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
