Posts filed under: Q and A

Test 7 Section 4 #36

Two important concepts in this question, closely related: trigonometry and Pythagorean triples.

First, the trig. The fact that \tan B=\dfrac{3}{4} means that the short legs of both triangles in the question are \dfrac{3}{4} the lengths of the longer legs.

Now the Pythagorean triples: 3-4-5 is the most important Pythagorean triple to know! It’s called a Pythagorean triple because it’s a case where three integers work in the Pythagorean theorem: 3^2+4^2=5^2. If you know the legs of a right triangle are in a 3-4 ratio, then you know that you’ve got a 3-4-5 triangle (or one of its bigger cousins)!

The question tells you that BC = 15. That’s the hypotenuse of the longer side, so your larger triangle is a 9-12-15 (AKA a 3-4-5 times 3). So AC = 9 and AB = 12.

If DA = 4, that means DB = 12 – 4 = 8. This means the smaller of the triangles you’re dealing with is a 6-8-10 (AKA a 3-4-5 times 2). You’re asked for the value of DE, which must therefore be 6.

Test 7 Section 4 Number 28

I love this question. See how the given line is going through the origin? That means that all the other points on that line will have the same ratio of t to s!

Remember that any two points on a line tell you its slope. When you know a line goes through the origin, then you know one of those points is (0, 0). So using the other point this question gives us, (3, 6), we can calculate the slope:


That will hold for all other points on the line!* If the point (s, t) is on this line, then the ratio of t to s will be 2 to 1.

* …other than the origin because of the impossibility of dividing by zero…

Test 7 Section 3 #19

Because you’re told that (1, 4) and (2, 0) are on the line, you can easily calculate its slope!


From there, just use the points you know and the slope to count steps back to the y-axis. It’s only one step! From (1, 4), a slope of –4 tells you that you go up 4 steps when you go left 1 step. That lands you on (0, 8). Therefore, b = 8.

Test 7 Section 3 #17

The first thing to do on any find-the-measure-of-a-certain-angle problem is complete any 180°s you can. In this case, triangle PQR can be completed (the measure of angle PRQ must be 50°) and \overline{MQ} can be completed (the measure of angle MPR must be 120°).

From there, you’re almost home. The question tells you that MPPR, so you know that triangle MPR is isosceles. That means that angles PMR and PRM must be congruent! Because that triangle already has a 120° angle in it, the two unknowns must add up to 60°. Because they must be congruent, they must each be 30°.

Therefore, the measure of angle QMR is 30°.

Test 8 Section 3 #14

The fastest way to figure out how many solutions a system of equations will have if you can’t just graph them (this is the no calculator section) is to set both equations equal to y, then substitute and solve.


Because the only value that makes that equation true is x = 1, the system of equations only has one solution.

Test 8 Section 4 #27 please

Yeah, this is a pretty tricky one! The key to getting through it is recognizing that whenever you have a point on a line, you can plug the coordinates into the line’s equation. Because the question wants to know about \dfrac{r}{p}, we should start by getting our equations in terms of r and p.

We know (pr) is on yxb, so we substitute:


Likewise, we know that (2p, 5r) is on y = 2xb, so we substitute:


From there, we can use the fact that both equations have a b in them to our advantage. Solve each for b, then set them equal to each other:


Now just combine like terms and solve for \dfrac{r}{p}:



Test 8 Section 4 #30

First, don’t be intimidated by all the visuals here. This one is not nearly as bad as it looks.

Start by identifying the maximum of the graph of f. Hopefully you agree with me that the greatest y-value the graph reaches is 3, which it reaches when x = 4. So we know that the maximum value of f is 3. Therefore, k = 3.

From there, all you need to do is read the table and find g(3). What does the table say in the g(x) column when it has a 3 in the x column? That’s right, g(3) = 6, so the answer is 6.

Test 8 Section 3 #18. Is there a fast way to get this?

Yes! Because the systems have –x and x terms, respectively, you can eliminate x super easily by adding the equations! This is called solving a system by elimination and this is a particularly nice example of the SAT letting you do it.


Test 8 Section 3 #20

This question is asking you to figure out the measure of a circle’s central angle from information about an arc length. This is a piece of cake if you remember that arc length, sector area (a sector is like a pizza slice shape), and central angle are all tied together via proportion:

    \begin{align*}\dfrac{\text{central angle measure}}{360^\circ}=\dfrac{\text{arc length}}{\text{circumference}}=\dfrac{\text{sector area}}{\text{circle area}}\end{align*}

Lots of circle questions (see some examples in the style of the old SAT here) will simply ask you to solve variations on that proportion.

