrote:
The average (arithmetic mean) of 14, 19 and 24 is 7 more than the average of 2, 12 and

A. 11
B. 19
C. 22
D. 33
E. 50


\(\frac{(14 + 19 + 24 )}{3} - 7 = \frac{( 2 + 12 + x)}{3}\)

Or, \((19 - 7 )3 = 14 + x\)

Or, \(x = 36 - 14\)

So, \(x = 22\), Answer must be (C)
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Re: The average (arithmetic mean) of 14, 19 and 24 is 7 more than the aver &nbs [#permalink] 29 Sep 2018, 09:48
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