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

**
Attempt all questions.
Show all work.
**

(1.)**ACT** If $a$ and $b$ are real numbers such that $a \gt 0$ and $b \lt 0$,
then which of the following is equivalent to $|a| - |b|$?

$ F.\:\: |a - b| \\[3ex] G.\:\: |a + b| \\[3ex] H.\:\: |a| + |b| \\[3ex] J.\:\: a - b \\[3ex] K.\:\: a + b $

We shall solve this question using two methods.

If you really do not know how to begin to solve this question, use the first method.

For faster solution because the ACT is a timed test, use the second method.

__First Method - Arithmetic method (Test Numbers)__

$ a\:\: and\:\: b\:\: are\:\: real\:\: numbers \\[3ex] a \gt 0, b \lt 0 \\[3ex] Let\:\: a = 1, b = -1 \\[3ex] |a| - |b| = |1| - |-1| = 1 - 1 = 0 \\[3ex] Test \\[3ex] |a - b| = |1 - (-1)| = |1 + 1| = |2| = 2 \:\: NO \\[3ex] |a + b| = |1 + (-1)| = |1 - 1| = |0| = 0 \:\: Maybe \\[3ex] |a| + |b| = |1| + |-1| = 1 + 1 = 2 \:\: NO \\[3ex] a - b = 1 - (-1) = 1 + 1 = 2 \:\: NO \\[3ex] a + b = 1 + (-1) = 1 - 1 = 0 \:\: Maybe \\[3ex] $ So, we are left with two options: $|a + b|$ and $a + b$

Specific to the question, $a$ and $b$ are real numbers.

Their absolute values are also real numbers.

The subtraction of their absolute values should give us real numbers, rather than the absolute value of real numbers.

In that sense, $a + b$ is a better answer.

__Second Method - Algebraic method__

$ a\:\: and\:\: b\:\: are\:\: real\:\: numbers \\[3ex] a \gt 0, b \lt 0 \\[3ex] $ Be definition

$|a|$ means that $a$ could be positive, zero, or negative.

However, from the question; $a \gt 0 \implies a$ is positive

$|b|$ means that $b$ could be positive, zero, or negative.

However, from the question; $b \lt 0 \implies b$ is negative

This means:

$ |a| - |b| \\[3ex] = (+a) - (-b) \\[3ex] = a - (-b) \\[3ex] = a + b $

$ F.\:\: |a - b| \\[3ex] G.\:\: |a + b| \\[3ex] H.\:\: |a| + |b| \\[3ex] J.\:\: a - b \\[3ex] K.\:\: a + b $

We shall solve this question using two methods.

If you really do not know how to begin to solve this question, use the first method.

For faster solution because the ACT is a timed test, use the second method.

$ a\:\: and\:\: b\:\: are\:\: real\:\: numbers \\[3ex] a \gt 0, b \lt 0 \\[3ex] Let\:\: a = 1, b = -1 \\[3ex] |a| - |b| = |1| - |-1| = 1 - 1 = 0 \\[3ex] Test \\[3ex] |a - b| = |1 - (-1)| = |1 + 1| = |2| = 2 \:\: NO \\[3ex] |a + b| = |1 + (-1)| = |1 - 1| = |0| = 0 \:\: Maybe \\[3ex] |a| + |b| = |1| + |-1| = 1 + 1 = 2 \:\: NO \\[3ex] a - b = 1 - (-1) = 1 + 1 = 2 \:\: NO \\[3ex] a + b = 1 + (-1) = 1 - 1 = 0 \:\: Maybe \\[3ex] $ So, we are left with two options: $|a + b|$ and $a + b$

Specific to the question, $a$ and $b$ are real numbers.

Their absolute values are also real numbers.

The subtraction of their absolute values should give us real numbers, rather than the absolute value of real numbers.

In that sense, $a + b$ is a better answer.

$ a\:\: and\:\: b\:\: are\:\: real\:\: numbers \\[3ex] a \gt 0, b \lt 0 \\[3ex] $ Be definition

$|a|$ means that $a$ could be positive, zero, or negative.

However, from the question; $a \gt 0 \implies a$ is positive

$|b|$ means that $b$ could be positive, zero, or negative.

However, from the question; $b \lt 0 \implies b$ is negative

This means:

$ |a| - |b| \\[3ex] = (+a) - (-b) \\[3ex] = a - (-b) \\[3ex] = a + b $

(2.) **ACT** If $x \lt y$ and $y \lt 4$, then what is the greatest possible integer value of
$x + y$

$ A.\:\: 0 \\[3ex] B.\:\: 3 \\[3ex] C.\:\: 4 \\[3ex] D.\:\: 7 \\[3ex] E.\:\: 8 $

We shall solve this question using the Arithmetic Method by testing numbers.

