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Posts filed under: Q and A

Test 6 calc section number 28

There are two equally valid ways to go here. If you’ve read enough of me, though, you won’t be surprised to learn that I’ll suggest the simpler, less mathy way first.

The question tells you that two points are both 3 units away from –4. You don’t need to make any equations to know what those points are! They’re –1 and –7! So just work through the answer choices until you find one that is true for both x = –1 and x = –7. Start with choice A:

    \begin{align*}|x+4|&=3\\\\|-1+4|&=3\\|3|&=3\\3&=3\\\\|-7+4|&=3\\|-3|&=3\\3&=3\end{align*}

Lookit that: already done!

Algebraically, what’s happening here is that you’re looking to write an equation for “the positive difference between x and –4 is 3.” The way you write “the positive difference” algebraically when you don’t know which of the values you’re subtracting is bigger is you put the subtraction in absolute value brackets. Therefore, “the positive difference between x and –4″ is written |x-(-4)|=|x+4|.

The reason I love the simpler way on this question is that you can just as easily write |-4-x| for the positive difference between x and –4. That’s equivalent because of the absolute value brackets, of course, but if that’s what you start out writing, you might not recognize the correct answer choice right away. If, on the other hand, you just start out by finding the answer choice that works for –7 and –1, you can’t go wrong!

Test 6 calc section number 35

Remember that at the x-intercept, y = 0. Therefore, all you need to do to find the x-intercept is substitute 0 for y and solve!

    \begin{align*}\dfrac{4}{5}x+\dfrac{1}{3}y&=1\\\dfrac{4}{5}x+\dfrac{1}{3}(0)&=1\\\dfrac{4}{5}x&=1\\4x&=5\\x&=\dfrac{5}{4}\end{align*}

How do you do #18 in Test 6 Section 3 without a calculator?

The key to this one is recognizing that you’ve got similar triangles. First, both triangle CDB and triangle CEA contain angle C. The question also tells you that \overline{BD}\parallel\overline{AE}, which means that angle CDB is also a right angle. Therefore, you have angle-angle similarity.

It’s also good to recognize a 6-8-10 triangle quickly. Know your Pythagorean triples! 🙂

From there, all you need to do is recognize that the length of \overline{AE} is 3 times the length of \overline{BD}, which means all the sides of triangle CEA are 3 times the length of their corresponding sides in triangle CDB. Therefore, because the length of \overline{CD} is 10, the length of \overline{CE} must be 30.

What’s the fastest way to do test 5 #18 in the no calculator section?

Because this is a no calculator question, it is worth your while to take a second to strategize before you dive in. In this case, a couple seconds of thinking will tell you two things:

  1. Those fractions will be easy to cancel out if you multiply by 2.
  2. You’re only asked to solve for x, so substitute accordingly.

First, cancel out the fraction:

    \begin{align*}\dfrac{1}{2}(2x+y)&=\dfrac{21}{2}\\2\left(\dfrac{1}{2}(2x+y)\right)&=2\left(\dfrac{21}{2}\right)\\2x+y&=21\end{align*}

Now take the second equation, y=2x, and substitute:

    \begin{align*}2x+y&=21\\2x+2x&=21\\4x&=21\\x&=\dfrac{21}{4}\end{align*}

You can grid in that fraction or its decimal equivalent: 5.25.

Can you do Test 2 Section 3 #14? Thanks!

The thing to know here is that the basic form for exponential growth or decay is this:

\text{Value after }n\text{ periods}=(\text{Starting Value})(1+\text{Growth Rate per Period})^n

When this question tells you that the substance decays at an annual rate of 13%, that means you have a growth rate of –0.13. The question also tells you that the starting amount is 325 grams and that the period in years is represented by t.

We can plug the given information into the standard form:

f(t)=325(1+(-0.13))^t

Simplify the growth rate and you have your answer:

f(t)=325(0.87)^t

Test 4, Section 4, Number 36 (Calculator Section)

Great question. Remember that arc length and central angle are related thusly:

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

You have a circle with radius 10, so its circumference is 2\pi r=20\pi, or about 62.83. If the arc formed by the central angle has a length between 5 and 6, that means it’s between \dfrac{5}{62.83} and \dfrac{6}{62.83} of the full circumference.

    \begin{align*}x_\text{smallest}=360\left(\dfrac{5}{62.83}\right)=28.65\end{align*}

    \begin{align*}x_\text{biggest}=360\left(\dfrac{6}{62.83}\right)=34.38\end{align*}

Therefore, x can be any integer between 28.65 and 34.38: 29, 30, 31, 32, 33, or 34.

