SAT Official Practice Tests 1-2: Hardest Math

The ten hardest questions with hints and solutions cross referenced to (even harder) CrushTheTest problems. Only CrushTheTest is meaner than SATAN!

www.DrMatthewKohler.com ~~~~~~~ DrMatthewKohler@gmail.com.

Numbering key: 1.3.14 means official test 1 section 3 (no calculator) question 14. Or you might see 2.4.15 which means test 2 section 4 (calculator allowed) question 15.

CTT Practice: Step 3 on the SAT Prep tab has a free pdf download. Each hard official question here is cross referenced to similar (harder) questions in the book. References are page/problem#. So 100/4 means go to page 100 in the book and give question 4 a shot.

**1.3.14**

If \( 3x-y=12 \) what is the value of \( \dfrac{8^x}{2^y} \)?

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8 is \(2^3\).

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\( {(2^a)}^b = 2^{ab} \).

A) \(2^{12}\)

B) \(4^{4}\)

C) \(8^{2}\)

D) The value cannot be determined from the given information.

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\( \dfrac{8^x}{2^y} = \dfrac{{(2^3)}^x}{2^y} = \dfrac{2^{3x}}{2^y}= 2^{3x-y} = 2^{12} \)

The answer is (A).

**1.3.15**

If \( (ax+2)(bx+7) = 15x^2+cx+14 \) for all values of \(x\) and \(a+b=8,\) what are the two possible values for \(c\)?

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\( ab=15 \).

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\( 7a+2b=c \).

A) 3 and 5

B) 6 and 35

C) 10 and 21

D) 31 and 41

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There is no special technique. SATAN knows you don't learn in class to do problems that require a little trial and error and SATAN takes advantage of this. Don't be afraid to just guess what a and b have to be. I once told a student that she was allowed to (and expected to) do this in the SAT. That's all I did for her. She went from a 700 to a 770 with this one piece of advice.

Once you multiply out the (FOIL) the two factors to get \( abx^2 + (7a+2b)x + 14 = 15x^2+cx+14 \) and match up the terms of the two polynomials to get \(ab=15\) and \(7a+2b = c\), you can just guess that (a,b) must be (3,5) or (5,3). That's the only way you are going to get them to add to 8 and multiply to 15.

Now what? You aren't used to being able to get two answers and SATAN is taking advantage of this. But there are two answers because you can plug in 3 and 5 for a and b or you can plug in 5 and 3 for a and b. When you do \(7a+2b=c\), you will get two answers for c. And that's it, you're done. Nothing fancy, just guessing and plugging in. If you try to solve this in some fancy, formal way, you will run out of time.

The answer is (D).

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147/3; 150/3; 83/2; 134/4

**1.4.18**

\[y<-x+a\] \[y>x+b\] In the

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The origin (0,0) is below a line with negative slope and intercept *a*.
Or plug (0,0) into the first inequality.

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The origin (0,0) is above a line with positive slope and intercept *b*. Or
plug (0,0) into the second inequality.

A) \(a>b\)

B) \(b>a\)

C) \(|a|>|b|\)

D) \(a=-b\)

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A quick way to do this one is to just plug (0,0) into both equations. Then you
get, right away, that \(0 < a\) and \( 0 > b\). A positive number (*a*) is
always greater than a negative number (*b*), so choice (A) looks pretty
good and is, in fact, the answer.

I didn't plug in (0,0) when I did it. Instead, I quickly sketched the two lines
even though I didn't know *a* and *b*. I just made sure the two lines
sandwiched the origin. And then I was done.

With inequalities, the "solution" is the region where the shading overlaps. Since, the first equation is a "less than" equation, I had to shade below the line. Since the second equation is a "greater than" equation, I had to shade above the line.

The first line has a slope of -1 and goes
through the y-axis at *a*. The second line has a slope of +1 and goes through the y-axis
at *b*. To sandwich the origin (0,0) I made sure
*a* was above the origin and *b* was below the origin.

