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For the 5 days shown in the graph, how many kilowatt-hours

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by VJesus12 » Sat Apr 20, 2019 3:19 am

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For the 5 days shown in the graph, how many kilowatt-hours greater was the median daily electricity use than the average (arithmetic mean) daily electricity use?

A) 1
B) 2
C) 3
D) 4
E) 5

[spoiler]OA=A[/spoiler]

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by Brent@GMATPrepNow » Sat Apr 20, 2019 4:40 am
VJesus12 wrote:Image

For the 5 days shown in the graph, how many kilowatt-hours greater was the median daily electricity use than the average (arithmetic mean) daily electricity use?

A) 1
B) 2
C) 3
D) 4
E) 5

[spoiler]OA=A[/spoiler]

Source: GMAT Prep
List the numbers in ASCENDING ORDER to get: 19, 24, 27, 29, 31.
So 27 is the median.

Average = (19 + 24 + 27 + 29 + 31)/5 = 130/5 = 26.

The median (27) is 1 greater than the average (26).

Answer: A

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Sat Apr 20, 2019 10:05 am
Median of \(19, 24, 27, 29\) and \(31\) is \(27\) (middle value in the list)

Mean \(= \frac{19+24+27+29+31}{5} = \frac{130}{5}=26\)

Difference \(=27-26=1\)
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by Scott@TargetTestPrep » Thu Apr 25, 2019 5:52 pm
VJesus12 wrote:Image

For the 5 days shown in the graph, how many kilowatt-hours greater was the median daily electricity use than the average (arithmetic mean) daily electricity use?

A) 1
B) 2
C) 3
D) 4
E) 5

[spoiler]OA=A[/spoiler]

Source: GMAT Prep
Listing the numbers from smallest to largest, we have: 19, 24, 27, 29 and 31. So 27 is the median. The average is (19 + 24 + 27 + 29 + 31)/5 = 130/5 = 26. Therefore, the median exceeds the average by 1.

Answer: A

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