In the CrushTheTest book, you do 10 questions in a row all
in the same category for concentrated practice. The sample questions below have been chosen from 12 of the 20 CrushTheTest
categories.
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This question from the Algebra:Fractions section is very
doable but is still not easy to do in under 2 minutes unless you've practiced a lot. In the old days, many people said
practicing wouldn't help for the SAT and they said it with a straight face even after it became obvious that practicing
was not only helpful but actually essential. Today, anyone saying "practicing won't help" would be laughed out of the room.
1. If you pour half of a full small bucket of water into an empty big bucket, the big
bucket will be three-tenths full. How many times bigger than the small bucket
is the big bucket?
You want b / s which is the ratio of big (b) to small (s). Don't think too much. The problem
says (in words): (1/2)s = (3/10)b. Now multiply both sides
by (10/3) and divide both sides by s to get b / s = 5/3 = 1.67 (B).
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This question is from the Algebra:Functions section. It involves the concept of intersection points
which is an SAT favorite. The good news is once you get the hang of "intersection point" problems
they are all pretty much the same. This problem is moderately difficult as CrushTheTest questions go
and on the hard side as real SAT questions go.
2. Two curves, y = px2 and y = qx2 intersect
rectangle ABCD at its four corners. Which of the following expressions gives the
area of the rectangle?
To get area you need length times width. The labeled point (2,0) shows you that the length is 4.
If you can get the y-coordinates of points C and B then you could subtract
to get CB. A point on a curve is ALWAYS (x, f(x)).
So point C is (2, 4p) and point
B is (2, 4q). Note that q is negative. Note also that the answers have p and q so you can pretend you know p and q.
We've got side CB = 4p - 4q and side AB = 4. If you multiply and pull out a 4 you get the area of the rectangle is 16·(p-q) or (E).
It might also help to
estimate p and q (the diagram is to scale since there's no warning). It looks like p is about 1/2 and q is about -1/4. So the area
is about 12 = 16·(3/4).
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This question from the Geometry:Areas section combines a number of circle and triangle skills that might
be tested over two or three questions on the real SAT.
3. Each circle in the figure above has radius r = 1. What is the area of the
shaded region surrounded by the three circles?
Each circle has area π (because the radius is 1). Since the angles of an
equilateral triangle are all 60 degrees, the area of each sector is one-sixth
of the whole area (60/360 = 1/6) or π/6.
Note that the answers all have a π in them (it is rare on the SAT that you have to plug in 3.14).
The area of the whole equilateral triangle is
√3 (see below). The area of all 3 sectors
is 3(π/6) = (π/2). Now subtract to get the shaded region: √3 - π/2.
You'll need a common denominator (2) to get the answer (D).
To get the area of the equilateral triangle do one-half base times
height. Half the base is equal to 1 and the height is √3 because it is
the long leg of a 30-60-90 and the short leg is 1. The area of the equilateral
triangle is √3.
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This problem from the Geometry:Lengths, Angles, Perimeters section has one of SATan's favorite tricks. Like all the
tricks, it's quite simple once you see it but totally baffling if you don't. It looks like you don't have
enough information to solve the problem, but really you do. Hint: Make sure you use everything they tell you as SATan
only very rarely gives you information you don't need.
Note: Figure not drawn to scale.
4. In the figure above, AC is parallel to EF, BC = CD, angle BCD = 100°,
and angle BDE = 70°. What is the measure in degrees of angle FED?
Since angle BCD is 100, the two base angles must each be 40 which means that
angle CDE = 70+40 = 110.
It's tempting to think you also need angles A and F individually but you really don't.
SATan loves this trick. All you need to know is that
they add up to 180 (they are same side interior
angles of parallel lines).
It's a five-sided figure so A + F + C + CDE + FED = 540. Now just plug in
180 + 100 + 110 + FED = 540 and get FED = 150. The answer is (D).
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Here is a classic units question from Units:Algebra.
You can think your way through a question like this, but the CrushTheTest mechanical
approach has some speed and accuracy advantages. Basically, you first write down what you
are given, then set up the units so the things you want to cancel cancel, then you replace the
units with letters and/or numbers.
5. It takes G gallons of water to fill 6 hot-tubs. If you have N gallons of water but use
80% of it to water the lawn, how many hot-tubs can you fill using the remaining water?
