Posts filed under: Q and A

Test 1 Calculator OK #16

The graph here is definitely a linear relationship between h and C, so we’re going to want to figure out the slope and the C-intercept. Then we’ll use the standard y=mx+b slope-intercept formula, only for this problem it’ll be C=mh+b.

Slope is rise over run, or the change in the vertical variable divided by the change in the horizontal variable. Let’s pick two points from the line and plug them into the slope formula. How about (1, 8) and (3, 14)?

    \begin{align*}\text{slope}=\dfrac{C_2-C_1}{h_2-h_1}=\dfrac{14-8}{3-1}=3\end{align*}

Once we have the slope, we’re actually done because whoever wrote this question wasn’t very tricky with the answer choices. The C-intercept of this graph is obviously not zero, and only choice C has the slope we just calculated plus a non-zero C-intercept. Sure enough, a quick glance at the graph confirms that the C-intercept is 5, meaning the equation should be C=3h+5.

Test 1 No Calculator #7

This is a really interesting question. The SAT is basically trying to psych you out with the complexity of the equation provided. However, note that all you’re asked to do is rearrange it to get P all by itself—that’s what it means to get P in terms of mr, and N. So let’s rewrite the original equation a bit more simply:

    \begin{align*}m&=\dfrac{\text{[apple][banana]}}{\text{[banana]}-1}P\end{align*}

See what I’m doing there? Because I can move those elements of the equation around in big chunks by multiplying and dividing, it doesn’t really matter what they are. The only question is: what do I need to do to get P by itself? And the answer to that is I need to move the big complicated fraction over to the other side of the equals sign.

    \begin{align*}m&=\dfrac{\text{[apple][banana]}}{\text{[banana]}-1}P\\\\(\text{[banana]}-1)m&=\text{[apple][banana]}P\\\\\dfrac{\text{[banana]}-1}{\text{[apple][banana]}}m&=P \end{align*}

That wasn’t so bad, right? The answer choice you seek is B—the one with the original fraction flipped and on the other side of the equals sign from P.

Test 2 Section 4 #7

When you get one of these “equivalent forms” questions, especially in a calculator section, you have lots of options. But before we get to those, let’s just make sure we understand what the question is asking.

“Which of the following equivalent forms of the equation…” means that ALL FOUR CHOICES should be equivalent—if you graph them all you’ll get the same parabola!

“…displays the x-intercepts of the parabola as constants or coefficients?” means that you’re looking for the form that has the actual numbers of both the parabola’s x-intercepts in the equation itself.

Now, here’s where your options come in.

First and fastest, if you know your parabola forms well, you know that the factored form, y=(x-a)(x-b) gives you the x-intercepts (they’re at a and b). Recognize that and you’re down to choice D right away.

If you don’t know that (or if you don’t trust your memory of arcane parabola forms), you can always graph the given equation, see its x-intercepts, and then look for those numbers!

Looks like the x-intercepts are at 2 and 4, so find the choice that has both a 2 and a 4 in it! Only choice D does, so that’s the answer.

Test 2 Section 3 #7

This is one of my favorite questions because there are a couple fun paths to take: one with pure exponent rules, and one with exponent rules and difference of two squares. So fun! (I understand if you don’t share my enthusiasm.)

First, the exponent-rules-only method. We need to take what we’re given and simplify it enough that we can stop worrying about x and focus exclusively on a and b. To do that, we first need to remember that when we raise a power to a power, we multiply the powers. In other words:

    \begin{align*}\dfrac{x^{a^2}}{x^{b^2}}&=x^{16}\\\\\dfrac{x^{2a}}{x^{2b}}&=x^{16}\end{align*}

From there, we need to remember that when we divide exponential expressions with the same base, we subtract the exponents.

    \begin{align*}\dfrac{x^{2a}}{x^{2b}}&=x^{16}\\\\x^{2a-2b}&=x^{16}\end{align*}

From there, we’re good to eliminate x and just say that 2a-2b=16. Since the question asks us for the value of a-b, just divide that equation by 2 to get a-b=8.

