How do we know which number is to be excluded ? Ans can change depending on the number we exclude

@GreenlightTestPrep

Here is an algebraic solution that allows us to examine all possible cases.

Let x = the smallest integer in the set
So, x + 1 = the next consecutive integer
x + 2 = the next integer
x + 3 = the next integer
x + 4 = the greatest integer

Average \(= \frac{x+(x+1)+(x+2)+(x+3)+(x+4)}{5}=\frac{5x+10}{5}=x+2\)

Key concept: In order to get the greatest DECREASE in the average, we must remove the biggest number in the set.
So we'll remove (x+4) from the set
The new average \(= \frac{x+(x+1)+(x+2)+(x+3)}{4}=\frac{4x+6}{4}=x+1.5\)

Percent decrease = (100)(old - new)/old

We get: percent decrease \(= \frac{(100)[(x+2)-(x+1.5)]}{x+2}\)

\(= \frac{(100)[0.5]}{x+2}\)

\(= \frac{50}{x+2}\)

Notice that the percent increase depends on the value of x.
For example, if \(x=8\), then the percent decrease \(= \frac{50}{8+2}= \frac{50}{10}=5%\), which is LESS THAN 20%

Also notice that, in order to maximize the percent decrease, we must minimize the value of x.
Since we're told x is a positive integer, the smallest possible value of x is 1.
When \(x=1\), then the percent decrease \(= \frac{50}{1+2}= \frac{50}{3}≈16.666...%\)
This means 16.666...% is the GREATEST possible value of Quantity A.

This means Quantity B will always be greater than Quantity A

Answer: B

Cheers,
Brent

Great question!
Notice that, if we remove the middle number, then the percent decrease in the average is zero.
So, in this case, Quantity B will be greater.
At this point our goal should be to MAXIMIZE the percent decrease. This is achieved by removing the biggest number in the set.
As I show in my post above, removing the biggest number in the set will always yield a percent decrease that is less than 20%.

Cheers,
Brent

Let the 5 numbers be: \(x, x+1, x+2, x+3, x+4\)

Since the numbers are in Arithmetic Progression (i.e. consecutive numbers have a constant gap), we have:
Mean = Median = 3rd term = \(x+2\)

(Note: You can always add up the terms and check the average)

One of the 5 numbers will be dropped

Maximum change in mean will occur if either of the 2 extreme terms, i.e. \(x\) or \(x+4\) is dropped. The average will decrease if \(x+4\) is dropped (Note: if \(x\) is dropped, the average would actually increase. Also, if the middle number, i.e. \(x+2\) is dropped, there will be no change in the mean)

If \(x+4\) is dropped: The 4 terms are: \(x, x+1, x+2, x+3\)

=> New Mean = Median = \([(x+1)+(x+2)]/2 = x+1.5\)

=> Percent decrease in mean = \([{(x+2)-(x+1.5)}/(x+2)] * 100\) = \([50/(x+2)]%\)

The above percent will be maximum if the value of \(x\) is minimum, i.e. \(x=1\)

=> Maximum percent decrease = \([50/(1+2)]% = 16.67%\)

Thus, Quantity B is greater than Quantity A

Answer B


Note: Some important results that come up here:

In an Arithmetic Progression i.e. consecutive terms having a constant difference of \(d\):

#1. Mean = Median
#2. The average remains unchanged if the middle term (or both middle terms) are removed
#3. The maximum change in mean occurs when one of the extreme terms is removed

#4. The maximum change in mean (when one of the extreme terms is removed) = \(d/2\), where \(d\) is the constant difference between consecutive terms

I don't understand why we have to decrease; I do not read the question in a way that states that as a rule. Is it possible to see a positive increase as a negative decrease? Would the GRE risk this issue? The entire micro-intention of the test is separate the people who know from the people who don't know and playing word games can produce a knower with a wrong answer and an Edwin J. Goodwin with the right one which doesn't serve GRE's purpose (or those of the admissions offices)

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