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To all students who just finished high school and are about to apply to a university and need SAT Test , this collection is good for you, it includes various and hard questions in math that helps you prepare for SAT test or any other math exam
Study Set Content:
SAT Math Hard Practice Quiz
Numbers and Operations
1.
A bag contains tomatoes that are either green or red.
The ratio of green tomatoes to red tomatoes in the bag
is 4 to 3. When five green tomatoes and five red tomatoes
are removed, the ratio becomes 3 to 2. How many red
tomatoes were originally in the bag?
(A)
12
(B)
15
(C)
18
(D)
24
(E)
30
2.
If each digit in an integer is greater than the digit to the
left, the integer is said to be “monotonic”. For example,
12 is a monotonic integer since 2
>
1. How many positive
two-digit monotonic integers are there?
(A)
28
(B)
32
(C)
36
(D)
40
(E)
44
a
, 2
a
−
1, 3
a
−
2, 4
a
−
3,
. . .
3.
For a particular number
a
, the first term in the sequence
above is equal to
a
, and each term thereafter is 7 greater
than the previous term. What is the value of the 16
th
term in the sequence?
4.
If
p
is a prime number, how many factors does
p
3
have?
(A)
One
(B)
Two
(C)
Three
(D)
Four
(E)
Five
5.
How many integers between 10 and 500 begin and end in
3 ?
6.
A particular integer
N
is divisible by two different prime
numbers
p
and
q
. Which of the following must be true?
I.
N
is not a prime number.
II.
N
is divisible by
pq
.
III.
N
is an odd integer.
(A)
I only
(B)
II only
(C)
I and II only
(D)
I and III only
(E)
I, II, and III
7.
A perfect square is an integer that is the square of an
integer. Suppose that
m
and
n
are positive integers such
that
mn >
15. If 15
mn
is a perfect square, what is the
least possible value of
mn
?
8.
M
is a set of six consecutive even integers. When the
least three integers of set
M
are summed, the result is
x
.
When the greatest three integers of set
M
are summed,
the result is
y
. Which of the following is true?
(A)
y
=
x
−
18
(B)
y
=
x
+ 18
(C)
y
= 2
x
(D)
y
= 2
x
+ 4
(E)
y
= 2
x
+ 6
erikthered.com/tutor
pg. 1
SAT Math Hard Practice Quiz
9.
A three-digit number,
XYZ
, is formed of three different
non-zero digits
X
,
Y
, and
Z
. A new number is formed by
rearranging the same three digits. What is the greatest
possible difference between the two numbers? (For ex-
ample, 345 could be rearranged into 435, for a difference
of 435
−
345 = 90.)
10.
An integer is subtracted from its square. The result could
be which of the following?
(A)
A negative integer.
(B)
An odd integer.
(C)
The product of two consecutive even integers.
(D)
The product of two consecutive odd integers.
(E)
The product of two consecutive integers.
erikthered.com/tutor
pg. 2
SAT Math Hard Practice Quiz
Algebra and Functions
1.
Let
m
be an even integer. How many possible values of
m
satisfy
√
m
+ 7
≤
3 ?
(A)
One
(B)
Two
(C)
Three
(D)
Four
(E)
Five
2.
Let
x
be defined by
x
=
x
+ 3
x
−
1
for any
x
such that
x
6
= 1. Which of the following is equivalent to
x
−
1 ?
(A)
x
+ 2
x
−
1
(B)
4
x
−
1
(C)
2
x
+ 4
x
−
1
(D)
2
x
−
1
(E)
x
+ 2
x
−
2
3.
Let
a
and
b
be numbers such that
a
3
=
b
2
. Which of the
following is equivalent to
b
√
a
?
(A)
b
2
/
3
(B)
b
4
/
3
(C)
b
2
(D)
b
3
(E)
b
4
4.
Let
m
and
n
be positive integers such that one-third of
m
is
n
less than one-half of
m
. Which of the following is
a possible value of
m
?
(A)
15
(B)
21
(C)
24
(D)
26
(E)
28
5.
If
a
and
b
are numbers such that (
a
−
4)(
b
+ 6) = 0, then
what is the smallest possible value of
a
2
+
b
2
?
6.
Let
f
(
x
) =
ax
2
and
g
(
x
) =
bx
4
for any value of
x
. If
a
and
b
are positive constants, for how many values of
x
is
f
(
x
) =
g
(
x
) ?
(A)
None
(B)
One
(C)
Two
(D)
Three
(E)
Four
7.
Let
a
and
b
be numbers such that 30
< a <
40 and 50
<
b <
70. Which of the following represents all possible
values of
a
−
b
?
(A)
−
40
< a
−
b <
−
20
(B)
−
40
< a
−
b <
−
10
(C)
−
30
< a
−
b <
−
20
(D)
−
20
< a
−
b <
−
10
(E)
−
20
< a
−
b <
30
erikthered.com/tutor
pg. 3
SAT Math Hard Practice Quiz
x
3
+
y
12
=
z
8.
In the equation shown above,
x
,
y
, and
z
are positive
integers. All of the following could be a possible value of
y
EXCEPT
(A)
4
(B)
6
(C)
8
(D)
12
(E)
20
√
72 +
√
72 =
m
√
n
9.
