If there is one prayer that you should

- Samuel Dominic Chukwuemeka
**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.

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

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 WASSCE Students

Any question labeled WASCCE is a question for the WASCCE General Mathematics

Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

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.

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.

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 $p$

$ \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 $k$

$ \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 $EC\$50.00$

(i) Calculate the cost of $5$ litres of gasoline in St. Vincent,

(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 $p$

$litres$ | $cost(EC\$)$ |
---|---|

$3$ | $10.4$ |

$5$ | $p$ |

$ \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 $k$

$litres$ | $cost(EC\$)$ |
---|---|

$3$ | $10.4$ |

$k$ | $50$ |

$ \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 $

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 $

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 $\$6.50$

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

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 $37.50$

$ \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 $\$12.60$

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

$ 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.

(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

$ \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 $

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

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.

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.

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 $

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.

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 = $B$

$ \dfrac{A}{1} = \dfrac{2.14}{150} \\[5ex] A = 0.0142666667 \\[3ex] A \approx \$0.01 \\[3ex] $

$ \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 $

Which of the jars shown above is the BETTER buy?

We can do this question in at least two ways...Without a Calculator and With a Calculator.

Use any method you prefer.

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 = $B$

$Amount (g)$ | $Cost (\$)$ |
---|---|

$150$ | $2.14$ |

$1$ | $A$ |

$ \dfrac{A}{1} = \dfrac{2.14}{150} \\[5ex] A = 0.0142666667 \\[3ex] A \approx \$0.01 \\[3ex] $

$Amount (g)$ | $Cost (\$)$ |
---|---|

$400$ | $6.50$ |

$1$ | $B$ |

$ \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.

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 $\$358.75$

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

$ \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 $

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

$ 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.

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?

The correct option is $A$

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°C.

On the 4th day, the noon temperature was 12°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 $

On the 1st day, the noon temperature was -4°C.

On the 4th day, the noon temperature was 12°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 $

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 $

$ 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 $

(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 $

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. $

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 $

$ 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 $

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.

(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 $

Some items in the table are missing.

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

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 (

Keep in mind that it is not just the common multiple of $16$ and $28$, but the

$ \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.

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] $

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.

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] $

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 $

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 $

(29.)

(30.) **ACT** The DigiPhone Company advertises the following calling plans:

Tanisha is on the $89.99 plan, and Suki is on the $119.99 plan.

To the nearest cent, how much more is the before-tax charge per minute for 600 minutes on Tanisha's plan than for 1,000 minutes on Suki's plan?

$ A.\;\; \$0.03 \\[3ex] B.\;\; \$0.12 \\[3ex] C.\;\; \$0.13 \\[3ex] D.\;\; \$0.15 \\[3ex] E.\;\; \$0.75 \\[3ex] $

$ \underline{Tanisha's\;\;plan} \\[3ex] time = 600\;\;minutes \\[3ex] cost = \$89.99 \\[3ex] charge\;\;per\;\;minute \\[3ex] = \dfrac{89.99}{600} \\[5ex] = \$0.1499833333/minute \\[5ex] \underline{Suki's\;\;plan} \\[3ex] time = 1000\;\;minutes \\[3ex] cost = \$119.99 \\[3ex] charge\;\;per\;\;minute \\[3ex] = \dfrac{119.99}{1000} \\[5ex] = \$0.11999\;\;per\;\;minute \\[5ex] \underline{Difference\;\;between\;\;the\;\;charges} \\[3ex] 0.1499833333 - 0.11999 \\[3ex] = 0.0299933333 \\[3ex] \approx \$0.03\;\;per\;\;minute $

DigiPhone Calling Plans |
---|

600 minutes* for $89.99** per month 1,000 minutes* for $119.99** per month 1,400 minutes * for $149.99** per month |

*The charge for each additional minute is $0.20** **Taxes are NOT included. |

Tanisha is on the $89.99 plan, and Suki is on the $119.99 plan.

To the nearest cent, how much more is the before-tax charge per minute for 600 minutes on Tanisha's plan than for 1,000 minutes on Suki's plan?

$ A.\;\; \$0.03 \\[3ex] B.\;\; \$0.12 \\[3ex] C.\;\; \$0.13 \\[3ex] D.\;\; \$0.15 \\[3ex] E.\;\; \$0.75 \\[3ex] $

$ \underline{Tanisha's\;\;plan} \\[3ex] time = 600\;\;minutes \\[3ex] cost = \$89.99 \\[3ex] charge\;\;per\;\;minute \\[3ex] = \dfrac{89.99}{600} \\[5ex] = \$0.1499833333/minute \\[5ex] \underline{Suki's\;\;plan} \\[3ex] time = 1000\;\;minutes \\[3ex] cost = \$119.99 \\[3ex] charge\;\;per\;\;minute \\[3ex] = \dfrac{119.99}{1000} \\[5ex] = \$0.11999\;\;per\;\;minute \\[5ex] \underline{Difference\;\;between\;\;the\;\;charges} \\[3ex] 0.1499833333 - 0.11999 \\[3ex] = 0.0299933333 \\[3ex] \approx \$0.03\;\;per\;\;minute $

(31.)