Going the second way involves the division exponent rule but not the power-to-a-power rule. Skip right to the division rule:

    \begin{align*}\dfrac{x^{a^2}}{x^{b^2}}&=x^{16}\\\\x^{a^2-b^2}&=x^{16}\end{align*}

From there, we can say that a^2-b^2=16 and use our difference of two squares skills.

    \begin{align*}a^2-b^2&=16\\(a+b)(a-b)&=16\end{align*}

The question tells us that a+b=2, so let’s substitute:

    \begin{align*}(a+b)(a-b)&=16\\2(a-b)&=16\\a-b&=8\end{align*}

So there you go—two fun ways to get this one done. Which do you prefer?

It might be a stupid question, but on exercise 6 about plugging in, I plugged in 3 for x, and none of the answer choices were correct. Letter c was the closest, but still, 3 raised to 6 equals 729; minus 9 equals 720. okay, but the result for letter c, if y=9, is 72.
So, I wanted to ask you if, on this specific exercise, the only option to get the right aswer is plugging in 2 for x.
(sorry if I said any terms in the wrong way, engish is not my first language)

Here’s the question for everyone else playing along:

I think the problem is that you’re squaring x instead of cubing it. If you’re plugging in x = 3, then you should have x^3=3^3=27 for y and x^6-x^3 should be 3^6-3^3=729-27=702.

So you’re looking for an answer choice that equals 702 given y = 27. Choice C works: 27(27 - 1) = 702.

For the question 8 about plugging in
I plugged in 30 for the number of desks, and so e=25 and c=48. But, it didn’t match any of the answer choices. I don’t understand what went wrong. I checked it over and over again. Pleease help aaaaa

Thanks for the question, but I think you mean #10? This one?

Let’s go step by step, starting with your choice of 30 for the number of desks. Five desks are not occupied, so you’re correct that there are 25 employees. So far so good.

All but two employees have two chairs. So 23 employees have 2 chairs each—46 chairs. Here’s the tricky part: all other desks, whether occupied or not, have one chair. That means the 2 employees that don’t have two chairs plus the 5 empty desks each have a chair at them. So there are 46 + 2 + 5 = 53 chairs.

Which expression is equal to 53? Choice D works: 2e + 3 = 2(25) + 3 = 53.

Can you do Test 7 Section 4 #30?

The manager projects that the percent increase from 2012 to 2013 will be double the increase from 2013 to 2014. You’re given the 2012 and 2013 numbers, so start by figuring out that percent increase.

There were 5600 subscriptions sold in 2012 and 5880 sold in 2013. Percent increase is change divided by original value times 100%:

    \begin{align*}\text{percent change}&=\dfrac{\text{change}}{\text{original value}}\times 100\%\\&=\dfrac{5880-5600}{5600}\times 100\%\\&=5\%\end{align*}

If there was a 5% increase from 2012 to 2013, and the manager believes that will be double the increase from 2013 to 2014, then he projects a 2.5% increase into 2014.

All we need to do from here is apply 2.5% growth to the 2013 number we know to arrive at the manager’s 2014 projection.

    \begin{align*}5880+\left(\dfrac{2.5}{100}\times 5880\right)=6027\end{align*}

So the answer is B.

Can you help with 33 in test 7 section 4?

Yep! This one is all about setting up the equations—translating words into math. The question tells us that the score in the game is calculated by subtracting the number of incorrect answers from twice the number of correct answers. Let’s say x is the number of correct answers and y is the number of incorrect answers. “Subtracting the number of incorrect answers” is easy enough: you’re going to have a ” – y” in your score equation. “Twice the number of correct answers” is pretty straightforward, too: if the number of correct answers is x, then twice that is 2x. So we can say that:

Score = 2xy

We’re not done though. We need to write another equation before we can solve. The question tells us the total number of questions answered; we need to use that information. The total number of questions answered must be the number of questions answered correctly plus the number answered incorrectly, right?