We shall use maximum numbers to test.

*
Student: Why do we have to use "maximum numbers" to test? *

Teacher: We shall use "maximum numbers" to test because the question is asking for the__greatest possible integer value__

Keep in mind that $x$ and $y$ do not have to be integer values.

But, we are interested in the integer value of their sum

**Arithmetic Method - Test Numbers**

$ y \lt 4 \\[3ex] Let\:\: y = 3.9 \\[3ex] x \lt y \\[3ex] Let\:\: x = 3.8 \\[3ex] x + y = 3.9 + 3.8 = 7.7 \\[3ex] $ Greatest possible integer value = $7$

*
Student: Why is the greatest possible integer value not equal to $8$? *

$ Teacher:\:\: 7.7 \ne 8 \\[3ex] 7.7 \gt 7 \\[3ex] 7.7 \lt 8 $

$ A.\:\: 0 \\[3ex] B.\:\: 3 \\[3ex] C.\:\: 4 \\[3ex] D.\:\: 7 \\[3ex] E.\:\: 8 $

We shall solve this question using the Arithmetic Method by testing numbers.

We shall use maximum numbers to test.

Teacher: We shall use "maximum numbers" to test because the question is asking for the

Keep in mind that $x$ and $y$ do not have to be integer values.

But, we are interested in the integer value of their sum

$ y \lt 4 \\[3ex] Let\:\: y = 3.9 \\[3ex] x \lt y \\[3ex] Let\:\: x = 3.8 \\[3ex] x + y = 3.9 + 3.8 = 7.7 \\[3ex] $ Greatest possible integer value = $7$

$ Teacher:\:\: 7.7 \ne 8 \\[3ex] 7.7 \gt 7 \\[3ex] 7.7 \lt 8 $

(3.) **ACT** What is the smallest integer greater than $\sqrt{85}$

$ A.\:\: 5 \\[3ex] B.\:\: 9 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 43 $

$ \sqrt{85} \approx 9.21954 \\[3ex] $ Next integer $\gt 9.21954 = 10$

$ A.\:\: 5 \\[3ex] B.\:\: 9 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 43 $

$ \sqrt{85} \approx 9.21954 \\[3ex] $ Next integer $\gt 9.21954 = 10$

(4.) **ACT** The text message component of each of Juan's monthly phone bills consists of
$\$10.00$ for the first $300$ text messages sent that month, plus $\$0.10$
for each additional text message sent that month. On Juan's most recent phone bill, he was charged a
total of $\$16.50$ for text messages. For how many text messages in total was Juan charged
on this bill?

We know the cost of the first $300$ messages. It is $\$10.00$

We do not know the number of "additional text messages". Let it be $x$

We do know that the cost of each additional text message = $\$0.10$

So, the cost of the additional messages = $0.1(x) = 0.1x$

We also know the total cost. It is $\$16.50$

Basic cost (for the first $300$ messages) + Cost of "additional" text messages = Total cost

$ 10 + 0.1x = 16.50 \\[3ex] 0.1x = 16.5 - 10 \\[3ex] 0.1x = 6.5 \\[3ex] x = \dfrac{6.5}{0.1} \\[5ex] x = 65 \\[3ex] $ Additional text messages = $65$

"Basic" text messages = $300$

Total text messages for that bill = $65 + 300 = 365$ messages

We know the cost of the first $300$ messages. It is $\$10.00$

We do not know the number of "additional text messages". Let it be $x$

We do know that the cost of each additional text message = $\$0.10$

So, the cost of the additional messages = $0.1(x) = 0.1x$

We also know the total cost. It is $\$16.50$

Basic cost (for the first $300$ messages) + Cost of "additional" text messages = Total cost

$ 10 + 0.1x = 16.50 \\[3ex] 0.1x = 16.5 - 10 \\[3ex] 0.1x = 6.5 \\[3ex] x = \dfrac{6.5}{0.1} \\[5ex] x = 65 \\[3ex] $ Additional text messages = $65$

"Basic" text messages = $300$

Total text messages for that bill = $65 + 300 = 365$ messages

(5.) **ACT** For every pair of real numbers $x$ and $y$ such that $xy = 0$ and $\dfrac{x}{y} = 0$, which of the
following statements is true?