(32.)

**ACT** Use the following information to answer Questions 33 - 35.

Many humans carry the gene Yq77.

The Yq test determines, with 100% accuracy, whether a human carries Yq77.

If a Yq test result is negative, the human does NOT carry Yq77

Sam designed a less expensive test for Yq77 called the Sam77 test.

It produces some incorrect results.

To determine the accuracy of the Sam77 test, both tests were administered to 1,000 volunteers.

The results from this administration are summarized in the table below.

Positive Yq test | Negative Yq test | |

Positive Sam77 test Negative Sam77 test |
590 25 |
10 375 |

(33.) It cost $2,500 to administer each Yq test and $50 to administer each Sam77 test.

What was the total cost to administer both tests to all the volunteers?

$ A.\;\; \$1,537,500 \\[3ex] B.\;\; \$1,556,750 \\[3ex] C.\;\; \$1,568,250 \\[3ex] D.\;\; \$2,500,000 \\[3ex] E.\;\; \$2,550,000 \\[3ex] $

$ Number\;\;of\;\;participants = 1000 \\[3ex] 1000\;\;Yq77\;\;tests\;\;@\;\;\$2500/test = 1000(2500) = 2,500,000 \\[3ex] 1000\;\;Sam77\;\;tests\;\;@\;\;\$50/test = 1000(50) = 50,000 \\[3ex] Total\;\;cost = \$2,500,000 + \$50,000 = \$2,550,000 $

What was the total cost to administer both tests to all the volunteers?

$ A.\;\; \$1,537,500 \\[3ex] B.\;\; \$1,556,750 \\[3ex] C.\;\; \$1,568,250 \\[3ex] D.\;\; \$2,500,000 \\[3ex] E.\;\; \$2,550,000 \\[3ex] $

$ Number\;\;of\;\;participants = 1000 \\[3ex] 1000\;\;Yq77\;\;tests\;\;@\;\;\$2500/test = 1000(2500) = 2,500,000 \\[3ex] 1000\;\;Sam77\;\;tests\;\;@\;\;\$50/test = 1000(50) = 50,000 \\[3ex] Total\;\;cost = \$2,500,000 + \$50,000 = \$2,550,000 $

(34.) What percent of the volunteers actually carry Yq77?

$ F.\;\; 57.5\% \\[3ex] G.\;\; 60.0\% \\[3ex] H.\;\; 60.5\% \\[3ex] J.\;\; 61.5\% \\[3ex] K.\;\; 62.5\% \\[3ex] $

$ Number\;\;of\;\;participants = 1000 \\[3ex] Number\;\;of\;\;positive\;\;Yq77\;\;test = 590 + 25 = 615 \\[3ex] \%\;\;of\;\;positive\;\;Yq77\;\;test \\[3ex] = \dfrac{615}{1000} * 100 \\[5ex] = 61.5\% $

$ F.\;\; 57.5\% \\[3ex] G.\;\; 60.0\% \\[3ex] H.\;\; 60.5\% \\[3ex] J.\;\; 61.5\% \\[3ex] K.\;\; 62.5\% \\[3ex] $

$ Number\;\;of\;\;participants = 1000 \\[3ex] Number\;\;of\;\;positive\;\;Yq77\;\;test = 590 + 25 = 615 \\[3ex] \%\;\;of\;\;positive\;\;Yq77\;\;test \\[3ex] = \dfrac{615}{1000} * 100 \\[5ex] = 61.5\% $

(35.) For how many volunteers did the Sam77 test give an incorrect result?

$ A.\;\; 10 \\[3ex] B.\;\; 25 \\[3ex] C.\;\; 35 \\[3ex] D.\;\; 385 \\[3ex] E.\;\; 400 \\[3ex] $

Yq77 test is 100% accurate but Sam77 test is not.

So, an incorrect result by the Sam77 test implies that the:

(a.) volunteer had a positive Sam77 test__but__ a negative Yq77 test (10 volunteers)

(b.) volunteer had a negative Sam77 test__but__ a positive Yq77 test (25 volunteers)

10 + 25 = 35

This means that 35 volunteers received an incorrect result from the Sam77 test.

$ A.\;\; 10 \\[3ex] B.\;\; 25 \\[3ex] C.\;\; 35 \\[3ex] D.\;\; 385 \\[3ex] E.\;\; 400 \\[3ex] $

Yq77 test is 100% accurate but Sam77 test is not.

