An exam is given in a certain class. The average (arithmetic mean) of

Question

An exam is given in a certain class. The average (arithmetic mean) of the highest score and the lowest score is equal to x. If the average score for the entire class is equal to y and there are z students in the class, where z > 5, then in terms of x, y, and z, what is the average score for the class excluding the highest and lowest scorers?

(A) (zy – 2x)/z
(B) (zy – 2)/z
(C) (zx – y)/(z – 2)
(D) (zy – 2x)/(z -2)
(E) (zy – x)/(z + 2)

(A) (zy – 2x)/z
(B) (zy – 2)/z
(C) (zx – y)/(z – 2)
(D) (zy – 2x)/(z -2) 
(E) (zy – x)/(z + 2)

Alchemist14 · Answer

Highest: H
Lowest: L
Number of students in the class: Z
Number of students in the class excluding the highest and lowest : Z-2

Average of Highest and Lowest: (H + L)/2=X => H+L=2X

Average of Entire Class : (H+L+Others)/Z=Y => Others= ZY-2X

Average of the others in the class: (ZY-2X)/(Z-2)

Ans: D

TeamGMATIFY · Answer

Given  : 
Average of highest and lowest = x
Hence total of highest and lowest = 2x

Average score of entire class = y
Total students = z
Hence, total score of entire class = yz

Required  : average score for the class excluding the highest and lowest?
Total score excluding the highest and the lowest = yz - 2x
Total students excluding the highest and the lowest scorer = z - 2

Average =  \frac{{yz - 2x}}{{z - 2}}

Option D

EMPOWERgmatRichC · Answer

Hi All,

While this question is wordy, it can be solved by TESTing VALUES and some proper note-taking. The answer choices also provide a great 'logic shortcut' that we can use to avoid doing some of that math.

To start, I want to keep the numbers 'easy' to work with, so let's say that...

Z = 6 students
The 6 scores are: 10, 20, 30, 40, 50 and 60
X = average of lowest and highest scores = (10+60)/2 = 35
Y = average of ALL scores = (10+20+30+40+50+60)/6 = 35

We're asked for the average of the 4 "middle" scores...(20+30+40+50)/4 = 35

So we're looking for an answer that equals 35, when Z=6, X=35 and Y=35. Since we're looking for an answer that focuses on the average of those 4 scores, there will have to be a "4" in the denominator. Looking at the answer choices, Answers A, B and E do NOT have a 4 in the denominator, so they cannot be correct. That leaves us with just 2 answers to check: Answers C and D...

(C) (zx – y)/(z – 2) =  /4 = 175/4 = not an integer
(D) (zy – 2x)/(z -2) =  /4 = 140/4 = 35

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

stonecold · Answer

Fantastic Question.
Here is what i did in this Question =>

Mean = \frac{Sum}{#} 
Sum(lowest+highest)= 2x
Sum(Entire class)=zy

Sum(Entire Class excluding the highest and lowest)=> zy-2x

Hence Mean => zy-2x/z-2
Hence D

bumpbot · Answer

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An exam is given in a certain class. The average (arithmetic mean) of

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