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For the set of positive, distinct integers \(v, w, x, y, z\)

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by BTGmoderatorLU » Wed Jul 31, 2019 4:49 pm

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Source: Princeton Review

For the set of positive, distinct integers \(v, w, x, y, z,\) the median is 10. What is the minimum value of \(v+w+x+y+z\)?

A. 25
B. 32
C. 36
D. 40
E. 50

The OA is C
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Source: — Problem Solving |

by Jay@ManhattanReview » Thu Aug 01, 2019 12:46 am
BTGmoderatorLU wrote:Source: Princeton Review

For the set of positive, distinct integers \(v, w, x, y, z,\) the median is 10. What is the minimum value of \(v+w+x+y+z\)?

A. 25
B. 32
C. 36
D. 40
E. 50

The OA is C
Since there are five (Odd numbers of) integers, one of \(v, w, x, y, z,\) is the median. Say \(v, w, x, y, z,\) are arranged in ascending order; thus, x = median = 10. Since we want the minimum value of \(v+w+x+y+z\), the minimum possible value of v = 1 and w = 2. Again, since x < y < z, the minimum possible values of y = 11 and z = 12.

Thus, the minimum value of \(v+w+x+y+z\) = 1 + 2 + 10 + 11 + 12 = 36.

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Sun Aug 04, 2019 10:28 am
BTGmoderatorLU wrote:Source: Princeton Review

For the set of positive, distinct integers \(v, w, x, y, z,\) the median is 10. What is the minimum value of \(v+w+x+y+z\)?

A. 25
B. 32
C. 36
D. 40
E. 50

The OA is C
We need to make the numbers as small as possible. Notice that the smallest possible values for the two integers smaller than the median are 1 and 2; and the smallest possible values for the two integers greater than the median are 11 and 12.

Thus, the minimum value is:

1 + 2 + 10 + 11 + 12 = 36

Answer: C

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