If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

Solved Examples: Word Problems on Numbers

Samuel Dominic Chukwuemeka (SamDom For Peace) 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.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Attempt all questions.
Use at least two (two or more) methods whenever applicable.
Show all work.

(1.) ACT Diego purchased a car that had a purchase price of $\$13,400$, which included all other costs and tax.
He paid $\$400$ as a down payment and got a loan for the rest of the purchase price.
Diego paid off the loan by making $48$ payments of $\$300$ each.
The total of all his payments, including the down payment, was how much more than the car's purchase price?

$ F.\:\: \$1,000 \\[3ex] G.\:\: \$1,400 \\[3ex] H.\:\: \$13,000 \\[3ex] J.\:\: \$14,400 \\[3ex] K.\:\: \$14,800 \\[3ex] $

Purchase price of car = $\$13,400$

Down payment = $\$400$

$48$ payments @ $\$300$ per payment = $48 * 300 = 14,400$

Total of all payments he made = $\$400 + \$14,400 = \$14,800$

This is the question:

$\$14,800$ is how much more than $\$13,400$

$14,800 - 13,400 = 1,400$

The total payments made by Diego is $\$1,400$ more than the car's purchase price.
(2.) ACT Ming purchased a car that had a purchase price of $\$5,400$, which included all other costs and tax.
She paid $\$1,000$ as a down payment and got a loan for the rest of the purchase price.
Ming paid off the loan by making $28$ payments of $\$200$ each.
The total of all her payments, including the down payment, was how much more than the car's purchase price?


Purchase price of car = $\$5,400$

Down payment = $\$1000$

$28$ payments @ $\$200$ per payment = $28 * 200 = 5,600$

Total of all payments she made = $\$1000 + \$5,600 = \$6,600$

This is the question:

$\$6,600$ is how much more than $\$5,400$

$6,600 - 5,400 = 1,200$

The total payments made by Ming is $\$1,200$ more than the car's purchase price.
(3.) CSEC In St. Vincent, $3$ litres of gasoline cost $EC\$10.40$

(i) Calculate the cost of $5$ litres of gasoline in St. Vincent, stating your answer correct to the nearest cent.

(ii) How many litres of gasoline can be bought for $EC\$50.00$ in St. Vincent?
Give your answer correct to the nearest whole number.


$ 10.40 = 10.4 \\[3ex] 50.00 = 50 \\[3ex] $ (i)
Let the cost of $5$ litres of gasoline be $p$

Proportional Reasoning Method
$litres$ $cost(EC\$)$
$3$ $10.4$
$5$ $p$

$ \dfrac{p}{5} = \dfrac{10.4}{3} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] 5 * \dfrac{p}{5} = 5 * \dfrac{10.4}{3} \\[5ex] p = \dfrac{5 * 10.4}{3} \\[5ex] p = \dfrac{52}{3} \\[5ex] p = 17.3333333 \\[3ex] p \approx EC\$17.33 \\[3ex] $ The cost of $5$ litres of gasoline is $EC\$17.33$

(ii)
Let the volume of gasoline that can be bought for $EC\$50.00$ be $k$

Proportional Reasoning Method
$litres$ $cost(EC\$)$
$3$ $10.4$
$k$ $50$

$ \dfrac{k}{3} = \dfrac{50}{10.4} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:3 \\[3ex] 3 * \dfrac{k}{3} = 3 * \dfrac{50}{10.4} \\[5ex] k = \dfrac{3 * 50}{10.4} \\[5ex] k = \dfrac{150}{10.4} \\[5ex] k = 14.4230769 \\[3ex] k \approx 14\:\:litres \\[3ex] $ About $14$ litres of gasoline can be bought with $EC\$50.00$
(4.) ACT The length of a rectangle is $12$ feet.

The width of the rectangle is $\dfrac{1}{2}$ the length.

What is the perimeter of the rectangle, in feet?

$ F.\:\: 18 \\[3ex] G.\:\: 24 \\[3ex] H.\:\: 30 \\[3ex] J.\:\: 36 \\[3ex] K.\:\: 72 \\[3ex] $

$ Length = 12 \\[3ex] Width = \dfrac{1}{2} * 12 = 1 * 6 = 6 \\[5ex] Perimeter = 2 * Length + 2 * Width \\[3ex] = 2(12) + 2(6) \\[3ex] = 24 + 12 \\[3ex] = 36\:\:feet $
(5.) ACT When Tyrone fell asleep one night, the temperature was $24^\circ F$.
When Tyrone awoke the next morning, the temperature was $-12^\circ F$.
Letting $+$ denote a rise in temperature and $-$ denote a drop in temperature, what was the change in temperature from the time Tyrone fell asleep until the time he awoke?

$ F.\:\: -36^\circ F \\[3ex] G.\:\: -12^\circ F \\[3ex] H.\:\: +6^\circ F \\[3ex] J.\:\: +12^\circ F \\[3ex] K.\:\: +36^\circ F \\[3ex] $

$ Initial\:\:temperature = 24^\circ F \\[3ex] Final\:\:temperature = -12^\circ F \\[3ex] Change\:\:in\:\:temperature = Final\:\:temperature - Initial\:\:temperature \\[3ex] = -12 - 24 \\[3ex] = -36^\circ F $
(6.) ACT Discount tickets to a basketball tournament sell for $\$4.00$ each.
Enrico spent $\$60.00$ on discount tickets, $\$37.50$ less than if he had bought the tickets at the regular price.
What was the regular ticket price?