Questions = xy

Now we can plug in he values we know and solve the system of equations.

50 = 2xy
40 = xy

The question asks for the number the player answered correctly, so we need to solve for x. Conveniently, we can solve for x by eliminating y—all we need to do is add the equations together!

    50 = 2x – y
+ (40 = xy)
    90 = 3x

Of course, if 90 = 3x, then 30 = x. 30 is the answer.

Can you help me with the Official Test 4 Question 21?

Yep! I actually think it’s helpful to do a little drawing here. Below is my very rough (sorry, had to draw it on my laptop trackpad) drawing of the axes as described in the question, with year 2000 data on the x-axis and year 2010 data on the y-axis. I’ve also drawn the yx line.

Now let’s pick a set of bars and try to plot it. Looks like the bars for wood were at about 2.25 quadrillion Btu in 2000 and a little less than 2 quadrillion Btu in 2010.

Biofuels, on the other hand, were fairly small in 2000 (around 0.25) and larger in 2010 (say around 1.8).

Do you see what’s going on here? When consumption of an energy source grew from 2000 to 2010 (like biofuels), its dot will be above the yx line. When consumption declined from 2000 to 2010 (like wood), that dot will below the yx line.

Now you can stop plotting and just count the energy sources for which the 2010 bar is higher than the 2000 bar. I’m seeing three: biofuels, geothermal, and wind. So 3 is your answer.

Can you do test 4 section 4 number 35 please?

This is a very SAT-ish question: the math is pretty trivial, but conceptually the question is still a bit tricky.

There are a couple things you need to have nailed down to get this one right.

First, you need to recognize that they’re asking you for a ratio of two dynamic pressures, so they’re asking you for a ratio of two q‘s. Let’s call them q_1 and q_2. We’ll say q_1 is the one that corresponds to a velocity of v and q_2 is the one that corresponds to a velocity of 1.5v.

Second, you need to make sure you’re providing the ratio that’s asked for: the q of the faster fluid to the q of the slower fluid. Which fluid is faster—the one with velocity v or the one with velocity 1.5v? 1.5v is always going to be a larger number than v, so that’s the faster fluid. Therefore, we need to calculate the ratio of q_2 to q_1.

    \begin{align*}\dfrac{q_2}{q_1}&=\dfrac{\frac{1}{2}n\left(1.5v\right)^2}{\frac{1}{2}nv^2}\\\\&=\dfrac{(1.5v)^2}{v^2}\\\\&=\dfrac{2.25v^2}{v^2}\\\\&=2.25\end{align*}

How do you do Test 2 Section 3 Number 18?

The key here is to recognize that you’re dealing with similar triangles (pro tip: similar triangle questions often take this “hourglass” form). The two angles at point B are vertical, so they must be congruent. And because segments AE and CD are parallel, you’ve got alternate interior angles for the rest. When all the angles in two triangles are congruent, those triangles are similar.

(It might be easier to see the alternate interior angles if you extend the lines…expand the image on the right.)

Anyway, now that you’ve established that these are similar triangles, you just need to use ratios to solve. If AB=10 and [latexBD=5[/latex], then each set of corresponding sides will be in the same ratio. So we can solve for BC thusly:

    \begin{align*}\dfrac{AB}{BD}&=\dfrac{BE}{BC}\\\\\dfrac{10}{5}&=\dfrac{8}{BC}\\\\2&=\dfrac{8}{BC}\\BC&=4\end{align*}

But wait—you’re not quite done! The question asks for CE, not BC!

CEBCBE = 4 + 8 = 12

There. Now you’re done.

Could you please explain Test 2 Calculator Active #25

Sure thing. The best way to see what’s going on with this one is to plug in. When they tell you that ab = 0 and a ≠ b, they’re telling you that a and b must be nonzero opposites (i.e., that a = –b and that a ≠ 0). So just pick any nonzero number and rock and roll!

