Mean & Median

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Mean & Median

by peelamedu » Sun May 22, 2011 11:30 pm
In a graduating class, the difference between the highest and lowest salaries is $100,000. The median salary is $50,000 higher than the lowest salary and the average salary is $20,000 higher than the median. What is the minimum number of students in the class?

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by Frankenstein » Mon May 23, 2011 12:09 am
Hi,
Lets the salaries be a1,a,2,...a2n+1 in the increasing order.
i.e. a1<=a2<=a3...<=a2n+1
So, a2n+1 = a1+100K
and an+1 = a1+50K
For the number to be minimum and the average to be a1+70K, we have to make sure we take the greatest possible values of the terms. So, we assign a2=a3=...an+1=a1+50K and an+2=an+3=....a2n+1 = a1+100K.
So, the mean is (a1+a2=...+a2n+1)/2n+1 = a1+n.(a1+50K)+n.(a1+100K)/(2n+1), which is given to be a1+70K
=>[(2n+1)a1+150nK]/(2n+1) = a1+70K
=> 150n = 70(2n+1) =>n=7
So 2n+1 =15 .
Hence, the minimum number of terms would be 15
You might wonder, why I have taken odd number of terms. But, if you take even number of terms. you are essentially adding a term so that the two middle most terms average is a1+50K, which is less than the mean of the series. So, in essence, employing a similar procedure we find that it would require more number of terms to satisfy the conditions.
sorry, this might be a bit complicated approach. But, I don't find any easier solution as of now. Will let you know if I end up getting a simpler solution.
Btw, my answer is 15 . May I know the OA

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by peelamedu » Mon May 23, 2011 12:37 am
Thanks Frankenstein. This was one my questions for my last week's test from my test center and I don't have the answer yet. Will post the answer once I get my answer key.

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by Frankenstein » Mon May 23, 2011 4:59 am
peelamedu wrote:Thanks Frankenstein. This was one my questions for my last week's test from my test center and I don't have the answer yet. Will post the answer once I get my answer key.
Hi,
I am not that good at expressing clearly. So, if you do not comprehend my post, feel free to express it

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by djiddish98 » Mon May 23, 2011 6:27 am
I spent way too long on this problem, but here's a solution I found.

I'm going to cut off some digits from the dollars to make the math easier.

Max = 10
Min = 0
Median = 5
Mean = 7

If we know the Max is 10 and the median is 5, we need to figure out a way to get the mean up to 7. The fastest way to do this is to add the maximum values to the set that still satisfy the conditions.

We know we can't add more than 10 to the set, or else we'll violate our RANGE condition. So that max number we can add is 10.

We also can't throw off the balance of the median (at 5). So for any number we add, we'll have to add a 5 to cancel out the adjustment of the median.

Therefore, if we want to increase our mean, we'll do so in increments by adding the values 5 and 10 to the set.

We can then setup an equation to try and figure out how many times we add the two values.

Let's pretend that we start off with the values 0,5,10 - this will be our case when x (the number of increments of adding 5 and 10) = 0.

With this set our mean is 15/3 or 5.

Now if we add 1 increment (x=1) of 5 and 10, we end up with a mean of 30/5, or 6.

If we add another increment of 5 and 10, we end up with 45/7.

If we keep repeating, we see a pattern, which is defined by this equation.

15(x+1)/(2x+3)

And we want to set this equation = 7, our desired mean.

15(x+1)/(2x+3) = 7 -> when we solve for x, we get 6. This is not the answer, but how many increments of 10 and 5 we need.

To figure out our answer, we take 6 * 2 (since we add 2 numbers) plus our original 3 numbers in our original case (0,5,10) and we get 15 total numbers.

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by GMATGuruNY » Mon May 23, 2011 8:11 am
peelamedu wrote:In a graduating class, the difference between the highest and lowest salaries is $100,000. The median salary is $50,000 higher than the lowest salary and the average salary is $20,000 higher than the median. What is the minimum number of students in the class?
An efficient approach would be to plug in values that satisfy the conditions given and to plug in the answer choices (which the GMAT would provide).

We should plug in small numbers that satisfy the relationships described in the problem. Let:
Lowest salary = 0.
Highest salary = lowest + 10 = 0+10 = 10.
Median salary = lowest + 5 = 0+5 = 5.
Mean = median + 2 = 5+2 = 7.

Since the mean must be higher than the median, and the number of values must be as small as possible, all the values in the list need to be as large as possible.
Thus, the list of values looks like this:
[0,5,5,5...median=5,10,10,10...]

Here are the answer choices, which represent the minimum number of values needed to achieve a mean of 7:

A) 11
B) 13
C) 14
D) 15
E) 17

Since we're looking for smallest number of values that will work, the smallest answer choice (A) is a little too obvious and unlikely to be correct.

Answer choice B: 13 values
Set of values = {0,5,5,5,5,5,5,10,10,10,10,10,10}
Mean = 90/13 = 6.92.
Too small. Eliminate B.

Answer choice C: 14 values
To maintain a median of 5, the maximum value that can be added to the list is 5.
Including another 5 will decrease the mean.
Eliminate C.

Answer choice D: 15 values
Set of values = {0,5,5,5,5,5,5,5,10,10,10,10,10,10,10}
Mean = 105/15 = 7.

The correct answer is D.
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by peelamedu » Mon May 23, 2011 8:58 pm
Thanks Mitch. Perfect illustration with examples.