In the equation above,
m
and
n
are integers such that
m > n
. Which of the following is the value of
m
?
(A)
6
(B)
12
(C)
16
(D)
24
(E)
48
t
0
1
2
N
(
t
)
128
16
2
10.
The table above shows some values for the function
N
.
If
N
(
t
) =
k
·
2
−
at
for positive constants
k
and
a
, what is
the value of
a
?
(A)
−
3
(B)
−
2
(C)
1
3
(D)
2
(E)
3
11.
Amy is two years older than Bill. The square of Amy’s
age in years is 36 greater than the square of Bill’s age in
years. What is the sum of Amy’s age and Bill’s age in
years?
1
y
1
x
y
=
f
(
x
)
12.
The function
f
is graphed in its entirety above. If the
function
g
is defined so that
g
(
x
) =
f
(
−
x
), then for what
value of
x
does
g
attain its maximum value?
(A)
−
3
(B)
−
2
(C)
0
(D)
2
(E)
3
13.
If (
x
+ 1)
2
= 4 and (
x
−
1)
2
= 16, what is the value of
x
?
(A)
−
3
(B)
−
1
(C)
1
(D)
3
(E)
5
erikthered.com/tutor
pg. 4
SAT Math Hard Practice Quiz
x
12
x
8
14.
On the number line above, the tick marks correspond to
consecutive integers. What is the value of
x
?
15.
The value of
y
increased by 12 is directly proportional to
the value of
x
decreased by 6. If
y
= 2 when
x
= 8, what
is the value of
x
when
y
= 16 ?
(A)
8
(B)
10
(C)
16
(D)
20
(E)
28
16.
Two cars are racing at a constant speed around a circu-
lar racetrack. Car A requires 15 seconds to travel once
around the racetrack, and car B requires 25 seconds to
travel once around the racetrack. If car A passes car B,
how many seconds will elapse before car A once again
passes car B ?
erikthered.com/tutor
pg. 5
SAT Math Hard Practice Quiz
Geometry
y
x
y
=
x
2
2
y
=
x
2
(
a, b
)
O
1.
The curve
y
=
x
2
/
2 and the line
y
=
x/
2 intersect at
the origin and at the point (
a, b
), as shown in the figure
above. What is the value of
b
?
(A)
1
8
(B)
1
4
(C)
1
2
(D)
1
(E)
2
6
8
A
B
C
60
◦
2.
In the figure above,
AB
= 6 and
BC
= 8. What is the
area of triangle
ABC
?
(A)
12
√
2
(B)
12
√
3
(C)
24
√
2
(D)
24
√
3
(E)
36
√
3
a
c
b
Note: Figure not drawn to scale.
3.
In the figure above, 3
< a <
5 and 6
< b <
8. Which of
the following represents all possible values of
c
?
(A)
0
< c <
3
(B)
1
< c <
3
(C)
0
< c <
13
(D)
1
< c <
13
(E)
3
< c <
13
4.
Line
l
goes through points
P
and
Q
, whose coordinates
are (0
,
1) and (
b,
0), respectively. For which of the fol-
lowing values of
b
is the slope of line
l
greater than
−
1
2
?
(A)
1
2
(B)
1
(C)
3
2
(D)
5
3
(E)
5
2
erikthered.com/tutor
pg. 6
SAT Math Hard Practice Quiz
A
B
C
D
5.
In the figure above,
AB
= 6 and
BC
= 8. What is the
length of segment
BD
?
(A)
2
(B)
12
5
(C)
4
(D)
24
5
(E)
6
6.
If four distinct lines lie in a plane, and exactly two of
them are parallel, what is the least possible number of
points of intersection of the lines?
(A)
Two
(B)
Three
(C)
Four
(D)
Five
(E)
More than five
7.
The perimeter of a particular equilateral triangle is
numerically equal to the area of the triangle. What is
the perimeter of the triangle?
(A)
3
(B)
4
(C)
4
√
3
(D)
12
√
3
(E)
18
√
3
8.
In the figure above, a square is inscribed in a circle. If
the area of the square is 36, what is the perimeter of the
shaded region?
(A)
6 +
3
√
2
2
π
(B)
6 + 3
π
(C)
6 + 3
√
2
π
(D)
36 + 6
√
2
π
(E)
9
2
π
−
9
A
B
C
7
Note: Figure not drawn to scale.
9.
In the figure above,
AC
= 7 and
AB
=
BC
. What is the
smallest possible integer value of
AB
?
erikthered.com/tutor
pg. 7
SAT Math Hard Practice Quiz
y
x
(2
, a
)
(10
,
0)
O
10.
In the figure above, two line segments in the
x
-
y
plane
form a right triangle with the
x
-axis. What is the value
of
a
?
(A)
2
√
2
(B)
4
(C)
5
(D)
4
√
2
(E)
5
√
2
11.
The perimeter of square
ABCD
is
x
, and the perimeter
of isosceles triangle
EF G
is
y
. If
AB
=
EF
=
F G
, which
of the following must be true?
(A)
0
< y <
x
4
(B)
x
4
< y <
x
2
(C)
x
2
< y < x
(D)
x < y <
2
x
(E)
2
x < y <
4
x
y
x
P
O
y
= 2
x
−
1
y
=
x
+
c
Note: Figure not drawn to scale.
