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
Devmitra Sen
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