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.
OG2015 PS The average (arithmetic mean)
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- lionsshare
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Let x = the missing (required number)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
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
Cheers,
Brent
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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: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
30 = 5 + (20 + 40 + n)/3
25 = (20 + 40 + n)/3
75 = 20 + 40 + n
15 = n
Answer: A
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