Quantity A
x
Quantity B
3. If 6 < 2
x
– 4 < 12, which of the following could be the value of
x
?
(A)
4
(B)
5
(C)
7
(D)
8
(E)
9
4. If
y
< 0 and 4
x
>
y
, which of the following could be equal to
?
(A)
0
(B)
(C)
(D)
1
(E)
4
3(
x
– 7) ≥ 9
0.25
y
– 3 ≤ 1
5.
Quantity A
x
Quantity B
y
6. If |1 –
x
| = 6 and |2
y
– 6| = 10, which of the following could be the value of
xy
?
Indicate all such values.
–40
–14
–10
56
7. If 2(
x
– 1)
3
+ 3 ≤ 19, which of the following must be true?
(A)
x
≥ 3
(B)
x
≤ 3
(C)
x
≥ –3
(D)
x
≤ –3
(E)
x
< –3 or
x
> 3
8. If 3
p
< 51 and 5
p
> 75, what is the value of the integer
p
?
(A)
15
(B)
16
(C)
24
(D)
25
(E)
26
9. A bicycle wheel has spokes that go from a center point in the hub to equally
spaced points on the rim of the wheel. If there are fewer than six spokes,
what is the smallest possible angle between any two spokes?
(A)
18°
(B)
30°
(C)
40°
(D)
60°
(E)
72°
|–
x
| ≥ 6
xy
2
< 0 and
y
is an integer.
10.
Quantity A
x
Quantity B
–4
11. If
> 5 and
x
< 0, which of the following could be the value of
x
?
Indicate all such values.
–6
–14
–18
|
x
3
| < 64
12.
Quantity A
–
x
Quantity B
–|
x
|
13. If |3
x
+ 7| ≥ 2
x
+ 12, then which of the following is true?
(A)
x
≤
(B)
x
≥
(C)
x
≥ 5
(D)
x
≤
or
x
≥ 5
(E)
≤
x
≤ 5
|3 + 3
x
| < –2
x
14.
Quantity A
|
x
|
Quantity B
4
15. If |
y
| ≤ –4
x
and |3
x
– 4| = 2
x
+ 6, what is the value of
x
?
(A)
–3
(B)
(C)
(D)
(E)
10
x
is an integer such that –
x
|
x
| ≥ 4.
16.
Quantity A
x
Quantity B
2
|x| < 1 and y > 0
17.
Quantity A
|
x
| +
y
Quantity B
xy
|
x
| > |
y
| and x + y > 0
18.
Quantity A
y
Quantity B
x
x
and
y
are integers such that |
x
|(
y
) + 9 < 0 and |
y
| ≤ 1.
19.
Quantity A
x
Quantity B
–9
p
+ |
k
| > |
p
| +
k
20.
Quantity A
p
Quantity B
k
|
x
| + |
y
| > |
x
+
z
|
21.
Quantity A
y
Quantity B
z
b
≠ 0
> 1
a
+
b
< 0
22.
Quantity A
a
Quantity B
0
23. If
f
2
g
< 0, which of the following must be true?
(A)
f
< 0
(B)
g
< 0
(C)
fg
< 0
(D)
fg
> 0
(E)
f
2
< 0
24.
and
. If
x
is an integer, which of the following
is the value of
x
?
(A)
2
(B)
3
(C)
4
(D)
5
(E)
6
|
x
|
y
>
x
|
y
|
25.
Quantity A
(
x
+
y
)
2
Quantity B
(
x
–
y
)
2
26. Which of the following could be the graph of all values of
x
that satisfy
the inequality 4 – 11
x
≥
?
(A)
(B)
(C)
(D)
(E)
–1 <
a
< 0 < |
a
| <
b
< 1
27.
Quantity A
Quantity B
x
> |
y
| >
z
28.
Quantity A
x
+
y
Quantity B
|
y
| +
z
29. The integers
k, l
, and
m
are consecutive even integers between 23 and 33.
Which of the following could be the average (arithmetic mean) of
k, l
, and
m
?
(A)
24
(B)
25
(C)
25.5
(D)
28
(E)
32
30.
The number line above represents which of the following inequalities?
(A)
x
< 1
(B)
–6 < 2
x
< 2
(C)
–9 < 3
x
< 6
(D)
1 < 2
x
< 3
(E)
x
> –3
31. For a jambalaya cook-off, there will be
x
judges sitting in a single row of
x
chairs. If
x
is greater than 3 but no more than 6, which of the following
could be the number of possible seating arrangements for the judges?
Indicate two such numbers.
6
25
120
500
720
32. Which of the following inequalities is equivalent to
for all
non-zero values of
a, b
, and
c
?
Indicate all such inequalities.
a
> –3
bc
|
x
+
y
| = 10
x
≥ 0
z
<
y
–
x
Quantity A
Quantity B
33.
z
10
0 <
a
< < 9
34.
Quantity A
9 –
a
Quantity B
–
a
For all values of the integer
p
such that 1.9 < |
p
| < 5.3,
the function
f
(
p
) =
p
2
.
35.
Quantity A
f
(
p
) for the greatest value of
p
Quantity B
f
(
p
) for the least value of
p
36. If
and
are reciprocals and
< 0, which of the following
must be true?
(A)
ab
< 0
(B)
< –1
(C)
< 1
(D)
(E)
37. If
mn
< 0 and
, which of the following must be true?
(A)
km
+
ln
< (
mn
)
2
(B)
kn
+
lm
< 1
(C)
kn
+
lm
> (
mn
)
2
(D)
k
+
l
>
mn
(E)
km
> –
ln
38. If the reciprocal of the negative integer
x
is greater than the sum of
y
and
z
, then which of the following must be true?