$ F.\:\: \$2.50 \\[3ex] G.\:\: \$6.40 \\[3ex] H.\:\: \$6.50 \\[3ex] J.\:\: \$7.50 \\[3ex] K.\:\: \$11.00 \\[3ex] $

Notice the wording of the question: regular ticket price.
This means that they want us to find the price of one regular ticket, NOT the price of all the regular tickets

First: Let us find the number of tickets that he bought by dividing the price of all the discount tickets by the price of one discount ticket

Second: We find the price of all the regular tickets
The price of all discount tickets is the price of all regular tickets minus $37.50$
This implies that the price of all the regular tickets is the price of all the discount tickets plus $37.50$

Third: We find the price of a regular ticket by dividing the price of all the regular tickets by the number of tickets

$ \underline{First} \\[3ex] Price\:\:of\:\:all\:\:discount\:\:tickets = 60.00 \\[3ex] Price\:\:of\:\:a\:\:discount\:\:ticket = 4.00 \\[3ex] Number\:\:of\:\:tickets = \dfrac{60}{4} = 15 \\[5ex] \underline{Second} \\[3ex] Price\:\:of\:\:all\:\:regular\:\:tickets = 60 + 37.50 = 97.50 \\[3ex] \underline{Third} \\[3ex] Price\:\:of\:\:a\:\:regular\:\:ticket = \dfrac{97.50}{15} = 6.50 \\[3ex] $ The price of a regular ticket is $\$6.50$
(7.) ACT Xuan sold $9$ used books for $\$9.80$ each.
With the money from these sales, she bought $4$ new books and had $\$37.80$ left over.
What was the average amount Xuan paid for each new book?

$ A.\:\: \$5.60 \\[3ex] B.\:\: \$9.45 \\[3ex] C.\:\: \$10.08 \\[3ex] D.\:\: \$22.05 \\[3ex] $

$ \underline{Sold} \\[3ex] Selling\:\:price = 9\:\:used\:\:books\:\:@\:\:\$9.80\:\:each \\[3ex] = 9 * 9.8 \\[3ex] = \$88.2 \\[3ex] \underline{Bought} \\[3ex] 4\:\:new\:\:books\:\:with\:\:\$37.80\:\:remaining \\[3ex] Cost\:\:of\:\:the\:\:4\:\:new\:\:books = 88.2 - 37.80 \\[3ex] = \$50.4 \\[3ex] Average\:\:cost\:\:of\:\:each\:\:new\:\:book \\[3ex] = \dfrac{50.4}{4} \\[5ex] = \$12.6 \\[3ex] $ The average cost of each new book is $\$12.60$
(8.) ACT Given today is Tuesday, what day of the week was it $200$ days ago?

$ A.\:\: Monday \\[3ex] B.\:\: Tuesday \\[3ex] C.\:\: Wednesday \\[3ex] D.\:\: Friday \\[3ex] E.\:\: Saturday \\[3ex] $

Let us look at this first:
If today is Tuesday,
Yesterday, (a day ago); it was Monday
$2$ days ago, it was Sunday
$3$ days ago, it was Saturday
$4$ days ago, it was Friday
$5$ days ago, it was Thursday
$6$ days ago, it was Wednesday
$7$ days ago (a week ago), it was Tuesday
$7$ days make a week.
$14$ days ago (2 weeks ago), it was Tuesday
So, let us find how many weeks and days there are in $200$ days.

$ 200\:\: days \\[3ex] 200 \div 7 = 28 + remainder \\[3ex] We\:\: have\:\: 28\:\: weeks \\[3ex] 28 * 7 = 196 \\[3ex] 200 - 196 = 4\:\: days \\[3ex] $ There are $28$ weeks and $4$ days in $200$ days
$196$ days ago (28 weeks ago), it was Tuesday
$197$ days ago, it was Monday
$198$ days ago, it was Sunday
$199$ days ago, it was Saturday
$200$ days ago, it was Friday
(9.) CSEC The table below shows a shopping bill prepared for Mrs Rowe.
The prices of some items are missing.

Shopping Bill
Item Unit Cost Price Total Cost Price
$3$ kg sugar $X$ $\$10.80$
$4$ kg rice $Y$ $Z$
$2$ kg flour $\$1.60$ $\$3.20$

(i) Calculate the value of $X$, the cost of $1\:kg$ of sugar.

(ii) If the cost price of $1\:kg$ of rice is $80$ cents MORE than for $1\:kg$ of flour, calculate the values of $Y$ and $Z$.

(iii) A tax of $10\%$ of the total cost price of the three items is added to Mrs Rowe's bill.
What is Mrs Rowe's TOTAL bill including the tax?


