Posts filed under: Q and A

In PWN p. 159 (p. 157 later printing) #8
In the xy-plane, where a and b are constants, the graphs …

The question does not specify that a and b are positive values. If one or both were negative, wouldn’t that change the answer?

Good question! It turns out, even though my solution assumes a and b are positive, that they don’t need to be. You have two parabolas with vertices on the x = 5 line, one higher than the other. As long as the leading coefficient of the higher one is greater than or equal to the leading coefficient of the lower one, the parabolas will not intersect. C is the answer.

Let’s look at the cases you’re asking about, where one or both of the coefficients is negative. First, say a = 1 and b = –1. In that case, ab, and one variable is negative.

We’re OK there—obviously those parabolas won’t ever intersect.

In the case where both are negative but ab, it turns out we’re STILL fine. Let’s say a = –2 and b = –3. It’s still true that ab, and both coefficients are negative.

Those also won’t ever intersect.

What’s going on here is that as long as ab and both signs are the same, the absolute value of the leading coefficient of the inner parabola will be higher—the inner parabola will be skinnier than the outer parabola!

That’s a lot to think about on one question though—maybe in the next edition I should just make them positive constants!

I came across a list of SAT facts to memorize, but I wonder if it’s dated.

Can you tell me if these geometry facts/formulas will be on the SAT?

1. Definition of a regular polygon.
2. The sum of the inside angles of an n-sided regular polygon is (n−2)·180◦.
3. Area of a parallelogram.
4. Definition and area of a rhombus.
5. Area of a rectangular solid.
6. Area of a right cylinder.

Thanks,
Kevin

That’s definitely not a list of the most important things to know for geometry on the SAT (which is at most 6 questions per test, often less). It’s missing important things like circle properties and the circle equation, which are common topics. Regular polygons do appear occasionally, and (n-2)\times 180^\circ could be useful (but certainly not on every test). The rest of the list is pretty weak.

I suppose a rhombus could appear, although the word “rhombus” doesn’t appear in any of the released tests, so I don’t think you need to know the definition (even though I bet you already do). I can’t think of a parallelogram area question on any tests I’ve seen.

You’re given the formulas for volume of a rectangular solid and a right cylinder in the beginning of every math section (where you’re also given much more important facts like the special right triangles and the Pythagorean Theorem). It’s good to know how to find their surface areas, too, but it’s probably sufficient just to know the basic concept of surface area—I wouldn’t recommend memorizing formulas to one of my students.

 

This no-calculator prompt is from the Daily Practice app:

Lisa gives her little brother Sam a 15-second (sec) head start in a 300-meter (m) race. During the race, Sam runs at an average speed of 5 m/sec and Lisa runs at an average speed of 8 m/sec, not including the head start.

Which of the following best approximates the number of seconds that had passed when Lisa caught up to Sam?

A. 5
B. 25
C. 40
D. 55

In 15 seconds at 5 m/sec, Sam will be 75 m ahead when Lisa starts running. Once she starts running, Lisa is going to gain 3 m/sec on Sam. How many seconds will it take to cover that 75 m gap at 3 m/sec?

    \begin{align*}\frac{75\text{ m}}{3\text{ m/sec}}=25\text{ sec}\end{align*}

The trick to this question is recognizing that you’re being asked (at least I think you’re being asked—the wording is not very precise) how many seconds have passed when Lisa catches up to Sam since the race started. You have to include Sam’s 15 second head start. Sam ran for 15 seconds, then both Lisa and Sam ran for 25 seconds, then Lisa caught up to Sam. Total elapsed time: 40 seconds.

PWN p. 151 #8 question

In PWN the SAT Math Guide (4th Ed, first printing p. 151), Polynomials chapter question #8, you explain the answer by graphing on calculator. Could you explain the answer to this question without graphing, in particular, why A, C and D are true and why B is false? Many thanks!

Sure! First, recognize that g(x) is factorable:

    \begin{align*}g(x)&=x^2+3x-10\\g(x)&=(x-2)(x+5)\end{align*}

From there, you can see that g(x) has two real zeros (2 and –5) so you can eliminate C.

Then, recognize that g(x) is a factor of f(x) as choice A says:

    \begin{align*}f(x)&=3x^3+9x^2-30x\\f(x)&=3x(x^2+3x-10)\\f(x)&=3x\left(g(x)\right)\end{align*}

Obviously that confirms that choice A is true, so you can eliminate A. However, that also confirms that choice B is false! Remember how we already factored g(x)!

    \begin{align*}f(x)&=3x\left(g(x)\right)\\f(x)&=3x(x-2)(x+5)\end{align*}

f(x) is divisible by (x + 5), but definitely not by (x – 5). You know from the way the question is constructed that only one choice will be false, so once you’ve found it, you can stop working and start bubbling. 🙂

Since you asked, though, we can also show that D is true from what we’ve already figured out using substitution. We already said f(x)=3x\left(g(x)\right), so we can substitute thusly:

    \begin{align*}h(x)&=f(x)+2g(x)\\h(x)&=3x\left(g(x)\right)+2g(x)\end{align*}

We can factor a g(x) out of each of those terms!

    \begin{align*}h(x)&=3x\left(g(x)\right)+2g(x)\\h(x)&=\left(g(x)\right)(3x+2)\end{align*}

Therefore, choice D is also true.

