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Posts filed under: Q and A

Can you do Test 6 Section 3 #15?

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*}

Test #4 Section 4 #23?

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.

Can you please do #16 in the Official Practice Test #4 NO CALCULATOR section?

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.

If the sum of the first six terms of an arithmetic sequence is 78 and the sixth term is 23…

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

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?

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.

Could you please explain question 8 of test #4 section 4 of the blue book?

Could you please explain question 8 of test #4 section 4 of the blue book?

Sure. It’s really helpful to see this one represented visually. The question says that the line in question goes through parts of quadrants II, III, and IV. That means it goes through each of the non-gray quadrants below.

The only way a line can be in all three quadrants is if it has a negative slope, like the lines sketched on the figure below.

Undefined slope means a vertical line, so that can’t be in three quadrants. Slope of 0 means a horizontal line; that can’t be in three quadrants, either. And try as you might, you can’t draw a line with a positive slope that doesn’t eventually get into quadrant I.

PWN The SAT Book – Volume Chapter – Question 10

PWN The SAT Book – Volume Chapter – Question 10

I am having hard time understanding how to differentiate if the question wants you to answer in feet or inches, since the last part asks : “How much does the water level rise, “in inches”, when Priya places the stone into the tank.

But the answer is 0.125 inches instead of 216 inches.

Shouldn’t the water displaced by a cube of sides 6 inches be 216 inches instead of 0.125 inches ?

Thanks

Here’s the question:

I think you’re getting confused between inches (a measure of length) and cubic inches (a measure of volume). A cube with 6 inch sides will have a volume of 6 inches × 6 inches × 6 inches = 216 cubic inches (also written \text{in}^3), but that doesn’t mean that placing a 6-inch-sided cube in a tank of water will make the water rise 216 inches (18 feet!).

To solve this question, you need to consider the volume of water in the tank and how much of that water will be displaced by the stone. Since the question is asking about inches of rise, let’s deal only in inches when we do our calculations, so instead of a 3-foot by 4-foot by 2-foot tank, we’re dealing with a 36-inch by 48-inch by 24-inch tank.

The tank begins with water filled up to the 12-inch mark, which means the volume of water in the tank to start is:

36\text{ in}\times 48\text{ in}\times 12\text{ in}=20,736\text{ in}^3

As we’ve already discussed, we’re displacing 216\text{ in}^3 when we add the stone. Since the cube will be completely submerged, the easier way to think about that is that now the volume in the tank is 20,736+216=20,952\text{ in}^3.

The width and length of the tank don’t change, so we can calculate the new height of the water in the tank, x, thusly:

36\text{ in}\times 48\text{ in}\times x\text{ in}=20,952\text{ in}^3

x\text{ in}=\dfrac{20,952\text{ in}^3}{36\text{ in}\times 48\text{ in}}

x\text{ in}=12.125\text{ in}

Since the height of the water before the stone was added was 12 inches, the water level only rises 0.125 inches.

PSAT/NMSQT practice test #1 section 4 question 18

PSAT/NMSQT practice test #1 section 4 question 18

This is a great one to plug in on. Say the height of the trapezoid is 2, the lower width is 5, and the upper one is 3. That way, assuming you don’t know the formula for the area of a trapezoid (which exists—see below—but which you don’t need to memorize for SAT purposes) you can break the trapezoid neatly into two right triangles and a rectangle thusly (note that my drawing is not to scale):

The rectangle has an area of 3\times 2=6, and the triangles each have an area of \dfrac{1}{2}(1)(2)=1. Total area of the trapezoid: 1+6+1=8.

Now if you do the manipulations the question asks you to do, the height gets cut in half and the bases are doubled:

The areas of the triangle are still \dfrac{1}{2}(2)(1)=1. The area of the rectangle is also unchanged: 6\times 1 = 6. Therefore, the area of the trapezoid is still 1+6+1=8.

The answer is C: the area does not change.

Since this post will be a reference point now, you can also do this using the area of a trapezoid formula:

A=\dfrac{b_1+b_2}{2}h

If you double both bases and cut the height in half, you get:

A=\dfrac{2b_1+2b_2}{2}\dfrac{h}{2}

A=\left(b_1+b_2\right)\dfrac{h}{2}

Of course, that’s equivalent to the original formula. Therefore, doubling the bases and halving the height won’t change the area of any trapezoid.

I want to stress, though, that the lesson you should take from this question is not that you need to memorize the trapezoid area formula. Rather, the lesson should be that you can 1) plug in, and 2) break more complex shapes into things like triangles and rectangles that you already know how to work with.

Test 3 Section 4 #34

Test 3 Section 4 #34

Remember that circle problems are often ratio problems, and this is definitely one of those cases. Generally, we can say that \dfrac{A_{\text{sector}}}{A_{\text{circle}}}=\dfrac{\text{central angle measure}}{360^\circ\text{ or }2\pi}.

Since we’re given the central angle in radians and no other information, we just need to figure out the ratio of the central angle to 2\pi.

    \begin{align*}\dfrac{\left(\dfrac{5\pi}{4}\right)}{2\pi}=\dfrac{5\pi}{8\pi}=\dfrac{5}{8}\end{align*}

You can either grid in that fraction or its decimal equivalent: .625.

