Posts filed under: new SAT

The five questions in this short quiz would all, I’m fairly sure, fit into the “Heart of Algebra” category that will be so heavily emphasized on the new SAT. Obviously, without even a full practice test released yet, it’s tough to know for sure if I’ve got the style right, but based on what I’ve seen, you’d better be able to cruise through these if you’re planning to take the SAT after January 2016.

(UPDATE: Try another set of questions in the style of the new SAT here.)

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How’s everyone else doing on this quiz?

36% got 5 right
29% got 4 right
14% got 3 right
12% got 2 right
7% got 1 right
2% got 0 right

College Board released a bunch of sample questions this week for the new PSAT and SAT, which will make their debuts in October 2015 and March 2016, respectively. Over the next few days, I’ll be making posts working through each question, a few at a time, and commenting on them when I feel like I have something insightful to say.

In this post, I’ll deal with questions 13 through 18 in the “calculator not permitted” section. For the “calculator permitted” section, see questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, 21 through 26 here, and 27 through 30 here. You can see questions 1 through 6 from the “calculator not permitted” section here, and 7 through 12 here.

Question 13 (link)

Difficulty: Hard
Is this new? Kinda. (This basic concept is tested on the current SAT, but the algebra in this question is a bit more involved.)

The thing you need to know here is that the only way two linear equations will have no solution is when they create parallel lines (and parallel lines have equal slopes). So put both equations into y=mx+b form, and then see what a needs to be to make the slopes the same:

\dfrac{1}{2}x-\dfrac{1}{4}y=5

-\dfrac{1}{4}y=-\dfrac{1}{2}x+5

y=2x-20

So the slope we’re looking for is 2. Proceeding to the next equation:

ax-3y=20

-3y=-ax+20

y=\dfrac{a}{3}x-\dfrac{20}{3}

So you know \dfrac{a}{3}=2. That means a must equal 6.

Question 14 (link)

Difficulty: Hard
Is this new? Yes. The current SAT generally stays away from work/rate problems. 

This is another one of the “new” questions that appeared in the first document we got about the new SAT months ago. I hated it then, and I still hate it now.

The basic idea is that you need to see, from the right-hand side of the equation, that the equation is telling you how much the printers will print in 1 hour, which is \dfrac{1}{5} of the job.

From there, you need to recognize that each fraction on the left represents one of the two printers. From there, hopefully it’s intuitive that the \dfrac{1}{x} represents the slow printer, and the \dfrac{2}{x} represents the fast one.

Question 15 (link)

Difficulty: Hard
Is this new? Not really—It’s basically a right triangle question. The only thing new is that the current SAT doesn’t have chord as part of its vocabulary. 

This question doesn’t immediately look like a right triangle question, but it’s totally a right triangle question—and one where plugging in will work, at that! Say r = 3, so that AB = 6 and CD = 4. Now draw some lines:

test_specifications_for_the_redesigned_sat_na3_pdf 8

Of course, since PD is a radius, it has a length of 3 just like AP and PB.

2^2+QP^2=3^2

4+QP^2 = 9

QP=\sqrt{5}

Only one answer choice has \sqrt{5} in it, so we can feel pretty good about choice D without even doing the last step of the plugging in process, which is to put 3 in for r in each answer choice. Of course, when you do, D is the only answer to give you \sqrt{5}.

Question 16 (link)

Difficulty: Hard
Is this new? Yes. There is no trigonometry on the current SAT.

A calculator would be nice here—plug in, graph, and you’re done. Without a calculator, you need to know the relationship between angles with opposite sines.

Since the sine function has a period of 2π, subtracting π from x inside the function will result in a negated sine. You might recognize that this really ends up being a graph translation problem. The red graph shifts the blue graph π units to the right (if we had added π and shifter π units to the left we would have had the same result).

Question 17 (link)

Difficulty: Hard
Is this new? Yes. There are no circle equation questions on the current SAT. This could appear on the current SAT Math Subject Tests, though.