$ A.\:\: x = 0 \:\:and\:\: y = 0 \\[3ex] B.\:\: x \ne 0 \:\:and\:\: y = 0 \\[3ex] C.\:\: x = 0 \:\:and\:\: y \ne 0 \\[3ex] D.\:\: x \ne 0 \:\:and\:\: y \ne 0 \\[3ex] $ $E.\:\:$ None of the statements is true for every such pair of real numbers $x$ and $y$

$xy = 0$ means that:

Either $x = 0$ OR $y = 0$ OR both $x = 0$ AND $y = 0$

This is known as the__Zero Product Property__

$\dfrac{x}{y} = 0$ means that $x = 0$

$0$ divided by anything is $0$

However, $y$ cannot be $0$.

Dividing anything by $0$ is undefined.

Combining the two conditions:

$x = 0 \:\:and\:\: y \ne 0$

$ A.\:\: x = 0 \:\:and\:\: y = 0 \\[3ex] B.\:\: x \ne 0 \:\:and\:\: y = 0 \\[3ex] C.\:\: x = 0 \:\:and\:\: y \ne 0 \\[3ex] D.\:\: x \ne 0 \:\:and\:\: y \ne 0 \\[3ex] $ $E.\:\:$ None of the statements is true for every such pair of real numbers $x$ and $y$

$xy = 0$ means that:

Either $x = 0$ OR $y = 0$ OR both $x = 0$ AND $y = 0$

This is known as the

$\dfrac{x}{y} = 0$ means that $x = 0$

$0$ divided by anything is $0$

However, $y$ cannot be $0$.

Dividing anything by $0$ is undefined.

Combining the two conditions:

$x = 0 \:\:and\:\: y \ne 0$

(6.) **ACT** A company that builds bridges used a pile driver to drive a post into the ground.
The post was driven $18$ feet into the ground by the first hit of the pile driver. On each hit after
the first hit, the post was driven into the ground an additional distance that was $\dfrac{2}{3}$ the
distance the post was driven in the previous hit. After a total of $4$ hits, the post was driven how
many feet into the ground?

We can solve this in two ways.

__First Method: Arithmetic Method__

This method is recommended for ACT.

However, you may use the Second Method if you feel comfortable with it.

$ First\:\: hit = 18 \\[3ex] Second\:\: hit = \dfrac{2}{3} * 18 = 2 * 6 = 12 \\[5ex] Third\:\: hit = \dfrac{2}{3} * 12 = 2 * 4 = 8 \\[5ex] Fourth\:\: hit = \dfrac{2}{3} * 8 = \dfrac{16}{3} \\[5ex] Total\:\: distance = 18 + 12 + 8 + \dfrac{16}{3} \\[5ex] = 38 + \dfrac{16}{3} \\[5ex] = \dfrac{114}{3} + \dfrac{16}{3} \\[5ex] = \dfrac{114 + 16}{3} \\[5ex] = \dfrac{130}{3} \\[5ex] = 43\dfrac{1}{3} \:feet \\[5ex] $__Second Method: Algebraic Method (Sum of a Geometric Sequence)__

This question is actually a geometric sequence.

You are asked to calculate the sum of the first four terms of a Geometric Sequence.

You can learn about Geometric Sequences here

The sequence is like this: $18, \dfrac{2}{3} \:\:of\:\: 18, \dfrac{2}{3} \:\:of\:\:\dfrac{2}{3}\:\:of\:\: 18, \dfrac{2}{3} \:\:of\:\:\dfrac{2}{3}\:\:of\:\:\dfrac{2}{3}\:\:of\:\: 18$

$ a = 18 \\[3ex] n = 4 \\[3ex] r = \dfrac{2}{3} \\[3ex] r \lt 1:\:\:So,\:\:use\:\: SGS_n = \dfrac{a(1 - r^n)}{1 - r} \\[5ex] SGS_4 = \dfrac{18\left(1 - \left(\dfrac{2}{3}\right)^4\right)}{1 - \dfrac{2}{3}} \\[7ex] = \dfrac{18\left(1 - \dfrac{2^4}{3^4}\right)}{\dfrac{3}{3} - \dfrac{2}{3}} \\[7ex] = \dfrac{18\left(1 - \dfrac{16}{81}\right)}{\dfrac{3 - 2}{3}} \\[7ex] = \dfrac{18\left(\dfrac{81}{81} - \dfrac{16}{81}\right)}{\dfrac{1}{3}} \\[7ex] = 18\left(\dfrac{81 - 16}{81}\right) \div \dfrac{1}{3} \\[7ex] = 18\left(\dfrac{65}{81}\right) * \dfrac{3}{1} \\[7ex] = 18 * \dfrac{65}{27} \\[5ex] = 2 * \dfrac{65}{3} \\[5ex] = \dfrac{130}{3} \\[5ex] = 43\dfrac{1}{3} \:feet $

We can solve this in two ways.

