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v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If the average (arithmetic mean) of the five numbers is 30 and their median is 25, which is the smallest possible value of z?

Re: v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If t [#permalink]

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20 Oct 2016, 01:35

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To minimize the value of Z we need to maximise the value of V and W. So the maximum value that we can assign to V and W is 25. So the first 3 number add up 75 and the total sum is 150. So the hightest two values will share the difference ie 150-75=75. To minimize the value of Z and also to keep Y<=Z the value that Z can take is 38.

Re: v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If t [#permalink]

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30 Nov 2017, 05:13

Bunuel wrote:

v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If the average (arithmetic mean) of the five numbers is 30 and their median is 25, which is the smallest possible value of z?

A. 30 B. 35 C. 37 D. 38 E. 40

Hi! Where is this question from? I did the Economist Contest (scholarship) test yesterday and this questions was in the test.

Do you have the rest of the questions? The company don't let you revise the test, so I can NOT review the questions!

v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If the average (arithmetic mean) of the five numbers is 30 and their median is 25, which is the smallest possible value of z?

A. 30 B. 35 C. 37 D. 38 E. 40

Hi! Where is this question from? I did the Economist Contest (scholarship) test yesterday and this questions was in the test.

Do you have the rest of the questions? The company don't let you revise the test, so I can NOT review the questions!

v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If t [#permalink]

Show Tags

30 Nov 2017, 11:05

Bunuel wrote:

v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If the average (arithmetic mean) of the five numbers is 30 and their median is 25, which is the smallest possible value of z?

A. 30 B. 35 C. 37 D. 38 E. 40

To minimize the value of \(z\), we need to maximize values of \(v\), \(w\), and\(y\). (Median, middle value, \(x\), is set.) And the "equal to" part of \(≤\) would be easy to miss (trap answer E).

Median = \(25 = x\) Maximize \(v\) and \(w\); let them equal \(x = 25\) Thus \((v≤w≤x)=> 25≤25≤25\) Total so far: \(75\)

Total sum of numbers: \((A*n)=S\) \(30*5 = 150\) Total remaining for \(y\) and \(z: (150 - 75) = 75\)

Maximize \(y\) to minimize \(z\). \((\frac{75}{2})=37.5\) Keep \(y\) as close as possible to \(z\). Closest integer "down" is 37, closest integer "up" is 38. \(y = (37.5 - 0.5) = 37\)

v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If the average (arithmetic mean) of the five numbers is 30 and their median is 25, which is the smallest possible value of z?

A. 30 B. 35 C. 37 D. 38 E. 40

To get the smallest value of z, we need to maximize the rest of the integers v,w,x and y.

The mean of the five numbers is 30. Thus, the sum total of 5 numbers = 30* 5 = 150.

We know that the median is 25.

Thus x is 25 (Assuming that to be the middle value).

Since we want to maximize all the values except z, we must assume that v and w are also equal to 25 each.

Thus v = w = x = 25

Now we need to distribute 75 between y and z. The best case would be if we divide is equally, that would give us 75/2 = 37.5.

But we know that all the number are integers and y is less than or equal to z.

v, w, x, y, and z are five positive integers for which v≤w≤x≤y≤z. If the average (arithmetic mean) of the five numbers is 30 and their median is 25, which is the smallest possible value of z?

A. 30 B. 35 C. 37 D. 38 E. 40

Since the average is 30, the sum is 150. To find the smallest possible value of z, which is the largest value by design, we need to maximize the other 4 numbers. We are given that x, the median, is 25. Since v ≤ w ≤ x, we can let v = w = x = 25. Since y ≤ z, y could be equal to z and hence we can let y = z. Therefore, we have:

25 + 25 + 25 + z + z = 150

75 + 2z = 150

2z = 75

z = 37.5

However, since z has to be an integer, then z must be 38 (and y will be 37).

Answer: D
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