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If x + z = 8, 2y + z =11, and 3x + y= 19, what is the

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by Vincen » Sat Apr 28, 2018 2:20 am
Hello gmat_mission.

This is how I'd solve it:

We need to find (x+y+z)/3.

Since
x + z = 8
2y + z =11

Then, we can get that x-2y=-3. If we derive x we will get: x=2y-3.

Now, we substitute it in the equation 3x + y= 19, hence $$3\left(2y-3\right)+y=19\ \Rightarrow\ \ 6y-9+y=19\ \Rightarrow\ \ 7y=28\ \Rightarrow\ \ y=4.$$ Therefore $$x=2\left(4\right)-3=5\ \ \ and\ \ \ \ 5+z=8\ \Rightarrow\ \ z=3.$$ This way we get that $$Average=\frac{5+4+3}{3}=\frac{12}{3}=4.$$ Therefore, the correct answer is the option [spoiler]C=4[/spoiler].

I hope it helps.
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by GMATGuruNY » Sat Apr 28, 2018 2:44 am
M7MBA wrote:If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithmetic mean) of x, y, and z?

A. 2
B. 3
C. 4
D. 5
E. 6
One approach:
Subtracting x+z = 8 from 2y+z = 11, we get:
(2y+z) - (x+z) = 11-8
2y - x = 3
2y - 3 = x.

Substituting x = 2y-3 into 3x+y = 19, we get:
3(2y-3) + y = 19
6y - 9 + y = 19
7y = 28
y = 4.

Substituting y=4 into 2y + z = 11, we get:
(2*4) + z = 11
z = 3.

Substituting z=3 into x+z = 8, we get:
x + 3 = 8
x = 5.

Average of x, y and z = (5+4+3)/3 = 4.

The correct answer is C.
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by Jeff@TargetTestPrep » Thu May 03, 2018 3:34 pm
M7MBA wrote:If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithmetic mean) of x, y, and z?

A. 2
B. 3
C. 4
D. 5
E. 6
Subtracting the first equation from the second, we have:

2y - x = 3

Multiplying the above equation by 3 and adding that to the third equation, we have:

7y = 28

So y = 4. Substituting y = 4 into the third equation, we have:

3x + 4 = 19

3x = 15

So x = 5. Substituting x = 5 into the first equation, we have:

5 + z = 8

z = 3

Thus, the average of x, y and z is (5 + 4 + 3)/3 = 4.

Answer: C

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