Inequalities

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For the Classic ACT exam:
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Also: please note that unless specified otherwise, geometric figures are drawn to scale. So, you can figure out the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.

These are the solutions to the ACT past questions on the topic: Inequalities.
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(1.) Which of the following number line graphs is that of the solution set to the inequality $-2x + 7 \ge 19$?

Number 1

$ -2x + 7 \ge 19 \\[3ex] -2x \ge 19 - 7 \\[3ex] -2x \ge 12 \\[3ex] x \le \dfrac{12}{-2} \\[5ex] x \le -6 \\[3ex] $ Check

$x \le -6$
LHS	RHS
$ Let\;\; x = -7 \\[3ex] -2x + 7 \\[3ex] -2(-7) + 7 \\[3ex] 14 + 7 \\[3ex] 21 $	19
21 ≥ 19

(2.) The inequality $7x^2y \lt 0$ is true for 2 fixed real numbers x and y.
Which of the following inequalities must be true?

$ F.\;\; x \gt 0 \\[3ex] G.\;\; y \gt 0 \\[3ex] H.\;\; x \lt 0 \\[3ex] J.\;\; y \lt 0 \\[3ex] K.\;\; xy \lt 0 \\[3ex] $

$7x^2y \lt 0$
Less than 0 implies a negative number

For $7x^2y$:
7 is a positive number
$x^2$ is a positive number because the square of any number is positive
The square of a positive number is positive
The square of a negative number is also positive
So, $7x^2$ is positive
⇒
$y$ must be negative because the product of positive and negative is negative
Option J. is the correct answer.

(3.) Fred’s cell phone service costs $25.00 per month and includes 100 minutes.
For any minutes used after the 100 minutes, he is charged $0.10 per minute.
Fred has budgeted $33.00 per month for his cell phone service.
What is the maximum number of minutes Fred could use in 1 month without exceeding his budget?

$ F.\;\; 108 \\[3ex] G.\;\; 132 \\[3ex] H.\;\; 180 \\[3ex] J.\;\; 330 \\[3ex] K.\;\; 430 \\[3ex] $

Let the number of minutes Fred could use in 1 month without exceeding his budget be x

$ \underline{Monthly} \\[3ex] \text{cell phone service cost} = \$25 \\[3ex] \text{over 100 minutes}, cost = \$0.1 \;\;per\;\;minute \implies 0.1(x - 100) \\[3ex] \text{budget cost} = \$33 \\[3ex] \implies \\[3ex] 25 + 0.1(x - 100) \le 33 \\[3ex] 25 + 0.1x - 10 \le 33 \\[3ex] 0.1x + 15 \le 33 \\[3ex] 0.1x \le 33 - 15 \\[3ex] 0.1x \le 18 \\[3ex] x \le \dfrac{18}{0.1} \\[5ex] x \le 180\;minutes \\[3ex] $ Fred can use up to 180 minutes in 1 month without exceeding his budget.

(4.) Marcy is making toys to sell at the local school fair.
Each toy costs Marcy $2.25 to make, and she will sell them for $4.05 each.
What is the minimum number of toys she can make and sell to earn a profit of at least $81.00?

$ F.\;\; 13 \\[3ex] G.\;\; 16 \\[3ex] H.\;\; 20 \\[3ex] J.\;\; 36 \\[3ex] K.\;\; 45 \\[3ex] $

Let the number of toys = n
Cost Price for n toys = $2.25n
Selling Price for n toys = $4.05n
Profit = Selling Price − Cost Price
At least $81.00 implies ≥$81.00
This implies that:

$ 4.05n - 2.25n \ge 81 \\[3ex] 1.8n \ge 81 \\[3ex] n \ge \dfrac{81}{1.8} \\[5ex] n \ge 45 \\[3ex] $ Marcy needs to make and sell at least 45 toys to earn a profit of at least $81.00

(5.) A quilt maker uses $25 worth of materials to produce a single quilt that sells for $100.
What is the minimum number of this type of quilt the quilt maker can produce and sell to earn a profit of at least $900?

$ F.\;\; 8 \\[3ex] G.\;\; 9 \\[3ex] H.\;\; 12 \\[3ex] J.\;\; 27 \\[3ex] K.\;\; 36 \\[3ex] $

For 1 quilt:
Cost Price = $25
Selling Price = $100
Profit = Selling Price − Cost Price
Profit = $100 − $25
Profit = $75

Units are in dollars: $
For 2 quilts:
Cost Price = 2(25)
Selling Price = 2(100)
Profit = 2(100) − 2(25)

For 3 quilts:
Cost Price = 3(25)
Selling Price = 3(100)
Profit = 3(100) − 3(25)

For x quilts:
Cost Price = x(25) = 25x
Selling Price = x(100) = 100x
Profit = 100x − 25x = 75x

At least $900 means ≥ 900
This implies that:

$ 75x \ge 900 \\[3ex] x \ge \dfrac{900}{75} \\[5ex] x \ge 12 \\[3ex] $ The minimum number of this type of quilt the quilt maker can produce and sell to earn a profit of at least $900 is 12 quilts.