 

This no-calc prompt is from the Daily Practice app:

The temp T in °C of a chilled drink after the drink has been sitting on a table for m min is represented by T(m) = 32 – 28*3^(-0.05m).

What is the best interpretation of the # 32 in this context?

A. The drink is originally 32°C.
B. Every 32 min, the temperature warms by 3°C.
C. After 32 min, the drink will fully warm to the ambient temp.
D. After the drink has been sitting for a very long time, the temp of the drink will approach 32°C.

I’m going to try to talk you through my thought process for this question rather than just do math.

First step for me here is to eliminate B and C—we know that m is minutes and that appears in the exponent in this function, far away from the 32.

From there,* your job is to figure out whether the expression will approach 32 as m increases (in which case the answer is D) or get farther away from 32 as m increases (in which case the answer is A).

Think about the equation this way: T(m)=32-\text{[something]}. Of course, that something is 28\times 3^{-0.05m}, but stay with me for a minute. If the [something] is going to grow, then A will be the answer. If the something is going to shrink, then D will be the answer.

Now, does 28\times 3^{-0.05m} grow or shrink?

We know a negative exponent is the same as the inverse of the positive exponent, so let’s look at it that way:

    \begin{align*}28\times \frac{1}{3^{0.05m}}\end{align*}

When m = 0, that’s going to just be 28:

    \begin{align*}28\times \frac{1}{3^{0.05(0)}}\\28\times \frac{1}{3^0}}\\28\times \frac{1}{1}\\28\end{align*}

When m grows to 20, however…

    \begin{align*}28\times \frac{1}{3^{0.05(20)}}\\28\times \frac{1}{3^1}}\\28\times \frac{1}{3}\\9.333...\end{align*}

When m grows, our [something] shrinks! Therefore, the function will get closer to 32 as m increases, and D is the answer!

 

* From the get-go, you might also be able to eliminate A if you know a few things about Celsius. 32° C is about 90° F, which is in no way a “chilled” drink. In that case, you really need not do any math at all!

How do you solve Practice test 6 Section 3 Number 16?

Remember that, for questions like this one, you only need to find one solution, not all solutions. Therefore, go for the simplest one you can!

Think of powers you know that land on 16. The first one that comes to mind for me is 4^2=16. The only restriction the question places on a and b is that they must both be positive integers, so let’s pick numbers that will make the given equation the same as 4^2=16.

Obviously we can make the base a = 4. How do we make the exponent equal to 2? Well, \dfrac{8}{4}=2, so let’s say b = 8.

    \begin{align*}a^\frac{b}{4}&=16\\\\4^\frac{8}{4}&=16\\\\4^2&=16\end{align*}

So there you go: 8 is one valid answer. For each of the other answers (1, 2, 4, 8, and 16 are all valid) there’s an exponential expression that works. While I find some of the following routes a little less obvious, all are valid:

    \begin{align*}2^4=16\qquad\Longrightarrow\qquad b=16\end{align*}

    \begin{align*}16^1=16\qquad\Longrightarrow\qquad b=4\end{align*}

    \begin{align*}256^\frac{1}{2}=16\qquad\Longrightarrow\qquad b=2\end{align*}

    \begin{align*}65536^\frac{1}{4}=16\qquad\Longrightarrow\qquad b=1\end{align*}

PWN the SAT Math Guide p. 122 (p. 120 in newer printing) #9

h = -4.9^2 + t + 1.5
The equation … After how many seconds will the coin land on the ground?

When the coin lands, y=0, so we just need to solve the quadratic. Using the quadratic formula, however, I get a messy repeating decimal, not .664, .665. Could you solve this for me using the quadratic formula?