So, an incorrect result by the Sam77 test implies that the:

(a.) volunteer had a positive Sam77 test

(b.) volunteer had a negative Sam77 test

10 + 25 = 35

This means that 35 volunteers received an incorrect result from the Sam77 test.

(36.)

(37.) It takes someone 52 seconds to walk from the first (ground) floor of a building to the third floor.

How long will it take the person to walk from the first floor to the sixth floor (at the same pace, assuming all floors have the same height)?

Walking from the 1st floor to the third floor: 52 seconds

There are two intervals between the first floor and the third floor

This means that from 1st floor to 2nd floor (1st interval)__and__ from 2nd floor to 3rd floor (2nd interval): 52 seconds

Assume equal proportion of times:

$\dfrac{52}{2} = 26$ seconds

1st floor to 2nd floor: 26 seconds

2nd floor to 3rd floor: 26 seconds

This means: 26 seconds per interval

At the same pace, how long will it take the person to walk from the first floor to the sixth floor?

first floor to sixth floor = 5 intervals (1st to 2nd, 2nd to 3rd, 3rd to 4th, 4th to 5th, and 5th to 6th)

Therefore, the time it will take the person to walk from the first floor to the sixth floor = 26(5) = 130 seconds.

How long will it take the person to walk from the first floor to the sixth floor (at the same pace, assuming all floors have the same height)?

Walking from the 1st floor to the third floor: 52 seconds

There are two intervals between the first floor and the third floor

This means that from 1st floor to 2nd floor (1st interval)

Assume equal proportion of times:

$\dfrac{52}{2} = 26$ seconds

1st floor to 2nd floor: 26 seconds

2nd floor to 3rd floor: 26 seconds

This means: 26 seconds per interval

At the same pace, how long will it take the person to walk from the first floor to the sixth floor?

first floor to sixth floor = 5 intervals (1st to 2nd, 2nd to 3rd, 3rd to 4th, 4th to 5th, and 5th to 6th)

Therefore, the time it will take the person to walk from the first floor to the sixth floor = 26(5) = 130 seconds.

(38.) **ACT** On the first day of school, Ms. Dubacek gave her third-grade students 6 new spelling words to learn.

On each day of school after that, she gave the students 3 new spelling words.

How many new spelling words had she given the students by the ends of the 21st day of school?

$ A.\;\; 60 \\[3ex] B.\;\; 63 \\[3ex] C.\;\; 66 \\[3ex] D.\;\; 69 \\[3ex] E.\;\; 72 \\[3ex] $

$ 1st\;\;day = 6\;\;new\;\;words \\[3ex] 2nd\;\;through\;\;21st\;\;day = 21 - 1 = 20\;\;days \\[3ex] 20\;\;days\;\;@\;\;3\;\;words\;\;per\;\;day = 20(3) = 60 \\[3ex] Total\;\;number\;\;of\;\;new\;\;words \\[3ex] = 6 + 60 = 66\;\;new\;\;words $

On each day of school after that, she gave the students 3 new spelling words.

How many new spelling words had she given the students by the ends of the 21st day of school?

$ A.\;\; 60 \\[3ex] B.\;\; 63 \\[3ex] C.\;\; 66 \\[3ex] D.\;\; 69 \\[3ex] E.\;\; 72 \\[3ex] $

$ 1st\;\;day = 6\;\;new\;\;words \\[3ex] 2nd\;\;through\;\;21st\;\;day = 21 - 1 = 20\;\;days \\[3ex] 20\;\;days\;\;@\;\;3\;\;words\;\;per\;\;day = 20(3) = 60 \\[3ex] Total\;\;number\;\;of\;\;new\;\;words \\[3ex] = 6 + 60 = 66\;\;new\;\;words $

(39.) **GCSE** Here is some information, by ticket type, about the number of people visiting a cinema one week.

(a) How many children visited the cinema?

(b) How many**more** students than adults visited the cinema?

(c) A bar chart is drawn to show the number of people visiting the cinema one month.

Give**one** criticism of the bar chart.

$ Key:\;\;1\;box = 40\;people \\[3ex] (a) \\[3ex] Number\;\;of\;\;children \\[3ex] = 4\;boxes \\[3ex] = 4(40) \\[3ex] = 160\;people \\[3ex] (b) \\[3ex] Number\;\;of\;\;students \\[3ex] = 6\;boxes \\[3ex] = 6(40) \\[3ex] = 240\;people \\[3ex] Number\;\;of\;\;adults \\[3ex] = 3\dfrac{1}{2}\;boxes \\[5ex] = \dfrac{7}{2}(40) \\[5ex] = 7(20) \\[3ex] = 140\;people \\[3ex] More\;\;students\;\;than\;\;adults \\[3ex] = students - adults \\[3ex] = 240 - 140 \\[3ex] = 100\;people \\[3ex] $ (c)

The scale on the vertical axis (*Number of people*) is incorrect because 2500 is missing.