Practice Test 4, Section 3, Number 11 (No Calc)

I love this question because the fastest way to go involves almost no math. You just have to know a little bit about the shapes of lines and parabolas.

First, think about the parabola. Its equation is y=(2x-3)(x+9). From that we know it’s a parabola that opens up (the x^2 term will be positive) and has x-intercepts at \dfrac{3}{2} and -9. You should figure out its y-intercept, too, by plugging in zero for x:

    \begin{align*}y&=(2(0)-3)(0+9)\\y&=(-3)(9)\\y&=-27\end{align*}

Do a very rough drawing of that on your paper (forgive my MS Paint skillz, but your drawing can be sloppier than mine and still be plenty good enough):

Now do a rough drawing of the line. To do that, put it in slope-intercept form:

    \begin{align*}x&=2y+5\\x-5&=2y\\y&=\dfrac{1}{2}x-\dfrac{5}{2}\end{align*}

The important detail there is that the y-intercept is -\dfrac{5}{2}, which is above the parabola’s y-intercept of -27, so you know the line will intersect the parabola twice. Like so:

Practice Test 6, number 17, no calc.

I find the way to get questions like this right without worrying too much about fractions is just to break everything down into the most simple steps. In other words, instead of trying to divide by a fraction all at once, first multiply by the denominator to kill the fraction, then solve as you normally would.

    \begin{align*}\dfrac{2}{3}t&=\dfrac{5}{2}\\\dfrac{2}{3}t\times 3&=\dfrac{5}{2}\times 3\end{align*}

That will eliminate the fraction on the left:

    \begin{align*}\\2t&=\dfrac{15}{2}\end{align*}

Now you just need to divide by 2 (or multiply by \dfrac{1}{2}) to get t all by itself:

    \begin{align*}2t\times \dfrac{1}{2}&=\dfrac{15}{2}\times\dfrac{1}{2}\\t&=\dfrac{15}{4}\end{align*}

Hi Mike, I find these Qs confusing and I lose valuable time trying to think my way through them (usually get them wrong anyway!) What’s a good stepwise approach? Thanks!

If 6 < |x-3| < 7 and x < 0, what is one possible value of |x| ?

I think it’s helpful on questions like this to plug in—to think in terms of actual numbers as much as possible. In this case, I’d look at the inequality, which says that |x-3| must be between 6 and 7, and I’d simplify that by saying that 6.5 is between 6 and 7, so why not just say |x-3|=6.5?

From there, you can say that x-3=6.5, in which case x=9.5, or x-3=-6.5, in which case x=-3.5. Since the question says that x<0, we must choose x=-3.5, which means |x|=3.5. So that’s what we grid.

Algebraically this is a bit trickier, but if you want to see a step-by-step solution, you have to begin by noting that removing the absolute value brackets from the original inequality range results in two possible inequality ranges:

    \begin{align*}6&<x-3<7\\&-or-\\-6&>x-3>-7\end{align*}

Again, we know that x<0, so we only need to deal with the second inequality. Solve for x by adding 3 to each part:

    \begin{align*}-6+3&>x>-7+3\\-3&>x>-4\end{align*}

So x must be between –3 and –4. That means the absolute value of x must be between 3 and 4.

Can you do Test 6 Section 3 #15?

Sure can. I’ll even do it two ways. First, I’ll plug in: let’s say a=3 and b=4. Then we can say that \left(a+\dfrac{b}{2}\right)^2=\left(3+\dfrac{4}{2}\right)^2=25. When you plug those values into the answer choices, which one gives you 25? Only choice D will: a^2+ab+\dfrac{b^2}{4}=3^2+(3)(4)+\dfrac{4^2}{4}=9+12+4=25.

