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Test Code 52, Section 6, Question 10

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by irfan_m1973 » Wed Dec 02, 2009 1:30 am

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Answers

A

B

C

D

E

Stats

Difficulty—

For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3
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by papgust » Wed Dec 02, 2009 1:36 am
The mean will always be 1 greater than the median in this list of numbers. Hence, B

Try to solve this by substituting values of n.
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by Stuart@KaplanGMAT » Wed Dec 02, 2009 2:36 am
irfan_m1973 wrote:For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3
Picking numbers is a great way to attack this question; we can also solve algebraically.

The median of a set with an odd number of terms is simply the middle term in the set, in this case n+2.

To solve for the mean, we use the average formula:

Average = sum of terms / # of terms

= (n + (n + 1) + (n + 2) + (n + 4) + (n+ 8)) / 5
= (5n + 15)/5
= n + 3

We want to know by how much the mean is greater than the median, so:

mean - median = (n + 3) - (n + 2) = 1... choose (B).
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by [email protected] » Sun Mar 18, 2018 7:05 pm
Hi All,

We're told that five positive numbers are N, (N+1), (N+2), (N+4) and (N+8). We're asked for the difference between the MEAN of the group and the MEDIAN of the group. This question can be solved in a couple of different ways, including by TESTing VALUES.

IF.... N=2, then the five numbers are....
2, 3, 4, 6 and 10
The median = 4
The average = (25/5) = 5
The difference is 5 - 4 = 1

Final Answer: B

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by Jeff@TargetTestPrep » Thu Apr 19, 2018 5:05 pm
irfan_m1973 wrote:For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3
Let's first calculate the mean (arithmetic average).

mean = sum/quantity

mean = (n + n + 1 + n + 2 + n + 4 + n + 8)/5

mean = (5n + 15)/5

mean = n + 3

Next, we determine the median. The median is the middle value when the terms are ordered from least to greatest. The terms ordered from least to greatest are as follows:

n, n + 1, n + 2, n + 4, n + 8

The median is n + 2.

Finally we are asked how much greater the mean is than the median. To determine the difference we can subtract the smaller value (the median) from the larger value (the mean) and we get:

n + 3 - (n + 2) = n + 3 - n - 2 = 1

Answer: B

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[email protected]

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Re: Test Code 52, Section 6, Question 10

by iammohakjain » Wed Jul 01, 2020 9:39 am
Average = sum of terms / # of terms

= (n + (n + 1) + (n + 2) + (n + 4) + (n+ 8)) / 5
= (5n + 15)/5
= n + 3

We want to know by how much the mean is greater than the median, so:

mean - median = (n + 3) - (n + 2) = 1... choose (B).
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Re: Test Code 52, Section 6, Question 10

by Scott@TargetTestPrep » Sat Jul 04, 2020 8:55 am
irfan_m1973 wrote: ↑
Wed Dec 02, 2009 1:30 am
 For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3
Solution:

Let’s first calculate the mean (arithmetic average).

mean = sum/quantity

mean = (n + n + 1 + n + 2 + n + 4 + n + 8)/5

mean = (5n + 15)/5

mean = n + 3

Next, we determine the median. The median is the middle value when the terms are ordered from least to greatest. The terms ordered from least to greatest are as follows:

n, n + 1, n + 2, n + 4, n + 8

The median is n + 2.

Finally we are asked how much greater the mean is than the median. To determine the difference we can subtract the smaller value (the median) from the larger value (the mean) and we get:

n + 3 – (n + 2) = n + 3 – n – 2 = 1

Answer: B

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