SOLUTION: If the average (arithmetic mean) of x and y is k, which of the following is the average of x, y, and z? (A) 2k + z/3 (B) 2k +z/2 (C) k+z/3 (D) k+z/2 (E) 2(k+z)/3

Algebra ->  Test -> SOLUTION: If the average (arithmetic mean) of x and y is k, which of the following is the average of x, y, and z? (A) 2k + z/3 (B) 2k +z/2 (C) k+z/3 (D) k+z/2 (E) 2(k+z)/3      Log On


   

Question 93017: If the average (arithmetic mean) of x and y is k, which of the following is the average of x, y, and z?
(A) 2k + z/3
(B) 2k +z/2
(C) k+z/3
(D) k+z/2
(E) 2(k+z)/3
Found 2 solutions by stanbon, bucky:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If the average (arithmetic mean) of x and y is k, which of the following is the average of x, y, and z?
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If "k" is the average of x and y then (x+y)/2 = k and x+y=2k
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Average of x,y,and z is (x+y+z)/3 = (2k+z)/3
=================
Cheers,
Stan H.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
None of the given answers are correct.
.
Here's how you do the problem.
.
You are told that the average of x and y is equal to k. But how do you find the average
of x and y? You add them together and divide by 2. So the average of x and y is:
.
%28x+%2B+y%29%2F2
.
and setting that equal to k you get:
.
%28x+%2B+y%29%2F2+=+k
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Get rid of the denominator by multiplying both sides of this equation by 2:
.
%282%2A%28x%2By%29%29%2F2+=+2k
.
Cancel the two in the numerator and the two in the denominator on the left side:
.
%28cross%282%29%2A%28x+%2B+y%29%29%2Fcross%282%29+=+2k
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and we end up with:
.
+x+%2B+y+=+2k
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Remember this result.
.
Now let's find the average (call it A) of x, y, and z. To average these three you add them
together and divide by 3. In equation form this becomes:
.
A+=+%28x+%2B+y+%2B+z%29%2F3
.
but recall that %28x+%2B+y%29+=+2k. So we can substitute 2k for x + y in the equation
for A. When we make that substitution, the answer for A is given by the equation:
.
A+=+%282k+%2B+z%29%2F3
.
The right side of this equation is the form of the answer. Notice that A equals a quantity
that is divided by 3. So in your answer list you can eliminate answers B and D because
they are divided by 2, not 3.
.
By the rules of algebraic combination for single line expressions, the other answers are:
.
(A) 2k%2Bz%2F3
.
(C) k+%2B+z%2F3
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(E) 2%2A%28k+%2B+z%29%2F3 and this multiplies out to %282k+%2B+2z%29%2F3
.
I suppose that (A) is supposed to be the correct answer, but it sure isn't written
that way. The way it is written is:
.
(A) 2k + z/3
.
This actually translates to 2k+%2B+z%2F3 because by the rules of algebraic combination
you do the division first and then you do the addition.
.
Instead, it should have been written:
.
(A) (2k + x)/3
.
In this form you divide everything inside the parentheses by 3 which translates to
%282k+%2B+z%29%2F3, and this is the correct form of the answer.
.
Hope this clarifies the problem for you and helps you to see your way through it.