This implies that the number of*Students* is not represented correctly on the bar chart.

(a) How many children visited the cinema?

(b) How many

(c) A bar chart is drawn to show the number of people visiting the cinema one month.

Give

$ Key:\;\;1\;box = 40\;people \\[3ex] (a) \\[3ex] Number\;\;of\;\;children \\[3ex] = 4\;boxes \\[3ex] = 4(40) \\[3ex] = 160\;people \\[3ex] (b) \\[3ex] Number\;\;of\;\;students \\[3ex] = 6\;boxes \\[3ex] = 6(40) \\[3ex] = 240\;people \\[3ex] Number\;\;of\;\;adults \\[3ex] = 3\dfrac{1}{2}\;boxes \\[5ex] = \dfrac{7}{2}(40) \\[5ex] = 7(20) \\[3ex] = 140\;people \\[3ex] More\;\;students\;\;than\;\;adults \\[3ex] = students - adults \\[3ex] = 240 - 140 \\[3ex] = 100\;people \\[3ex] $ (c)

The scale on the vertical axis (

This implies that the number of

(40.) **ACT** A weeklong summer camp is held in June for children in Grades 3 - 6

Parents and guardians who enrolled their children for camp by May 15 received a 20% discount off the regular enrollment fee for each child enrolled.

For each grade, te table below gives the number of children enrolled by May 15 as well as the*regular*
enrollment fee oer child.

The grade of any child is that child's grade in school as of May 15.

Parents and guardians who enrolled their children for camp by May 15 received a 20% discount off the regular enrollment fee for each child enrolled.

For each grade, te table below gives the number of children enrolled by May 15 as well as the

The grade of any child is that child's grade in school as of May 15.

Grade | Enrollment by May 15 | Regular enrollment fee |

3 4 5 6 |
20 15 28 18 |
$350 $400 $450 $500 |

(41.)

(42.) **ACT** One caution sign flashes every 4 seconds, and another caution sign flashes every 10 seconds.

At a certain instant, the 2 signs flash at the same time.

How many seconds elapse until the 2 signs next flash at the same time?

$ A.\;\; 6 \\[3ex] B.\;\; 7 \\[3ex] C.\;\; 14 \\[3ex] D.\;\; 20 \\[3ex] E.\;\; 40 \\[3ex] $

The question is asking for the Least Common Multiple (LCM) of 4 and 10

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 4, 10 \\[3ex] 4 = \color{black}{2} * 2 \\[3ex] 10 = \color{black}{2} * 5 \\[5ex] LCM = \color{black}{2} * 2 * 5 \\[3ex] LCM = 20 \\[3ex] $ 20 seconds will elapse until the 2 signs next flash at the same time

At a certain instant, the 2 signs flash at the same time.

How many seconds elapse until the 2 signs next flash at the same time?

$ A.\;\; 6 \\[3ex] B.\;\; 7 \\[3ex] C.\;\; 14 \\[3ex] D.\;\; 20 \\[3ex] E.\;\; 40 \\[3ex] $

The question is asking for the Least Common Multiple (LCM) of 4 and 10

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 4, 10 \\[3ex] 4 = \color{black}{2} * 2 \\[3ex] 10 = \color{black}{2} * 5 \\[5ex] LCM = \color{black}{2} * 2 * 5 \\[3ex] LCM = 20 \\[3ex] $ 20 seconds will elapse until the 2 signs next flash at the same time

(43.)

(44.) **ACT** One neon sign flashes every 6 seconds.

Another neon sign flashes every 8 seconds.

If they flash together and you begin counting seconds, how many seconds after they flash together will they next flash together?

$ F.\;\; 48 \\[3ex] G.\;\; 24 \\[3ex] H.\;\; 14 \\[3ex] J.\;\; 7 \\[3ex] K.\;\; 2 \\[3ex] $

Similar to Question (42.); the question is asking for the Least Common Multiple (LCM) of 6 and 8

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 6, 8 \\[3ex] 6 = \color{black}{2} * 3 \\[3ex] 8 = \color{black}{2} * 2 * 2 \\[5ex] LCM = \color{black}{2} * 3 * 2 * 2 \\[3ex] LCM = 24 \\[3ex] $ 24 seconds after they flash together, they will next flash together

Another neon sign flashes every 8 seconds.

If they flash together and you begin counting seconds, how many seconds after they flash together will they next flash together?