(i)
Unit cost implies the price for $1$ unit

Proportional Reasoning Method
Item (kg) Cost ($\$$)
$3$ $10.80$
$1$ $X$

$ \dfrac{X}{1} = \dfrac{10.80}{3} \\[5ex] X = 3.6 \\[3ex] X = \$3.60 \\[3ex] $ The cost of $1$ kg of sugar is $\$3.60$

(ii)
The cost price for $1$ kg of rice is Y, and it is $80$ cents more than for $1$ kg of flour
The unit cost (the cost for $1$ kg) of flour is $\$1.60$
This means that $Y$ is $80$ cents more than $\$1.60$

$ 80\:\:cents = \dfrac{80}{100} = 0.8 \\[5ex] Y = 0.8 + 1.6 \\[3ex] Y = 2.4 \\[3ex] Y = \$2.40 \\[3ex] $ The cost of $1$ kg of rice is $\$2.40$

$ Z = cost\:\:of\:\:4\:\:kg\:\:of\:\:rice \\[3ex] Z = 4 * 2.40 \\[3ex] Z = 9.6 \\[3ex] Z = \$9.60 \\[3ex] $ Therefore, the cost of $4$ kg of rice will be $\$9.60$

$ (iii) \\[3ex] Total\:\:cost\:\:price\:\:of\:\:all\:\:three\:\:items \\[3ex] = 10.80 + Z + 3.20 \\[3ex] = 10.80 + 9.60 + 3.20 \\[3ex] = 23.6 \\[3ex] 10\%\:\:tax\:\:of\:\:23.6 \\[3ex] = \dfrac{10}{100} * 23.6 \\[5ex] = 0.1 * 23.6 \\[3ex] = 2.36 \\[3ex] Total\:\:bill \\[3ex] = 23.6 + 2.36 \\[3ex] = 25.96 \\[3ex] = \$25.96 $
(10.) ACT Mr. Dietz is a teacher whose salary is $\$22,570$ for this school year, which has $185$ days.
In Mr. Dietz's school district, substitute teachers are paid $\$80$ per day.
If Mr. Dietz takes a day off without pay and a substitute teacher is paid to teach Mr. Dietz's classes, how much less does the school district pay in salary by paying a substitute teacher instead of paying Mr. Dietz for that day?

$ A.\:\: \$42 \\[3ex] B.\:\: \$80 \\[3ex] C.\:\: \$97 \\[3ex] D.\:\: \$105 \\[3ex] E.\:\: \$122 \\[3ex] $

$ \underline{Mr.\:\:Dietz} \\[3ex] School\:\:Year\:\:salary = \$22,570 \\[3ex] School\:\:Days\:\: = 185\:\:days \\[3ex] Average\:\:daily\:\:pay = \dfrac{22570}{185} = \$122 \\[5ex] \underline{Substitute\:\:teacher} \\[3ex] Daily\:\:pay = \$80 \\[3ex] For\:\:one\:\:day: \\[3ex] Difference = 122 - 80 = 42 \\[3ex] $ The school district saved $\$42$ by paying the substitute teacher on the day Mr. Dietz was absent.
(11.) ACT A construction company builds $3$ different models of houses (A, B, and C).
They order all the bathtubs, shower stalls, and sinks for the houses from a certain manufacturer.
Each model of house contains different numbers of these bathroom fixtures.
The tables below give the number of each kind of these fixtures required for each model and the cost to the company, in dollars, or each type of fixture.

Fixture Model
A B C
Bathtubs
Shower stalls
Sinks
$1$
$0$
$1$
$1$
$1$
$2$
$2$
$1$
$4$

Fixture Cost
Bathtubs
Shower stalls
Sinks
$\$250$
$\$150$
$\$120$

The company plans to build $3$ A's, $4$ B's, and $6$ C's.
What will be the cost to the company of exactly enough of these bathroom fixtures to put the required number in all of these houses?

$ F.\:\: \$1,940 \\[3ex] G.\:\: \$2,070 \\[3ex] H.\:\: \$8,940 \\[3ex] J.\:\: \$9,180 \\[3ex] K.\:\: \$10,450 \\[3ex] $

Model A requires $1$ bathtub, no shower stall, and $1$ sink

Model B requires $1$ bathtub, $1$ shower stall, and $2$ sinks

Model C requires $2$ bathtubs, $1$ shower stall, and $4$ sinks

A bathtub costs $\$250$

A shower stall costs $\$150$

A sink costs $\$120$

$ Cost\:\:of\:\:Model\:\:A \\[3ex] = 1(250) + 0(150) + 1(120) \\[3ex] = 250 + 0 + 120 \\[3ex] = \$370 \\[3ex] Cost\:\:of\:\:3\:\:A's \\[3ex] = 3(370) \\[3ex] = \$1110 \\[3ex] Cost\:\:of\:\:Model\:\:B \\[3ex] = 1(250) + 1(150) + 2(120) \\[3ex] = 250 + 150 + 240 \\[3ex] = \$640 \\[3ex] Cost\:\:of\:\:4\:\:B's \\[3ex] = 4(640) \\[3ex] = \$2560 \\[3ex] Cost\:\:of\:\:Model\:\:C \\[3ex] = 2(250) + 1(150) + 4(120) \\[3ex] = 500 + 150 + 480 \\[3ex] = \$1130 \\[3ex] Cost\:\:of\:\:6\:\:C's \\[3ex] = 6(1130) \\[3ex] = \$6780 \\[3ex] Total\:\:Cost \\[3ex] = 1110 + 2560 + 6780 \\[3ex] = \$10,450 $
(12.) CSEC The diagram below, not drawn to scale, shows two jars of peanut butter of the same brand.

Application of Numbers


Which of the jars shown above is the BETTER buy?
Show ALL working to support your answer.


We can do this question in at least two ways...Without a Calculator and With a Calculator.
Use any method you prefer.

First Method: Proportional Reasoning Method
This method uses a calculator.

