3 variables 1 equation..tearing my hair out

[faraz_jeddah]

First post here.

This question does not make any sense to me. I'd appreciate a simple explanation.

if 30(x�(x+y)) + 60(y�(x+y)) = z, what could be the value of z?


A)30
B)40
C)45
D)50
E)60

[Anju@Gurome]

If 30(x�(x+y)) + 60(y�(x+y)) = z, what could be the value of z?

The problem is missing some information like x > y or x < y and whether x and y are positive or not.

Anyway, the standard procedure is as follows...

30*(x/(x + y)) + 60*(y/(x + y))
= 30*(x/(x + y)) + 30*(y/(x + y)) + 30*(y/(x + y))
= 30*(x/(x + y) + y/(x + y)) + 30*(y/(x + y))
= 30*((x + y)/(x + y)) + 30*(y/(x + y))
= 30 + 30*(y/(x + y))

Now, if x and y are both positive, 0 < y < (x + y)
So, 0 < y/(x + y) < 1
--> 0 < 30*(y/(x + y)) < 30

Hence, the given expression is greater than (30 + 0) = 30 but less than (30 + 30) = 60

Now, if it is mentioned that x < y, then (x + y) < (y + y)
So, 2y > (x + y)
--> y/(x + y) > 1/2
--> 30*(y/(x + y)) > 30*(1/2) = 15

Hence, the given expression will be greater than (30 + 15) = 45
So, only possible value of the expression is 50.

[Anju@Gurome]

The solution becomes a lot easier if you identify that z is nothing but the weighted average of 30 and 60, x and y being the weights of 30 and 60 in that average.

Hence, z will lie between 30 and 60.

Now, if x < y, i.e. the weight of 30 in the average is less than the weight of 60 in the average.
Hence, the average will be more close to 60 than 30.

Only such option is 50.

Similarly, if x = y : the average will be equidistant from 30 and 60 ---> z = (30 + 60)/2 = 45
And if x > y : the average will be more close to 30 than 60 --> only possible option is 40.

Hope that helps.

[srcc25anu]

This is a classic weighted average problem where X and y are two elements and X is weighted by 30 and Y is weighted by 60. we are asked to find the weighted average.

Wtd Avg (Z) = [(30*x) � (x+y)] + [(60*y) � (x+y)]

since Y is weighted more heavily, the wtd avg should be closer to Y's weight than X's weight. In this case z should be closer to 60 than 30. Only option that suggests so is 50.

if X and Y were weighted equally by say 50 each, z would have been 50
if x was given a 0 weight, and Y was given 70 weight, the wtd avg would be 70
If y was given 0 weight and X was given a weight of 40, wtd avg would be 40

however, usual practice in weighted average problems is that SUM of weights is equal to 1.

[Anju@Gurome]

This is a classic weighted average problem where X and y are two elements and X is weighted by 30 and Y is weighted by 60

The case is exactly the other way around.
To get weighted average,

• 1. We multiply the quantities with some weights. Now, they become weighted quantities.
  2. Add the weighted quantities
  3. Divide the sum of weighted quantities by the sum of the weights.

If it was weighted average of X and Y with 30 and 60 as the respective weights, then the weighted average should be : (30*X + 60*Y)/(30 + 60)

Hope that helps.

[GMATGuruNY]

x and y are positive integers such that y>x. If 30* ( x / (x+y) ) + 60 ( y / (x+y) ) = z, what could be the value of z?

A)30
B)40
C)45
D)50
E)60

I believe that the added conditions (in red above) reflect the intention of the problem.

Putting the sum over a common denominator, we get:
(30x + 60y) / (x+y) = z.

Let x = the number of $30 shirts purchased at a certain store.
Let y = the number of $60 shirts purchased at a certain store.
Total cost of the $30 shirts = 30x.
Total cost of the $60 shirts = 60y,
Total number of shirts purchased = x+y.
Thus, the AVERAGE cost per shirt is equal to the following:

(30x + 60y) / (x+y).

In the problem above, z represents the average cost per shirt.
Since each shirt costs either $30 or $60, the average cost per shirt must be BETWEEN 30 and 60.
Since y>x, the number of $60 shirts purchased is GREATER than the number of $30 shirts purchased, with the result that the average cost per shirt must be CLOSER TO 60 than to 30.
Of the answer choices, the only viable option is z=50.

The correct answer is D.