$ F.\;\; 48 \\[3ex] G.\;\; 24 \\[3ex] H.\;\; 14 \\[3ex] J.\;\; 7 \\[3ex] K.\;\; 2 \\[3ex] $

Similar to Question (42.); the question is asking for the Least Common Multiple (LCM) of 6 and 8

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 6, 8 \\[3ex] 6 = \color{black}{2} * 3 \\[3ex] 8 = \color{black}{2} * 2 * 2 \\[5ex] LCM = \color{black}{2} * 3 * 2 * 2 \\[3ex] LCM = 24 \\[3ex] $ 24 seconds after they flash together, they will next flash together

(45.)

(46.)

(47.)

(48.) **ACT** Karen invested $2,000 in a special savings account.

The balance of this special savings account will double every 5 years.

Assuming that Karen makes no other deposits and no withdrawals, what will be the balance of Karen's investment at the end of 40 years?

$ A.\;\; \$80,000 \\[3ex] B.\;\; \$256,000 \\[3ex] C.\;\; \$400,000 \\[3ex] D.\;\; \$512,000 \\[3ex] E.\;\; \$1,024,000 \\[3ex] $

double__means__ multiplication by 2

$ initial\;\;amount = 2000 \\[5ex] At\;\;the\;\;end\;\;of\;\;5\;\;years: \\[3ex] updated\;\;amount = 2000(2) = 4000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(10\;\;years): \\[3ex] updated\;\;amount = 4000(2) = 8000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(15\;\;years): \\[3ex] updated\;\;amount = 8000(2) = 16000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(20\;\;years): \\[3ex] updated\;\;amount = 16000(2) = 32000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(25\;\;years): \\[3ex] updated\;\;amount = 32000(2) = 64000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(30\;\;years): \\[3ex] updated\;\;amount = 64000(2) = 128000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(35\;\;years): \\[3ex] updated\;\;amount = 128000(2) = 256000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(40\;\;years): \\[3ex] updated\;\;amount = 256000(2) = 512000 \\[3ex] $ The balance of Karen's investment at the end of 40 years is $512,000

The balance of this special savings account will double every 5 years.

Assuming that Karen makes no other deposits and no withdrawals, what will be the balance of Karen's investment at the end of 40 years?

$ A.\;\; \$80,000 \\[3ex] B.\;\; \$256,000 \\[3ex] C.\;\; \$400,000 \\[3ex] D.\;\; \$512,000 \\[3ex] E.\;\; \$1,024,000 \\[3ex] $

double

$ initial\;\;amount = 2000 \\[5ex] At\;\;the\;\;end\;\;of\;\;5\;\;years: \\[3ex] updated\;\;amount = 2000(2) = 4000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(10\;\;years): \\[3ex] updated\;\;amount = 4000(2) = 8000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(15\;\;years): \\[3ex] updated\;\;amount = 8000(2) = 16000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(20\;\;years): \\[3ex] updated\;\;amount = 16000(2) = 32000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(25\;\;years): \\[3ex] updated\;\;amount = 32000(2) = 64000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(30\;\;years): \\[3ex] updated\;\;amount = 64000(2) = 128000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(35\;\;years): \\[3ex] updated\;\;amount = 128000(2) = 256000 \\[5ex] At\;\;the\;\;end\;\;of\;\;another\;\;5\;\;years\;(40\;\;years): \\[3ex] updated\;\;amount = 256000(2) = 512000 \\[3ex] $ The balance of Karen's investment at the end of 40 years is $512,000

(49.)

(50.)

(51.)

(52.) **ACT** In 2011, the U.S Mint in Philadelphia produced 10,334,590,000 1-cent pieces, commonly called pennies.

These pennies were then bagged, with $50 in pennies per bag.

This process resulted in how many bags of pennies?

**A.** 2,066,918

**B.** 206,691,800

**C.** 5,167,295,000

**D.** 20,669,180,000

**E.** 51,672,950,000,000

$ \$1 = 100\;\;pennies \\[3ex] \implies \\[3ex] \$50 = 5000\;\;pennies = 1\;bag \\[3ex] \implies \\[3ex] Number\;\;of\;\;bags \\[3ex] = \dfrac{10,334,590,000}{5000} \\[5ex] = 2,066,918\;bags $

These pennies were then bagged, with $50 in pennies per bag.

This process resulted in how many bags of pennies?

$ \$1 = 100\;\;pennies \\[3ex] \implies \\[3ex] \$50 = 5000\;\;pennies = 1\;bag \\[3ex] \implies \\[3ex] Number\;\;of\;\;bags \\[3ex] = \dfrac{10,334,590,000}{5000} \\[5ex] = 2,066,918\;bags $

(53.)

(54.)

(55.)