To determine the better buy, we need to calculate the cost of a unit gram (cost of 1 gram) of peanut butter for each jar.
The jar that has the least cost for a unit gram of peanut butter is the better buy.
Let the unit cost of peanut butter in Jar A = $A$
Let the unit cost of peanut butter in Jar B = $B$

Proportional Reasoning Method (Jar A)
$Amount (g)$ $Cost (\$)$
$150$ $2.14$
$1$ $A$

$ \dfrac{A}{1} = \dfrac{2.14}{150} \\[5ex] A = 0.0142666667 \\[3ex] A \approx \$0.01 \\[3ex] $
Proportional Reasoning Method (Jar B)
$Amount (g)$ $Cost (\$)$
$400$ $6.50$
$1$ $B$

$ \dfrac{B}{1} = \dfrac{6.5}{400} \\[5ex] B = 0.01625 \\[3ex] B \approx \$0.02 \\[3ex] $ Jar $A$ costs about a cent for a unit gram of peanut butter
Jar $B$ costs about two cents for a unit gram of peanut butter
Jar $A$ is the better buy because it costs less for a unit gram of peanut butter.

Second Method: Quantitative Reasoning Method
This method does not need a calculator.

$ \underline{Jar\:A} \\[3ex] 150\:g\:\:for\:\:\$2.14 \\[3ex] 2(150)\:\:g\:\:for\:\:2(2.14) \rightarrow 300\:g\:\:for\:\:\$4.28 \\[3ex] 3(150)\:\:g\:\:for\:\:3(2.14) \rightarrow 450\:g\:\:for\:\:\$6.42 \\[3ex] 3\:\:jars\:\:of\:\:A = 450g\:\:for\:\:\$6.42 \\[3ex] \underline{Jar\:\:B} \\[3ex] 400\:g\:\:for\:\:\$6.50 \\[3ex] Compare: \\[3ex] 450\:g\:\:for\:\:\$6.42\:\:versus\:\:400\:g\:\:for\:\:\$6.50 \\[3ex] 450\:g\:\:for\:\:\$6.42\:\:much\:\:better...get\:\:more\:\:for\:\:less \\[3ex] \therefore Jar\:A\:\:is\:\:the\:\:better\:\:buy $
(13.) ACT The total cost of renting a car is $\$35.00$ for each day the car is rented plus $42.5¢$ for each mile the car is driven.
What is the total cost of renting the car for $6$ days and driving $350$ miles?
(Note: No sales tax is involved.)

$ A.\:\: \$154.75 \\[3ex] B.\:\: \$224.88 \\[3ex] C.\:\: \$358.75 \\[3ex] D.\:\: \$420.00 \\[3ex] E.\:\: \$1,697.50 \\[3ex] $

The answer options are in dollars.

Therefore, we need to convert the $42.5$ cents to dollars.

It is required that we work in the same unit.

$ \underline{Cost\:\:of\:\:renting} \\[3ex] 6\:\:days\:\:@\:\:\$35.00\:\:per\:\:day = 6(35) = \$210 \\[3ex] \underline{Cost\:\:of\:\:driving} \\[3ex] 100¢ = \$1 \\[3ex] 42.5¢ = \dfrac{42.5}{100} = \$0.425 \\[3ex] 350\:\:miles\:\:@\:\:\$0.425\:\:per\:\:mile = 350(0.425) = \$148.75 \\[3ex] \underline{Total\:\:Cost} \\[3ex] Total\:\:Cost = \$210 + \$148.75 = \$358.75 \\[3ex] $ The total cost of renting the car for $6$ days and driving $350$ miles is $\$358.75$
(14.) ACT Taho earns his regular pay of $\$11$ per hour for up to $40$ hours of work per week.
For each hour over $40$ hours of work per week, Taho earns $1\dfrac{1}{2}$ times his regular pay.
How much does Taho earn in a week in which he works $50$ hours?

$ F.\:\: \$550 \\[3ex] G.\:\: \$605 \\[3ex] H.\:\: \$625 \\[3ex] J.\:\: \$750 \\[3ex] K.\:\: \$825 \\[3ex] $

$ Work\:\:hours = 50 \\[3ex] Regular\:\:work\:\:hours = 40 \\[3ex] Overtime\:\:work\:\:hours = 50 - 40 = 10 \\[3ex] \underline{Regular\:\:Work\:\:Hours} \\[3ex] Regular\:\:pay = \$11\:\:per\:\:hour \\[3ex] 40\:hours\:\:@\:\:\$11\:\:per\:\:hour = 40(11) = \$440 \\[3ex] \underline{Overtime\:\:Work\:\:Hours} \\[3ex] 1\dfrac{1}{2} = \dfrac{2 * 1 + 1}{2} = \dfrac{2 + 1}{2} = \dfrac{3}{2} = 1.5 \\[5ex] Overtime\:\:pay = 1.5(11) = \$16.5\:\:per\:\:hour \\[3ex] 10\:hours\:\:@\:\:\$16.5\:\:per\:\:hour = 10(16.5) = \$165 \\[3ex] \underline{Total\:\:Pay} \\[3ex] Total\:\:pay\:\:for\:\:50\:\:hours\:\:of\:\:work \\[3ex] = \$440 + \$165 = \$605 $


ACT Use the following information to answer questions $15 - 17$
A large theater complex surveyed $5,000$ adults.
The results of the survey are shown in the tables below.

Age groups Number
$21 - 30$
$31 - 40$
$41 - 50$
$51$ or older
$2,750$
$1,225$
$625$
$400$

Moviegoer category Number
Very often
Often
Sometimes
Rarely
$830$
$1,650$
$2,320$
$200$

Tickets are $\$9.50$ for all regular showings and $\$7.00$ for matinees.