 To solve this one, you need to know the general equation of a circle, which is this:

A circle with radius r and a center of (ab) has the equation (x-a)^2+(y-b)^2=r^2.

You also need to know how to complete the square, because otherwise you’re not going to be able to wrangle what you’re given into that form.

x^2+y^2-6x+8y=144

Do a little rearranging:

(x^2-6x)+(y^2+8y)=144

Now, what binomial square begins with x^2-6x? What binomial square begins with y^2+8y?

  • x^2-6x is the beginning of the (x-3)^2=x^2-6x+9 binomial square
  • y^2+8y is the beginning of the (y+4)^2=y^2+8y+16 binomial square.

So, to complete those squares, we need to add 9 and 16 to both sides of the circle equation!

(x^2-6x+9)+(y^2+8y+16)=144+9+16

(x-3)^2+(y+4)^2=169

(x-3)^2+(y+4)^2=13^2

So we have a circle centered at (3, –4) with a radius of 13. The question asked for the diameter, though, so the answer is 26.

Question 18 (link)

Difficulty: Hard
Is this new? Not really. The algebra is trickier than you’d see on most current SAT questions, but I would not be surprised to see this as a #18 in the current SAT grid-in section.

\dfrac{24}{x+1}-\dfrac{12}{x-1}=1

This is just algebra. Let’s do it. First, get a common denominator:

\dfrac{24(x-1)}{(x+1)(x-1)}-\dfrac{12(x+1)}{(x+1)(x-1)}=1

Hopefully you recognize that the denominator is a difference of two squares:

\dfrac{24x-24}{x^2-1}-\dfrac{12x+12}{x^2-1}=1

\dfrac{24x-24-(12x+12)}{x^2-1}=1

\dfrac{12x-36}{x^2-1}=1

Now you can move that denominator over:

12x-36=x^2-1

And set everything equal to 0:

0=x^2-12x+35

And finally, factor:

0=(x-7)(x-5)

x could equal 7 or 5.

College Board released a bunch of sample questions this week for the new PSAT and SAT, which will make their debuts in October 2015 and March 2016, respectively. Over the next few days, I’ll be making posts working through each question, a few at a time, and commenting on them when I feel like I have something insightful to say.

In this post, I’ll deal with questions 7 through 12 in the “calculator not permitted” section. For the “calculator permitted” section, see questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, 21 through 26 here, and 27 through 30 here. You can see questions 1 through 6 from the “calculator not permitted” section here, and questions 13 through 18 here

Question 7 (link)

Difficulty: Medium
Is this new? Kinda. The vertex form of a parabola has shown up (rarely) on the current SAT.

And here we see that the new SAT will expect you to know the vertex form of a parabola. Only one of the answer choices is in vertex form, so if you know what that looks like, you don’t even need to confirm that it actually has the right numbers in it—you just need to recognize the structure. Not sure how I feel about that.

Anyway, the vertex form of a parabola is this:

For constants ah, and k, the parabola y=a(x-h)^2+k will have its vertex at (h, k). If you’re nervous about whether y=(2x-4)(x-4) and y=2(x-3)^2+(-2) are equivalent, you can do some algebra (CB walks you through that in the answer explanation) but I’m sticking to my guns—if they’re going to make it so that only one choice is actually in the right form, then you don’t need to bother checking.

Question 8 (link)

Difficulty: Medium
Is this new? Very much so.

Imaginary. Numbers. On the SAT. My whole world is turned upside-down.

No, seriously though, this stuff is easy if you know how to FOIL. All you need to do is remember, at the end, that i^2=-1.

(14-2i)(7+12i)
(14)(7) + (14)(12i) + (-2i)(7) +(-2i)(12i)
98 + 168i -14i -24i^2
98 + 154i -24i^2

Now remember that i^2=-1

98 + 154i +24
122 + 154i

See? Easy.

