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