(15.) ACT Based on the survey results, what was the average number of moviegoers for each of the $4$ categories?

$ A.\:\: 610 \\[3ex] B.\:\: 1,060 \\[3ex] C.\:\: 1,240 \\[3ex] D.\:\: 1,250 \\[3ex] E.\:\: 1,985 \\[3ex] $

$ Average = Mean = \bar{x} \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \Sigma x = 830 + 1650 + 2320 + 200 = 5000 \\[3ex] n = 4 \\[3ex] \bar{x} = \dfrac{5000}{4} \\[5ex] \bar{x} = 1250 \\[3ex] $ The average number of moviegoers for each of the $4$ categories is $1250$ moviegoers
(16.) ACT Suppose all the adults surveyed happened to attend $1$ movie each in one particular week.
The total amount spent on tickets by those surveyed in that week was $\$44,000.00$
How many adults attended matinees that week?

$ F.\:\: 500 \\[3ex] G.\:\: 1,400 \\[3ex] H.\:\: 2,500 \\[3ex] J.\:\: 3,600 \\[3ex] K.\:\: 4,500 \\[3ex] $

For more examples of similar questions, please review Solved Examples on Linear Systems

$ Let\:\:the: \\[3ex] Number\:\:of\:\:adults\:\:that\:\:attended\:\:matinees = m \\[3ex] Number\:\:of\:\:adults\:\:that\:\:attended\:\:regular\:\:showings = r \\[3ex] Adults:\:\: m + r = 5000...eqn.(1) \\[3ex] Cost:\:\: 7m + 9.5r = 44000...eqn.(2) \\[3ex] To\:\:find\:\:m, \:\:eliminate\:\: r \\[3ex] 9.5 * eqn.(1) \implies 9.5m + 9.5r = 47500...eqn.(3) \\[3ex] eqn.(3) - eqn.(2) \implies \\[3ex] (9.5m - 7m) + (9.5r - 9.5r) = 47500 - 44000 \\[3ex] 2.5m = 3500 \\[3ex] m = \dfrac{3500}{2.5} \\[5ex] m = 1400 \\[3ex] $ $1400$ adults attended matinees that week.
(17.) ACT One of the following circle graphs represents the proportion by age group of the adults surveyed. Which one?
Numbers Application Question 17


Age groups Number Percentage
$21 - 30$ $2,750$ $ \dfrac{2750}{5000} * 100 = 0.55 * 100 = 55\% $
$31 - 40$ $1,225$ $ \dfrac{1225}{5000} * 100 = 0.245 * 100 = 24.5\% $
$41 - 50$ $625$ $ \dfrac{625}{5000} * 100 = 0.125 * 100 = 12.5\% $
$51$ or older $400$ $ \dfrac{400}{5000} * 100 = 0.08 * 100 = 8\% $

The correct option is $A$
(18.) ACT Pablo recorded the noon temperature, in degrees Celsius, on $4$ consecutuve days as part of a science project.
On the $1st$ day, the noon temperature was $-4^\circ C$.
On the $4th$ day, the noon temperature was $12^\circ C$.
What was the change in the noon temperature from the $1st$ day to the $4th$ day?

$ F.\:\: -16^\circ C \\[3ex] G.\:\: -4^\circ C \\[3ex] H.\:\: 4^\circ C \\[3ex] J.\:\: 8^\circ C \\[3ex] K.\:\: 16^\circ C \\[3ex] $

$ Initial\:\:temperature = temperature\:\:on\:\:the\:\:1st\:\:day = -4^\circ C \\[3ex] Final\:\:temperature = temperature\:\:on\:\:the\:\:4th\:\:day = 12^\circ C \\[3ex] Change\:\:in\:\:temperature \\[3ex] = Final\:\:temperature - Initial\:\:temperature \\[3ex] = 12 - (-4) \\[3ex] = 12 + 4 \\[3ex] = 16^\circ C $


ACT Use the following information to answer questions $19 - 21$
Kyla purchased $25$ pieces of candy from Laszko's Candy Shop.
Her purchase consists of $7$ lollipops, $4$ candy bars, $10$ licorice sticks, and $4$ gumballs.
The unit price of each candy item is shown in the table below.
Kyla's purchase, without sales tax, totaled $\$10.30$.
Laszko's charges an $8\%$ sales tax on each purchase, which is calculated by multiplying the purchase total by $0.08$ and rounding to the nearest $\$0.01$.

Candy Item Unit price
Lollipop
Candy bar
Licorice stick
Gumball
$\$0.90$
$\$0.60$
$\$0.10$
$\$0.15$


(19.) ACT Kyla gave the shop clerk $\$15.00$.
How much change should Kyla have received?