Question 9 (link)

Difficulty: Medium
Is this new? Not really. The current SAT doesn’t usually ask questions that require this kind of equation solving, but it’s not unheard of.

Good ol’ fashioned algebra, here. Finally!

\dfrac{5(k+2)-7}{6}=\dfrac{13-(4-k)}{9}

First, simplify the numerators…

\dfrac{5k+3}{6}=\dfrac{9+k}{9}

Now cross multiply!

(9)(5k+3)=(6)(9+k)
45k+27=54+6k
39k=27

k=\dfrac{27}{39}=\dfrac{9}{13}

Question 10 (link)

Difficulty: Medium
Is this new? Kinda. Systems of equations on the current SAT can almost always be solved easily in one or two steps. This is not that kind of systems question.

More algebra! I’m so happy! Buckle up…

First, get the first equation into a useful form:

4x-y=3y+7
4x-4y=7

Now multiply it by 2:

8x-8y=14

Now add that to the second equation to eliminate the y terms and find x!

8x-8y=14
+(x+8y=4)
9x=18

So you know x = 2. From there, it’s easy to find y:

2+8y=4
8y=2
y=\dfrac{1}{4}

So what’s xy? Well, it’s \left(2\right)\left(\dfrac{1}{4}\right)=\dfrac{1}{2}.

Question 11 (link)

Difficulty: Medium
Is this new? No.

After the last two questions, I kinda feel like they’re joking with this one.

\dfrac{1}{2}x + \dfrac{1}{3}y = 4

Multiply that all by 6 and, well, you’re done.

6\left(\dfrac{1}{2}x + \dfrac{1}{3}y\right) = 6(4)

3x + 2y = 24

Question 12 (link)

Difficulty: Hard
Is this new? Yes. Trig doesn’t appear at all on the current SAT.

And we get to finish this post up with some trig! Wahoo!

There are a few ways to go here. One is to know your unit circle well enough to know that choices A, B, and D don’t make sense. Honestly, that’s how I thought through this when I first looked at it.

There’s a better way, though, which I’m ashamed not to have recognized at first. It’s what makes this an SAT question, and not just a math test question. The graphs of sine and cosine are in phase, such that \sin x =\cos \left(\dfrac{\pi}{2}-x\right)! Therefore:

\sin\dfrac{\pi}{5}=\cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{5}\right)

\sin\dfrac{\pi}{5}=\cos\left(\dfrac{5\pi}{10}-\dfrac{2\pi}{10}\right)

\sin\dfrac{\pi}{5}=\cos\dfrac{3\pi}{10}

College Board released a bunch of sample questions this week for the new PSAT and SAT, which will make their debuts in October 2015 and March 2016, respectively. Over the next few days, I’ll be making posts working through each question, a few at a time, and commenting on them when I feel like I have something insightful to say.

In this post, I’ll deal with questions 1 through 6 in the “calculator not permitted” section. For the “calculator permitted” section, see questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, 21 through 26 here, and 27 through 30 here. You can see questions 7 through 12 from the “calculator not permitted” section here, and questions 13 through 18 here.

Math_Sample_Question__1___SAT_Suite_of_AssessmentsQuestion 1 (link)

Difficulty: Easy
Is this new? No.

This is an interesting question, because it’s really just testing whether you know that when a line is translated, its slope doesn’t change at all! The slope of the line in the figure is -\dfrac{3}{2}, which is easy to see from the two darkened points—the line moves down 3, and right 2. Therefore, the translated line will also have a slope of -\dfrac{3}{2}.

Question 2 (link)

Difficulty: Easy
Is this new? Kinda. The content is familiar, but the way the question is asked is new.

Heart of Algebra questions (the easy ones, anyway) will usually look like this, apparently. The thing to pay attention to in this equation is where the variable, x, is. If x represents the number of years since 2004, we can conclude that the average must have been 27.227 in 2004, and has increased slightly every year since then, at a roughly constant rate. The 0.8636 number, which is multiplied by x, must be the number that’s added to the average every year—the estimated annual increase in the average number of students per classroom.