12.
In the
x
-
y
plane, the lines
y
= 2
x
−
1 and
y
=
x
+
c
inter-
sect at point
P
, where
c
is a positive number. Portions
of these lines are shown in the figure above. If the value
of
c
is between 1 and 2, what is one possible value of the
x
-coordinate of
P
?
erikthered.com/tutor
pg. 8
SAT Math Hard Practice Quiz
Data, Statistics, and Probability
1.
The first term of a sequence is the number
n
, and each
term thereafter is 5 greater than the term before. Which
of the following is the average (arithmetic mean) of the
first nine terms of this sequence?
(A)
n
+ 20
(B)
n
+ 180
(C)
2
n
(D)
2
n
+ 40
(E)
9
n
+ 180
2.
The average (arithmetic mean) of a particular set of
seven numbers is 12. When one of the numbers is re-
placed by the number 6, the average of the set increases
to 15. What is the number that was replaced?
(A)
−
20
(B)
−
15
(C)
−
12
(D)
0
(E)
12
3.
Let
a
,
b
, and
c
be positive integers. If the average (arith-
metic mean) of
a
,
b
, and
c
is 100, which of the following
is NOT a possible value of any of the integers?
(A)
1
(B)
100
(C)
297
(D)
298
(E)
299
4.
M
is a set consisting of a finite number of consecutive
integers. If the median of the numbers in set
M
is equal
to one of the numbers in set
M
, which of the following
must be true?
I.
The average (arithmetic mean) of the
numbers in set
M
equals the median.
II. The number of numbers in set
M
is odd.
III. The sum of the smallest number and the
largest number in set
M
is even.
(A)
I only
(B)
II only
(C)
I and II only
(D)
I and III only
(E)
I, II, and III
erikthered.com/tutor
pg. 9
SAT Math Hard Practice Quiz Answers
Numbers and Operations
1.
B
(Estimated Difficulty Level: 5)
The number of green and red tomatoes are 4
n
and 3
n
,
respectively, for some integer
n
. In this way, we can be
sure that the green-to-red ratio is 4
n/
3
n
= 4
/
3. We
need to solve the equation:
4
n
−
5
3
n
−
5
=
3
2
.
Cross-multiplying, 8
n
−
10 = 9
n
−
15 so that
n
= 5.
There were 3
n
, or 15, red tomatoes in the bag.
Working with the answers may be easier. If answer A
is correct, then there were 16 green tomatoes and 12
red tomatoes, in order to have the 4 to 3 ratio. But
removing five of each gives 11 green and 7 red, which is
not in the ratio of 3 to 2. If answer B is correct, then
there were 20 green tomatoes and 15 red tomatoes, since
20
/
15 = 4
/
3. Removing five of each gives 15 green and
10 red, and 15
/
10 = 3
/
2, so answer B is correct.
2.
C
(Estimated Difficulty Level: 4)
From 10 to 19, 12 and up (eight numbers) are mono-
tonic. Among the numbers from 20 to 29, seven (23 and
up) are monotonic. If you can see a pattern in counting
problems like this, you can save a lot of time. Here, the
30s will have 6 monotonic numbers, the 40s will have 5,
and so forth. You should find 8 + 7 + 6 + 5 + 4 + 3 +
2 + 1 + 0 = 36 total monotonic numbers.
3.
113
(Estimated Difficulty Level: 5)
Since the second term is 7 greater than the first term,
(2
a
−
1)
−
a
= 7 so that
a
= 8. The sequence is 8,
15, 22,
. . .
You can either continue to write out the
sequence until the 16
th
term, or realize that the 16
th
term is 16
a
−
15 = 16(8)
−
15 = 128
−
15 = 113.
4.
D
(Estimated Difficulty Level: 4)
The answer must be true for any value of
p
, so plug in
an easy (prime) number for
p
, such as 2. The factors of
2
3
= 8 are 1, 2, 4, and 8, so answer D is correct.
In general, since
p
is prime, the only numbers that go
into
p
3
without a remainder are 1,
p
,
p
2
, and
p
3
.
5.
11
(Estimated Difficulty Level: 4)
For the two-digit numbers, only 33 begins and ends in
3. For three-digit numbers, the only possibilities are:
303, 313,
. . .
, 383, and 393. We found ten three-digit
numbers, and one two-digit number, for a total of 11
numbers that begin and end in 3.
Yes, this was a counting problem soon after another
counting problem. But this one wasn’t so bad, was it?
6.
C
(Estimated Difficulty Level: 4)
This type of SAT math question contains three separate
mini-problems. (This kind of question is also known as
“one of those annoying, long, SAT math questions with
roman numerals”). Let’s do each mini-problem in order.
First, recall that a prime number is only divisible by
itself and 1, and that 1 is not a prime number. So,
statement I must be true, since a number that can be
divided by two prime numbers can’t itself be prime.
Next, recall that every number can be written as a prod-
uct of a particular bunch of prime numbers. Let’s say
that
N
is divisible by 3 and 5. Then,
N
is equal to
3
·
5
·
p
1
·
p
2
· · ·
, where
p
1
,
p
2
, etc. are some other primes.
So,
N
is divisible by 3
·
5 = 15. Statement II must be
true.
