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)

A set has exactly five consecutive integers

Quantity A	Quantity B
The percentage change in the average of the numbers when one of the numbers is dropped from the set	20 %

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

I believe this only works when the first integer in the set isn't zero, right?

If it is 0, then the average of the five consecutive integers would be 10 and the average would be 2.
If you dropped 0, the average of the remaining 4 would be 2.5 and the percent change in the averages would be (2.5-2)/2 = 25%, which is greater than 20%.
Likewise, if you dropped 4 (the 5th consecutive integer), the average would be 1.5 and the percent change in the averages would also be 25%.

What if it was -2, -1, 0, 1, 2 rather than 0, 1, 2, 3, 4? What could we say?

Hey Carcass,

In the case of the set being -2, -1, 0, 1, 2 with 0 dropped, the change would have been 0% that is less than 20%. However, if it was -2 dropped out of the series, it would add up to 0.5 which stands for 50% increase that is greater than 20%.

Is there any other information needed that is missing in the question?..

I think there may be a mistake here?

the original values can be represented as x, x+1, x+2, x+3, x+4

Thus, the average is (5x + 10) / 5 = x + 2.

For the largest possible changes to the average, take the ends.

Remove x: average = (4x + 10) / 4 = x + 2.5. If x is 0, then the change is 25%, which is greater than 20%. If x were 3, then the change would be less than 20%. Am I missing something?

Merged similar topics.

Please, refer to the explanations provided above

1) https://gre.myprepclub.com/forum/a-set- ... tml#p24923
2) https://gre.myprepclub.com/forum/a-set- ... tml#p44919
3) https://gre.myprepclub.com/forum/a-set- ... tml#p44923

i did like 98 99 100 101 102, tried to remove 102 average 99.5 , tried to remove 98 , average 101.5 , tried to remove 101 average smth else does not matter, always smaller than %20.