Question 3 (link)

Difficulty: Easy
Is this new? Not really, although they’re calling it easy here and this would not be an “easy” question on the current test.

In this “Passport to Advanced Math” question (I promise that I’ll get over these crazy category names soon, but I’m not quite there yet) we’re being tested on whether we recognize the opportunity to complete a binomial square. That a^2+14a bit is most of what we get when we square the quantity a+7. What’s missing?

(a+7)^2=a^2+14a+49

Do you see the opportunity to substitute? We already know thata^2+14a=51!

(a+7)^2=51+49

(a+7)^2=100

Because the question tells us that a is positive, we can ignore the negative square root and just say:

a+7=10

(If this weren’t the “calculator not permitted” section, we also could have solved this by graphing very quickly.)

Question 4 (link)

Difficulty: Medium
Is this new? No.

Just do a little cross-multiplying here:

\dfrac{2}{a-1}=\dfrac{4}{y}

2y=4a-4

y=2a-2

Question 5 (link)

Difficulty: Medium
Is this new? No.

They’re calling this one a medium difficulty question, but I just don’t buy it, unless they think the simple presence of a cubic term is going to send kids running for the hills.*

If y=x^3+2x+5 and z=x^2+7x+1, then:

2y+z=2\left(x^3+2x+5\right)+x^2+7x+1  =2x^3+4x+10+x^2+7x+1  =2x^3+x^2+11x+11

*With the elimination of the guessing penalty on the new SAT, there’s less reason than ever before to go running for the hills.

Question 6 (link)

Difficulty: Medium
Is this new? No.

I have to say that so far I’m liking the “calculator not permitted” questions a lot better. I don’t really know why, since most of the “calculator permitted” questions weren’t really calculator questions either, but it seems like we’ll get to do our fun algebraic manipulation work mostly in the “calculator not permitted” section.

A few ways to go here, but what I usually do with a question like this is just take it one step at a time until I’ve isolated the thing I want to isolate. In this case, that means I’m going to first raise both sides to the –1 power, and then square both sides.

a^{-\frac{1}{2}}=x

\left(a^{-\frac{1}{2}}\right)^{-1}=x^{-1}

a^{\frac{1}{2}}=x^{-1}

\left(a^{\frac{1}{2}}\right)^{2}=\left(x^{-1}\right)^2

a=x^{-2}

That’s not an answer choice, so let’s apply what we know about negative exponents:

a=\dfrac{1}{x^{2}}

College Board released a bunch of sample questions this week for the new PSAT and SAT, which will make their debuts in October 2015 and March 2016, respectively. Over the next few days, I’ll be making posts working through each question, a few at a time, and commenting on them when I feel like I have something insightful to say.

In this post, I’ll deal with questions 21 through 26 in the “calculator permitted” section. See questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, and 27 through 30 here.

Question 21 (link)

This one’s just about as easy as a medium question can come, but it requires careful reading of both the question and the graph. The thing you need not to miss in the question is that the dishes are 10 square centimeters in area. So when bacteria cover 7 square centimeters of the dish, they also cover 70% of the 10 square centimeter dish.

Let’s just look at each choice one at a time:

  • A: Nope. At time t = 0, neither dish is even close to being 100% covered.
  • B: Yes, this is the answer. Dish 2 has 2 square centimeters covered, and Dish 1 has 1 square centimeter covered. Because the dishes are 10 square centimeters in area, those percentages are correct.
  • C: Nope. Dish 2 covers twice as much as Dish 1 at t = 0. That’s not 50% more, that’s 100% more.
  • D: Nope. The opposite is true—Dish 1 grows much more quickly in the first hour than does Dish 2.

