Posts tagged with: algebra

If x and y are positive and sqrtx = y, which of the following must be equal 0?

A) x-y
B) x- sqrty
C) y-2x
D) y- x^2
E) y^2 -x

Square both sides:

From there, all you need to do is subtract!

You could also plug in! Say x = 49 and y = 7. Then it’s true that the square root of x equals y, and you can try each answer choice to see which equals 0. Only E will.

If 13≤31-2x≤39, which of the following represents all possible values of x?

(A) -9≤x≤4
(B) -4≤x≤4
(C) -4≤x≤9
(D) 9≤x≤22
(E) 22≤x≤35

How should I go about doing this question, and what’s the quickest way? Thanks for your help!

Treat this just like it’s any equation, only instead of remembering to do any operations on both sides of the equals sign, you have to remember to do every operation to ALL THREE PARTS of the inequality.

First, subtract 31 from each part:

13 – 31 ≤ 31 – 2x – 31 ≤ 39 – 31
–18 ≤ –2x ≤ 8

Now divide by –2, remembering that when you multiply or divide a negative through an inequality, you need to flip the direction of the inequality:

9 ≥ x ≥ –4

Which of the following must be negative whenever x>y>0?

A) x^2 -4xy + y^2
B) x^2 -2xy – y^2
C) x^2 – xy -2y^2
D) y^2 -x^2 -xy
E) y^2 -x^2 +xy

First, get rid of (A) and (E) because they have + signs in them. Those aren’t gonna work.

Now look at what you have left. Can you factor any of those in a useful way?

Hmm…(D) has a difference of two squares in it.

If x is bigger than y, then y – x is always going to be negative. Since x and y are both positive, y + x will always be positive, and xy will always be positive. Negative times positive is always negative, so  is always (positive)(negative) – positive. That’s always going to be negative.

if x^2 – y^2 = x- y and x<y, what is the average (arithmetic mean) of x and y?

A) 0
B) 1/2
C) 1
D) x/2
E) y/2

Factor:

Now divide by :

The arithmetic mean of 2 numbers is their sum divided by 2. Since you know the sum of x and y is 1, you know their arithmetic mean is .

In a certain apartment building, residents are allowed to own exactly one dog, exactly one cat, or one dog and one cat. No other pets are allowed. There are 22 cats and 17 dogs in the building. If there are 25 apartments that contain only one dog or one cat, how many apartments, in total, contain pets?

(A) 27
(B) 32
(C) 35
(D) 39
(E) 42

Thank you. 🙂

Hey, I know this question! I wrote it!

Start by drawing a Venn diagram:

The equations you can write are as follows:

xy = 17
yz = 22
xz = 25

That last one comes from what the question tells you about how many apartments contain only one dog or only one cat.

Anyway, the number you want is xyz. The easiest way to get that is to add ALL THREE equations together:

(x + y) + (y + z) + (x + z) = 17 + 22 + 25
2x + 2y + 2z = 64

Now just divide everything by 2 to get your answer!

x + y + z = 32

Which of the following equations, together with 2x + y= 8, results in a system of equations with exactly one solution?

A) -2x – y= -8
B) x+2y=0
C) 2x+y=0
D) 2x+y=14
E) 4x+2y= 16

These are all linear equations, so they’ll either have one solution (if the lines intersect), no solutions (if the lines are parallel), or infinite solutions (if the lines are actually just the same line).

Since you want one solution, you want lines that will intersect, which means you want lines that don’t have the same slopes. Put the original equation into slope-intercept () form:

I won’t put all the answer choices into that form (you can do that yourself) but I will point out that the correct answer choice, B, looks like this in slope-intercept form:

If two lines have slopes of  and , then they aren’t parallel, so they intersect.

College Board released a bunch of sample questions this week for the new PSAT and SAT, which will make their debuts in October 2015 and March 2016, respectively. Over the next few days, I’ll be making posts working through each question, a few at a time, and commenting on them when I feel like I have something insightful to say.