Also, I tried to search the Q&A here to see if you had already answered this question. Could you make an index for PWN the SAT Math Gu

Sure. In the book, I solve this by graphing—which, as you know from reading my book, I love to do. 🙂 But of course we should also land on the same result using the quadratic formula.

    \begin{align*}0=-4.9t^2+t+1.5\end{align*}

    \begin{align*}t&=\dfrac{-1\pm\sqrt{1^2-4(-4.9)(1.5)}}{2(-4.9)}\\\\t&=\dfrac{-1\pm\sqrt{30.4}}{-9.8}\\\\t&=\dfrac{-1}{-9.8}\pm\dfrac{\sqrt{30.4}}{-9.8}\end{align*}

At this point you’re going to have to enter everything into your calculator (in fact, you probably did that a few steps ago!) to see where everything lands. You’ll get a negative number (t = –0.46…) that you can ignore, and the positive number you want: = 0.664655…

As for an index for PWN questions, I don’t get all that many so I haven’t bothered with an index. Between the solutions at the end of the book and the videos in the Owners Area (for the chapters that have them) I guess people are mostly OK! Of course, I’m quite happy to post solutions here, too, so you should feel free to submit without checking. If, on the off chance, I’ve posted a solution before, I’ll link to it.

Can you solve #18 from practice test 5 calculator section? Thank you.

Sure. We can approximate the age of each tree using the table provided (the paragraph under the table says so). So a white birch tree with a diameter of 12 inches is approximately 12\times 5.0=60 years old and a pin oak tree with a diameter of 12 inches is approximately 12\times 3.0=36 years old. (You actually don’t need to do this for this question because on the test you would have figured these mechanics out on question 16 but I’m including it here for completeness’s sake.)

Now, note that the concept of a “growth factor” as provided in the table only works if the trees grow at roughly a constant rate. That means that we can incorporate units into the multiplication we just did, which helps clarify how everything is working. If we’re multiplying a diameter in inches by a growth factor and getting an output in years, what must be the units of the growth factor?

12\text{ inches}\times 5.0\text{ [???]}=60\text{ years}

The only thing you can multiply inches by to get years is \frac{\text{years}}{\text{inches}}!

12\text{ inches}\times 5.0\:\dfrac{\text{years}}{\text{inches}}=60\text{ years}

What the “growth factor” really tells us, then, is how many years it takes a species of tree to grow an inch: it takes a white birch tree 5.0 years to grow an inch and it takes a pin oak tree 3.0 years to grow an inch.We can DIVIDE a number of years by the growth factor to see how many inches a tree would grow in that many years!

10\text{ years}\div 5.0\:\dfrac{\text{years}}{\text{inches}}=2\text{ inches}

10\text{ years}\div 3.0\:\dfrac{\text{years}}{\text{inches}}=3.33...\text{ inches}

So in 10 years a white birch tree will have grown 2 inches and a pin oak tree will have grown about 3.3 inches. Since the two trees we were starting with both had a diameter of 12 inches, their new diameters will be about 12+2=14 inches and 12+3.3 =15.3 inches, respectively. That’s a difference of about 1.3 inches.

PWN the SAT Math Guide p. 112 (p. 110 in later printing) #3:

Question relating to the exponent in the answer: It is a little unclear to me that the Value column indicates the value at the END of each year. The text states that the values represent “the value of his account in each of the next three years.” Must we presume that “in each” means at the END of each of those years, at which point the interest would have been accrued for that full year?

Your question is a good one—I think I could have been a bit more careful with the wording on this one and I will tighten that up in the 5th edition. The key to getting this, though, is looking at the table and the answer choices. The answer choices show you that your starting value must be 3000 and that you’re either increasing by 10% per year (choices A and B), 1% per year (choice C), or -90% per year (choice D). Obviously choice D is right out, and a quick look at the year 1 and year 2 values (3000 to 3300) shows that 10% is what you’re looking for.

From there, you need to figure out whether year 7 will be 3000(1.1)^6 or 3000(1.1)^7. The way I’d recommend thinking about this is looking, again, at the first two values. The year 1 value is 3,000, which suggests that the exponent in year 1 is zero (i.e., 3000(1.1)^0=3000). Likewise, the exponent in year 2 should be 1 (i.e., 3000(1.1)^1=3300). Following that pattern suggests that the exponent in year 7 should be 6. Does that help?

How would you solve this one? Please see the link to the pic below.

I removed your picture link because it was a question I knew—Test 2 Section 4 #19. See my solution to that one here. (See all my solutions to the Official Tests here, btw.)

I don’t understand a question from your book. Question #2 on functions. Could you explain it in more detail?

The question:

I like to approach this by thinking about functions abstractly. A function is a process by which the thing in the parenthesis (AKA the argument) is transformed into something else using only mathematical operations. In this case, for all values of x, some mathematical operation or operations are done to x-1 to transform it into x+1.

What could those operations be? You can rule out multiplication or division—the leading coefficient of both x-1 and x+1 is 1. Likewise, you can easily rule out exponents or roots. But what about addition/subtraction? Could something be added to or subtracted from x-1 to transform it into x+1? Sure! If you add 2 to x-1, you get x+1.