Question 22 (link)

Pay SUPER close attention to units here. They’re your lifeblood. First, figure out what 11.2 gigabits is in megabits:

11.2 gigabits × 1024 megabits per gigabit = 11,468.8 megabits

Now figure out how many seconds are in 11 hours:

11 hours × 60 minutes per hour × 60 seconds per minute = 39,600 seconds

Now figure out how many megabits can be transmitted in that period based on the 3 megabits per second rate:

3 megabits per second × 39,600 seconds  = 118,800 megabits

Finally, figure out how many images that’ll let the tracking station receive:

118,800 megabits / 11,468.8 megabits per image ≈ 10.4

That’s 10 full images.

Question 23 (link)

This is one of the most old-school SAT questions we’ve seen–all you need to do is make one quick substitution and solve, but make sure you don’t go one step too far! The question asks for x^2, not x!

x^2+y^2=153

y = -4x

x^2 + (-4x)^2 = 153

17x^2 = 153

x^2 = 9

Question 24 (link)

I hope you don’t mind me plagiarizing myself here—the next three question were floated by College Board way back when they announced the new SAT, and I solved them in a couple posts then. This section and the following two are reprints from here and here.

To get this, first find the area of the hexagon, and then subtract the area of the circle. Once you’ve done that, you can multiply by 1 centimeter to get the volume. Then you can use the given density to find the mass. Sound like fun???

A regular hexagon can be divided into 6 equilateral triangles, like so:

The area of an equilateral triangle is \left(\dfrac{1}{2}\right)\left(b\right)\left(\dfrac{b\sqrt{3}}{2}\right). (Derive that for yourself—it’s good practice! Or see here.) For each of those equilateral triangles, then, the area is \left(\dfrac{1}{2}\right)\left(2\right)\left(\dfrac{2\sqrt{3}}{2}\right)=\sqrt{3}. There are 6 of those, so the area of the hexagon is 6\sqrt{3} square centimeters.

The hole drilled through the middle has a diameter of 2, so it has a radius of 1, and therefore an area of π square centimeters. So the area of the hexagonal face of the nut, minus the hole drilled through it, is 6\sqrt{3}-\pi square centimeters.

Multiply that area by the height of the nut, 1 centimeter, and you get the volume:6\sqrt{3}-\pi cubic centimeters.

Now, you’re told that density = mass/volume, you’re given the density 7.9 grams per cubic centimeter), and we just found the volume. We’re asked for the mass to the nearest gram.

7.9=\dfrac{m}{6\sqrt{3}-\pi}

7.9\left(6\sqrt{3}-\pi\right)=m

57.2706… = m

Since we’re asked for the answer to the nearest gram, we write 57.

Question 25 (link)

Yikes, right?

If the bank converts Sara’s purchase to dollars, and adds a 4% charge, then the $9.88 she’s charged doesn’t convert directly to rupees. First we need to get that 4% fee out of there. If x = the dollar cost of her purchase without the fee, then 1.04x = 9.88.

x = \dfrac{9.88}{1.04} = 9.5

So her purchase was worth $9.50, which means the exchange rate that day, in rupees per dollar, is:

\dfrac{602\text{ rupees}}{9.5\text{ dollars}}\approx 63.36\text{ rupees per dollar}

Round that to the nearest whole number and the answer is 63.

Question 26 (link)

If Sara buys the prepaid card, she doesn’t have to pay the 4% fee her credit card charges her, but she could lose money if she doesn’t spend the whole card. The exchange rate doesn’t actually matter anymore—we can actually forget about dollars entirely at this point.

The question is really how much Sara has to spend on her credit card, with the 4% fee, before she’d save money by buying the prepaid card, even if she doesn’t spend it all. Say y = the amount of rupees Sara spends:

7,500 = 1.04y
7,211.54 = y

If Sara spends 7,211.54 rupees on her credit card, it costs her the same as buying a 7,500 rupee prepaid card. Therefore, if Sara spends 7212 rupees or more on her prepaid card, she gets a better deal than if she spends the same amount on her credit card.