(A)
x
>
y
+
z
(B)
y
and
z
are positive.
(C)
1 >
x
(
y
+
z
)
(D)
1 <
xy
+
xz
(E)
>
z
–
y
39. If
u
and –3
v
are greater than 0, and
, which of the following
cannot be true?
(A)
< –
v
(B)
> –3
(C)
(D)
u
+ 3
v
> 0
(E)
u
< –3
v
40. In the figure above, an equilateral triangle is inscribed in a circle. If the
arc bounded by adjacent corners of the triangle is between 4
π
and 6
π
long,
which of the following could be the diameter of the circle?
(A)
6.5
(B)
9
(C)
11.9
(D)
15
(E)
23.5
Inequalities and Absolute Values Answers
1.
(D).
Solve the inequality algebraically:
7
y
– 3 ≤ 4
y
+ 9
3
y
– 3 ≤ 9
3
y
≤ 12
y
≤ 4
Because
y
could be less than or equal to 4, the relationship cannot be
determined from the information given.
2.
(D).
Solve the inequality by first dividing both sides by 3 to isolate the
absolute value. Then solve for the positive and negative possibilities of (
x
–
4), using the identity that |
a
| =
a
when
a
is positive or zero and |
a
| = –
a
when
a
is negative:
Thus,
x
could be
or
, making the two quantities equal or Quantity B
greater, respectively. The relationship cannot be determined from the
information given.
3.
(C).
When manipulating a “three-sided” inequality, perform the same
operations on all “sides.” Therefore, the first step to simplify this inequality
would be to add 4 to all three sides to get: 10 < 2
x
< 16. Next, divide all three
sides by 2. The result is 5 <
x
< 8. The only answer choice that fits within the
parameters of this inequality is 7.
4.
(A).
If
y
is negative, then dividing both sides of the second inequality by
y
yields
< 1. Remember to switch the direction of the inequality sign
when multiplying or dividing by a negative (whether that negative is in
number or variable form). Next, dividing both sides by 4 changes the
inequality to
<
. The only answer choice less than
is 0.
5.
(D).
Solve each inequality algebraically:
3(
x
– 7) ≥ 9
x
– 7 ≥ 3
x
≥ 10
0.25
y
– 3 ≤ 1
0.25
y
≤ 4
y
≤ 16
Since the ranges for
x
and
y
overlap, either quantity could be greater. For
instance,
x
could be 11 and
y
could be 15 (
y
is greater), or
x
could be 1,000
and
y
could be –5 (
x
is greater). The relationship cannot be determined from
the information given.
6. –
40,
–
14, and 56 only.
Solve each absolute value, using the identity that |
a
|
=
a
when
a
is positive or zero and |
a
| = –
a
when
a
is negative:
|1 –
x
| = 6
+ (1 –
x
) = 6
or
– (1 –
x
) = 6
(1 –
x
) = 6
(1 –
x
) = –6
–
x
= 5
–
x
= –7
x
= –5
x
= 7
x
= –5 or 7
|2
y
– 6| = 10
+ (2
y
– 6) = 10
or
– (2
y
– 6) = 10
(2
y
– 6) = 10
(2
y
– 6) = –10
2
y
= 16
2
y
= –4
y
= 8
y
= –2
y
= 8 or –2
Since
x
= –5 or 7 and
y
= 8 or –2, calculate all four possible combinations for
xy
:
(–5)(8)
= –40
(–5)(–2) = 10
(7)(8)
= 56
(7)(–2)
= –14
Select –40, –14, and 56. (Do
not
pick –10, as
xy
could be 10, but not –10.)
7.
(B).
2(
x
– 1)
3
+ 3 ≤ 19
2(
x
– 1)
3
≤ 16
(
x
– 1)
3
≤ 8
Taking the cube root of an inequality is permissible here, because cubing a
number, unlike squaring it, does not change its sign.
x
– 1 ≤ 2
x
≤ 3
8.
(B).
Dividing the first inequality by 3 results in
p
< 17. Dividing the second
inequality by 5 results in
p
> 15. Therefore, 15 <
p
< 17. Because
p
is an
integer, it must be 16.
9.
(E).
In this scenario, if there are
n
spokes, there are
n
angles between them.
Thus, the measure of the angle between spokes is
. Since
n
< 6, rewrite
this expression as
. Dividing by a “less than” produces a
“greater than” result. Therefore,
= greater than 60°. The only
answer that is greater than 60° is (E). To verify, note that
n
can be at most 5,
because
n
must be an integer. Because there are 360° in a circle, a wheel with
5 spokes would have
= 72° between adjacent spokes.
10.
(B).
First, solve the inequality for
x
, remembering the two cases that must
be considered when dealing with absolute value: |
a
| =
a
when
a
is positive or
zero and |
a
| = –
a
when
a
is negative:
|–
x
| ≥ 6
+(–
x
) ≥ 6
or
–(–
x
) ≥ 6
–
x
≥ 6
x
≥ 6
x
≤ –6
x
≤ –6 or
x
≥ 6
Because
xy
2
< 0, neither
x
nor
y
equals zero. A squared term cannot be
negative, so
y
2
must be positive. For
xy
2
to be negative,
x
must be negative.
This rules out the
x
≥ 6 range of solutions for
x
. Thus,
x
≤ –6 is the only range
of valid solutions. Since all values less than or equal to –6 are less than –4,
Quantity B is greater.
11. –
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