**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

$\rightarrow 2h - 200 \gt 0$

Purchasing hot dogs and other supplies costs $200 per month.

The solution of which of the following inequalities models the numbers of hot dogs,

$ 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.

Julie wants to make

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

(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

The maximum score on the next test is also 100 points.

What score does she need to score in the next test to have

In the United States academic scale, the range 80 – 89 denotes a

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

(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.

Naomi's mother wants her to make an

Can Naomi make an

In the United States academic scale, the range 90 – 100 denotes an

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

Determine the value of

$ 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

(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.

Determine the length for which the area of the rectangle is

$ 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

(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__ $\$25.00$

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

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

(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 $

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.

The cost of a partial hour is charged the same rate as a full hour.

Determine the

$ 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

(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.

The cost of a partial hour is charged the same rate as a full hour.

Determine the

$ 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

(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: 5*x* + 2*y* ≤ 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.

(a) Pam must buy

(i) Write an inequality to represent this information.

The TOTAL number of pens and pencils must

(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: 5

(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

(c) Pam sells the

(i) Write an expression in

(ii) Calculate the maximum profit Pam makes.

(iii) If Pam buys 4 pens, show

$ (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

(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 $115$

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

What is the

$ A.\:\: 15 \\[3ex] B.\:\: 16 \\[3ex] C.\:\: 101 \\[3ex] D.\:\: 115 \\[3ex] E.\:\: 116 \\[3ex] $

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

What is the

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.

$ \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.

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.

(14.)

(15.)

(16.)

(17.)

(18.)