Applications of Linear Inequalities



Samuel Dominic Chukwuemeka (SamDom For Peace) Pre-requisites: Expressions and Equations
For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Show all work.
Interpret your solutions.

(1.) ACT At her hot dog stand, Julie sells hot dogs for $2 each.
Purchasing hot dogs and other supplies costs $200 per month.
The solution of which of the following inequalities models the numbers of hot dogs, h, Julie can sell per month and make a profit?

$ A.\:\: h - 200 \gt 0 \\[3ex] B.\:\: h - 200 \lt 0 \\[3ex] C.\:\: h + 200 \gt 0 \\[3ex] D.\:\: 2h - 200 \lt 0 \\[3ex] E.\:\: 2h - 200 \gt 0 \\[3ex] $

The number of hot dogs = $h$

$ profit = selling\:\: price - cost\:\: price \\[3ex] cost\:\: price = 200 \\[3ex] selling\:\: price = 2 * h = 2h \\[3ex] profit = 2h - 200 \\[3ex] $ To make a profit, the profit must be greater than zero

$\rightarrow 2h - 200 \gt 0$
(2.) Based on Question (1.)
Julie wants to make at least a thousand dollars profit by the end of the month.
How many hot dogs must she sell to make that profit?


Based on the Answer to Question $1$

$ profit = 2h - 200 \\[3ex] profit \ge 1000 \\[3ex] \rightarrow 2h - 200 \ge 1000 \\[3ex] 2h \ge 1000 + 200 \\[3ex] 2h \ge 1200 \\[3ex] h \ge \dfrac{1200}{2} \\[5ex] h \ge 600 \\[3ex] $ Julie must sell at least $600$ hot dogs by the end of the month to make at least a thousand dollars profit for that month.
(3.) Naomi scored 72 out of 100 on a Statistics test.
The maximum score on the next test is also 100 points.
What score does she need to score in the next test to have at least a B grade average?
In the United States academic scale, the range 80 – 89 denotes a B.


At least a $B$ means $\ge 80$
Let her score on the next test be $p$

$ 1st\:\:test = 72 \\[3ex] 2nd\:\:test = p \\[3ex] Average = \dfrac{72 + p}{2} \\[5ex] \rightarrow \dfrac{72 + p}{2} \ge 80 \\[5ex] LCD = 2 \\[3ex] 2\left(\dfrac{72 + p}{2}\right) \ge 2(80) \\[5ex] 72 + p \ge 160 \\[3ex] p \ge 160 - 72 \\[3ex] p \ge 88 \\[3ex] $ Naomi needs to make at least an $88$ on the next test to earn a $B$ average
(4.) Based on Question (3.)
Naomi's mother wants her to make an A average.
Can Naomi make an A average based on the two tests? Justify your answer.
In the United States academic scale, the range 90 – 100 denotes an A.


At least an $A$ means $\ge 90$
Let her score on the next test be $p$

$ 1st\:\:test = 72 \\[3ex] 2nd\:\:test = p \\[3ex] Average = \dfrac{72 + p}{2} \\[5ex] \rightarrow \dfrac{72 + p}{2} \ge 90 \\[5ex] Cross\:\:Multiply \\[3ex] 72 + p \ge 2(90) \\[3ex] 72 + p \ge 180 \\[3ex] p \ge 180 - 72 \\[3ex] p \ge 108 \\[3ex] $ Naomi cannot make an $A$ average based on the two tests.
(5.) The two sides of a rectangle are x and x + 4 meters.
Determine the value of x for which the perimeter of the rectangle is less than 116 meters.


$ Let\:\:the\:\:width\:\:of\:\:the\:\:rectangle = x \\[3ex] Let\:\:the\:\:length\:\:of\:\:the\:\:rectangle = x + 4 \\[3ex] Perimeter = 2x + 2(x + 4) \\[3ex] \rightarrow 2x + 2(x + 4) \lt 116 \\[3ex] 2x + 2x + 8 \lt 116 \\[3ex] 4x \lt 116 - 8 \\[3ex] 4x \lt 108 \\[3ex] x \lt \dfrac{108}{4} \\[5ex] x \lt 27 \\[3ex] $ $x$ must be less than $27$ meters for the perimeter of the rectangle to be less than $116$ meters
(6.) A rectangle has a fixed width of 7 meters.
Determine the length for which the area of the rectangle is more than 147 square meters.