You want to get "tubs." They gave us 6/G "tubs/gallons" (it's good to have tubs on top because that's what you're
trying to get). They also gave us N/5 "gallons" (80% = 4/5 was used to water the lawn so you have 1/5 left).
To get "tubs" you multiply "tubs/gallons" by "gallons" so that
"gallons" cancels. Now replace the units with numbers and letters. The answer is (6/G) times (N/5) which is
6N/(5G). The answer is (A).
Note that if you needed "gallons/tubs" you would just use G/6. We call this
the "upside down rule."
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This question from the Numbers:Divisibility section is easy IF you know the trick. This trick will solve
many classic SAT questions in the "number theory" category: prime numbers,
remainders, divisibility, and inequalities.
6. The quantity 12100 is NOT divisible by which of the following?
The number 12 is 2·2·3 (factoring a number into its primes is often useful on SAT questions). If you repeat this 100 times you get
a long chain of 200 2's and 100 3's.
Now let's look at the answers: 18 = 2·3·3 which would all cancel so 12100/18 is a whole number;
24, 27, and 36 are also all 2's and 3's. But 30 has a 5 in it which would never cancel no matter
how many 2's and 3's you have on top of the fraction.
So 12 to the 100th power is not divisible by 30 and the answer is (D).
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Remainders (technically, modular math) aren't usually emphasized in school. That's probably why
SATan loves them so much. If you struggle with and finally understand all 10 CrushTheTest
Numbers:Remainders questions, you'll be an expert.
7. When positive integer n is divided by 5, the remainder is x. When 2n is divided by
5, the remainder is y. Which pair (x, y) is not possible?
You could use the "play around" technique for this one.
Let's try 12/5 = 2R2 and 24/5 = 4R4. Now we know that the pair of
remainders (2,4) is possible so A is NOT the answer.
Now do 13/5 = 2R3 and 26/5 = 5R1 which means (3,1) is possible.
Next is 14/5 = 2R4 and 28/5 = 5R3 so (4,3) works too.
Obviously (0,0) will work as well. So the answer must be (E).
Here's another, fancier way. If you divide n by 5, there are 5 possible remainders: 0, 1, 2, 3, or 4. If you divide
2n by 5 you get twice as many 5's AND you also double the remainder. So your remainders for 2x are
0, 2, 4, 6, and 8. BUT you can't really have 6 and 8 as remainders: 6 is 1R1 and 8 is 1R3 if you are dividing by 5.
So the real set of remainders for 2n is 0, 2, 4, 1, and 3.
If you match this list up with 0, 1, 2, 3, 4 you get choices A-D but not choice E.
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It's not so much a trick that works for this
Statistics:Probability question as it is
a certain approach. This particular way of dealing with an apparently hard problem (SATan loves problems that
are apparently hard) is among the most useful for ALL types SAT questions.
Hint: CrushTheTest calls the approach that works for this problem "concreteness."
8. If the probability of picking a yellow marble out of a bag is 1/n and n is an
integer greater than 1, what is the ratio of the number of yellow marbles in the
bag to the number of other marbles?
They didn't say what n is so you can pick any n you want. It has to work for all n or they
can't ask the question. We call this trick "concreteness."
Suppose we have n = 5 marbles total and 1 yellow marble. The probability of picking a yellow marble is 1/5.
So now we've restated the question with nice clear numbers. Who needs letters?
We've got 1 yellow marble and 4 other marbles.
The ratio of yellow marbles to other marbles is 1 to 4 which is 1/(n–1) or (D).
If you try some fancy equation for getting this one, you might end up just chasing your tail around! "Don't
be too fancy" is a good SAT rule.
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Occasionally, SAT questions are really just the kind of puzzle you would have seen many times before if
you happened to be a logic puzzle fanatic. This one from the Logic:Thinking section
is a fun example. BTW, a good puzzle book to start with is "Games for the Superintelligent."
9. Bob and Carol and Ted and Alice smuggle treasure from a Caribbean island to the
United States. The FBI is onto them. There are four possible outcomes and each
of our four heroes will experience one outcome and each will experience a different
outcome.
Bob or Carol or Ted will go to jail.
Ted or Alice will get rich.
Carol or Alice will change her identity.
Bob or Ted or Alice will go into hiding.
Which of the following is true?
