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The average (arithmetic mean) of x, y, and z is 50. What is

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by BTGmoderatorDC » Tue Aug 13, 2019 4:42 pm

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C

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The average (arithmetic mean) of x, y, and z is 50. What is the sum of (4x + y), (3y + z), and (3z)?

A. 150
B. 200
C. 600
D. 800
E. It cannot be determined from theinformation given.

OA C

Source: Princeton Review
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by Jay@ManhattanReview » Tue Aug 13, 2019 8:54 pm
BTGmoderatorDC wrote:The average (arithmetic mean) of x, y, and z is 50. What is the sum of (4x + y), (3y + z), and (3z)?

A. 150
B. 200
C. 600
D. 800
E. It cannot be determined from theinformation given.

OA C

Source: Princeton Review
Given that the average (arithmetic mean) of x, y, and z is 50, we have (x + y + z) = 150.

Sum of (4x + y), (3y + z), and (3z) = 4x + y + 3y + z + 3z = 4x + 4y + 4z = 4(x = y + z) = 4*150 = 600.

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Sun Aug 18, 2019 6:13 pm
BTGmoderatorDC wrote:The average (arithmetic mean) of x, y, and z is 50. What is the sum of (4x + y), (3y + z), and (3z)?

A. 150
B. 200
C. 600
D. 800
E. It cannot be determined from theinformation given.

OA C

Source: Princeton Review
We see that x + y + z = 50(3) = 150. Therefore,

(4x + y) + (3y + z) + (3z) = 4x + 4y + 4z = 4(x + y + z) = 4(150) = 600

Answer: C

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