$ A.\:\: \$3.88 \\[3ex] B.\:\: \$4.35 \\[3ex] C.\:\: \$4.62 \\[3ex] D.\:\: \$4.70 \\[3ex] E.\:\: \$5.08 \\[3ex] $

$ Kyla's\:\:purchase\:\:without\:\:sales\:\:tax = \$10.30 \\[3ex] 8\%\:\:sales\:\:tax = 0.08(10.30) = 0.824 \\[3ex] Checkout\:\:price = 10.30 + 0.824 = 11.124 \approx \$11.12 \\[3ex] Gave\:\: \$15.00 \\[3ex] Change = 15.00 - 11.12 = \$3.88 $
(20.) ACT Without sales tax, what was the average price Kyla paid per piece of candy, to the nearest $\$0.01?$

$ F.\:\: \$0.21 \\[3ex] G.\:\: \$0.25 \\[3ex] H.\:\: \$0.32 \\[3ex] J.\:\: \$0.41 \\[3ex] K.\:\: \$0.44 \\[3ex] $

$ Average\:\:price = \dfrac{\$10.30}{25} \\[5ex] = \$0.412 \:\:per\:\: candy \\[3ex] \approx \$0.41 \:\:per\:\: candy $




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(21.) ACT Kyla offers to sell $2$ of the $25$ pieces of candy to her brother Virgil.
She lets Virgil have a choice of $2$ pieces of the same candy item or $1$ piece each of $2$ different candy items.
Kyla will have Virgil pay the same total cost for the $2$ pieces that he would pay for the $2$ pieces at Laszko's.
How many different total costs (in dollars) are possible for Virgil's choice of $2$ pieces?

$ A.\:\: 8 \\[3ex] B.\:\: 10 \\[3ex] C.\:\: 12 \\[3ex] D.\:\: 14 \\[3ex] E.\:\: 16 \\[3ex] $

We can do this in at least two ways.
The first method involves Quantitative Literacy
The second method is Combinatorics.
Use any method you prefer.

$ \underline{First\:\:Method: - Quantitative\:\:Reasoning} \\[3ex] Combinations\:\:are \\[3ex] 1\:Lollipop \:\:and\:\: 1\:Candy\:\:bar \\[3ex] 1\:Lollipop \:\:and\:\: 1\:Licorice\:\:stick \\[3ex] 1\:Lollipop \:\:and\:\: 1\:Gumball \\[3ex] 1\:Candy\:\:bar \:\:and\:\: 1\:Licorice\:\:stick \\[3ex] 1\:Candy\:\:bar \:\:and\:\: 1\:Gumball \\[3ex] 1\:Licorice\:\:stick \:\:and\:\: 1\:Gumball \\[3ex] 2\:Lollipops \\[3ex] 2\:Candy\:\:bars \\[3ex] 2\:Licorice\:\:sticks \\[3ex] 2\:Gumballs \\[3ex] = 10\:\:combinations $
(22.) ACT Marietta purchased a car that had a purchase price of $\$10,400$, which included all other costs and tax.
She paid $\$2,000$ as a down payment and got a loan for the rest of the purchase price.
Marietta paid off the loan by making $48$ payments of $\$225$ each.
The total of all her payments, including the down payment, was how much more than the car's purchase price?

$ A.\:\: \$400 \\[3ex] B.\:\: \$2,400 \\[3ex] C.\:\: \$8,400 \\[3ex] D.\:\: \$10,800 \\[3ex] E.\:\: \$12,800 \\[3ex] $

$ Purchase price of car = $\$10,400$

Down payment = $\$2000$

$48$ payments @ $\$225$ per payment = $48 * 225 = 10,800$

Total of all payments she made = $\$2000 + \$10,800 = \$12,800$

This is the question:

$\$12,800$ is how much more than $\$10,400$

$12,800 - 10,400 = 2,400$

The total payments made by Marietta is $\$2,400$ more than the car's purchase price. $


ACT Use the following information to answer questions $23 - 24$
The table below gives the price per gallon of unleaded gasoline at Gus's Gas Station on January $1$ for $5$ consecutive years in the $1990s$.
At Gus's, a customer can purchase a car wash for $\$4.00$.

Year Price
$1$
$2$
$3$
$4$
$5$
$\$1.34$
$\$1.41$
$\$1.41$
$\$1.25$
$\$1.36$


(23.) ACT What is the mean price per gallon, to the nearest $\$0.01$, on January $1$ for the $5$ years listed in the table?

$ F.\:\: \$3.88 \\[3ex] G.\:\: \$4.35 \\[3ex] H.\:\: \$4.62 \\[3ex] J.\:\: \$4.70 \\[3ex] K.\:\: \$5.08 \\[3ex] $

$ Mean = \dfrac{1.34 + 1.41 + 1.41 + 1.25 + 1.36}{5} \\[5ex] = \dfrac{6.77}{5} \\[5ex] = 1.354 \\[3ex] \approx \$1.35 $
(24.) ACT On January $1$ of Year $5$, Anamosa bought gas and a car wash at Gus's.
She put $11.38$ gallons of gas in her car and $1.85$ gallons of gas in a container for her snowblower.
To the nearest $\$0.01$, how much did Anamosa pay for the gas for her car and snowblower, and a car wash?

$ F.\:\: \$15.48 \\[3ex] G.\:\: \$17.23 \\[3ex] H.\:\: \$17.99 \\[3ex] J.\:\: \$19.48 \\[3ex] K.\:\: \$21.99 \\[3ex] $

$ \underline{January\:\:1\:\:of\:\:Year\:\:5} \\[3ex] Total\:\:gallons\:\:of\:\:gas\:\:purchased = 11.38 + 1.85 = 13.23 \\[3ex] 13.23\:\:gallons\:\:@\:\:\$1.36\:\:per\:\:gallon = 13.23(1.36) = \$17.9928 \\[3ex] Car\:\:wash = \$4.00 \\[3ex] Total\:\:cost = \$17.9928 + \$4.00 = \$21.9928 \\[3ex] \approx \$21.99\:\:to\:\:the\:\:nearest\:\:\$0.01 $
(25.) CSEC The table below shows the number of tickets sold for a bus tour.
Some items in the table are missing.
Tickets Sold for Bus Tour
Category Number of Tickets Sold Cost per Ticket in $\$$ Total Cost in $\$$
Juvenile $5$ $P$ $130.50$
Youth $14$ $44.35$ $Q$
Adult $R$ $2483.60$

(i) Calculate the value of $P$
(ii) Calculate the value of $Q$
(iii) An adult ticket is TWICE the cost of a youth ticket. Calculate the value of $R$
(iv) The bus company pays taxes of $15\%$ on each ticket sold. Calculate the taxes paid by the bus company.