I. If Bob goes to jail, Alice gets rich.
II. If Ted goes to jail, Bob goes into hiding.
III. If Carol goes to jail, Ted gets rich.
Suppose Bob goes to jail. You could have Ted get rich, Carol change her
identity, and Alice go into hiding. So (I) isn't true.
Suppose Ted goes to jail.
This means Alice has to get rich. If Alice gets rich, then Carol has to change
her identity. This means Bob goes into hiding. So (II) is true.
Finally, suppose
Carol goes to jail. This means Alice has to change her identity. If Alice changes
her identity then Ted must get rich. So (III) is true. The answer is (D).
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This medium-difficulty question from Sequences:Counting illustrates
a method that works again and again for all kinds of "counting" problems: how many routes, how many ways to arrange N
objects, how many possible teams/outfits/sundaes etc. etc. Even rather hard SAT counting questions may
be solved using this one technique; no fancy equations are necessary.
10. From the (fictional) secret headquarters of the Standardized Testing Service (STS),
there are three underground roads that each lead directly
to the big boss's office. From SATan's office, three
more roads each go deeper underground to the ETC (Eternal Testing Center)
where bad people spend eternity taking the SAT. How many different routes are there
from STS headquarters to the ETC and back if no route can use the same road twice?
To get from STS to SATan's office
you have 3 roads to choose from. From the boss's office to the Eternal
Testing Center you have another 3 choices. Coming back from the ETC to the office
you have 2 choices because you
can't repeat roads. To get from the office back to STS HQ you also have 2 choices.
When you have a series of choices
you multiply the number of choices for each pick to get the total number of possibilities.
So the total number of routes is 3·3·2·2 = 36 or (D).
On the other hand, one might say the answer should be "none of these" because there is
no returning from the ETC.
Note: with some "series of choices" problems you have to divide (often by 2) because order
doesn't matter and you don't want to over-count. For example, if you are picking a team of 2 from
6 people, there are 6·5/2 = 15 possible teams because the team of Bob and Joe is the same as the team
of Joe and Bob.
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This is the bonus round. These last two questions are somewhat harder than anything the SAT will
throw at you. About 10% of CrushTheTest questions fall into this "extra-hard" category. A few questions
like this will give you some crucial "over-preparation" hopefully without undue agony.
This first one is from the Statistics: Mean, Median, Mode section.
11. Four positive numbers a, b, c, and d have mean
x and median y. If a < b < c < d then which of the following is possible?
I. x < y/2
II. y = b
III. x > c
The numbers are in order so the median y is between b and c:
y = (b+c)/2 and
y/2 = (b+c)/4
The mean of four numbers is always x = (a+b+c+d)/4.
There
is a y/2 hidden in the equation for the mean:
x = (a+d)/4 + (b+c)/4 = (a+d)/4 + y/2.
So there's no way x can be less than y/2 and (I) is not possible.
Since b and c are different, (II) is not possible.
In general, averages can be gigantic without affecting the median (the classic example is Bill Gates walks into
a room full of ordinary people – the median income in the room hardly changes but the average income goes right up into
the millions).
If you have 1, 2, 3, 1000 the average will be bigger than the third number so
(III) is possible. The answer is (C).
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Many SAT questions are most efficiently and accurately solved by "playing around" a bit.
In other words, you use trial and error. It's not fancy, but it is effective. This problem, from the
Statistics:Averages section is, as far as we know, impervious to fancy techniques: it can
only be solved by using the "playing around" technique.
12. The average of 5 different integers is 20. The largest of the 5 integers is 23.
The smallest of the five integers is N. How many possible values of N are there?
You can pick 23, 22, 21, 20, and 14. You have a total of 1+2+3=6 above 20 and the 14 is 6 below 20
so the average is 20. The smallest number is 14 and that's as low as
you can go because the other four numbers are as big as they can be.
You could also do 23, 22, 21, 19, and 15.
You could do 23, 22, 21, 18, and 16.
You could do 23, 22, 20, 18, and 17.
Now try to make the lowest one 18. You "run
out of room." The 23 is 3 above 20 and an 18 and a 19 make 3 below 20. But you still
have two more numbers to pick. There's no way to keep the average equal to 20 and still keep 18 as
your lowest number and have all the numbers different.
So there
are four possible smallest numbers: 14, 15, 16, and 17. The answer is (B).
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