Obviously, from my sketch, I could see that *a* was greater than *b*.
Note that choice (C) could be true if *a* happened to be a large positive
number. But only choice (A) is guaranteed to be true.

The answer is (A).

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6/8; 1/1; 13/5

**1.4.20**

Alma bought a laptop computer at a store that gave a 20 percent discount off its original price. The total amount she paid to the cashier was

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First multiply the original price by something that removes 20 percent.

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Next, multiply the original price by something that adds 8 percent.

A) 0.88*p*

B) \( \dfrac{p}{0.88} \)

C) (0.8)(1.08)*p*

D) \( \dfrac{p}{(0.8)(1.08)} \)

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You can do 30 percent off by figuring out 30 percent of the price and then subtracting. NEVER do this. To do 30 percent off, multiply by 0.70. To do 10 percent off, multiply by 0.90. This method is MUCH better and is almost always the best way.

Same deal when increasing a price. To add 30 percent to a price, do NOT figure out 30 percent and then add it even though this seems to be the most obvious way. Instead, multiply by 1.30. To increase something by 5 percent, multiply by 1.05.

SATAN knows people often don't know this trick, so he
gives problems that can't be done without it (evil, remember?).
In this case, since you are knocking 20 percent off the original price,
you multiply by 0.80.
Then, since you are charging 8 percent sales tax, you multiply by 1.08. Here
is the equation:
\[ 0.8 * p_{original} * 1.08 = p_{paid} = p. \]
Order doesn't matter when multiplying: *abc* = *cba* = *bca* *etc.*

Note that SATAN uses the symbol *p*, defines it as
the price paid, and asks for the *original* price in terms
of *p*.

On every question on the math part, ALWAYS ask yourself, "Is my answer the one
SATAN is asking for?" Often, you'll correctly solve for *x* but SATAN
actually wants *y*. This is one of SATAN's favorite tricks and it can fool
anyone, including me! On this question, if you are going too fast and/or don't
read carefully, you can easily pick choice (C) and get it wrong.

In SATAN's answer explanation on page 442 of his precious book, he acts as if this problem can be done without knowing the trick for discounting or taxing: 32 percent increase means multiply by (1 + 0.32); 12 percent decrease means multiply by (1 - 0.12).

SATAN figures you can derive this technique on the fly and he shows you how.
SATAN is being his usual absurd self. Most people aren't going
to develop a new problem solving technique *while they take a speeded test.*
But hey, that's just my opinion. What do I know? It's not like I'm lord of
the underworld or anything — my cat won't even obey me.

One other point: Don't let the phrase "in terms of" bother you. It sounds fancy,
but all it means in this case is that you can have a *p* in your answer.

The answer is (D).

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67/3; 69/1; 69/2; 28/5; 33/1

**1.4.28**

If a system of inequalities \( y ≥ 2x+1 \) and \( y > \tfrac{1}{2}x-1 \)
is graphed in the *xy*-plane above, which quadrant contains no solutions to the system?

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Imagine shading the region above the line that has a slope of 2 and crosses
the *y*-axis at 1.

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Imagine shading the region above the line that has a slope of 1/2 and crosses
the *y*-axis at -1.

A) Quadrant II

B) Quadrant III

C) Quadrant IV

D) There are solutions in all four quadrants.

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Draw or sketch the lines and shade the region that is above BOTH lines. There will be no shading in quadrant IV. You can also shade above one line with blue and above the other line with red. The solutions to the system are in the region that has both blue and red. Or you can shade using lines and look for cross-hatching to find the solution. However you do it, there is no part of quadrant IV that is above both lines.

Note that there is some part of quadrant IV that is above the second line, but this region is BELOW the first line and so isn't part of the solution.

The answer is (C).

**1.4.29**

For a polynomial

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We know *p*(*x*) is not divisible by \(x-3\) because if it were,
*p*(3) would be zero.

