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.
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