Answer is (D). The average of x, y, and z is . Calculated similarly, the average of 0.5x, 0.5y, and
0.5z is exactly half that. If the sum of the variables is positive, Quantity A is greater. However, if the
sum of the variables is negative, Quantity B is greater. If the sum of the variables is zero, the two
quantities are equal.

plug in 10, 20,30

(10+20+30)/3= 60/3= 20

do the same for Quantity B and we get

(5+10+15)/3= 30/3 =10

So here, quantity A looks greater

then plug in the negative versions for Quantity A

(-10+-20+-30)/3= -20

Plug in the negative versions for Quantity B

(-5+-10+-15)/3= -10

-10 is greater than -20, which is the opposite of when the numbers are positive.

since A is greater when the numbers are positive, and B is greater when the numbers are negative,
we have an inconsistent solution, and choose D

The average (arithmetic mean) of x, y, and z => A=\(\frac{x+y+z}{3}\)
The average (arithmetic mean) of 0.5x, 0.5y, and 0.5z => B=\(\frac{0.5x+0.5y+0.5z}{3}=0.5*\frac{x+y+z}{3}=0.5A\)

so we are comparing A and 0.5A..
when A = 0, both are equal that is A=B
when A>0, A>0.5A
when A<0, A<0.5A

hence the relation cannot be detremined..
D
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