Finally, remember that 2 is a prime number. So,
N
could be 6, since 6 = 2
·
3. Statement III isn’t always
true, making C the correct answer.
erikthered.com/tutor
pg. 10
SAT Math Hard Practice Quiz Answers
7.
60
(Estimated Difficulty Level: 5)
First, note that 15
mn
= 3
·
5
·
mn
. We need
√
3
·
5
·
mn
to be an integer. We could have, for example,
m
= 3
and
n
= 5 since
√
3
·
5
·
3
·
5 =
√
3
2
·
5
2
= 15, except
that the problem requires that
mn >
15. (This is a hard
problem for a reason, after all!) If
m
= 3
·
2 and
n
= 5
·
2
then 15
mn
= (3
·
5)(3
·
2)(5
·
2) = 3
2
·
5
2
·
2
2
. That way,
√
15
mn
=
√
3
2
·
5
2
·
2
2
= 30 is still an integer, making
the least possible value of
mn
equal to 6
·
10 = 60.
8.
B
(Estimated Difficulty Level: 4)
A good opportunity to plug in real numbers! For ex-
ample, suppose set
M
consists of the integers: 2, 4, 6,
8, 10, and 12. The sum of the least three is 12 and the
sum of the greatest three is 30, so answer B is correct.
You say you want an algebraic solution? Suppose that
n
is the first even integer. The remaining integers are then
n
+2,
n
+4,
n
+6,
n
+8, and
n
+10. The sum of the least
three of these integers is
x
=
n
+(
n
+2)+(
n
+4) = 3
n
+6,
and the sum of the greatest three of these integers is
y
= (
n
+6)+(
n
+8)+(
n
+10) = 3
n
+24. So,
y
−
x
= 18,
or
y
=
x
+ 18.
9.
792
(Estimated Difficulty Level: 5)
To get the greatest difference, we want to subtract a
small number from a large one, so we will need the
digit 9 and the digit 1, in order to make a number in
the 100’s and a number in the 900’s. The large number
will look like 9
N
1 and the small number will look like
1
N
9, where
N
is a digit from 2 to 8. You will find that,
no matter what you make
N
, the difference is 792.
10.
E
(Estimated Difficulty Level: 5)
Suppose that the integer is
n
. The result of subtract-
ing
n
from its square is
n
2
−
n
=
n
(
n
−
1), which is
the product of two consecutive integers, so answer E is
correct.
Notice that if you multiply any two consecutive integers,
the result is always even, since it is the product of an
even integer and an odd integer. To win an Erik The
Red Viking Hat, see if you can determine why the result
is never a negative integer.
erikthered.com/tutor
pg. 11
SAT Math Hard Practice Quiz Answers
Algebra and Functions
1.
E
(Estimated Difficulty Level: 5)
The answers suggest that there aren’t that many pos-
sibilities. So, make up some even integers, plug them
in for
m
, and see if they work! Since we can’t take the
square root of a negative number,
m
can’t be less than
−
6. Also, if
m
= 2, then
√
m
+ 7 = 3, but any larger
value of
m
won’t work. So, the possible values are
−
6,
−
4,
−
2, 0, and 2. (Don’t forget that zero is a perfectly
good even integer.)
2.
B
(Estimated Difficulty Level: 4)
Plug in real numbers for
x
! You can plug in anything
other than 1. If you set
x
= 2, then
x
−
1 = 5
−
1 = 4.
Now go through the answers, plugging in 2 for
x
. You
will find that answers A and B are both equal in value
to 4. If this happens, simply plug in another number.
(You don’t need to retry the answers that were wrong.)
If
x
= 0, then
x
−
1 =
−
3
−
1 =
−
4, and only answer
B is also
−
4, so that is the correct answer.
If you love algebra, here is how to do it:
x
−
1 =
x
+ 3
x
−
1
−
1
=
x
+ 3
x
−
1
−
x
−
1
x
−
1
=
x
+ 3
−
(
x
−
1)
x
−
1
=
4
x
−
1
Be still my beating heart!
3.
B
(Estimated Difficulty Level: 5)
Solve the first equation given for
a
:
a
=
b
2
/
3
. Then
√
a
=
a
1
/
2
=
b
1
/
3
. (You really need to know your ex-
ponent rules for this one.) So,
b
√
a
=
b
·
b
1
/
3
=
b
4
/
3
.
Using real numbers also works here, but it may be hard
to come up with two that work for
a
and
b
(such as
a
= 4 and
b
= 8).
4.
C
(Estimated Difficulty Level: 4)
Translate the words into an algebraic equation:
m
3
=
m
2
−
n.
Multiplying both sides by 6 (the common denominator)
gives 2
m
= 3
m
−
6
n
, or
m
= 6
n
. So,
m
must be a
positive multiple of 6, which means that answer C is
correct.
5.
16
(Estimated Difficulty Level: 4)
Since (
a
−
4)(
b
+6) = 0, the possible solutions are:
a
= 4
and
b
is anything, or
b
=
−
6 and
a
is anything. Now,
the expression
a
2
+
b
2
is made smallest by choosing
a
and
b
to be close to zero as possible. So,
a
= 4 and
b
= 0 will give us the smallest value of
a
2
+
b
2
, namely,
16. Using the other solution would give
a
2
+ 36, which
will always be bigger than 16.