$ Let\:\:the\:\:width = W \\[3ex] W = 7 \\[3ex] Let\:\:the\:\:length = L \\[3ex] Let\:\:the\:\:area = A \\[3ex] A = L * W \\[3ex] A \gt 147 \\[3ex] \rightarrow L * 7 \gt 147 \\[3ex] L \gt \dfrac{147}{7} \\[5ex] L \gt 21 \\[3ex] $ The length of the rectangle must be greater than $21$ meters for the area to be more than $147$ square meters.
(7.) ACT Koji had $50.00 to spend on school supplies.
He bought a schoolbag for $18.50, box of pencils for $2.50, and a package of pens for $4.00
Koji still must buy b binders at $3.50 each and n spiral notebooks at $0.75 each.
Which of the following inequalities represents how many binders and spiral notebooks he can buy with the remaining money?
(Note: No tax is charged on school supplies.)

$ F.\:\: 3.5b + 0.75n \le 75 \\[3ex] G.\:\: 3.5b + 0.75n \lt 50 \\[3ex] H.\:\: 3.5b + 0.75n \le 50 \\[3ex] J.\:\: 3.5b + 0.75n \lt 25 \\[3ex] K.\:\: 3.5b + 0.75n \le 25 \\[3ex] $

$ Total = 50 \\[3ex] \underline{Bought} \\[3ex] Schoolbag = 18.50 \\[3ex] Pencils = 2.50 \\[3ex] Pens = 4.00 \\[3ex] Sum = 18.50 + 2.50 + 4.00 = 25 \\[3ex] \underline{Balance} \\[3ex] 50 - 25 = 25 \\[3ex] b\:\:binders\:@\:3.50\:\:each = 3.50(b) = 3.5b \\[3ex] n\:\:spiral\:\:notebooks\:@\:0.75\:\:each = 0.75n \\[3ex] \rightarrow 3.5b + 0.75n \le 25 \\[3ex] $ The cost of the binders and spiral notebooks should be at most $\$25.00$
(8.) JAMB In a racing competition, Musa covered a distance of $5x\:km$ in the first hour and $(x +10)\:km$ in the next hour.
He was second to Ngozi who covered a total distance of $118\:km$ in the two hours.
Which of the following inequalities is correct?

$ A.\:\: 0 \lt -x \lt 15 \\[3ex] B.\:\: -3 \lt x \lt 3 \\[3ex] C.\:\: 15 \lt x \lt 18 \\[3ex] D.\:\: 0 \lt x \lt 15 \\[3ex] E.\:\: 0 \lt x \lt 18 \\[3ex] $

Musa was second only to Ngozi
This implies that his total distance for the two hours was less than Ngozi's distance of $118\:km$

$ 5x + (x + 10) \lt 118 \\[3ex] 5x + x + 10 \lt 118 \\[3ex] 6x \lt 118 - 10 \\[3ex] 6x \lt 108 \\[3ex] x \lt \dfrac{108}{6} \\[5ex] x \lt 18\:km \\[3ex] \therefore 0 \lt x \lt 18 $
(9.) Assume it costs $2 for the first hour and $1.75 for each hour thereafter to park in the public lot in the village of Sandwich, Massachusetts
The cost of a partial hour is charged the same rate as a full hour.
Determine the longest time that James can park in the public lot for $9.


$ Let\:\:the\:\:number\:\:hours = p \\[3ex] longest\:\:time\:\:\implies at\:\:most \\[3ex] 2 + 1.75p \le 9 \\[3ex] 1.75p \le 9 - 2 \\[3ex] 1.75p \le 7 \\[3ex] p \le \dfrac{7}{1.75} \\[5ex] p \le 4 \\[3ex] $ James can park in the public lot for at most $4$ hours.
(10.) Assume it costs $2 for the first half hour and $1.75 for each hour thereafter to park in the public lot in the village of Sandwich, Massachusetts
The cost of a partial hour is charged the same rate as a full hour.
Determine the longest time that James can park in the public lot for $$9.


$ Let\:\:the\:\:number\:\:hours = p \\[3ex] longest\:\:time\:\:\implies at\:\:most \\[3ex] 2 + 1.75p \le 9 \\[3ex] 1.75p \le 9 - 2 \\[3ex] 1.75p \le 7 \\[3ex] p \le \dfrac{7}{1.75} \\[5ex] p \le 4 \\[3ex] $ This means that the $\$9$ can be used to park for $4$ hours.
But, the $\$9$ can also be used to park for $4$ and a half hours.
This is because it cost $\$2$ to park for the first half hour.
It also cost $\$2$ to park for the first hour because a partial hour is charged the same rate as a full hour.
So, we have a balance of a half-hour to use that $\$9$

$ half-hour = 0.5\:hour \\[3ex] 4 + 0.5 = 4.5\:hours \\[3ex] $ James can park in the public lot for at most $4.5$ hours.
(11.) CSEC Pam visits the stationery store where she intends to buy x pens and y pencils.
(a) Pam must buy at least 3 pens.
(i) Write an inequality to represent this information.

