Posts tagged with: proportions

Test 3 Section 4 #26

Fun fact and major shortcut: when a line passes through the origin, the coordinates of each point are in the same ratio. For example, if you knew a line passed through (3, 6) and (30, 60), you’d know that line passed through the origin. That’s because when a line passes through the origin, its equation will simply be ymx + 0, or simply ymx. Put another way, y will be directly proportional to x.

We can use that fact to solve quickly:

    \begin{align*}\frac{2}{k}&=\frac{k}{32}\\\\k^2&=64\\k&=\pm 8\end{align*}

Of course, because there are no negative answer choices, we choose C for 8.

The other way to go here would be to backsolve by just trying to draw the points. You can cross off A right away because any line that goes through (2, 0) and (0, 32) can’t also go through the origin! Even a rough drawing should convince you that the line containing(2, 16) and (16, 32) doesn’t go through the origin. Things get a little trickier with (2, 4) and (4, 32) because it might look CLOSE, but hopefully the right answer looks even closer.

Screenshot_3_17_16__9_07_PM

The table above lists the results of a survey of a random sample of 250 high school seniors and freshmen. Each student selected one subject that was his or her favorite.If the sample is representative of a high school with 2,000 freshmen and seniors, then based on the table, what is the predicted number of freshmen at the high school who would select math as their favorite high school subject? A) 520 B) 240 C) 65 D) 30

If the survey is of a true random sample, then you would expect the proportion of freshmen who like math in the sample to be pretty close to the proportion of freshmen who like math in the whole school’s population. \dfrac{30}{130}=\dfrac{3}{13} of freshmen chose math as their favorite subject, so roughly 3/13 of the freshman population probably feels the same way.

In my opinion, this question is trying to get a little too cute by using freshmen and seniors (instead of just including all grades). I don’t think a real question would do that. But what the question is getting at when it says that there are 250 total in the sample and 2000 freshmen and seniors overall is that you can just multiply every number by 8 and have a decent approximation of the preferences of the full population because \dfrac{250}{2000}=\dfrac{1}{8}.

So, just multiply! 8\times 30=240, so the answer is B.

The volume of water in a small pool is directly proportional to the depth of water. When there are 10 cubic feet of water in the pool, the depth of the water is 5 inches. How many cubic feet of water are in the pool when the depth of the water is 2 feet?

When two variables are directly proportional, they always make the same fraction. Just make sure you convert 2 feet to 24 inches before you set up your proportion!

\dfrac{10 \text{ ft}^3}{5 \text{ in}}=\dfrac{x\text{ ft}^3}{24\text{ in}}

x=48\text{ ft}^3

Someone requested recently at my Q&A page that I post a challenge question about direct and inverse variation. I quickly agreed, but it’s taken me longer than I wanted it to because I’ve been having a hard time hitting the right level of difficulty for a challenge question. I’m still not sure I’ve nailed it, but regardless, the wait is over. First correct response, as usual, gets a Math Guide shipped to their home (in the US) for free. Full contest rules, as always, apply.

  • p and q are directly proportional to each other; their proportionality constant is k
  • x and y are inversely proportional to each other; their proportionality constant is j
  • k = j
  • a is a constant greater than 1

According to the conditions above, what is one possible value of y in terms of a when p = a3, q = a, and x = a?

Put your answers in the comments. If you’re not registered with this site and your comment doesn’t appear immediately, don’t panic. I get comments in the order they’re submitted, but not everything shows up right away because I’m trying to prevent spam.

UPDATE: Rushil won the book, and Peter gave a nice explanation in the comments. My solution is below the cut.

To begin work on this, get all the information in the bullet points into equations.

If p and q are directly proportional, with proportionality constant k, there are actually two equations you could write:

This is because all that really needs to be true is that p and q always make the same fraction. I’m going to deal with the first one first, and then I’ll give a quick treatment to the second one. I would have accepted the answer that results from either one.

If x and y are inversely proportional, with proportionality constant j, there’s only one equation you can write:

Of course, if k and j are equal, then we can set up one equation:

And now we’re ready to substitute in our values. Remember, p = a3, q = a, and x = a. So here we go:

And now we solve for y:

Had you used the second version of the p and q equation, you’d instead get y = a–3 here instead. I would have accepted that as an answer as well. Here’s how that’d work:

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 – 32+ 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.

There are two kinds of proportionality (some call these problems “variation” problems, but I’m sticking with proportionality) problems that you might see on the SAT: direct and inverse. I’m going to cover both here since I’m in the business of preparing you for any eventuality, but you should know that the the former is much more prevalent than the latter. Don’t sweat inverse proportions all that much.

Direct Proportions

There are a few ways to represent direct proportionality mathematically. The Blue Book likes to say that when x and y are directly proportional, y = kx for some constant k. This definition is correct, of course, but I find it to be less useful since it introduces an extra value into the mix and doesn’t lend itself as easily to the kinds of questions you’ll usually be asked on the SAT. I much prefer to say:

When x and y are directly proportional:
 

Note that k is still in there, but we don’t have to deal with it directly anymore. I like to streamline.

In a direct proportion, as one value gets bigger, the other gets bigger by the same factor. As one gets smaller, the other gets smaller by the same factor*. Observe:

p and q are directly proportional
p = 4, q = 10
= 8 (), q = 20 ()
= 40 (), q = 100 ()
= 2 (), q = 5 ()
= 1 (), q = 2.5 ()

For easy direct proportion questions, all you’ll need to do is plug values into the proportion above, and solve. And the “hard” direct proportion questions won’t actually be much harder.

Let’s see an example
  1. If y is directly proportional to x2, and y = 8 when x = 4, what is y when x = 5?
     
    (A) 5.12
    (B) 10
    (C) 12.5
    (D) 14
    (E) 25

What makes this question tricky is that y is proportional to x2, but we’re given values of x. Don’t get freaked out. Just square your x values before you plug them into the proportion. 42 = 16, and 52 = 25. Watch:

No big deal, right? You should get y2 = 12.5, which is answer choice (C).

Inverse Proportions

Inverse proportions are much less common on the SAT, but as I said above, they are fair game and so you should know what to do if you actually do encounter one. Again, the Blue Book’s definition of inverse proportions (y = k/x for some constant k) involves a constant and is therefore not the most expedient definition for test-day deployment. Instead, think of it this way:

When x and y are inversely proportional:
 
 

Again, note that k hasn’t disappeared; k is equal to both sides of the equation. I’ve just given you one less value to keep track of and have a name for.

In an inverse proportion, as one value gets bigger, the other gets smaller, and vice versa*:

p and q are inversely proportional
= 4, q = 10
= 8 (), q = 5 ()
= 10 (), q = 4 ()
= 2 (), q = 20 ()
= 1 (), q = 40 ()

So…how about another example?
  1. If u and w are inversely proportional, and u = 11 when w = 5, what is u when w = 110?
     
    (A) 242
    (B) 50
    (C) 5
    (D) 1
    (E) 0.5

Drop your values into the formula, and you’re good to go:

Solve, and you’ll find that u2 (uns…dos…tres…CATORCE!) is 0.5, so choice (E) is correct. Note that choice (A) is there in case you misread the question and set up a direct proportion instead. Note also that in the first example above, choice (A) would have been the solution to an inverse proportion. I’m not doing that to be a jerk. I’m doing that because the SAT will. Read the question carefully.

* Of course, these statements assume a positive k. If k is negative (in the realm of possibility I suppose, but I’ve never seen it on the SAT) you’d have to revise them to be about absolute values.

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