OG2015 PS The average (arithmetic mean)

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OG2015 PS The average (arithmetic mean)

by lionsshare » Wed Sep 13, 2017 2:37 am
The average (arithmetic mean) of 10, 30, and 50 is 5 more than the average of 20, 40, and

(A) 15
(B) 25
(C) 35
(D) 45
(E) 55

OA: A

Hello, Experts. Please explain the solution to this. Thanks.

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by Brent@GMATPrepNow » Wed Sep 13, 2017 8:01 am
lionsshare wrote:The average (arithmetic mean) of 10, 30, and 50 is 5 more than the average of 20, 40, and

(A) 15
(B) 25
(C) 35
(D) 45
(E) 55

OA: A
Let x = the missing (required number)
Average of 10, 30, and 50 = (10 + 30 + 50)/3 = 90/3 = 30
Average of 20, 40, and x = (20 + 40 + x)/3 = (60 + x)/3

The average (arithmetic mean) of 10, 30, and 50 is 5 more than the average of 20, 40, and
So, 30 = (60 + x)/3 + 5
Subtract 5 from both sides to get: 25 = (60 + x)/3
Multiply both sides by 3 to get: 75 = 60 + x
Solve: x = 15

Answer: A

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by Scott@TargetTestPrep » Wed Sep 20, 2017 2:38 pm
lionsshare wrote:The average (arithmetic mean) of 10, 30, and 50 is 5 more than the average of 20, 40, and

(A) 15
(B) 25
(C) 35
(D) 45
(E) 55
We'll use the average formula: average = sum/number. The average of 10, 30, and 50 is 90/3 = 30. If we let n = the unknown number, then we have:

30 = 5 + (20 + 40 + n)/3

25 = (20 + 40 + n)/3

75 = 20 + 40 + n

15 = n

Answer: A

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