6.
D
(Estimated Difficulty Level: 5)
This is a tough one. For
f
(
x
) to be equal to
g
(
x
) for
all
x
, we need
ax
2
=
bx
4
. First, notice that if
x
= 0,
both sides are zero, so
x
= 0 is a solution. If
x
is not
zero, we can divide both sides of the equation by
x
2
to
get:
a
=
bx
2
. Solving for
x
results in
x
=
±
p
a/b
. This
makes three solutions total, so answer D is correct. It
may help to plug in numbers for
a
and
b
to make this
problem more concrete.
7.
B
(Estimated Difficulty Level: 5)
To make
a
−
b
as large as possible, we need to make
a
as
large as possible and
b
as small as possible. So,
a
−
b
has
to be less than 40
−
50 =
−
10. To make
a
−
b
as small
as possible, we need to make
a
as small as possible and
b
as large as possible. So,
a
−
b
has to be greater than
30
−
70 =
−
40. The expression that gives all possible
values of
a
−
b
is then
−
40
< a
−
b <
−
10.
erikthered.com/tutor
pg. 12
SAT Math Hard Practice Quiz Answers
8.
B
(Estimated Difficulty Level: 5)
Solve the equation for
y
. You should get:
y
= 12
z
−
4
x
.
Factor out a 4 from the right-hand side:
y
= 4(3
z
−
x
).
Since
z
and
x
are integers, 3
z
−
x
is an integer, so that
y
is a multiple of 4. Only answer B is not a multiple of
4, so it can’t be a possible value of
y
.
You could also combine the fractions on the left-hand
side of the equation to get:
4
x
+
y
12
=
z.
For
z
to be an integer, 4
x
+
y
must be a multiple of
12. Try plugging in various integers for
x
and
y
to get
multiples of 12; you should find that
y
can take on all
of the values in the answers except for 6.
9.
B
(Estimated Difficulty Level: 4)
Combining the two terms on the left-hand side of the
equation gives us 2
√
72, but that doesn’t give us what
we need, which is
m > n
.
For the SAT test, you should know how to rewrite and
simplify radicals. Here,
√
72 =
√
36
·
2 = 6
√
2, so the
left-hand side is equal to 12
√
2, making
m
= 12 and
n
= 2.
10.
E
(Estimated Difficulty Level: 5)
Hint: if you see a zero in a table problem like this one,
try to use it first! When you plug in 0 for
t
, you get
N
(0) =
k
·
2
−
0
=
k
·
1 =
k
, which means that
k
= 128.
Next, try plugging in 1 for
t
:
N
(1) = 128
·
2
−
a
. From
the table,
N
(1) = 16, so that 128
·
2
−
a
= 16, or 2
−
a
=
1
/
2
a
= 1
/
8. Since 8 = 2
3
,
a
= 3. (Your calculator
may also help here, but try to understand how to do it
without it.)
11.
18
(Estimated Difficulty Level: 4)
Let
a
be Amy’s age and
b
be Bill’s age. The problem
tells us that:
a
=
b
+ 2 and
a
2
=
b
2
+ 36. One way to
do this is to plow ahead, substitute for one variable and
solve for the other (a bit messy!). But SAT questions
are designed to be solved without tedious calculations
and/or messy algebra. Let’s try doing the problem us-
ing the SAT way, not the math teacher way.
First, notice that the second equation can be written as:
a
2
−
b
2
= 36. This is a difference of two squares, and is
the same as: (
a
+
b
)(
a
−
b
) = 36. The first equation can
be written as:
a
−
b
= 2. This means that the second
equation is just (
a
+
b
)
·
2 = 36, so that
a
+
b
= 18. We
don’t know what
a
and
b
are, and we don’t even care!
12.
B
(Estimated Difficulty Level: 4)
You can obtain the graph of
y
=
f
(
−
x
) by “flipping”
the graph of
y
=
f
(
x
) across the
y
-axis. For exam-
ple, if the point (3
,
1) is on the graph of
f
(
x
), then
the point (
−
3
,
1) must be on the graph of
f
(
−
x
), since
f
(
−
(
−
3)) =
f
(3) = 1.
The figure below tells the story. Here, the function
g
(
x
) =
f
(
−
x
) is shown as a dashed line:
1
y
1
x
f
(
x
)
g
(
x
)
From the graph,
g
(
x
) is maximum when
x
=
−
2.
erikthered.com/tutor
pg. 13
SAT Math Hard Practice Quiz Answers
13.
A
(Estimated Difficulty Level: 4)
A good “skip-the-algebra” way to do this problem is to
use the answers by plugging them into
x
until the two
given equations work. Using answer A, you should find
that (
−
3+1)
2
= (
−
2)
2
= 4 and (
−
3
−
1)
2
= (
−
4)
2
= 16,
so answer A is correct.
You must have the algebraic solution, you say? Try
taking the square root of both sides of the equations,
but don’t forget that there are two possible solutions
when you do this. The first equation gives:
x
+ 1 =
±
2
so that
x
= 1 or
x
=
−
3. The second equation gives
x
−
1 =
±
4 so that
x
= 5 or
x
=
−
3. The only solution
that works for both equations is
x
=
−
3.