In this post, I’ll deal with questions 27 through 30 in the “calculator permitted” section. See questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, and 21 through 26 here.

Awesome question. You could spend some time wrestling this into ymxb form, but you don’t need to if you’re paying attention. Look at the choices! One of them is a parabola. There’s no term in , so you know you’re not going to get a parabola. Cross off D. The other thing you should notice is that A and C have non-zero y-intercepts. Since the given equation is doesn’t have any constant terms, it’s going to go through the origin. You can cross off A and C, and you’re left with B. This is one of the short-cuttiest questions yet!

OK, don’t be intimidated here. All you need to do is take some of the information you’re given, and use it correctly. Put the point (–4, 0) into the given equation, and you’ll be able to solve for c. Simple as that!

Is it weird that they tell you that c is a constant but don’t feel the need to tell you that p is a constant? Yes, I suppose. But do you need p at all to solve for c? Nope. Not at all. Is this question slightly harder because you had to puzzle with that for a minute before getting to work? You bet.

This question is kinda fun, although I bet a lot of students who encounter it will disagree with me. When the question says that those two expressions are equivalent forms, what that means is that you can set them equal to each other.

So let’s do some algebra to get A in terms of x!

Do you see the difference of two squares up top there?

Ah, finally some trig!

You’re going to want to drop an altitude segment down from the top vertex. That’s going to end up being a perpendicular bisector and make two right triangles. (You knew that, because you know those congruent base angles mean this is an isosceles triangle, right? RIGHT???)

Remember your SOH-CAH-TOA. We’re dealing with cosine here, which is the adjacent leg over the hypotenuse.

The cosine of x, then, is just .

How’s everyone else doing on this quiz?  …

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How’s everyone else doing on this quiz?

14% got 5 right
39% got 4 right
23% got 3 right
10% got 2 right
14% got 0 or 1 right

Thanks to Hurricane Sandy, this has been a rough week for a lot of people. If you’re feeling charitable, please note that I am still giving away Math Guides for donations to the Red Cross.

Let me note, before we get into this, that these challenge questions are WAY harder than anything you’d see on the SAT. That’s why I call them challenges. They test your knowledge of concepts that appear on the SAT, sure, but my challenge questions are meant to stump you for at least a few minutes.  Please don’t freak out about these, especially if you’re taking the November SAT tomorrow. Anyway, let’s get to it.

Amy is putting together an epic 7-song playlist that she can listen to while she’s working out at the gym. She’s a bit particular about how it’s constructed. The second song must be 25% longer than the first. The third song must be 12.5% shorter than the second. The fourth song must be half a minute shorter than the third. The fifth song must be 1.5 times as long as the fourth, and the sixth song must be 30% shorter than the fifth. The seventh song must be her current favorite song, “Gangnam Style,” the mp3 for which she recently purchased from Amazon. The total playlist must last exactly 25 minutes, with no pauses between songs. How long must the first song on Amy’s playlist be?

As usual, the first correct answer in the comments will win a copy of the Math Guide.

[Full contest rules.]

UPDATE: It’s taken me a long time to write the solution to this, because, well, I kinda started thinking I had already done it. Commenter Yeana got it right first, and the commenter Peter posted a good short explanation. My full explanation below the cut.

The first thing I’d recommend doing is deciding whether you want to work with minutes in decimal form, or in seconds. My preference is for seconds (and I went to some trouble to make sure none of these songs had fractional second lengths). So yeah. My explanation will use seconds.

A 25 minute playlist will be 25 minutes × 60 seconds per minute = 1500 seconds long. And “Gangnam Style” (I specified the Amazon version because I found other online versions that differed by a second and I didn’t that to frustrate anyone) is 3 minutes and 39 seconds, or 219 seconds long. Everything else you’re going to have to figure out by reading through the problem.

Let’s work through this step by step.