(56.) **ACT** Ricardo started a savings account for his daughter Ruth by depositing $500 into the
account for her 1st birthday.

For each successive birthday, Ricardo deposits $200 more than the amount deposited for the previous birthday.

This is the only money deposited into the account.

What is the total amount of money Ricardo will have deposited into the account for Ruth up to and including her 6th birthday?

$ F.\;\; \$4,000 \\[3ex] G.\;\; \$4,200 \\[3ex] H.\;\; \$4,700 \\[3ex] J.\;\; \$4,900 \\[3ex] K.\;\; \$6,000 \\[3ex] $

$ \underline{1st\;\;birthday} \\[3ex] Deposit = 500 \\[5ex] \underline{2nd\;\;birthday} \\[3ex] Deposit = 200 + 500 = 700 \\[5ex] \underline{3rd\;\;birthday} \\[3ex] Deposit = 200 + 700 = 900 \\[5ex] \underline{4th\;\;birthday} \\[3ex] Deposit = 200 + 900 = 1100 \\[5ex] \underline{5th\;\;birthday} \\[3ex] Deposit = 200 + 1100 = 1300 \\[5ex] \underline{6th\;\;birthday} \\[3ex] Deposit = 200 + 1300 = 1500 \\[5ex] \underline{Total\;\;Deposited} \\[3ex] Total\;\;deposit \\[3ex] = 500 + 700 + 900 + 1100 + 1300 + 1500 \\[3ex] = 6000 \\[3ex] $ The total amount of money Ricardo will have deposited into the account for Ruth up to and including her 6th birthday is $6,000

For each successive birthday, Ricardo deposits $200 more than the amount deposited for the previous birthday.

This is the only money deposited into the account.

What is the total amount of money Ricardo will have deposited into the account for Ruth up to and including her 6th birthday?

$ F.\;\; \$4,000 \\[3ex] G.\;\; \$4,200 \\[3ex] H.\;\; \$4,700 \\[3ex] J.\;\; \$4,900 \\[3ex] K.\;\; \$6,000 \\[3ex] $

$ \underline{1st\;\;birthday} \\[3ex] Deposit = 500 \\[5ex] \underline{2nd\;\;birthday} \\[3ex] Deposit = 200 + 500 = 700 \\[5ex] \underline{3rd\;\;birthday} \\[3ex] Deposit = 200 + 700 = 900 \\[5ex] \underline{4th\;\;birthday} \\[3ex] Deposit = 200 + 900 = 1100 \\[5ex] \underline{5th\;\;birthday} \\[3ex] Deposit = 200 + 1100 = 1300 \\[5ex] \underline{6th\;\;birthday} \\[3ex] Deposit = 200 + 1300 = 1500 \\[5ex] \underline{Total\;\;Deposited} \\[3ex] Total\;\;deposit \\[3ex] = 500 + 700 + 900 + 1100 + 1300 + 1500 \\[3ex] = 6000 \\[3ex] $ The total amount of money Ricardo will have deposited into the account for Ruth up to and including her 6th birthday is $6,000

(57.)

(58.)

(59.)

(60.) **ACT** Mary takes 2 medications throughout the day and night.

One medication is to be taken every 6 hours and the other is to be taken every 4 hours.

If Mary begins taking both medications at 7:00 A.M. and takes both medications on schedule, how many hours later will it be when she next takes both medications at the same time?

$ F.\;\; 6 \\[3ex] G.\;\; 9 \\[3ex] H.\;\; 10 \\[3ex] J.\;\; 12 \\[3ex] K.\;\; 24 \\[3ex] $

We need to find the Least Common Multiple (LCM) of 6 and 4

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 6, 4 \\[3ex] 6 = \color{black}{2} * 3 \\[3ex] 4 = \color{black}{2} * 2 \\[5ex] LCM = \color{black}{2} * 3 * 2 \\[3ex] LCM = 12 \\[3ex] $ 12 hours later, Mary will take both medications at the same time.

*
If we are asked to find the actual time, then it will be: *

7:00 AM + 12 hours

= 19

= 7:00 PM

At 7:00 PM, Mary will take both medications at the same time.

One medication is to be taken every 6 hours and the other is to be taken every 4 hours.

If Mary begins taking both medications at 7:00 A.M. and takes both medications on schedule, how many hours later will it be when she next takes both medications at the same time?

$ F.\;\; 6 \\[3ex] G.\;\; 9 \\[3ex] H.\;\; 10 \\[3ex] J.\;\; 12 \\[3ex] K.\;\; 24 \\[3ex] $

We need to find the Least Common Multiple (LCM) of 6 and 4

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 6, 4 \\[3ex] 6 = \color{black}{2} * 3 \\[3ex] 4 = \color{black}{2} * 2 \\[5ex] LCM = \color{black}{2} * 3 * 2 \\[3ex] LCM = 12 \\[3ex] $ 12 hours later, Mary will take both medications at the same time.