14.
96
(Estimated Difficulty Level: 5)
Since the tick marks correspond to consecutive integers,
and it takes four “steps” to go from
x/
12 to
x/
8, we
know that
x/
8 is four greater than
x/
12. (Or, think of
the spaces between the tick marks: there are four spaces
and each space is length 1, so the distance from
x/
12
to
x/
8 is 4.) In equation form:
x
8
=
x
12
+ 4
.
Multiplying both sides by 24 gives: 3
x
= 2
x
+ 4
·
24 so
that
x
= 96.
15.
B
(Estimated Difficulty Level: 5)
First, recall that if
y
is proportional to
x
, then
y
=
kx
for some constant
k
. So, “
y
increased by 12 is directly
proportional to
x
decreased by 6” translates into the
math equation:
y
+ 12 =
k
(
x
−
6). Plugging in
y
= 2
and
x
= 8 gives 14 =
k
·
2 so that
k
= 7. Our equation
is now:
y
+ 12 = 7(
x
−
6). Plugging in 16 for
y
gives
28 = 7(
x
−
6) so that
x
−
6 = 4, or
x
= 10.
16.
75
/
2 or 37
.
5
(Estimated Difficulty Level: 5)
To make this problem more concrete, make up a number
for the circumference of the racetrack. It doesn’t really
matter what number you use; I’ll use 75 feet. Since
speed is distance divided by time, the speed of car A
is 75
/
15 = 5 feet per second, and the speed of car B is
75
/
25 = 3 feet per second. (I picked 75 mostly because
it is divided evenly by 15 and 25.) Every second, car A
gains 2 feet on car B. To pass car B, car A must gain
75 feet on car B. This will require 75
/
2 = 37
.
5 seconds.
You may be thinking, “Whoa, tricky solution!” Here
is the mostly straightforward but somewhat tedious al-
gebraic solution. Once again, I’ll use 75 feet for the
circumference of the track. Suppose that you count
time from when car A first passes car B. Then, car A
travels a distance (75
/
15)
t
= 5
t
feet after
t
seconds.
(Remember that distance = speed
×
time.) For ex-
ample, after 15 seconds, car A has traveled a distance
5
·
15 = 75 feet, and after 30 seconds, car A has traveled
a distance 5
·
30 = 150 feet. Similarly, car B travels a
distance (75
/
25)
t
= 3
t
feet after
t
seconds. When the
two cars pass again, car A has traveled 75 feet more
than car B: 5
t
= 3
t
+ 75. Solving for
t
gives: 2
t
= 75,
or
t
= 75
/
2 = 37
.
5 seconds.
erikthered.com/tutor
pg. 14
SAT Math Hard Practice Quiz Answers
Geometry
1.
C
(Estimated Difficulty Level: 4)
To determine where two curves intersect, set the equa-
tions equal to one another and solve for
x
. (Hint: know
this for the SAT!) In this question, we need to figure
out which values of
x
satisfy:
x
2
/
2 =
x/
2. If
x
= 0,
this equation works, but we need the solution when
x
6
= 0. Dividing both sides of the equation by
x
gives:
x/
2 = 1
/
2 so that
x
= 1. By plugging in
x
= 1 to either
of the two curves, you should find that
y
= 1
/
2. So, the
point of intersection is (1
,
1
/
2), making answer C the
correct one.
2.
B
(Estimated Difficulty Level: 5)
This is the kind of problem that would be too hard
and/or require things you aren’t expected to know for
the SAT (such as trigonometry),
unless
you draw a con-
struction line in the figure. (This is a very hard question
anyway.) In this case, you want to draw a line from
A
perpendicular to the opposite side of the triangle:
6
8
A
B
C
60
◦
30
◦
This forms a 30-60-90 triangle whose hypotenuse has
length 6. Now, use the 30-60-90 triangle diagram given
to you at the beginning of each SAT math section: The
length of the side opposite the 30
◦
angle is 3, and the
length of the side opposite the 60
◦
angle (the dashed
line) is 3
√
3.
So finally, if the base of the triangle is segment
BC
,
then the dashed line is the height of the triangle, and
the area of the triangle is (1
/
2)
·
8
·
3
√
3 = 12
√
3.
3.
D
(Estimated Difficulty Level: 5)
You need to know the “third-side rule” for triangles to
solve this question:
The length of the third side of a
triangle is less than the sum of the lengths of the other
two sides and greater than the positive difference of the
lengths of the other two sides
. Applied to this question,
the first part of the rule says that the value of
c
must be
less than
a
+
b
. Since we are interested in
all
possible
values of
c
, we need to know the greatest possible value
of
a
+
b
. With
a <
5 and
b <
8,
a
+
b <
13 so that
c
must be less than 13.
For the second part of the rule,
c
must be greater than
b
−
a
. (Note that
b
is always bigger than
a
, so that
b
−
a
is positive.) We are interested in
all
possible values of
c
, so we need to know the least possible value of
b
−
a
.
The least value occurs when
b
is as small as possible and
a
is as large as possible:
b
−
a >
6
−
5 = 1. Then,
c
must
be greater than 1. Putting this together, 1
< c <
13,
making answer D the correct one.
4.