• Since the question asks about the first song, let’s call its length x.
• The second song’s length is 25% greater than that of the first. So it’s going to be 1.25x.
• The third song is 12.5% shorter than the second. Another way to think of this is that it’s 87.5% as long as the second. So the third song will be 0.875(1.25x) seconds long.
• The fourth song is an easy one, but at the same time continues to complicate the question. It’s 30 seconds shorter than the third song, so its length is 0.875(1.25x) – 30.
• The fifth song is 1.5 times as long as the fourth, so it’s 1.5(0.875(1.25x) – 30) seconds long.
• The sixth song must be 30% shorter than the fifth (in other words, 70% as long as the fifth). So its length is 0.7(1.5(0.875(1.25x) – 30))
• And of course, as already stated, the seventh song is 219 seconds long.
All of those add up to 1500. So let’s solve for x!
…On second thought, let’s let Wolfram Alpha do it for us. 🙂
x = 192
So the last thing we need to do is convert that to minutes and seconds. Note that seconds-to-minutes conversions are one place other than the SAT and 3rd grade where remainders are useful. The closest minute mark is 3 minutes (180 seconds), and there are 12 seconds left over. In other words, 192/60 = 3 remainder 12. So the first song is 3:12 long.

When a student asks me how to solve a math problem, my default response is to show, if possible, how to solve it by plugging in, backsolving, or guesstimating. I do this because I figure if the “math way” was obvious, the student wouldn’t be asking me for help in the first place. Besides, problem solving—in life, or on the SAT—isn’t about following a circumscribed set of procedures. It’s about creativity and flexibility. I’ve written before about the importance of being nimble. Consider this post a sequel.

It’s fun to be good at math, and it’s nice to understand how the underlying algebra on a tough word problem works. But if you’re aiming for top scores, it’s imperative that you cast a critical eye on your own ability to tease the “math way” of solving a problem out of the problem during the fairly tight time constraints imposed by the SAT.

If x + y = p and x – y = q, what is p2 + q2 in terms of x and y?

(A) 2(x + y)2
(B) 4xy
(C) 2x2 – 2y2
(D) 2(x2 + y2)
(E) 2(x2 – 4xy + y2)

Like all questions, there’s a “math way” to do this, but unlike all questions, this one is a prime candidate for plugging in. There will be some students who can breeze through the algebra in their head and identify the correct answer almost instantly. If that’s you, then great. You needn’t plug in. But if that’s not you, or if you only kinda think that’s you, then you should probably just plug in. It’s fast, it’ll get you the right answer, and then, later on you can go home, make an awesome couch fort, and figure out the algebra when you’re not pressed for time.

The plugging in solution

Say x = 3 and y = 2. Then 3 + 2 = p = 5, and 3 – 2 = q = 1. 52 + 12 = 26, so you’re looking for an answer choice to give you 26. Type the answer choices into your calculator carefully, substituting 3 for x and 2 for y, and you’ll be done in a hot second:

(A) 2(x + y)2 = 2(3 + 2)2 = 50
(B) 4xy = 4(3)(2) = 24
(C) 2x2 – 2y= 2(3)2 – 2(2)2 = 10
(D) 2(x2 + y2) = 2(32 + 22) = 26
(E) 2(x2 – 4xy + y2) = 2(32 – 4(2)(3) + 22) = –22

The algebra
Now add ’em up:
Not impossible, right? Totally doable. But arguably more involved than the plug-in solution above.
The bottom line

Look, I really just want you to be happy. If you want the algebra, I’ll give you the algebra. But I really think it’s a good idea for you to know how to plug in, too. Because if you have to ask me for the algebra on a question like this, that means it wasn’t obvious to you right away when you encountered it on the test. And that means there’s a good chance that when you sit down for the real thing, the algebra isn’t going to be obvious to you for every single question. And if, when the algebra isn’t obvious, you don’t have a backup plan, then you’re doing yourself a disservice.

Try the algebra first, if that’s your bent. But you should have a few other tricks up your sleeve for the questions where the “math way” isn’t jumping off the page onto your lap.

…Say what now?

This isn’t tested on the SAT all that often, but it has appeared (you’ll find an example in the Blue Book: Test 3 Section 5 Number 8) and I’ve had a bunch of kids tell me lately that they don’t remember ever learning it in school.