(6.) Given a − b > a + b for real numbers a and b, which of the inequalities below must be true?

I. a > 0
II. b < 0
III. b < a

F. I only
G. II only
H. I and III only
J. II and III only
K. I, II, and III

$ For\;\; a - b \gt a + b \\[3ex] a - a \gt b + b \\[3ex] 0 \gt 2b \\[3ex] 2b \lt 0 \\[3ex] b \lt \dfrac{0}{2} \\[5ex] b \lt 0 \\[3ex] $ This is a necessary condition.
The correct answer is Option G.

(8.) Brian had $12.80 to spend at the flea market.
He bought a portable CD player for $5.50 and wants to buy some CDs that are $0.75 each.
Brian can determine n, the number of CDs he can buy, using which of the following inequalities?

$ F.\;\; 0.75n \le 12.80 \\[3ex] G.\;\; n - 5.50 \le 12.80 \\[3ex] H.\;\; n + 5.50 \le 12.80 \\[3ex] J.\;\; 0.75n - 5.50 \le 12.80 \\[3ex] K.\;\; 0.75n + 5.50 \le 12.80 \\[3ex] $

$ \text{Total amount to spend} = \$12.80 \\[3ex] \text{Spent: } \$5.50 \\[3ex] \text{To spend: } 0.75n \\[3ex] \implies \\[3ex] 0.75n + 5.50 \le 12.80 $

(9.) Kate is going to buy meat (beef and chicken) for a party.
Beef costs $3.00 per pound, chicken costs $2.00 per pound, and Kate has a budget of $90.00 to spend on meat for the party.
The shaded region in one of the following graphs represents all and only the possible combinations of beef and chicken Kate can buy while staying within her budget.
Which one?
(Note: Do not consider tax.)

Number 9

Let:
the amount of chicken = C
the amount of beef = B

Amount of beef @ $3.00 per pound = B × 3 = 3B
Amount of chicken @ $2.00 per pound = C × 2 = 2C

Kate has a budget of $90.00 to spend on meat for the party.
This implies that: 3B + 2C ≤ 90

$3B + 2C \le 90$
Beef, B	Chicken, C	Point (B, C)
$0$	$ 3B + 2C = 90 \\[3ex] 3(0) + 2C = 90 \\[3ex] 2C = 90 \\[3ex] C = \dfrac{90}{2} \\[5ex] C = 45 $	$(0, 45)$
$ 3B + 2C = 90 \\[3ex] 3B + 2(0) = 90 \\[3ex] 3B = 90 \\[3ex] B = \dfrac{90}{3} \\[5ex] B = 30 $	$0$	$(30, 0)$

Indicate the two points on the graph
Join the points with a solid line because of the equality sign in ≤
Shade the region below the solid line because of the less than sign in ≤
This implies that the correct answer is Option G

(10.) Given that x ≤ 2 and x + y ≥ 6, what is the LEAST value that y can have?

$ A.\;\; -8 \\[3ex] B.\;\; -4 \\[3ex] C.\;\; 0 \\[3ex] D.\;\; 4 \\[3ex] E.\;\; 8 \\[3ex] $

$ x + y \ge 6 \\[3ex] y \ge 6 - x \\[3ex] $ The least value that y can have is the maximum (most) value that x can have.

$ x \le 2 \\[3ex] \text{most value of } x = 2 \\[3ex] \implies \\[3ex] y \ge 6 - 2 \\[3ex] y \ge 4 \\[3ex] \text{least value of } y = 4 $

(14.) Given that c ≠ d, what are all the real values of b that make the inequality $\dfrac{bc - bd}{6c - 6d} \lt 0$ true?

$ A.\;\; 6\;\;only \\[3ex] B.\;\; \dfrac{1}{6}\;\;only \\[5ex] C.\;\; -\dfrac{1}{6}\;\;only \\[5ex] D.\;\; \text{All positive real numbers} \\[3ex] E.\;\; \text{All negative real numbers} \\[3ex] $

$ \dfrac{bc - bd}{6c - 6d} \lt 0 \\[5ex] \dfrac{b(c - d)}{6(c - d)} \lt 0 \\[5ex] \dfrac{b}{6} \lt 0 \\[5ex] b \lt 0(6) \\[3ex] b \lt 0 \\[3ex] $ All negative real numbers are less than 0

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