E
(Estimated Difficulty Level: 4)
First, calculate the slope of line
l
using the given points:
slope =
rise
run
=
0
−
1
b
−
0
=
−
1
b
.
At this point, a good approach is to work with the an-
swers by plugging them into the expression for slope
above until you get a value greater than
−
1
/
2. For ex-
ample, using answer A gives a slope of
−
1
/
(1
/
2) =
−
2,
which is not greater than
−
1
/
2, so answer A is incor-
rect. You should find that answer E is the correct one,
since
−
1
/
(5
/
2) =
−
2
/
5 is greater than
−
1
/
2. (Know-
ing the decimal equivalents of basic fractions will really
help speed this process up.)
Here is the algebraic solution:
−
1
b
>
−
1
2
⇒
1
b
<
1
2
⇒
b >
2
.
(Remember to flip the inequality when multiplying by
negative numbers or when taking the reciprocal of both
sides.) Only answer E makes
b >
2.
erikthered.com/tutor
pg. 15
SAT Math Hard Practice Quiz Answers
5.
D
(Estimated Difficulty Level: 5)
Since
ABC
is a right triangle, the length of segment
AC
is
√
6
2
+ 8
2
= 10. (Hint: you will see 3-4-5 and 6-8-10
triangles a lot on the SAT.) The area of triangle
ABC
is (1
/
2)
·
b
·
h
, where
b
is the base and
h
is the height of
triangle
ABC
.
The key thing to remember for this problem is that the
base can be
any
of the three sides of a triangle, not just
the side at the bottom of the diagram. If the base is
AB
, then the height is
BC
and the area of the triangle
is (1
/
2)(6)(8) = 24. If the base is
AC
, then the height is
BD
and the area of the triangle is still 24. This means
that (1
/
2)(
AC
)(
BD
) = (1
/
2)(10)(
BD
) = 24 so that
BD
= 24
/
5.
6.
B
(Estimated Difficulty Level: 4)
Draw a diagram for this problem! With exactly two
parallel lines, the other two lines cannot be parallel to
themselves or to the first two lines. Your diagram may
seem to suggest five points of intersection; however, the
point of intersection of the two non-parallel lines can
overlap with a point of intersection on one of the parallel
lines:
From the figure, the least possible number of intersec-
tion points is then three.
7.
D
(Estimated Difficulty Level: 5)
One formula that a good math student such as yourself
may want to memorize for the SAT is the area of an
equilateral triangle. If the length of each side of the
triangle is
s
, then the area is
√
3
s
2
/
4. The perimeter of
this triangle is 3
s
.
Now, since the perimeter equals the area for this tri-
angle, we have: 3
s
=
√
3
s
2
/
4 so that 3 =
√
3
s/
4 and
s
= 12
/
√
3 = 4
√
3. The perimeter is then 12
√
3, mak-
ing answer D the correct one. (Did you get
s
= 4
√
3
and then choose answer C? Sorry about that.)
8.
A
(Estimated Difficulty Level: 5)
For many difficult SAT questions, it can be very helpful
to know some “extra” math along with the “required”
math. First, when a square is inscribed in a circle,
the diagonals are diameters of the circle. Second, the
diagonals of a square meet at right angles. Third, a
diagonal of a square is
√
2 times as long as the length
of one of the sides. (A diagonal of a square makes a 45-
45-90 triangle with two sides.) For this question, the
length of each side of the square is 6 (since the area is
6
2
= 36), and the length of a diagonal is 6
√
2, so the
radius of the circle is 3
√
2, as shown below:
6
6
6
6
3
√
2
3
√
2
A final piece of needed math: the arc length of a portion
of a circle is the circumference times the central angle of
the arc divided by 360
◦
. Here, the central angle is 90
◦
,
so the needed arc length (shown darkened in the figure
above) is just 1
/
4 times the circle’s circumference. The
arc length is then 2
πr/
4 = 2
π
·
3
√
2
/
4 = 3
π
√
2
/
2 and
the perimeter of the shaded region is 6 + 3
π
√
2
/
2.
erikthered.com/tutor
pg. 16
SAT Math Hard Practice Quiz Answers
9.
4
(Estimated Difficulty Level: 5)
You need to know half of the “third-side rule” for trian-
gles to solve this question:
The length of the third side
of a triangle is less than the sum of the lengths of the
other two sides.
For this question, we will make
AC
the
third side.
Now, suppose that the length of each of the other two
sides of the triangle is
x
, so that
AB
=
BC
=
x
. Then,
the third-side rule says that
AC
is less than the sum of
AB
and
BC
: 7
< x
+
x
. Simplifying gives: 2
x >
7 so
that
x >
3
.
5. The smallest possible integer value for
x
is 4.
10.
B
(Estimated Difficulty Level: 5)
One way to do this question is to use the fact that the
product of the slopes of two perpendicular lines (or line
segments) is
−
1. The slope of the line segment on the
left is (
a
−
0)
/
(2
−
0) =
a/
2. The slope of the line
segment on the right is (0
−
a
)
/
(10
−
2) =
−
a/
8. The
two slopes multiply to give
−
1:
a
2
·
−
a
8
=
−
a
2
16
=
−
1
.