So let’s think about how that works in function notation. If we want to say that the function adds 2 to its input, we can write:

    \begin{align*}f(x-1)=(x-1)+2=x+1\end{align*}

That tells us what f(x) would be:

    \begin{align*}f(x)=x+2\end{align*}

And of course from there, we can see what f(x+1) would be:

    \begin{align*}f(x+1)=(x+1)+2=x+3\end{align*}

Does that help?

College Board Test 4 Section 3 #9:
____
√x-a = x-4
If a=2 what is the solution set of the preceding equation?
A. {3, 6}
B. {2}
C. {3}
D. {6}

Is there another way to solve this quickly besides plugging in the answer choices?

Backsolve is far and away my preferred method on a question like this. The answer choices only show you 2, 3, and 6 as possible solutions, so you have to try, at most, three numbers.

However, of course there is an algebraic solution! If you go that way, you must be careful because you will generate extraneous solutions. To begin, square both sides, combine like terms, and solve the resulting quadratic:

    \begin{align*}\sqrt{x-2}&=x-4\\\left(\sqrt{x-2}\right)^2&=(x-4)^2\\x-2&=x^2-8x+16\\0&=x^2-9x+18\\0&=(x-6)(x-3)\end{align*}

From there, you must check that both solutions you generated actually work in the original equation, because when you square both sides of an equation you can cause extraneous solutions.

    \begin{align*}\sqrt{3-2}&=3-4\\1&\ne -1\end{align*}

    \begin{align*}\sqrt{6-2}&=6-4\\2&=2\end{align*}

As you can see, 3 doesn’t work in the original equation, so it’s an extraneous solution. 6 DOES work in the original equation, so it’s the one we go with.

Hopefully this reinforces the appeal of backsolving here: even if you do the algebra, you still need to try the answers you find in the original equation anyway. Why not just start by trying the answers provided!? 🙂

Test 8 Section 4 #23

I think the easiest way to get this one is to plug in some numbers. We know that every 4 quarters is a year, so the correct answer will be the one that gives us the same result for 1 year as 4 quarters, 2 years as 8 quarters, 3 years as 12 quarters, etc.

At 1 year, the original equation works out thusly:

    \begin{align*}M&=\num{1800}(1.02)^t\\M&=\num{1800}(1.02)^1\end{align*}

Which answer choice(s) result in the same expression when q = 4? Only choice A does:

    \begin{align*}M&=1800(1.02)^\frac{q}{4}\\M&=1800(1.02)^\frac{4}{4}\\M&=1800(1.02)^1\end{align*}

If you’re concerned that another one might also work, just check with your calculator!

    \begin{align*}1800(1.02)^1&=1836\\1800(1.02)^\frac{4}{4}&=1836\\1800(1.02)^{4(4)}&\approx 2471\\1800(1.005)^{4(4)}&\approx 1950\\1800(1.082)^4&\approx 2467\end{align*}

Sure enough, only choice A gives you the same result after 4 quarters that the original equation gives you for 1 year.

Test 8 Section 4 #8

This is a good one to backsolve! Note that you’re asked for a value of x + 1 (as opposed to a value for x) so backsolving is really easy. Start with the easiest ones to try.

    \begin{align*}x+1&=\dfrac{2}{x+1}\\\\\text{C)     }\ \ \qquad 2&=\dfrac{2}{2}\\\\2&\ne1\\\\\text{D)     }\ \ \qquad 4&=\dfrac{2}{4}\\\\4&\ne 0.5\\\\\text{B)     }\qquad\sqrt{2}&=\dfrac{2}{\sqrt{2}}\\\\\sqrt{2}&=\dfrac{2}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sqrt{2}&=\dfrac{2\sqrt{2}}{2}\\\\\sqrt{2}&=\sqrt{2}\end{align*}

Choice B looks good to me!

Test 8 Section 3 #13

You’re translating from words into math here. Key words to note: “at a constant rate” means you’re dealing with something linear, that has a slope. And “decreased at a constant rate” means you’re looking for a negative slope.

Because the function is a function of t, years are your x-variable and and millions of barrels is your y-variable. Calculate the slope from the numbers the question provides: 4 million barrels in 2000 and 1.9 million barrels in 2013.

    \begin{align*}\text{Slope}&=\dfrac{4-1.9}{2000-2013}\\&=\dfrac{2.1}{-13}\end{align*}

Note that while no answer choice has that slope in that exact form, choice C is equivalent. \dfrac{2.1}{-13}=-\dfrac{21}{130}.

Because all the answer choices have the same y-intercept of 4, you need not think much about it, but it does make sense given the question. There were 4 million barrels in 2000 and t represents years after 2000, so t = 0 when the y-variable is 4.