 

 

College Board released a bunch of sample questions this week for the new PSAT and SAT, which will make their debuts in October 2015 and March 2016, respectively. Over the next few days, I’ll be making posts working through each question, a few at a time, and commenting on them when I feel like I have something insightful to say.

In this post, I’ll deal with questions 16 through 20 in the “calculator permitted” section. See questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 21 through 26 here, and 27 through 30 here.

Question 16 (link)

Math_Sample_Question__16___SAT_Suite_of_AssessmentsWell, this is embarrassing for CB, but now there have been 2 questions in a row where an answer choice (and the correct answer choice, no less) is printed incorrectly. (See Question 15 in my previous post.) In this case, choices A and C, screengrabbed at right, should have || symbols, not blank squares. Someone didn’t check his or her font library carefully enough before choosing a math font!

Typo aside, this is a good question. The thing you need to know about similar triangles is that their corresponding angles are congruent. Since you have vertical angles at point C, which must be congruent to each other, you know how the corresponding angles are laid out. I’ve color coded them below.

 

Do you see what’s going on at points B and D, and points A and E? Those are alternate interior angles of lines AB and DE, so if they’re congruent, then those lines must be parallel!

Question 17 (link)

Another Heart of Algebra question, another scenario described that we need to identify an equation for. For this one, it’ll be helpful to use units. You know you want to end up with gallons, so make sure all the miles and hours units in your equation cancel out!

Hopefully it’s obvious that you need to start with “17 –” something. That right there eliminates choices C and D, amazingly.

Each remaining choice has a 50t in it. The 50, we know, is a speed, in miles per hour. The t is time, in hours. So the units of 50t is miles per hour × hours = miles. Remember that we want our final units to be gallons.

The units of 21 is miles per gallon. So between choices A and B, we’re choosing between \dfrac{\text{miles per gallon}}{\text{miles}} and \dfrac{\text{miles}}{\text{miles per gallon}}. Which of those simplifies to gallons? B does.

Question 18 (link)

They’re not even asking you to solve systems of equations anymore—they’re just asking you to be able to write them. This is just the first step on medium and hard questions on the current test, but on this medium question it’s the only step. There are x cars and y trucks, so you xy should equal the total number of vehicles, 187. You also know how much each kind of vehicle has to pay in tolls. Cars pay $6.50 and trucks pay $10. So 6.5x + 10y should tell you the total amount collected in tolls, $1338.

Your equations, then, are xy = 187, and 6.5x + 10y = 1338. That’s choice C.

Question 19 (link)

OK. There a few ways we might go here. One is to just plug the (d, p) coordinates we know (9, 18.7) and (14, 20.9) into each answer choice and see which one gives us a true equation. Since calculators are allowed on this question, that might be the fastest way. You might also enter each equation into your graphing calculator and use the table function to see which equation gives you the points you want, like so:

See how Y2 has the points you want? That means the equation entered into Y2 is the right equation!

Of course, you can also use the points you’re given to find the equations algebraically. First you find the slope:

\dfrac{20.9-18.7}{14-9}=0.44

…then you find the intercept using one of the points:

18.7 = 0.44(9) + b
14.74 = b

That’ll tell you that the equation you want is p = 0.44d + 14.74. If you ask me, though, using the calculator is the way to go here.

Question 20 (link)

This question is essentially asking for the slope of the line of best fit, so our job is to find 2 points on the graph, and then use the slope formula: slope = \dfrac{y_2-y_1}{x_2-x_1}.

Math_Sample_Question__20___SAT_Suite_of_Assessments 2Remember that it’s fine to approximate here! The answer choices are really far apart, so even if our approximations aren’t very good, we’ll know the right answer. Using my numbers, here’s the slope calculation:

\dfrac{3000-2000}{2003-1996}=\dfrac{1000}{7}\approx 143

143 isn’t an answer choice, but it’s awfully close to 150, and not very close at all to any other answer choice. 150, therefore, is the right answer.