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If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is

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Bunuel
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If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink] New post   22 Apr 2021, 09:00
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If the average of the numbers 18, 15, x, 22, 28 and 31 is 21, what is the value of x?

(A) 12
(B) 14
(C) 17
(D) 16
(E) 18
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Re: If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink] New post   22 Apr 2021, 09:13
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I dont understand, for the avg to be 31, x has to be greater than 31, and there is not option for that. Am I missing something?
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If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink] New post   Updated on: 23 Apr 2021, 05:36
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Bunuel wrote:
If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is the value of x?

(A) 12
(B) 14
(C) 17
(D) 16
(E) 18


Average of {18, 15, x, 22, 28 and 31} = 21


CONCEPT: If the Average is added/subtracted in average of numbers then the average doesn't change
e.g. If you add/subtract 10 in average of several numbers with average 10 then the new average will also be 10


i.e. AVerage of {18, 15, x, 22, 28, 31} = 21

i.e. \(\frac{18+15+x+22+28+31}{6} = 21\)

i.e. \(114+x = 21*6=126\)

i.e. \(x = 12\)

Answer: Option A

Originally posted by GMATinsight on 22 Apr 2021, 11:10.
Last edited by GMATinsight on 23 Apr 2021, 05:36, edited 1 time in total.
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Re: If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink] New post   22 Apr 2021, 11:27
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GMATinsight wrote:
Bunuel wrote:
If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is the value of x?

(A) 12
(B) 14
(C) 17
(D) 16
(E) 18


Average of {18, 15, x, 22, 28 and 31} = 31


CONCEPT: If the Average is added/subtracted in average of numbers then the average doesn't change
e.g. If you add/subtract 10 in average of several numbers with average 10 then the new average will also be 10


i.e. AVerage of {18, 15, x, 22, 28} = 31

i.e. \(\frac{18+15+x+22+28}{5} = 31\)

i.e. \(83+x = 31*5=155\)

i.e. \(x = 72\)

Bunuel: PLease check options or numbers.

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Fixed the issue. Thank you!
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Re: If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink] New post   22 Apr 2021, 12:52
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Bunuel wrote:
If the average of the numbers 18, 15, x, 22, 28 and 31 is 21, what is the value of x?

(A) 12
(B) 14
(C) 17
(D) 16
(E) 18


Method 1:
Average = Total of the terms ÷ number of terms

=> (18+15+x+22+28+31)/6 = 21
=> 114+x = 126
=> x = 12

Method 2:

If we compute the deviations of each of the terms about the average, the SUM of those deviations MUST be ZERO
(This is a very useful concept)

Let us use that:
Deviation of 18 about 21 = 18-21 = -3
Deviation of 15 about 21 = 15-21 = -6
Deviation of 22 about 21 = 22-21 = +1
Deviation of 28 about 21 = 28-21 = +7
Deviation of 31 about 21 = 31-21 = +10

Sum of the above deviations = 9
Hence, deviation of the last number, x, must be -9
Thus, the number x = 21 + (-9) = 12

Answer A

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Re: If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink] New post   23 Apr 2021, 03:32
Bunuel wrote:
If the average of the numbers 18, 15, x, 22, 28 and 31 is 21, what is the value of x?

(A) 12
(B) 14
(C) 17
(D) 16
(E) 18


sum of numbers = total numbers * avg
in this case we have
x= (21*6)-(18+15+22+28+31)
x=126-114
x=12
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If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink] New post   23 Apr 2021, 04:49
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Solution:

Difference of the numbers from the mean are (18-21) , (15-21), (22-21),(28-21),(31-21)

= -3 , -6 , +1 , +7 , +10

Sum of the differences = -9 +18 = +9

The sum of differences of the elements from the mean is always 0.

{For ex- If we have 1,2,3,4,5 and we replace every element with the mean of these 5 numbers that is 3 then the differences are -2,-1,0,1,2 and the sum of the differences is 0}

=> Thus x must be creating a difference of -9 from 21 in order to make the sum 0

=> x = 21-9

=12 (option a)

Devmitra Sen
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If the average of the numbers 18, 15, x, 22, 28 and 31 is 31, what is [#permalink]
23 Apr 2021, 04:49
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