When you have two polynomials that equal each other, their corresponding coefficients equal each other.

IF:

ax2 + bx + c = mx2 + nx + p

THEN:

a = m
b = n
c = p

You might find it useful, in fact, when you’re presented with polynomials that equal each other, to stack them on top of each other and put circles (use your imagination because I can’t figure out how to circle things in HTML) around the corresponding coefficients:

ax2 + bx + c
=
mx2 + nx + p

Example (Grid-in)

(x + 9)(x + k) = x2 + 4kx + p

1. In the equation above, k and p are constants. If the equation is true for all values of x, what is the value of p?

Alllllright. First, foil the left hand side:

(x + 9)(x + k)
= x2 + 9x + kx + 9k

Might look a little better like this:

x2 + (9 + k)x + 9k

Now stack up the two sides, and see what equals what:

x2 + (9 + k)+ 9k
=
x2 + 4kx + p

So we know that:

9 + k = 4k
9k = p

From here, this is cake, no?

9 + k = 4k
9 = 3k
3 = k

9(3) = p
27 = p

Math is fun!

Try a few more, won’t you?

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Also check out this brutal old weekend challenge testing the same concept.

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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 Leonardo da Vinci totally <3’d parabolas.

The parabola is actually a hugely important mathematical concept with tons of forms, properties, and even its own history. It can open up, down, left, right, or any other direction. It can be used to graph the trajectory of my last AT&T cell phone that I threw in a lake when it dropped one too many calls. But if you’re interested in that stuff, you should go to the Wikipedia parabola article, and tattoo the word NERD on your forehead while you’re at it.

On the SAT, there’s actually not much you need to know about parabolas. So let’s keep it simple, huh?

Parabolas Are Symmetrical

This is the most important thing to remember about parabolas, because this is the key that unlocks most of the SAT’s most difficult parabola questions. The awesome thing is that you probably already knew this. The not awesome thing is that the SAT still finds ways to make you miss these questions. Let’s look at an example:

1. The graph above represents the parabolic function f(x). If the function’s minimum is at f(–3), and f(0) = 0, which of the following is also equal to 0?

(A) f(3)
(B) f(–1)
(C) f(–4)
(D) f(–5)
(E) f(–6)

Right. So let’s translate this into English first (if you’re having trouble with the function notation, have a look at this post for some help). What they’re saying here is that the line of symmetry for this parabola is at x = -3, and that the function goes right through the origin: f(0) = 0 means that this graph contains the point (0,0). They’re basically asking us to find the other x-intercept.

Here’s where they whole symmetry thing really comes in. If we know the line of symmetry, and we know one of the x-intercepts, it’s CAKE to find the other. Put very simply, they have to be the exact same distance from the line of symmetry as each other. Since (0,0) is a distance of 3 from the line of symmetry at x = -3, our other x-intercept has to be a distance of 3 away as well!

So we’re looking for the point (-6,0). Choice (E) is the one that does that for us: f(-6) = 0.

Can they find ways to make symmetry questions difficult? You bet. Will you be ready for them? Yarp.

The equation of a parabola:

y = ax2 + bx + c

Again, let me say that there are many more equations of parabolas. If you go on to do advanced math in college, you’ll need to learn some of them. If you’re doing three-dimensional math now maybe you know some of them. But you won’t need any of them for the SAT. Know this one (and what the coefficients signify), and you’re good to go.

In this equation:

• a tells you whether the parabola opens up or down. If a is positive, it’s a smiley face. If a is negative, it’s a frowny face. Easy to remember, no?
• b is pretty useless for you, as far as the SAT is concerned.
• c is your y-intercept. If there is no c, that means your parabola has a y-intercept of 0 (which is to say, it goes through the origin).

Ready for some practice problems?

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For a little bit more practice, see this previously posted parabola problem (with explanation).

