Source: Nature’s Graffiti.

Do you guys do the Black Friday thing? One time, about 7 years ago, I got up at 3 in the morning and met some friends to wait outside Best Buy. It was complete pandemonium, and I was not fast or ruthless enough to get the TV I wanted. Then, I got the same TV for the same price a week later online. I still have it. I’m staring at it right now, actually. And that’s the story of why I hate Black Friday. The end.

As you may or may not know, the Beta for the Math Guide is closed since the book is almost done. I haven’t really figured out what to give away as a prize for these challenges now that I don’t have the Beta. I might come up with something good in the future, but this one’s just for good old fashioned bragging rights. So, I dunno, go on College Confidential and tell everyone you’re super smart. You’ll fit right in! (I kid, I kid. That site is a fantastic resource.)

x is directly proportional to y
x is inversely proportional to z

If the two statements above are true, and y = 10 and z = 5 when x = 2, what are x, y, and z when x + y + z = 32 and
x < y < z?

Good luck, troops. I’ll post the solution Monday (probably late).

UPDATE: Jeffery nailed it. Bragging rights awarded. Go lord it over everyone you know. 🙂

Solution below.

Jeffery did a nice job of laying out the pure algebra solution, so I’m basically just going to add some commentary. Let it be known that this question doesn’t really lend itself to much trickery–you’re going to have to do some equation solvin’. Remember, these Weekend Challenges are supposed to be harder than regular SAT questions!

If x is directly proportional to y, then y will always be the same quotient, no matter what. We know y = 10 when x = 2, so we know that, conveniently, y will always be 5 times bigger than x. Write the equation: y = 5x.

You can figure out the relationship between x and z in a similar way, although when we’re talking about inverse proportionality, we know that the product xz will always equal the same thing. In this case, z = 5 when x = 2, so xz = 10, always. z = 10/x.

We can now rewrite the equation from the question:

x + y + z = 32
x + 5x + 10/x = 32
6x + 10/x = 32
Now we’re going to have to solve. First, multiply everything by x to get that one out of the denominator:
6x2 + 10 = 32x
6x2 – 32x + 10 = 0
It’s not trivial to factor this, so from here you can either put it in the quadratic equation to find your roots, or you can graph it on your calculator and (depending on your calculator) figure it out that way.
The roots are x = 1/3 and x = 5.
When x = 5, y = 25, and z = 2. That adds up to 32, but breaks the x < y < z rule.
When x = 1/3, y = 5/3, and z = 10/(1/3) = 30. Yep, that adds up to 32, AND x < y < z. Nice.

Comments (4)

x is directly proportional to y:  x = ay
x is inversely proportional to z: x = b/z
when x = 2,
y = 10
z = 5

x = ay;
2 = 10a;
a = 1/5;
x = 1/5y;
y = 5x

x = b/z
2 = b/5
10 = b
x = 10/z
z = 10/x

x + y + z = 32;
x + 5x + 10/x = 32;
x² + 5x² + 10 = 32x;
6x² – 32x + 10 = 0;
6(x² – 16/3x + 5/3) = 0;
x² – 16/3x + 5/3 = 0;
3x² – 16x + 5 = 0;
(3x-1)(x-5) = 0;
x = 5 or 1/3

If x = 5:
x = 5
y = 25
z = 2
5 + 25 + 2 = 32, but this set of numbers does not meet the requirement of x < y < z.

If x = 1/3:
x = 1/3
y = 5/3
z = 30
1/3 + 5/3 + 30 = 32, and this set meets the requirement of x < y < z.

I don't think this was exactly the "best" solution, and I would love to see other approaches.. I like algebra very much, so most of my approaches are algebraic.. 🙂

Hello, here is a similar way of looking at it:

Follow whatever Jeffrey wrote so we have: 

y/x = 5        (1)
 
xz = 10        (2)

x + y +z =32   (3)

so here is the idea, substitute 5x = y into equation three and we get

6x + z =32 

add 6xz +1 to both sides

6xz + 6x + z + 1 = 93

(6x+1)(z+1) = 93

93 is factored as 3*31 so clearly z=30 and x=1/3 which implies y=5/3. 

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