$ Number\;\;of\;\;tickets\;\;sold * Cost\;\;per\;\;ticket = Total\;\;cost \\[3ex] (i) \\[3ex] \underline{Juvenile} \\[3ex] 5 * P = 130.50 \\[3ex] P = \dfrac{130.5}{5} \\[5ex] P = \$26.10 \\[3ex] (ii) \\[3ex] \underline{Youth} \\[3ex] 14 * 44.35 = Q \\[3ex] 620.9 = Q \\[3ex] Q = \$620.90 \\[3ex] (iii) \\[3ex] \underline{Adult} \\[3ex] Cost\;\;per\;\;ticket = 2 * 44.35 \\[3ex] Cost\;\;per\;\;ticket = \$88.70 \\[3ex] R * 88.70 = 2483.60 \\[3ex] R = \dfrac{2483.6}{88.7} \\[5ex] R = 28\;tickets \\[3ex] (iv) \\[3ex] Total\;\;Cost\;\;of\;\;all\;\;tickets \\[3ex] = 130.50 + Q + 2483.60 \\[3ex] = 130.50 + 620.90 + 2483.60 \\[3ex] = \$3235.00 \\[3ex] 15\%\;\;tax \\[3ex] = \dfrac{15}{100} * 3235 \\[5ex] = 0.15(3235) \\[3ex] = \$485.25 $
(26.) ACT Rya and Sampath start running laps from the same starting line at the same time and in the same direction on a certain indoor track.
Rya completes one lap in $16$ seconds, and Sampath completes the same lap in $28$ seconds.
Both continue running at their same respective rates and in the same direction for $10$ minutes.
What is the fewest number of seconds after starting that Rya and Sampath will again be at their starting line at the same time?

$ F.\:\: 88 \\[3ex] G.\:\: 112 \\[3ex] H.\:\: 120 \\[3ex] J.\:\: 220 \\[3ex] K.\:\: 448 \\[3ex] $

The question is simply asking for the LCM (least common multiple) of $16$ and $28$ because of fewest number of seconds
Keep in mind that it is not just the common multiple of $16$ and $28$, but the least common multiple

$ \underline{Prime\;\;Factorization\;\;Method} \\[3ex] 16 = \color{black}{2} * \color{darkblue}{2} * 2 * 2 \\[3ex] 28 = \color{black}{2} * \color{darkblue}{2} * 7 \\[3ex] LCM = \color{black}{2} * \color{darkblue}{2} * 2 * 2 * 7 \\[3ex] LCM = 112 \\[3ex] $ The next time Rya and Sampath will again be at their starting line is $112$ seconds
(27.) ACT Jen is doing an experiment to determine whether a high-protein food affects the ability of white mice to find their way through a maze.
The mice is the experimental group were given the high-protein food; the mice in the control group were given regular food.
Jen then timed the mice as they found their way through the maze.
The table below shows the results.
Mouse number Experimental group Control group
$1$ $1$ min $46$ sec $2$ min $13$ sec
$2$ $2$ min $2$ sec $1$ min $49$ sec
$3$ $2$ min $20$ sec $2$ min $28$ sec
$4$ $1$ min $51$ sec $2$ min $7$ sec
$5$ $1$ min $41$ sec $1$ min $58$ sec

The average time the mice in the experimental group took to find their way through the maze was how many seconds less than the average time taken by the mice in the control group?

$ A.\;\; 8 \\[3ex] B.\;\; 11 \\[3ex] C.\;\; 13 \\[3ex] D.\;\; 16 \\[3ex] E.\;\; 19 \\[3ex] $

We can solve this question in at least two ways.
Use any method you prefer.

$ 1\;min = 60\;sec \\[3ex] 2\;min = 2 * 60 = 120\;sec \\[3ex] 1\;min \;\; 46\;sec = 60 + 46 = 106\;sec \\[3ex] 2\;min \;\; 2\;sec = 120 + 2 = 122\;sec \\[3ex] 2\;min \;\; 20\;sec = 120 + 20 = 140\;sec \\[3ex] 1\;min \;\; 51\;sec = 60 + 51 = 111\;sec \\[3ex] 1\;min \;\; 41\;sec = 60 + 41 = 101\;sec \\[3ex] 2\;min \;\; 13\;sec = 120 + 13 = 133\;sec \\[3ex] 1\;min \;\; 49\;sec = 60 + 49 = 109\;sec \\[3ex] 2\;min \;\; 28\;sec = 120 + 28 = 148\;sec \\[3ex] 2\;min \;\; 7\;sec = 120 + 7 = 127\;sec \\[3ex] 1\;min \;\; 58\;sec = 60 + 58 = 118\;sec \\[3ex] $
1st Method: By Seconds
Mouse number Experimental group Experimental group (seconds) Control group Control group (seconds)
$1$ $1$ min $46$ sec $106$ $2$ min $13$ sec $133$
$2$ $2$ min $2$ sec $122$ $1$ min $49$ sec $109$
$3$ $2$ min $20$ sec $140$ $2$ min $28$ sec $148$
$4$ $1$ min $51$ sec $111$ $2$ min $7$ sec $127$
$5$ $1$ min $41$ sec $101$ $1$ min $58$ sec $118$
Sum: $580$ $635$
Average: $\dfrac{580}{5} = 116$ $\dfrac{635}{5} = 127$
Difference: $127 - 116 = 11$

The average time the mice in the experimental group took to find their way through the maze is $11$ seconds less than the average time taken by the mice in the control group.