I’ll pick 3 for a, which means I’m picking –3 for b. Therefore, the graph passes through (3, 0) and (0, –3). What do we know about its slope? (Don’t just imagine it in your head—draw it in your test book!)

That’s a positive slope! Conveniently, finding a positive slope for even one set of values of a and b is enough to eliminate choices B, C, and D: a positive slope is not negative, zero, or undefined!

If you’re nervous though, maybe see what happens if you set a = –3 and b = 3 instead. Then your line would go through (–3, 0) and (0, 3). That’s still a positive slope!

Test 7 Section 3 #20 please

This is a funny question because unlike most #20s, there’s not much going on conceptually. However, it’s still an easy one to miss. The best advice I can give about questions like this is to write out every step.

    \begin{align*}&(7532+100y^2)+10(10y^2-110)\\=&7532+100y^2+(100y^2-1100)\\=&100y^2+100y^2+7532-1100\\=&200y^2+6432\end{align*}

So in ay^2+b, we now know that a = 200 and b = 6432. Therefore, a+b=200+6432=6632.

Test 7 Section 3 #18 please

To convert between radians and degrees, you just need to remember that a full circle, 360°, is the same as 2π radians. You can use that fact to solve any degree/radian conversion question with a simple ratio:

    \begin{align*}\dfrac{\text{angle in degrees}}{360^\circ}&=\dfrac{\text{angle in radians}}{2\pi}\end{align*}

In this case, we take the 720° we’re given and solve thusly:

    \begin{align*}\dfrac{720^\circ}{360^\circ}&=\dfrac{\text{angle in radians}}{2\pi}\\\\2&=\dfrac{\text{angle in radians}}{2\pi}\\\\4\pi&=\text{angle in radians}\end{align*}

That tells you that the a in  is equal to 4.

Hi Mike, Can you work the solution for Test 7, Section 3, #13? Thanks!

Absolutely. First of all, this is a great one to plug in on. Equivalent means they’ll be the same for all values of x, so pick a value of x that’s easy to work with and get crackin’. I’m going to use x = 4, because that’ll make the denominator (x – 3) equal to 1.

The original expression:

    \begin{align*}&\dfrac{x^2-2x-5}{x-3}\\\\=&\dfrac{4^2-2(4)-5}{4-3}\\\\=&\dfrac{16-8-5}{1}\\\\=&3\end{align*}

Now, which answer choice gives you 3 when you plug in 4 for x? You can cross off A and B right away because they’re going to be negative—they start with x – 5! So try C and D:

    \begin{align*}\text{C)}\qquad &x+1-\dfrac{8}{x-3}\\\\=&4+1-\dfrac{8}{4-3}\\\\=&-3\\\\\text{D)}\qquad &x+1-\dfrac{2}{x-3}\\\\=&4+1-\dfrac{2}{4-3}\\\\=&3\end{align*}

Boom! D it is.

If you don’t want to plug in, you’re going to need to do polynomial division. (In other words, you should want to plug in!)

Polynomial long division is a real pain to render in text, so please forgive the handwriting.

Step 1: You’re always looking at the lead terms in long division. The leading term of the divisor, x, goes into the leading term of the dividend, x^2, exactly x times. Therefore, put an x at the top, and then subtract the product of x and x-3 from the dividend:

Step 2: x goes into x exactly 1 time, so put a + 1 on top and then subtract the product of x-3 and 1 from the x-5 we’re working with:

Step 3: Now you’re done, because x can’t go into –2. What’s on top is your quotient, and what’s left on bottom is your remainder.

The SAT will often show quotients and remainders as it does in the answer choices of this question: in the \text{QUOTIENT}+\dfrac{\text{REMAINDER}}{\text{DIVISOR}} form. In other words, x+1-\dfrac{2}{x-3} is the answer we’re looking for.