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The simplest polynomial is a constant.

A) \(x - 5\) is a factor of *p*(*x*).

B) \(x - 2\) is a factor of *p*(*x*).

C) \(x + 2\) is a factor of *p*(*x*).

D) The remainder when *p*(*x*) is divided by \(x - 3\) is -2.

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In the answers, SATAN derives the remainder theorem as if you are going to do that on the test. Again, this is absurd. If you don't happen to remember the remainder theorem, you'll need some sort of trick. Here's one of my favorites: concreteness.

\[ \text{Simplest possible polynomial: }p(x)=-2 \implies \text{A, B, and C all wrong, D okay.}\]

Is that scary or what? All we did is
assume *p*(*x*) is a ridiculously simple function. They said
*p*(3) was -2 so why not make *p*(*x*) always -2?
As long as you don't contradict
anything in the question, you can get away with this.
If you're worried, you could look at a couple of other possible polynomials that give you
the right answer when you plug in \(x=3,\) such as \(p(x)=x-5\)
and \(p(x)=x^2-11\), but it isn't necessary in this case. Remember, a polynomial
is just numbers and powers of *x* added and multiplied and we are free to
make up the simplest one that fits the problem. Who needs *x* at all?
Take that SATAN!

Once you have your super-simple concrete example, you can go through the answers and see how many you can eliminate. Your polynomial is a constant and so it isn't divisible by anything. So choices A, B, and C can all be eliminated. When you check choice (D) and divide your simple polynomial by \(x - 3\), you get zero with remainder -2 and so the concreteness method worked and you have your answer. If we had used \(p(x)=x-5\) as the polynomial, that would fit with choices A and D and so we would have to test a different polynomial to narrow it down. But we were lucky since the simplest version of a polynomial allowed us to eliminate 3 of the choices.

You can also remember the remainder theorem for polynomials which says that
any polynomial *p*(*x*) can be written as follows:
\[p(x) = q(x)*(x-3) + p(3)\]
and you can simply choose the number to be 3 (or 10 or whatever) if you want to.
Note that this example is guaranteed to work when \(x=3\) and also note that
*p*(3) is automatically the remainder when you divide by \(x - 3.\)

Numbers work the same way as polynomials (but are a little easier).
So any number *N* can be divided by anything. Suppose we
divide *N* by 5. We don't know that *N* is divisible by 5, but that's okay because
that's where remainders come in. We get the following:
\[ N = q*5 + R\]
where *q* is however many 5's there are in *N* while *R*
is the remainder when you divide *N* by 5.

SATAN has always had an unnatural fondness for remainders, but CrushTheTest has 18 hard remainder questions all grouped together. Do them and SATAN won't be able to touch you.

The answer is (D).

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123/1; 123/2; 124/5; 124/6; 124/7

**2.3.8**

\[ nA=360 \] The measure

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Translation: How big can you make *n* if you aren't allowed to let *A*
go below 50?

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If *n* is 2, *A* is 180. If *n* is 3, *A* is 120.
If *n* is 4, *A* is 90. Keep going.

A) 5

B) 6

C) 7

D) 8

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SATAN knows you are used to getting exact numbers using a set of clearly
defined steps. Most people aren't used to answering "greatest possible number"
kinds of questions because they involve a little trial and error and there's
no clearly defined method. For this one, since you don't have a calculator, you
won't even get an exact value for *A*. Don't worry. Just try a few values
for *n*.

If *n* is 6, *A* is 60. So we're close. If *n* is 7,
*A* is 360/7 which is hard to do without a calculator. But I know 50*7=350,
so 360/7 is above 50. However, if *n* is 8, *A* is 45. So even
thought I don't know exactly what the exterior angle comes out to for a seven-
sided polygon, I know it's just a bit above 50 and 7 sides is the most I can
have if my exterior angle has to be above 50.

The answer is (C).

