The average (arithmetic mean) of the 5 positive integers, u, w ,x, y,

Given:

\(\quad u \quad w \quad x \quad y \quad z \)

Median = x

\(\quad u \quad w \quad x \quad y \quad 26 \)

The positive-negative difference of each term of the set should cancel out with respect to the mean

To minimize the value of x, we need to maximize y

\(\quad\) u \(\quad\) w \(\quad\) x \(\quad\) (14+11) \(\quad\) (14+12)

Hence, we have a value of 23 above the mean, the value of u, w, and x must cancel the excess of 23 to keep the mean at 14. We can have the set as shown below

\(\quad\) (14-9) \(\quad\) (14-8) \(\quad\) (14-6) \(\quad\) (14+11) \(\quad\) (14+12)

Min value of x = 14-6 = 8

Option C

Hi!
In this question-try thinking of which strategy could you use instead of setting any equation.
Deep think on the conditions provided.
The solution's rationale is based on the approach of maximizing and minimizing values around the median,
All values are positive integers with a constant sum of 14*5=70 for all the 5 integers and they are arranged in an ascending order.
Median is the third variable x. On the number line u and w are on the left side of x whereas y and z are on the right.
How can you make x very very small when the sum of the 5 terms is constant(which is 70) with z= 26?
You could do so if you maximise the value to the right of x and minimise the ones which are to the left of x.

Lets use the choices-
A. 3 If x =3 and z=26, the values to the left of x that is u and w can be 1 and 2.
Then the sum of u+w+x+y+z (which is 70) would be 1+2+3+y + 26=32+y.
This implies y would then be 70 - 32 =38. However the condition mentions y<z but 38 is not less than 26. Eliminate.

B.If x =7 and z=26, the values to the left of x that is u and w can be picked up as 5 and 6.
Then the sum of u+w+x+y+z (which is 70) would be 5 + 6 + 7 + y + 26 = 44 + y .
This implies y would then be 70 - 44 = 26. But y < z and not equal to z. Eliminate.
If you decrease the values of u and w to anything that is lesser than 5 and/or 6, the overall integer sum of u + w+ x+z would then decrease and y would then be greater than 26 which would also imply the answer to be eliminated.
In other words, any reduction in u and w would violate the given conditions.

C. If x = 8 and z = 26,the values to the left of x that is u and w can be picked up as 6 and 7.
Then the sum of u+w+x+y+z (which is 70) would be 6 + 7 + 8 + y + 26 = 47 + y
This implies y would then be 70 -47 = 23 which is less than z=26. So there is a set of 5 positive integers 6,7,8,23,26 with sum 70.

The objective is to find the "least possible value" of the median.
So, any option higher than the value of 8 would not be the minimum, even if they fit the other conditions.
Eliminate D and E.

Option C

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­KarishmaB MartyMurray chetan2u
It took me a while (~3 mins) not because of concept but because of plugging answer choices in. Would you be able to provide a faster solution?
After finding out u+w+x = 19, I went onto plug E through B and it took me time. 
Let me know. Thank you!

 ­

Once you know that u+w+x = 19 and the largest of them x, which is the median, should be the lowest possible, you have to get all three closest.

The middle value therefore, should be equal or almost equal to average­. 19/3 = 6.33
Hence, w is 6 or 7.
if 6, u will have the largest possible value as 5, so that x is the least possible = 19-5-6 = 8
If 7, no need to calculate, as the least will be 7+1 or 8.

Another way would be that if they could be equal u+w+x = 19 =6+6+7, but as they cannot be equal you have to shift one from lowest to largest => 6+6+7 = 5+6+8

Hence 8

houston1980

Given: The average (arithmetic mean) of the 5 positive integers, u, w ,x, y, and z is 14, and u < w <x < y < z.
Asked: If z is 26, what is the least possible value of the median of the 5 integers?

u + w + x + y + z = 14*5 = 70
z=26
Median = x

Since x is to be minimized; u,w,y have to be maximized

x-2 + x-1 + x + 25 + 26 = 70
3x = 70 - 48 = 22
x = 22/3 = 7 1/3 : Not feasible since 22/3 is not an integer

We can not increase 48, but we can decrease it by decreasing either of u, w or y or any combination of them.
3x = 70 - 47 = 23; Not feasible since x = 23/3 is not an integer

3x = 70 - 46 = 24; x = 24/3 = 8: feasible

The least possible value of the median of the 5 integers = 8

IMO C
­

 

Since we're looking for the least possible value of the median, one improvement on your approach would have been to start with (B) rather than with (E) since (B) was the lowest value you thought may work. Then, once you found that (C) worked, you would have been done.