7:00 AM + 12 hours

= 19

= 7:00 PM

At 7:00 PM, Mary will take both medications at the same time.

(61.)

(62.)

(63.)

(64.) **ACT** Two warning signs begin flashing at the same time.

One sign flashes every 3 seconds, and the other sign flashes every 8 seconds.

How many seconds elapse from the moment the 2 signs flash at the same time until they next flash at the same time?

$ F.\;\; 5 \\[3ex] G.\;\; 5.5 \\[3ex] H.\;\; 11 \\[3ex] J.\;\; 12 \\[3ex] K.\;\; 24 \\[3ex] $

The question is asking for the Least Common Multiple (LCM) of 3 and 8

$ Numbers = 3, 8 \\[3ex] 3 = 3 \\[3ex] 8 = 2 * 2 * 2 \\[5ex] LCM = 3 * 2 * 2 * 2 \\[3ex] LCM = 24 \\[3ex] $ 24 seconds elapse from the moment the 2 signs flash at the same time until they next flash at the same time

One sign flashes every 3 seconds, and the other sign flashes every 8 seconds.

How many seconds elapse from the moment the 2 signs flash at the same time until they next flash at the same time?

$ F.\;\; 5 \\[3ex] G.\;\; 5.5 \\[3ex] H.\;\; 11 \\[3ex] J.\;\; 12 \\[3ex] K.\;\; 24 \\[3ex] $

The question is asking for the Least Common Multiple (LCM) of 3 and 8

$ Numbers = 3, 8 \\[3ex] 3 = 3 \\[3ex] 8 = 2 * 2 * 2 \\[5ex] LCM = 3 * 2 * 2 * 2 \\[3ex] LCM = 24 \\[3ex] $ 24 seconds elapse from the moment the 2 signs flash at the same time until they next flash at the same time

(65.)

(66.)

(67.)

(68.) **ACT** One welcome sign flashes every 8 seconds, and another welcome sign flashes every 12 seconds.

At a certain instant, the 2 signs flash at the same time.

How many seconds elapse until the 2 signs next flash at the same time?

$ A.\;\; 4 \\[3ex] B.\;\; 10 \\[3ex] C.\;\; 20 \\[3ex] D.\;\; 24 \\[3ex] E.\;\; 96 \\[3ex] $

We need to find the Least Common Multiple (LCM) of 8 and 12

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 8, 12 \\[3ex] 8 = \color{black}{2} * \color{darkblue}{2} * 2 \\[3ex] 12 = \color{black}{2} * \color{darkblue}{2} * 3 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{2} * 2 * 3 \\[3ex] LCM = 24 \\[3ex] $ 24 seconds elapse until the 2 signs next flash at the same time

At a certain instant, the 2 signs flash at the same time.

How many seconds elapse until the 2 signs next flash at the same time?

$ A.\;\; 4 \\[3ex] B.\;\; 10 \\[3ex] C.\;\; 20 \\[3ex] D.\;\; 24 \\[3ex] E.\;\; 96 \\[3ex] $

We need to find the Least Common Multiple (LCM) of 8 and 12

The colors besides red indicate the common factors that should be counted only one time.

Begin with them in the multiplication for the LCM.

Then, include the rest.

$ Numbers = 8, 12 \\[3ex] 8 = \color{black}{2} * \color{darkblue}{2} * 2 \\[3ex] 12 = \color{black}{2} * \color{darkblue}{2} * 3 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{2} * 2 * 3 \\[3ex] LCM = 24 \\[3ex] $ 24 seconds elapse until the 2 signs next flash at the same time

(69.)

(70.)

**ACT** Use the following information to answer questions 71 - 73

The Dow Jones Industrial Average (DJIA) is an index of stock values.

The chart below gives the DJIA closing values from August 24 through September 30 of a certain year and the
change in
the closing value from the previous day.

A minus sign indicates a *decline* (a closing value less than the previous day's closing value).

A plus sign indicates an *advance* (a closing value greater than the previous day's closing value).

Date | Closing value | Change | Date | Closing value | Change |

8/24 8/25 8/26 8/27 8/30 8/31 9/01 9/02 9/03 9/07 9/08 9/09 9/10 |
8,600 8,515 8,160 8,050 7,540 7,825 7,780 7,680 7,640 8,020 7,860 8,045 7,795 |
-85 -355 -110 -510 +285 -45 -100 -40 +380 -160 +185 -250 |
9/13 9/14 9/15 9/16 9/17 9/20 9/21 9/22 9/23 9/24 9/27 9/28 9/29 9/30 |
7,945 8,020 8,090 7,870 7,895 7,930 7,900 8,150 8,000 8,025 8,110 8,080 7,845 7,630 |
+150 +75 +70 -220 +25 +35 -30 +250 -150 +25 +85 -30 -235 -215 |

(71.) Which of the following is closest to the percent of decrease from the August 24 closing value to the September 30
closing value?