$ \underline{2nd\;\;Method} \\[3ex] \underline{Experimental\;\;Group} \\[3ex] Total\;\;time \\[3ex] = 1\;min \;\; 46\;sec \\[3ex] + 2\;min \;\; 2\;sec \\[3ex] + 2\;min \;\; 20\;sec \\[3ex] + 1\;min \;\; 51\;sec \\[3ex] + 1\;min \;\; 41\;sec \\[3ex] = ...\;min \;\; 160\;sec \\[3ex] /* \\[3ex] 60\;sec = 1\;min \\[3ex] 160\;sec = \dfrac{160 * 1}{60} = 2.666666667\;sec \\[3ex] Integer\;\;part = 2 \\[3ex] 2 * 60 = 120 \\[3ex] 160 - 120 = 40 \\[3ex] \therefore 160\;sec = 2\;min \;\; 40\;sec \\[3ex] */ \\[3ex] = 9\;min \;\; 40\;sec \\[3ex] Average\;\;time \\[3ex] = \dfrac{Total\;\;time}{sample\;\;size} \\[5ex] = \dfrac{9\;min \;\; 40\;sec}{5} \\[5ex] /* \\[3ex] 5\;\;divide\;\;9\;min = 1\;min \;\;remaining\;\; 4\;min \\[3ex] 5\;\; cannot\;\; divide\;\; 4\;min \\[3ex] 4\;min = 4 * 60 = 240\;sec \\[3ex] 240\;sec + 40\;sec = 280\;sec \\[3ex] 5\;\;divide\;\;240\;sec = 56\;sec \\[3ex] */ \\[3ex] = 1\;min \;\; 56\;sec \\[3ex] \underline{Control\;\;Group} \\[3ex] Total\;\;time \\[3ex] = 2\;min \;\; 13\;sec \\[3ex] + 1\;min \;\; 49\;sec \\[3ex] + 2\;min \;\; 28\;sec \\[3ex] + 2\;min \;\; 7\;sec \\[3ex] + 1\;min \;\; 58\;sec \\[3ex] = ...\;min \;\; 155\;sec \\[3ex] /* \\[3ex] 60\;sec = 1\;min \\[3ex] 155\;sec = \dfrac{155 * 1}{60} = 2.583333333\;sec \\[3ex] Integer\;\;part = 2 \\[3ex] 2 * 60 = 120 \\[3ex] 155 - 120 = 35 \\[3ex] \therefore 155\;sec = 2\;min \;\; 35\;sec \\[3ex] */ \\[3ex] = 10\;min \;\; 35\;sec \\[3ex] Average\;\;time \\[3ex] = \dfrac{Total\;\;time}{sample\;\;size} \\[5ex] = \dfrac{10\;min \;\; 35\;sec}{5} \\[5ex] /* \\[3ex] 5\;\;divide\;\;10\;min = 2\;min \\[3ex] 5\;\;divide\;\;35\;sec = 7\;sec \\[3ex] */ \\[3ex] = 2\;min \;\; 7\;sec \\[3ex] \underline{Difference} \\[3ex] Average\;\;time\;\;for\;\;Control\;\;Group - Average\;\;time\;\;for\;\;Experimental\;\;Group \\[3ex] = 2\;min \;\; 7\;sec - 1\;min \;\; 56\;sec \\[3ex] 7\;sec - 56\;sec \;\;results\;\;in\;\;a\;\;negative\;\;value \\[3ex] Borrow\;\;1\;min\;\;from\;\;2\;min \\[3ex] Remaining\;\; 1\;min \\[3ex] 1\;min = 60\;sec \\[3ex] Add\;\;60\;sec \;\;to\;\; 7\;sec \rightarrow 67\;sec \\[3ex] 67\;sec - 56\;sec = 11\;sec \\[3ex] 1\;min - 1\;min = 0\;min \\[3ex] $ The average time the mice in the experimental group took to find their way through the maze is $11$ seconds less than the average time taken by the mice in the control group.
(28.) ACT A retail sales associate's daily commission during $1$ week was $\$30$ on Monday and Tuesday and $\$70$ on Wednesday, Thursday, and Friday.
What was the associate's average daily commission for these $5$ days?

$ F.\:\: \$50 \\[3ex] G.\:\: \$51 \\[3ex] H.\:\: \$54 \\[3ex] J.\:\: \$55 \\[3ex] K.\:\: \$56 \\[3ex] $

$ \underline{Daily\;\;Commission} \\[3ex] Monday \rightarrow \$30 \\[3ex] Tuesday \rightarrow \$30 \\[3ex] Wednesday \rightarrow \$70 \\[3ex] Thursday \rightarrow \$70 \\[3ex] Friday \rightarrow \$70 \\[3ex] Average\;\;daily\;\;commission \\[3ex] = \dfrac{30 + 30 + 70 + 70 + 70}{5} \\[5ex] = \dfrac{270}{5} \\[5ex] = \$54.00 $