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

Expert replies
by Max@Math Revolution » Thu Apr 26, 2018 12:12 am
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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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Source: — Problem Solving |

by Keith@ThePrincetonReview » Thu Apr 26, 2018 10:21 am
Max@Math Revolution 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
Hi Max,

The least of three integers is 10. Let's let x = 10.

The question stem provides the average of the three integers, so we can write the equation (10 + y + z)/3 = 11.
Simplify the equation, so that 10 + y + z = 33, and y + z = 23.

The question asks for the greatest possible value of the greatest integer.
To maximize the value of one of the remaining integers, minimize the value of the other.
Since the least of the three integers is 10, neither y nor z can have a value less than 10.
If y = 10, and y + z = 23, then z = 13.

The correct answer is choice D.
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by Scott@TargetTestPrep » Fri Apr 27, 2018 9:30 am
Max@Math Revolution 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
Since the average is 11, the sum of the three integers is 11 x 3 = 33.

Since the smallest possible integer in the set is 10, we can let two of the integers equal 10, so the maximum integer is 33 - (2 x 10) = 13.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
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by Max@Math Revolution » Sun Apr 29, 2018 5:23 pm
=>

Assume x ≤ y ≤ z.
( 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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by swerve » Mon Apr 30, 2018 9:58 am
Given minimum integer is = 10

Average of 3 integers x, y, and z =11

Therefore Total =11∗3=33

ie; (x + y + z = 33)

The greatest value is possible if the other two values are minimum.

Let xx and y= 10

Therefore the greatest possible value of their maximum;

x + y + z = 33

10 + 10 + z = 33

20 + z = 33

z = 33 - 20 = 13. Option D.

Regards!
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by Brent@GMATPrepNow » Mon Apr 30, 2018 10:21 am
Max@Math Revolution 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
Brent Hanneson - Creator of GMATPrepNow.com
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