If x, y, and K are positive numbers such that

[bduke66]

If x, y, and k are positive numbers such that [x/(x+y)](10) + [y/(x+y)] (20)=k and if x<y, which of the following could be the value of k?

(A) 10
(B) 12
(C) 15
(D) 18
(E) 30

The answer is (D), but I'm not sure about all the steps in the process to get that answer. Some of the steps to get that answer confuses me.

[tim]

Before we help with this question, we need you to show some effort of your own. What did you try on this question? Where did you get stuck?

[bduke66]

I've got as far as finding K, which is equal to 10 + [10y/(x+y)]. And I know 0 < x <y, but I'm not sure on the next steps after this.

[tim]

okay, your calculations appear to be wrong, because k shouldn't simplify to that. this is actually a classic weighted averages problem. you should recognize this because of the x+y in the denominator and the weighted version (10x+20y) in the numerator. think of it this way:

let's say we're creating a new blend of gasoline from brands X and Y. brand X costs $10 a gallon and brand Y costs $20 a gallon. let's combine x gallons of brand X and y gallons of brand Y. how much should we sell the blend for? well, let's divide the total cost by how many gallons there are.

total cost
x gallons at $10 a gallon plus y gallons at $20 a gallon
10x + 20y

number of gallons
x + y

this gives us a cost per gallon of [x/(x+y)](10) + [y/(x+y)] (20), which is exactly what k is according to the problem. well, what values of k make sense now that we know k is the cost of a blend of $10 and $20 brands? first, there is no way a blend of $10 and $20 gasoline can cost any less than $10 a gallon or more than $20 a gallon. but we also know that x < y, which means we have more of brand Y than brand X. this means the cost should be closer to $20 than $10, so k must be more than 15. this means k can be anything greater than 15 but less than or equal to 20, and 18 is the only answer choice that works.

please note that you would likely NEVER come up with this approach unless the concept of weighted averages was already on your radar. so make sure you learn this concept and more importantly how to recognize it in similar situations, so that next time you see a weighted averages problem you'll recognize it as such.

[bduke66]

Never thought of it that way but thanks for your help.

[jlucero]

Glad it helped.

[bduke66]

There's usually more than 1 way to work out a problem. If I wanted to approach this problem from another angle, any examples of how i would do this, and\or tips.

[jlucero]

Personally, I'm a big fan of plugging in numbers in a VIC question like this. It's something where you have to be quick with picking numbers and making calculations in order for it to be efficient. But I pick very simple numbers to see what sort of values I can get. What I notice about the two fractions is that x+y must be greater than zero and either x or y is in the numerator. So my first instinct is to pick:

x=1
y=0
k=10

and

x=0
y=1
k=20

It takes a bit of logic to work through this, but I've basically found my two extremes for all values of x and y. Take a look at the two fractions they've given us: x/(x+y) + y/(x+y) = 1. That means that as one goes up, the other goes down. So let's re-examine the equation:

10 (some%) + 20 (100-some%) = k

10 (100%) + 20 (0%) = 10
or
10 (0%) + 20 (100%) = 20
or (when x=y)
10 (50%) + 20 (50%) = 15

But since x<y, that means the second fraction must be larger than the first. Which means that the range of values must be between x=y (50/50 split) to x is 0 (0/100 split). Notice from the work above, the answer must be between 15-20.

Notice that this takes some time to think about (and is hard to type and explain), but it's the process of quickly picking smart numbers that can make this problem not so awful. Hope that helps!

[shinyms]

hi Joe,
The question says that the nos are positive, so I am guessing we cant assume y=0?

~SS

[RonPurewal]

shiny, thanks, yes, the numbers have to be positive. (if the numbers didn't have to be positive, there would actually be more than one possible correct answer.)
i have edited the original post. the original poster put the restriction in the title, but omitted it from the question itself (!)

There's usually more than 1 way to work out a problem. If I wanted to approach this problem from another angle, any examples of how i would do this, and\or tips.

here's another way requiring less intuition -- i.e., this method is still accessible if you don't quite grasp the whole "percentage" or "weighted average" interpretation.

namely:
multiply both sides of the original equation by (x + y). this gives:
10x + 20y = kx + ky

now try plugging in the choices, and see which one works with the other stated condition x < y.

(a)
10x + 20y = 10x + 10y
20y = 10y
y = 0
... not ok, since the numbers have to be positive.

(b)
10x + 20y = 12x + 12y
8y = 2x
4y = x
not good; this means that x is four times as big as y. since the numbers are positive, that contradicts the requirement x < y.

(c)
10x + 20y = 15x + 15y
5y = 5x
y = x
oops, we're supposed to have x < y, not x = y.

(d)
10x + 20y = 18x + 18y
2y = 8x
y = 4x
winner!

since the problem says "could", we are technically done at this point (they can't write the problem so that more than one answer would work). still, since there's only one choice left, we may as well check it.

(e)
10x + 20y = 30x + 30y
0 = 20x + 10y
10y = -20x
y = -2x
not good, since the numbers are supposed to be positive.

(d) wins

[shinyms]

thanks Ron. I would never have thought of that approach.

[bduke66]

That approach is the simplest. Thanks for the reply.

[tim]

:)