Can you please do #20 from the no calculator section in test 5?

Can you please do #20 from the no calculator section in test 5?

Sure. Here’s (basically) the image you’re given:

To find the value of x, you need to remember that the angles that make up straight lines and the angles that make up triangles will both sum to 180°. You’ve got a straight line (marked in red below) that helps you calculate that the upper left angle in the triangle is 180° – 106° = 74°.

From there, you can calculate the measure of the third angle in the triangle: 180° – 23° – 74° = 83°. Just one more 180° straight line to find the value of x:

= 180º – 83°
= 97°

A note since I’m sure someone would bring it up in the comments if I didn’t: another way to go here is the Exterior Angle Theorem, which states that the measure of an exterior angle of a triangle (the 106° angle in this case) equals the sum of the measures of the nonadjacent interior angles (the 23° and the 83°). You can use that theorem as a “shortcut” to find the 83° angle in one fewer step if you like. 🙂

Pwn the sat 4th edition page 36 question 3

Pwn the sat 4th edition page 36 question 3

I am tutoring my sophomore on her SAT math and we are going through your book a page at a time.

This problem is under back solving.
Given the amount of time you need to spend multiplying the left hand side of the equation. Wouldn’t it be faster to solve for a than plugging in the values?

How would you approach this differently?
Thanks

The question:

I get the “Wouldn’t it be faster…” question a lot about both the plug in and backsolve techniques, and my stock answer is maybe. In this case, I agree with you that FOILing the left hand side until you get the right answer (as many as 3 times—if you fail 3 times then you know the 4th choice is correct) might take a minute. I also agree that if you know what you’re doing with complex numbers, solving for a is fairly straightforward. To make a stronger case for backsolving here, I maybe should have made the question a tiny bit tougher. Like this, perhaps:

(9+ai)(1-i)=12-2ai

As I say at the top of that drill, though, I think it’s worth practicing the techniques, even if they feel slower or actually are slower, because if you really internalize the technique, it can be an escape hatch for you on test day when you encounter a question you can’t figure out the math way. If you never practice backsolving because it always feels too slow, then it probably won’t be there for you in a pinch when you need it most.

Here’s how backsolving would look for me on this question. I’d start with C, since I pretty much always start with C:

    \begin{align*}(9+2i)(1-i)&=12-6i\\9-9i+2i-2i^2&=12-6i\\9-7i+2&=12-6i\\11-7i&=12-6i\end{align*}

That’s obviously false, but it’s CLOSE! Being 1 off would almost certainly clue me in to trying choice B next, and I’d follow the pattern of the FOILing I just did, so I’d be a little faster the 2nd time.

    \begin{align*}(9+3i)(1-i)&=12-6i\\9-9i+3i-3i^2&=12-6i\\9-6i+3&=12-6i\\12-6i&=12-6i\end{align*}

 

An 84 meter length of fencing is attached to the side of a barn in order to fence in a rectangular area…

An 84 meter length of fencing is attached to the side of a barn in order to fence in a rectangular area, as shown in the figure above. If the length of the side of the fence running perpendicular to the barn is half the length of the side of the fence that is running parallel to the barn, what is the area of the fenced off land?

You know that 84 meters of fencing is used to make 3 sides of a rectangle (the barn forms the fourth side). You also know that one side of the rectangle is twice as long as the other. Let’s call the short side x and the long side 2x.

You know that x + x + 2x = 84, so you can solve for x. 4x = 84; x = 21.

That means the rectangle’s width and length will be 21 meters and 42 meters. Its area will therefore be (21 meters)(42 meters) = 882 square meters.

 

Test 2 calculator section #19

Test 2 calculator section #19. I guessed and got it right but how do you do it?

The key to figuring out the median in a question like this is remembering that the median is the middle number in a set of numbers, once that set is sorted (or the average of the two middle numbers in a set with an even number of numbers).

In this case, 600 students were surveyed in total (300 from each of two schools) so we need to figure out the middle number, which will be the average of the 300th and 301st value, although a quick look at the answer choices tells us we won’t have to worry about taking any averages.

Use the table to figure out where the middle value will fall. There were a total of 120 + 140 = 260 students surveyed who said they had 0 siblings. From there, another 80 + 110 = 190 students said they had 1 sibling.

260 was less than 300, but 260 + 190 = 450 is greater than 300, so we know that we hit the 300th and 301st value somewhere in the one-sibling crowd. Therefore, the median number of siblings is 1.

Test 1 #35 (calculator section)

Test 1 #35 (calculator section)

You can find the formula for the volume of a right circular cylinder at the beginning of the section; it’s V=\pi r^2h. Since the question tells you the volume (72π cubic yards) and the height (8 yards), you can solve for the radius:

    \begin{align*}72\pi&=\pi r^2(8)\\9&=r^2\\3&=r\end{align*}

Now, be careful! The question asks for the diameter, not the radius. Even though they underline the word diameter, I bet a lot of people still miss this question by putting the radius.

If the radius is 3, then the diameter is 6.