Solving for
a
gives
a
2
= 16 so that
a
= 4. A messier
way to do this problem is to use the distance formula
and the Pythagorean theorem. The length of the line
segment on the left is
√
2
2
+
a
2
, and the length of line
segment on the right is
p
(10
−
2)
2
+ (0
−
a
)
2
. Then,
the Pythagorean theorem says that:
p
2
2
+
a
2
2
+
p
(10
−
2)
2
+ (0
−
a
)
2
2
= 10
2
.
Simplifying the left-hand side gives: 2
a
2
+ 68 = 100 so
that 2
a
2
= 32. Then,
a
2
= 16, making
a
= 4.
11.
C
(Estimated Difficulty Level: 5)
Make a diagram, and fill it in with the information that
is given. (You should do this for any difficult geome-
try question without a figure.) Since the perimeter of
square
ABCD
is
x
, each side of the square has length
x/
4, so your figure should look something like this:
A
B
C
D
E
F
G
x
4
x
4
x
4
x
4
x
4
x
4
Now, use the third-side rule for triangles:
The length
of the third side of a triangle is less than the sum of
the lengths of the other two sides and greater than the
positive difference of the lengths of the other two sides
.
When the rule is applied to
EG
as the third side, we
get: 0
< EG < x/
2. If
y
is the perimeter of the triangle,
then
y
=
x/
4 +
x/
4 +
EG
=
x/
2 +
EG
. Solving for
EG
gives
EG
=
y
−
x/
2. Substituting into the inequality
gives 0
< y
−
x/
2
< x/
2 so that
x/
2
< y < x
, mak-
ing answer C the correct one. To make this problem
less abstract, it may help to make up a number for the
perimeter of the square. (A good choice might be 4 so
that
x
= 1. You’ll find 1
/
2
< y <
1, the same as answer
C when
x
= 1.)
12.
2
< x <
3
(Estimated Difficulty Level: 5)
In order to determine at what point two lines intersect,
set the equations of the lines equal to one another. In
this case, we have: 2
x
−
1 =
x
+
c
so that
x
=
c
+ 1.
In other words,
x
=
c
+ 1 is the
x
-coordinate of
P
, the
point where the lines intersect. Now, if
c
is between 1
and 2, then
c
+ 1 is between 2 and 3. Any value for the
x
-coordinate of
P
between 2 and 3 is correct.
erikthered.com/tutor
pg. 17
SAT Math Hard Practice Quiz Answers
Data, Statistics, and Probability
1.
A
(Estimated Difficulty Level: 4)
The first nine terms of the sequence are:
n
,
n
+ 5,
n
+ 10,
n
+ 15,
. . .
,
n
+ 40
.
(You should probably write all nine terms out to avoid
mistakes.) Adding these terms up gives: 9
n
+ 180. The
average is the sum (9
n
+ 180) divided by the number of
terms (9). The average is then: (9
n
+ 180)
/
9 =
n
+ 20.
2.
B
(Estimated Difficulty Level: 5)
The average of a set of numbers is the sum of the num-
bers divided by the number of numbers:
average =
sum
N
.
We can solve this equation for the sum:
sum = average
×
N.
Here, since there are 7 numbers and the average is 12,
the sum of the numbers is 7
×
12 = 84. The sum of the
new set of numbers is 7
×
15 = 105. Now, suppose that
the seven numbers are
a
,
b
,
c
,
d
,
e
,
f
, and
g
, and that
g
gets replaced with the number 6. Then, we have:
a
+
b
+
c
+
d
+
e
+
f
+
g
= 84
,
and
a
+
b
+
c
+
d
+
e
+
f
+ 6 = 105
.
The second equation says that
a
+
b
+
c
+
d
+
e
+
f
= 99.
Substituting into the first equation gives 99 +
g
= 84 so
that
g
=
−
15.
3.
E
(Estimated Difficulty Level: 4)
Using the definition of average gives:
a
+
b
+
c
3
= 100
so that
a
+
b
+
c
= 300. Since
a
,
b
, and
c
are all positive,
the smallest possible value for any of the numbers is 1.
The largest possible value of one of the three numbers
then occurs when the other two numbers are both 1. In
this case, the numbers are 1, 1, and 298, so that the
largest possible value is 298. Answer E can not be a
possible value, so it is the correct answer.
4.
E
(Estimated Difficulty Level: 5)
Plug in real numbers for set
M
to make this problem
concrete. For example, if
M
is the set of consecutive
integers from 1 to 5, then the median and average are
both 3. If
M
is the set of consecutive integers from 1
to 4, then the median and average are both 2
.
5. From
these examples, we can see that the number of numbers
in set
M
needs to odd, otherwise the median is not an
integer. Choice II must be true.
Also, if the number of numbers in a set of consecutive
integers is odd, then when the first number is odd, the
last number is odd. Or, when the first number is even,
the last number is even. This is because the difference
of the largest number and the smallest number will be
even when the number of numbers is odd. Choice III
must then be true, since the sum of two odd numbers
or two even numbers is an even number.
At this point, the only answer with choices II and III is
answer E, so that must be the correct answer. Why is
choice I also correct? The average of a set of consecutive
integers is equal to the average of the first and the last
integers in the set. The average of two integers that are
both odd or both even is the integer halfway between
the two, which is also the median of the set. Whew!
erikthered.com/tutor
pg. 18