The TOTAL number of pens and pencils must NOT be more than 10
(ii) Write an inequality to represent this information.

EACH pen costs $5.00 and EACH pencil costs $2.00.
More information about the pens and pencils is represented by: 5x + 2y ≤ 35
(iii) Write the information represented by this inequality as a sentence in your own words.

(b) (i) On the answer sheet provided, draw the graph of the TWO inequalities obtained in (a)(i) and (a)(ii) above.

(ii) Write the coordinates of the vertices of the region that satisfies the four inequalities (including y ≥ 0).

(c) Pam sells the x pens and y pencils and makes a profit of $1.50 on EACH pen and $1.00 on EACH pencil.
(i) Write an expression in x and y to represent the profit Pam makes.

(ii) Calculate the maximum profit Pam makes.

(iii) If Pam buys 4 pens, show on your graph the maximum number of pencils she can buy.


$ (a)\:\:(i) \\[3ex] x \ge 3 \\[3ex] (ii) \\[3ex] x + y \le 10 \\[3ex] (iii) \\[3ex] $ The cost of the total number of pens and pencils is at most thirty five dollars.
(12.) ACT The marketing students at Fort Link Business Academy took a placement test for an accounting class.
The mean of the test scores is 100 and the standard deviation is 15
In order to be accepted into the accounting class, a student must have attained a test score that is at least 1 standard deviation above the mean.
What is the lowest score a student could attain and still be accepted into the accounting class?

$ A.\:\: 15 \\[3ex] B.\:\: 16 \\[3ex] C.\:\: 101 \\[3ex] D.\:\: 115 \\[3ex] E.\:\: 116 \\[3ex] $ But guess what?
Can we paraphrase this question?
What if this question was worded like this?
To enroll in an accounting class, a student must make a test score of at least $15$ more than the class average of $100$
What is the lowest score a student could attain and still be accepted into the accounting class?

Do you see what I mean about lengthy word problems?
It is important you note the advice I gave about word problems



Mean = $100$

Standard deviation = $15$

$1$ standard deviation = $15$

$1$ standard deviation above the mean = $15 + 100 = 115$

at least $1$ standard deviation above the mean is $\ge 115$

This means $115, 116, 117, 118, ...$

The lowest score a student could attain and still be accepted into the accounting class is $115$
(13.) ACT Sani's course grade in his chemistry class is based on 3 tests and 1 final exam.
Each of the 3 test scores is weighted as 20% of the course grade, and the final exam score is weighted as 40% of the course grade.
Sani's 3 test scores are 78, 86, and 82, respectively.
What is the minimum score that Sani will have to earn on the final exam in order to receive a course grade of at least 86?

$ F.\;\; 82 \\[3ex] G.\;\; 84 \\[3ex] H.\;\; 90 \\[3ex] J.\;\; 92 \\[3ex] K.\;\; 98 \\[3ex] $

Let that minimum score be: $p$
Let us represent this information in a table
You may do it any way you prefer.

Assessment Weight (%) Score Weighted Score
Test 1 20 78 1560
Test 2 20 86 1720
Test 3 20 82 1640
Final Exam 40 $p$ $40p$
$\Sigma Weights = 100$ $\Sigma Weighted\;\;Scores = 4920 + 40p$

$ \dfrac{\Sigma Weighted\;\;Scores}{\Sigma Weights} = Course\;\;Grade \\[5ex] At\;\;least\;\;86 \;\;means\;\; \ge 86 \\[3ex] \implies \\[3ex] \dfrac{4920 + 40p}{100} \ge 86 \\[5ex] 4920 + 40p \ge 86(100) \\[3ex] 4920 + 40p \ge 8600 \\[3ex] 40p \ge 8600 - 4920 \\[3ex] 40p \ge 3680 \\[3ex] p \ge \dfrac{3680}{40} \\[5ex] p \ge 92 \\[3ex] $ Sani will have to earn 92% on the final exam in order to receive a course grade of at least 86.
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