In this case, when the question tells you that the length of arc BC is \dfrac{2}{5} of the circle’s circumference, it’s giving you the middle fraction! So we can solve for the value of x easily:

    \begin{align*}\dfrac{\text{central angle measure}}{360^\circ}&=\dfrac{\text{arc length}}{\text{circumference}}\\\\\dfrac{x^\circ}{360^\circ}&=\dfrac{2}{5}\\\\x^\circ&=144^\circ\end{align*}

On page 224, I have a question about exercise #2 (Designing and Interpreting Experiments and Studies chapter in the Problem Solving and Data Analysis section). Is Ms. Carlisle’s class considered a random selection? If so, can you generalize the results from her study to only students in her class (like future ones as well) or also to all math students?

No, I don’t think you can generally assume that any school class is a random selection. For example, it could be an honors class that only students with high grades in previous math classes could have gotten into.

In that question, though, there really aren’t even any causal results to generalize. Students weren’t randomly assigned to extra help; they self-selected. You can only conclude that an intervention (extra help) is causally responsible for some outcome (higher grades) if the intervention is randomly assigned.

Test 4 Section 4 Number 17. Could you give me the explanation for this problem? Thank you!

Sure! For questions like this, it’s really helpful to remember that the slope of any line represents the line’s rate of change—it represents how much the y-variable changes when the x variable changes. Applied to this question, the slope tells you how much the total cost of buying materials and renting tools (the y-variable) changes when the number of days (the x-variable) changes.

Because the materials cost is constant, only the tool rental costs change from day to day. Therefore, the slope tells you the daily rental cost of the tools.

You can also see this algebraically if that conceptual explanation falls flat. The question tells you that the equation for total cost y in terms of days x is:

yM + (WK)x

At Store C, which is the store question 17 is concerned with, we know from the table that M = 700, W = 20, and K = 70. Therefore:

y = 700 + (20 + 70)x
= 90x + 700

That’s the standard slope-intercept form of a line, ymxb, where m is the slope and b is the y-intercept. So the slope of the line will be 90, which is the total daily rental cost of the wheelbarrow and cement mixer.

Can you explain #9 on pg 273 in the PWN book please?

Sure. Here it is:

Break this into a few steps. First, note that when a wheel rolls without slipping, it travels a distance of one circumference when it makes one complete revolution. Since this wheel has a radius of 12 cm, it would travel 24π cm if it made one complete revolution. Of course, it’s not going to go quite as far.

Now, note that the wheel needs only to turn 30° for the line to be parallel to the ground. (Not obvious? Draw a line parallel to the ground through the wheel’s center.)

Once you turn that first 30°, you only need to turn 90° more to get the line to be perpendicular to the ground. So in total, the wheel will turn 30° + 90° = 120°.

120° is only one third of 360°, so the wheel only makes one third of a complete revolution. Therefore, it only travels one third of its complete circumference of 24π. One third of 24π is 8π, so the answer you’d grid in would be 8.

Can you give me some tips on managing my time in Section 4? What clues as to when using your calculator might be the fastest way to solve the problem in Section 4?

This depends a lot on your level of comfort and speed with your calculator—you have to know what you can do faster by hand and what you can do faster with a calculator. I’m pretty fast with my calculator, so I might decide to graph a system of equations to find its solution instead of solving the system algebraically, while someone else may not save much time that way.

Just scanning through PSAT #2 Section 4 (since I was just answering some questions from this test) I’d use my calculator for the questions listed below. Most of them are pretty mundane and obvious (like using the calculator to multiply or divide large numbers), which I suppose is my point. The opportunities to use the calculator for non-obvious things are not all that common; your decisions about calculator use alone will probably not make a huge difference in your speed over the whole section.

#3: I can do the proportion faster and more accurately with my calculator

#7: This kind of multiplication should definitely be done on a calculator.

#10: Graph both sides of the equation and find the intersection.

#12: Definitely using the calculator to do the division here.

#13: Yep, using the calculator for unit conversion.

#16 and #17: Would certainly use the calculator for division.

#20: I might graph just to make sure I’m doing my factoring right.

#24: I’d plug in values for radii here and use the calculator for all my calculations.

#27: Would definitely graph these equations to find their intersection.

#29: Would maybe graph these equations.

#30 and #31: Would use a calculator for all the arithmetic here.

what is the difference between the maximum value of y=-x^4-2x^3+5 and the minimum value of y = x^4+x^2-4

This doesn’t seem like an SAT question—maybe Subject Test? Anyway, use your graphing calculator to find the maximum of the first function (it’s 107/16) and the minimum of the second function (it’s –4). Now subtract!


PSAT #2, Section 4, #29

I’d get this one by substituting. Since both equations are given to you in y = notation, set them equal to each other.


That tells you that x is either 1 or 4, so pick one and solve for what y must be in that case. Use the second equation because it’s simpler! For example, if x = 1, then the second equation tells you that y = 1 – 1 = 0. So in that case, the product of x and y is (1)(0) = 0.

The other possibility is that x = 4 and y = 3. In that case, the product of x and y would be (4)(3) = 12.

Note that although my first instinct here was to solve algebraically, this is also a good one to use your graphing calculator (if you have one). Just graph both equations, find either of the intersections, and multiply their coordinates to find the product of x and y.