Alternatively, we could start with (C) and work up or down depending on whether (C) works, rather than starting at the high end and putting ourselves in the position of having to work our way down until we get to the answer.

Also, I think we can intuitively see that 10 probably won't be correct since using 10 for x leaves just 9 for u and w. So, even if we are going to test from high to low, we could start with (D), see that it works and then try (C) and (B), at least avoiding testing 10.

Another thought is that x would be minimized if the three integers u, w, and x were as close as possible. Without any other constraints, the three integers are as close as possible if u = x - 2 and w = x - 1.

In that case, 19 = x - 1 + x - 2 + x = 3x - 3. That case doesn't work because, in that case, 22 = 3x. So, x = 7.33, which is not an integer. At the same time, having explored that case, we've discovered that x can be just over 7, meaning x can be 8.­

­It's certainly a good question, not difficult but could be a bit confusing. Think conceptually. 

Given 5 distinct integers and Mean has to be 14. The greatest is 26. I need the median to be the least i.e the 3rd number to be the least possible. 

When mean has to be kept at a  certain value, how can we minimize a number? By maximizing other that we can maximise. If I want a lot of deficit, I should have a lot of excess too. So I am going to going to maximize the 4th number too and make it 25 (1 less than greatest number)

__ __ __ 25, 26

This gives me an excess of 11 + 12 = 23. So now I can have a maximum deficit of 23. So I should try to distribute this 23 evenly (as much as possible) among the first 3 numbers so that my greatest number out of these three (the median) does not increase too much.
23 divided by 3 is 7.something so perhaps I can have 6, 7, 8 as the first 3 numbers (as close to each other as possible but still distinct). But their sum will be 21 so the fourth number will be 23, not 25.

6, 7, 8, 23, 26 will be the numbers. 

My median cannot be any less i.e. it cannot be 7 or lower because 5, 6, 7 will create a deficit of 24 but we cannot have excess more than 23. 

Answer (C)

Here is a post on the concept of deviations: 
https://anaprep.com/arithmetic-usefulness-of-deviations/
 

­Hey GMAT guys! I'd like your help since I've read all the posts and I cannot precisely spot my mistake. My reasoning was as follows:

We're looking for the least possible value of the median which based on the order of the variables is the x. We're also given the average and the value on of the unknowns so we can tell that: u+w+x+y = 44 => x = 44 - y - (u+w).
To minimize the median we'll maximize the other 3 variables, so y = 25 (since is the one before z=26) and x = 19 -(u+w) similarly we'll insert the biggest possible value for the (u+w) ≤ 19 [in order for x to be positive].
Based on the options, the max(u+w) = 16 which results in x = median = 3 which also is the least possible.

Bunuel KarishmaB GMATCoachBen ScottTargetTestPrep MartyMurray manasp35 chetan2u­

 

Did you consider ­u < w <x
If x = 3, then w = 2, u = 1

we are getting u + w = 3 , not 16

so x=3 cannot be the answer. 

if x = 7, 
then w can be 6, u can be 5
u+w+x+y = 44

y = 44 - 7 -6 - 5 = 26 , which does not satisfy inequality u < w <x < y < z
 

You are wrong after selecting value of y.
x=19-(u+w) but u and w have to be less than x. So, their largest value can be x-1 and x-2.
So x=19-(x-1+x-2)=19-2x+3 ….3x=22 or x=7.33
We cannot x lesser than 7.33, so x is higher integer that is 8 and other two are 7 and 4 or 6 and 5.

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Thank you chetan2u and manasp35 ! I thought it as a group indeed. chetan2u you mentioned earlier that in the step of u+w+x=19 we'll find the average of these as to use the best-fit values (and I agree). When we work with w=7 wouldn't it be right to put u=6 (the largest possible) and as a result x = 19-7-6= 6 ?

That's the part I'm talking about: ­Once you know that u+w+x = 19 and the largest of them x, which is the median, should be the lowest possible, you have to get all three closest.

The middle value therefore, should be equal or almost equal to average­. 19/3 = 6.33
Hence, w is 6 or 7.
if 6, u will have the largest possible value as 5, so that x is the least possible = 19-5-6 = 8
If 7, no need to calculate, as the least will be 7+1 or 8.

omg// i was stuck with "least possible value" in the exam...
least means the smallest, i thought its asking the value that is least possible LOL.
love GMAT day by day 😂