$ A.\;\; 7.9\% \\[3ex] B.\;\; 8.9\% \\[3ex] C.\;\; 11.3\% \\[3ex] D.\;\; 12.7\% \\[3ex] E.\;\; 88.7\% \\[3ex] $

$ 8/24:\;\; initial = 8600 \\[3ex] 9/30:\;\; new = 7,630 \\[3ex] change = new - initial \\[3ex] change = 7630 - 8600 = -970 \\[3ex] The\;\;change\;\;is\;\;a\;\;decrease \\[3ex] \%\;decrease = \dfrac{change}{initial} * 100 \\[5ex] \%\;decrease \\[3ex] = \dfrac{970}{8600} * 100 \\[5ex] = 11.27906977\% \\[3ex] \approx 11.3\% $

$ A.\;\; 7.9\% \\[3ex] B.\;\; 8.9\% \\[3ex] C.\;\; 11.3\% \\[3ex] D.\;\; 12.7\% \\[3ex] E.\;\; 88.7\% \\[3ex] $

$ 8/24:\;\; initial = 8600 \\[3ex] 9/30:\;\; new = 7,630 \\[3ex] change = new - initial \\[3ex] change = 7630 - 8600 = -970 \\[3ex] The\;\;change\;\;is\;\;a\;\;decrease \\[3ex] \%\;decrease = \dfrac{change}{initial} * 100 \\[5ex] \%\;decrease \\[3ex] = \dfrac{970}{8600} * 100 \\[5ex] = 11.27906977\% \\[3ex] \approx 11.3\% $

(72.) The chart shows 4 more declines than advances.

All of the following statements are true.

Which one best explains why the decline from the August 24 closing value to the September 30 closing value was relatively large?

**F.** The greatest change in the chart was a decline.

**G.** The least change in the chart was an advance.

**H.** The greatest number of consecutive declines was greater than the greatest number of consecutive advances.

**J.** The first change was a decline.

**K.** The average of the declines was much greater than the average of the advances.

In other words, which of the options**best explains** why the decline from the August 24 closing value to the
September 30 closing value was relatively large (-970)?

All of the options are correct. However, which option is the most correct?

As observed and also written, there are more declines than advances.

But, the main reason for the large decline is because the average of the declines is much greater than the average of the advances.

*
Ask students to **verify* that statement...the last option...Option **K.** by calculating the:

(i.) average of the declines

(ii.) average of the advances

All of the following statements are true.

Which one best explains why the decline from the August 24 closing value to the September 30 closing value was relatively large?

In other words, which of the options

All of the options are correct. However, which option is the most correct?

As observed and also written, there are more declines than advances.

But, the main reason for the large decline is because the average of the declines is much greater than the average of the advances.

(i.) average of the declines

(ii.) average of the advances

(73.) What is the average closing value for the 5-day period from September 13 through September 17?

$ A.\;\; 7,895 \\[3ex] B.\;\; 7,920 \\[3ex] C.\;\; 7,964 \\[3ex] D.\;\; 7,980 \\[3ex] E.\;\; 8,090 \\[3ex] $

$ 9/13:\;\; 7945 \\[3ex] 9/14:\;\; 8020 \\[3ex] 9/15:\;\; 8090 \\[3ex] 9/16:\;\; 7870 \\[3ex] 9/17:\;\; 7895 \\[3ex] Average\;\;closing\;\;value\;\;for\;\;the\;\;5-day\;\;period \\[3ex] = \dfrac{7945 + 8020 + 8090 + 7870 + 7895}{5} \\[5ex] = \dfrac{39820}{5} \\[5ex] = 7964 $

$ A.\;\; 7,895 \\[3ex] B.\;\; 7,920 \\[3ex] C.\;\; 7,964 \\[3ex] D.\;\; 7,980 \\[3ex] E.\;\; 8,090 \\[3ex] $

$ 9/13:\;\; 7945 \\[3ex] 9/14:\;\; 8020 \\[3ex] 9/15:\;\; 8090 \\[3ex] 9/16:\;\; 7870 \\[3ex] 9/17:\;\; 7895 \\[3ex] Average\;\;closing\;\;value\;\;for\;\;the\;\;5-day\;\;period \\[3ex] = \dfrac{7945 + 8020 + 8090 + 7870 + 7895}{5} \\[5ex] = \dfrac{39820}{5} \\[5ex] = 7964 $

(74.)

(75.)

(76.)