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smallest number.

by j_shreyans » Tue Oct 21, 2014 12:41 am
List R contains five number that have an average value of 55 . If the median of the numbers in the list is equal to the mean and the largest number is equal to 20 more than two times the smallest number, what is the smallest possible value in the list.

A)35
B)30
C)25
D)20
E)15
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by GMATGuruNY » Tue Oct 21, 2014 4:18 am
j_shreyans wrote:List R contains five number that have an average value of 55 . If the median of the numbers in the list is equal to the mean and the largest number is equal to 20 more than two times the smallest number, what is the smallest possible value in the list.

A)35
B)30
C)25
D)20
E)15
Let the smallest number = x.
Since the largest number is equal to 20 more than 2 times the smallest number, the largest number = 2x + 20.
Median = mean = 55.
Let the other two numbers be y and z.
The 5 numbers are as follows:

x, y, 55, z, 2x+20

To MINIMIZE the value of x, we must MAXIMIZE y and z.
Since y cannot be greater than the median, the greatest possible value for y is 55.
Since z cannot be greater than 2x+20, the greatest possible value for z is 2x+20.
The 5 numbers are as follows:

x, 55, 55, 2x+20, 2x+20

Since the average is 55, the sum of the five numbers = 5*55 = 275.
Thus:
x + 55 + 55 + (2x+20) + (2x+20) = 275.
5x + 150 = 275
5x = 125
x = 25.

The correct answer is C.

Similar problem:
https://www.beatthegmat.com/set-median-t73013.html
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by [email protected] » Tue Oct 21, 2014 10:33 am
Hi J_shreyans,

While this question can be solved with Algebra (as Mitch's approach shows), it can also be solved by TESTing THE ANSWERS.

We're given a lot of little pieces of information to work with, so we have to stay organized:
1) 5 numbers with an average of 55. This means that 5(55) = 275 is the SUM of these 5 numbers.
2) The question does NOT state that the numbers need to be distinct, so DUPLICATES ARE ALLOWED.
3) The median = mean = 55. The 3rd value of the 5 numbers MUST be 55.
4) The largest number = 2(smallest number) + 20.

The question asks us for the smallest POSSIBLE value in the list. Let's start by TESTing the smallest answer....

E: 15

If the smallest number was 15, then the largest would be 50. This is NOT possible (the average and median are both 55).

D: 20

If the smallest number was 20, then the largest would be 60. This is NOT possible (the sum of the other 3 values = 195; there's no way to get to a total of 195 with the 3 remaining values and the above restrictions).

C: 25

If the smallest number was 25, then the largest would be 70. This IS POSSIBLE. The sum of the other 3 values would be 180 (with a 70, a 55 and another 55, we would get a total of 180).

Final Answer: C

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by gmat_for_life » Mon Mar 14, 2016 2:07 am
Hello Rich and Mitch,

Could you please let me know what's wrong with my approach?

If the median of the numbers in the list is equal to the mean, I assumed the 5 numbers as a,a+x,a+2x,a+3x,a+4x.

Now, (a+a+x+a+2x+a+3x+a+4x)/5=55-----(1)
and a+4x=20+2x -----(2)

Solving these 2 simultaneous equations, I arrived at a=30.

Regards,
Amit
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by GMATGuruNY » Mon Mar 14, 2016 4:57 am
gmat_for_life wrote:Hello Rich and Mitch,

Could you please let me know what's wrong with my approach?

If the median of the numbers in the list is equal to the mean, I assumed the 5 numbers as a,a+x,a+2x,a+3x,a+4x.
The expression in red is valid only if the integers evenly spaced.
For the median to be equal to the mean, the integers do NOT have to be evenly spaced:
{1, 9, 10, 15, 15}
Here, median = mean = 10, but the integers are not evenly spaced.
Since the prompt above does not require that the integers be evenly spaced, the expression in red is not valid.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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