This method is recommended for ACT.

However, you may use the Second Method if you feel comfortable with it.

$ First\:\: hit = 18 \\[3ex] Second\:\: hit = \dfrac{2}{3} * 18 = 2 * 6 = 12 \\[5ex] Third\:\: hit = \dfrac{2}{3} * 12 = 2 * 4 = 8 \\[5ex] Fourth\:\: hit = \dfrac{2}{3} * 8 = \dfrac{16}{3} \\[5ex] Total\:\: distance = 18 + 12 + 8 + \dfrac{16}{3} \\[5ex] = 38 + \dfrac{16}{3} \\[5ex] = \dfrac{114}{3} + \dfrac{16}{3} \\[5ex] = \dfrac{114 + 16}{3} \\[5ex] = \dfrac{130}{3} \\[5ex] = 43\dfrac{1}{3} \:feet \\[5ex] $

This question is actually a geometric sequence.

You are asked to calculate the sum of the first four terms of a Geometric Sequence.

You can learn about Geometric Sequences here

The sequence is like this: $18, \dfrac{2}{3} \:\:of\:\: 18, \dfrac{2}{3} \:\:of\:\:\dfrac{2}{3}\:\:of\:\: 18, \dfrac{2}{3} \:\:of\:\:\dfrac{2}{3}\:\:of\:\:\dfrac{2}{3}\:\:of\:\: 18$

$ a = 18 \\[3ex] n = 4 \\[3ex] r = \dfrac{2}{3} \\[3ex] r \lt 1:\:\:So,\:\:use\:\: SGS_n = \dfrac{a(1 - r^n)}{1 - r} \\[5ex] SGS_4 = \dfrac{18\left(1 - \left(\dfrac{2}{3}\right)^4\right)}{1 - \dfrac{2}{3}} \\[7ex] = \dfrac{18\left(1 - \dfrac{2^4}{3^4}\right)}{\dfrac{3}{3} - \dfrac{2}{3}} \\[7ex] = \dfrac{18\left(1 - \dfrac{16}{81}\right)}{\dfrac{3 - 2}{3}} \\[7ex] = \dfrac{18\left(\dfrac{81}{81} - \dfrac{16}{81}\right)}{\dfrac{1}{3}} \\[7ex] = 18\left(\dfrac{81 - 16}{81}\right) \div \dfrac{1}{3} \\[7ex] = 18\left(\dfrac{65}{81}\right) * \dfrac{3}{1} \\[7ex] = 18 * \dfrac{65}{27} \\[5ex] = 2 * \dfrac{65}{3} \\[5ex] = \dfrac{130}{3} \\[5ex] = 43\dfrac{1}{3} \:feet $

(7.) **ACT** An integer is *abundant* if its positive integer factors, excluding the integer itself, have
a sum that is greater than the integer. How many of the integers $6, 8, 10, \:\:and\:\: 12$ are
abundant?

Positive integer factors (exclude $6$) of $6 = 1, 2, 3$

$ 1 + 2 + 3 = 6 \\[3ex] 6 \ngtr 6 \\[3ex] $ Positive integer factors (exclude $8$) of $8 = 1, 2, 4$

$ 1 + 2 + 4 = 7 \\[3ex] 7 \ngtr 8 \\[3ex] $ Positive integer factors (exclude $10$) of $10 = 1, 2, 5$

$ 1 + 2 + 5 = 8 \\[3ex] 8 \ngtr 10 \\[3ex] $ Positive integer factors (exclude $12$) of $12 = 1, 2, 3, 4, 6$

$ 1 + 2 + 3 + 4 + 6 = 16 \\[3ex] 16 \gt 12 \\[3ex] $ Only one of those integers, $12$ is abundant.

Positive integer factors (exclude $6$) of $6 = 1, 2, 3$

$ 1 + 2 + 3 = 6 \\[3ex] 6 \ngtr 6 \\[3ex] $ Positive integer factors (exclude $8$) of $8 = 1, 2, 4$

$ 1 + 2 + 4 = 7 \\[3ex] 7 \ngtr 8 \\[3ex] $ Positive integer factors (exclude $10$) of $10 = 1, 2, 5$

$ 1 + 2 + 5 = 8 \\[3ex] 8 \ngtr 10 \\[3ex] $ Positive integer factors (exclude $12$) of $12 = 1, 2, 3, 4, 6$

$ 1 + 2 + 3 + 4 + 6 = 16 \\[3ex] 16 \gt 12 \\[3ex] $ Only one of those integers, $12$ is abundant.

(8.) **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?

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.

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.

(9.) **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 $

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 $

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

(10.) **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.

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.