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The minimum of the integers x, y, and z is 10 and their avera

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MathRevolution
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The minimum of the integers x, y, and z is 10 and their avera [#permalink] New post   26 Apr 2018, 01:22
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[GMAT math practice question]

The minimum of the integers \(x, y\), and \(z\) is \(10\) and their average is \(11\). What is the greatest possible value of their maximum?

\(A. 10\)
\(B. 11\)
\(C. 12\)
\(D. 13\)
\(E. 14\)
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The minimum of the integers x, y, and z is 10 and their avera [#permalink] New post   26 Apr 2018, 01:34
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MathRevolution wrote:
[GMAT math practice question]

The minimum of the integers \(x, y\), and \(z\) is \(10\) and their average is \(11\). What is the greatest possible value of their maximum?

\(A. 10\)
\(B. 11\)
\(C. 12\)
\(D. 13\)
\(E. 14\)


Given minimum integer is \(= 10\)

Average of \(3\) integers \(x\), \(y\), and \(z\) \(= 11\)

Therefore Total \(= 11*3 = 33\)

ie; \(x + y + z = 33\)\(\)

Greatest value is possible if the other two values are minimum.

Let \(x\) and \(y\) \(= 10\)

Therefore greatest possible value of their maximum;

\(x + y + z = 33\)

\(10 + 10 + z = 33\)

\(20 + z = 33\)

\(z = 33-20 = 13\)

Answer D
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The minimum of the integers x, y, and z is 10 and their avera [#permalink] New post   26 Apr 2018, 12:25
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MathRevolution wrote:
[GMAT math practice question]

The minimum of the integers \(x, y\), and \(z\) is \(10\) and their average is \(11\). What is the greatest possible value of their maximum?

\(A. 10\)
\(B. 11\)
\(C. 12\)
\(D. 13\)
\(E. 14\)

I could not understand the question. (Maybe that's the point?)

REWRITE:
Each of three integers \(x, y,\) and \(z\) has a minimum value of \(10\). The integers' average is \(11.\) What is the greatest possible value of any one variable?

We know the sum, \(x + y + z\):
\(A*n = S\)
\(11*3 = 33\) = Sum of all three

In order to maximize one value, minimize the other two. Each integer must = at least 10
Let \(x=10, y=10\)

\(10, 10,\) z? . . .They sum to \(33\):
\(10 + 10 + z = 33\)
\(z = (33 - 20) = 13\)
Maximum value of the third integer = \(13\)

Answer D
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Re: The minimum of the integers x, y, and z is 10 and their avera [#permalink] New post   29 Apr 2018, 18:32
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=>

Assume \(x ≤ y ≤ z.\)
\(\frac{( x + y + z )}{3} = 11\) and \(x = 10\)
We have \(10 + y + z = 33\) or \(y + z = 23.\)
In order to have the greatest maximum number, y must be the minimum which is \(10\).
\(10 + z = 23.\)
\(z = 13\).

Therefore, D is the answer.

Answer : D
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Re: The minimum of the integers x, y, and z is 10 and their avera [#permalink] New post   30 Apr 2018, 11:26
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MathRevolution wrote:
[GMAT math practice question]

The minimum of the integers \(x, y\), and \(z\) is \(10\) and their average is \(11\). What is the greatest possible value of their maximum?

\(A. 10\)
\(B. 11\)
\(C. 12\)
\(D. 13\)
\(E. 14\)


Key concept: If we know the sum of a set of numbers, and we want to MAXIMIZE the biggest number in the set, we must MINIMIZE all of the other numbers.

GIVEN: Average of x, y, and z is 11
So, (x + y + z)/3 = 33
This means x + y + z = 33
Great! We know the sum of the values.

In order to MAXIMIZE the biggest number in the set, we must MINIMIZE all of the other numbers.
We're told that 10 is the MINIMUM value in the set.
So, let's let TWO of the values equal 10
Say x = 10 and y = 10
We have now MINIMIZED two of the three values.

Since we know that x + y + z = 33, we can now write 10 + 10 + z = 33
Solve to get: z = 13
So, the MAXIMUM value is 13.

Answer: D

Cheers,
Brent
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Re: The minimum of the integers x, y, and z is 10 and their avera [#permalink]
30 Apr 2018, 11:26
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