Not all function questions have weird symbols, some are just vanilla f(x) type things. You’ve probably been working with the f(x) notation in school for some time now, but let’s review some of the things you’ll see over and over again on the SAT:

Interpreting function notation

One thing you’re definitely going to need to be able to do is interpret function notation. For some questions, it’s enough to remember that saying f(x) = x3, for example, is basically the same as saying y = x3.

For other questions, you’re going to need to take that a bit further and identify points on a graph using function notation. Here’s a quick cheat for you (with colors!). When you have f(x) = y, that’s the same as an ordered pair (xy). For example, if you know that f(4) = 5, then you know that the graph of the function f contains the point (4, 5). Likewise, if you know that h(c) = p, you know that the graph of function h contains the point (cp). And so on. Basically, whatever is inside the parentheses (that’s called the argument, if you care) is your x-value, and whatever’s across the equal sign is your y-value. This is important. If you don’t understand yet, read it over and over until you do. Might help to write it down. Just sayin’.

To make sure you’ve got this, think about what the following things mean. Once you’ve thought about them, hold your mouse over them (don’t click, just hover…follow directions) to see what they’re about.

• f(0) = 3
• p(12) = 0
• s(3) = r(3)
Nested functions (functions in functions)
 source

Have you ever seen those dolls where you open them up and there are smaller ones inside? If you haven’t, I’ve put a picture of some really artsy-fartsy ones there on the right. They’re called Russian nesting dolls. Welcome to the world. Can I take your coat?

Anyway, sometimes the SAT will put a function inside another function to try to bamboozle you. Don’t let yourself get stymied here. All you need to do is follow the instructions, same as you do with all other function questions. Let’s look at an example:

1. If f(x) = x2 – 10 and g(x) = 2f(x) + 3, what is g(√2)?

(A) 7
(B) 5
(C) –5
(D) –7
(E) –13

Let’s just take this one step at a time. First, let’s take the √2 we’re given and put it in for every x we see in g(x).

g(√2) = 2f(√2) + 3

Not so bad, right? Now let’s replace the f(√2) with an expression we can actually work with. Remember that f(x) = x2 – 10, so we write:

g(√2) = 2[(√2)2 – 10] + 3

See how this is working? Now just simplify:

g(√2) = 2[–8] + 3
g(√2) = –16 + 3
g(√2) = –13

So our answer is (E). Awesome, right?

Interpreting Graphs and Tables

Say the picture above shows the function g(x) over a given range. Note that, although we have no equation for g(x), we still know a lot about it. We know, for example, that g(4) = –2. We also know that the y-intercept, g(0), is about –1. Think about the following, again, mousing over them once you know what’s up to see if you’re right:

• If g(a) = 0, how many possible values are there for a in the given range?
• When, in the given range, is g(x) < –4?

For some focused practice with function notation and graph reading, click here.

You could also be presented with function information in table form. Peep this:

 x 2 3 4 5 6 f(x) 13 18 25 34 45

Just like in the graph above, we can use this table to find points. For example, f(5) = 34. See if you can do the following:

• If f(p) = 18, p =
• What is f(6 – 4)?
• What is f(6) – f(4)?
• Holy crap those are different?
• Can you figure out what the function f(x) is?
Graph Translation

Sometimes the SAT likes to test you on whether you can figure out where a graph will move based on some manipulation of its equation. Usually, though, they won’t give you the equation. They’ll draw some crappy squiggly like the one above, call it g(x), and then ask what will happen to g(x+1).

I’ll give you the rules for this, but I highly recommend reminding yourself of them with your calculator if you should need them on your SAT. It’s very easy to set a simple function (like f(x) = x2, which you’ve seen a million times) as your starting point, and experiment with your graphing calculator to see how graphs will behave based on various modifications. In fact, why don’t you play with this widget a bit right now to see what happens?

Here are those rules:

• f(x)+1 ⇒ (graph moves UP one)
• f(x)–1 ⇒ (graph moves DOWN one)
• f(x+1) ⇒ (graph moves LEFT one)
• f(x–1) ⇒ (graph moves RIGHT one)
Sample Questions!

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