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51/2; 65/2; 116/7; 153/2

**2.3.9**

The graph of a line in the

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The first line also goes through (0,6) and (-1, 4).

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The second line has a slope of -1 and goes through (2,1) and (1,2) and (0,3).

A) 4

B) 3

C) -1

D) -4

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I didn't use any equations for this one. I just sketched the lines and plotted points. Since I know the slope of both lines and I know a point for both lines, I can plot easy points and connect the dots and hope the lines share an easy point. On the SAT, questions like this often come out even. If they don't intersect in an easy point, I can always estimate and this is often plenty good enough and much faster than doing it in a formal way.

Since the first line has a slope of 2, you can back up from (1,8) to (0,6) to (-1,4) to (-2,2). Since the second line has a slope of –1, you can back up from (2,1) to (1,2) to (0,3) to (-1,4) to (-2,5).

Just a rough sketch and a little backing up got me (-1,4) for the intersection
point (*a, b*). Since they wanted the sum, the answer is 3.

The answer is (B).

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36/6; 36/8; 38/4; 61/2

**2.3.20**

\[ ax+by=12 \] \[ 2x+8y=60\] In the system of equations above

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An infinite number of solutions means the two equations have to be multiples of each other.

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Since 12*5=60, the first equation would have to be the second equation divided by 5.

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If *a* is 2/5 and *b* is 8/5, then the two equations would be
multiples of each other. That is, you could multiply the first equation by 5 and
it would become the second equation. So really, you would have just the one
equation. With one equation and two unknowns, *x* and *y* could be
anything. For example, if I say *x* + *y* = 6 and
2*x* + 2*y* = 12, there's really just one equation and there are an infinite
number of solutions including (0,6), (1,5), (1.1,4.9) *etc.*

For this question, you've got 2/5 and 8/5 for *a* and *b* and their
ratio is still just 2/8 or 1/4.

The answer is 1/4 or .250. On the SAT, if you put in a fraction, always put it in lowest terms. If you put in a decimal, always use 3 places.

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1/4; 2/7; 2/8

**2.4.30**

The figure above shows a regular hexagon with sides of length *a* and a square
with sides of length *a*. If the area of the hexagon is 384√3 square inches, what
is the area, in square inches, of the square.

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If the base of an equilateral triangle is *a*, the height is a*√3/2.

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There are six equilateral triangles in the hexagon.

A) 256

B) 192

C) 64√3

D) 16√3

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The six equilateral triangles have areas equal to one-half base times height.
The base of each triangle is *a* and the height is *a**√3/2.
Note that an equilateral
triangle consists of two 30-60-90 triangles which have these ratios:

hypotenuse : short leg : long leg = x : x/2 : x√3/2.

Definitely know 45-45-90, 30-60-90, and equilateral triangles cold for the SAT.

The total area of the six equilateral triangles that make the hexagon is: \( 6 \cdot \tfrac{1}{2} \cdot a \cdot \tfrac{\sqrt{3}}{2} \cdot a \) which they told us is \( 384\cdot\sqrt{3}. \) If we solve this for \(a^2\), we get 384*2/3 which is the area of the square and we're pretty much done.

You have to remember that you don't actually need *a*. SATAN is mean; he wants
you to try to get *a* without a calculator.
But all you need is \(a^2\) which
we already have because we wrote the equation for the area of the hexagon in terms
of *a*.

In the old days, this level of multi-step problem wouldn't be found on the SAT, but SATAN is apparently trying to reduce the number of perfect scores on the test and has upped the ante so to speak.

We still have to calculate 384*2/3 without a calculator. Of course, 384 is divisible by 3 which gets you 128 and then you multiply by 2 to get 256. Long problem. CrushTheTest (CTT) problems are still harder than SATAN's, but he is catching up! Try a few CTT problems and SATAN's worst will seem easy.

The answer is (A).

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43/4; 45/1; 47/5; 49/1