To get this algebraically, all you need to do is FOIL. Just be super careful with the fraction.

    \begin{align*}&\left(a+\dfrac{b}{2}\right)\left(a+\dfrac{b}{2}\right)\\\\=&a^2+\dfrac{ab}{2}+\dfrac{ab}{2}+\left(\dfrac{b}{2}\right)^2\\\\=&a^2+\dfrac{2ab}{2}+\dfrac{b^2}{4}\\\\=&a^2+ab+\dfrac{b^2}{4}\end{align*}

Dear Mike,

Can you do Khan Academy Test #4 Section 4 #23? I was wondering, how do you find Standard Deviation on your calculator (I have the normal TI-84 Plus)?

Pro tip: while your calculator certainly can do it, you will never need to calculate standard deviation to answer a standard deviation question on the SAT. All you need to do is look at the data and figure out which set is more spread out from the mean. In this case, you see that the mean for City A must be right around 79° degrees, since 14 of the 21 days were 79°. All that data is very clustered together around 79°, so the standard deviation will be low.

On the other hand, the mean for City B will be around 78°, even though there were 6 days at 80° and 6 days at 76°. That data is more spread out, so the standard deviation will be higher. Therefore, the answer is B.

It might help to see this data a bit more visually (although you probably wouldn’t do this during your test).

Again, see how the data for City A is really tightly clustered and the data for City B is really spread out? The more spread out the data, the higher the standard deviation.

Dear Mike,

Can you please do #16 in the Official Practice Test #4 NO CALCULATOR section
(aka Section 3)?

You bet!

This is a question about similar triangles. Here’s a simplified figure we can use to work through it.

Note that I’ve taken the step of drawing a vertical line from the top vertex to the base. Because the shelves are parallel, we know the angles they form with the vertical segment I drew and with the sides of the triangle will all be the same. That tells us that we have angle-angle similarity, so we know that the 2:3:1 ratio we have on the left side will also apply down the middle, like so:

Since we know that the height of the unit is 18 inches, we can solve for y:

y + 3y + 2y = 18
6y = 18
y = 3

From there, we know that the shampoo needs to fit on the middle shelf, so it needs to fit into a space that’s 3y = 3(3) = 9 inches tall.

Hey Mike, Is there a way do to this sequence problem without using the
“S=n[a1 + a…]” formula?
In an arithmetic sequence, each term after the first is found by adding the same number to the preceding term. If the sum of the first six terms of an arithmetic sequence is 78 and the sixth term is 23, what is the second term?
A) 3 B) 4 C) 7 D) 10 (The correct answer is 7)

Sure! (Although I don’t think you’d see this on the SAT—maybe a Subject Test.) You know that every term in the sequence is the same distance from its adjacent terms. Call that common difference n. Then you can say that the first six terms are:

23-5n,23-4n,23-3n,23-2n,23-n,23

You know those  terms add to 78, so solve:

    \begin{align*}(23-5n)+(23-4n)+(23-3n)+(23-2n)+(23-n)+23&=78\\6(23)-15n&=78\\138-15n&=78\\-15n&=-60\\n&=4\end{align*}

If n=4, then you can substitute into the expression we already have for the second term to get the answer. The second term is 23-4n, so 23-4(4)=7 is the answer.

Test 3 Section 4 #36

This is the calculator section; graphing will probably help you visualize what’s going on here. Don’t worry about the inequalities for now—just graph the lines. You’ll have to zoom out a bunch to see the intersection, but here’s what it looks like on my calculator:

The inequalities tell you that y is less than or equal to both of those lines, and you’re looking for the greatest possible y-value in the solution set. In other words, you’re looking for the highest point that’s BELOW OR ON both lines. With the graph in front of you, it should be easier to see that the point you need has to be the intersection of the two lines. So use your calculator to find that:

There you have it—the answer is 750.

Pro-tip: if you get another question like this, with two linear inequalities where you’re asked for the greatest (or least) possible value of one of the coordinates in the solution set, you’re looking for the intersection of the lines.

Once you know that, you may choose to just solve a problem like this algebraically next time:

–15x + 3000 = 5x
3000 = 20x
150 = x

If x = 150, at the intersection, then y ≤ 5(150); y ≤ 750.

Do you have a breakdown for the PSAT test #2 by math question topic. I didn’t see one in the help area and my book is in the mail. Maybe there is one in the book. thank you

A breakdown for PSAT/NMSQT Practice Test #1 is in the book, but I actually don’t have one for the new Test 2 yet, so thanks for asking! I will try to have it done by the end of this week, and I will post a link in the Math Guide Owners Area when it’s complete.