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

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

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

# Solved Examples: Numbers For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

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

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

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

Attempt all questions.
Use at least two (two or more) methods whenever applicable.
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 $|a| - |b|$?

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

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:
$|a| - |b| \\[3ex] = (+a) - (-b) \\[3ex] = a - (-b) \\[3ex] = a + b$
(2.) ACT If $x \lt y$ and $y \lt 4$, then what is the greatest possible integer value of $x + y$

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

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

$\underline{Arithmetic\:\:Method:\:\:Test\:\:Numbers} \\[3ex] y \lt 4 \\[3ex] Let\:\: y = 3.9 \\[3ex] x \lt y \\[3ex] Let\:\: x = 3.8 \\[3ex] x + y = 3.9 + 3.8 = 7.7 \\[3ex]$ Greatest possible integer value = $7$

Student: Why is the greatest possible integer value not equal to $8$?
$Teacher:\:\: 7.7 \ne 8 \\[3ex] 7.7 \gt 7 \\[3ex] 7.7 \lt 8$
(3.) Simplify the expression using the order of operations

$\dfrac{(6 - 7)^2 - 3|4 - 9|}{128 - 2 * 6^2} \\[5ex]$

$\dfrac{(6 - 7)^2 - 3|4 - 9|}{128 - 2 * 6^2} \\[5ex] \underline{Numerator} \\[3ex] (6 - 7)^2 - 3|4 - 9| \\[3ex] = (-1)^2 - 3|-5| \\[3ex] = (-1)(-1) - 3(5) \\[3ex] = 1 - 15 \\[3ex] = -14 \\[3ex] \underline{Denominator} \\[3ex] 128 - 2 * 6^2 \\[3ex] = 128 - 2 * 36 \\[3ex] = 128 - 72 \\[3ex] = 56 \\[3ex] \therefore \dfrac{(6 - 7)^2 - 3|4 - 9|}{128 - 2 * 6^2} = \dfrac{-14}{56} \\[5ex] = -\dfrac{1}{4}$
(4.) Evaluate $||-2| - |-10||$

$||-2| - |-10|| \\[3ex] |-2| = 2 \\[3ex] |-10| = 10 \\[3ex] \rightarrow |2 - 10| \\[3ex] = |-8| \\[3ex] = 8$
(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:
$x = 0 \:\:and\:\: y \ne 0$
(6.) CMAT $\sqrt{188 + \sqrt{51 + \sqrt{169}}} = ?$

$1.\:\: 16.4 \\[3ex] 2.\:\: 14.4 \\[3ex] 3.\:\: 16 \\[3ex] 4.\:\: 14 \\[3ex]$

$\sqrt{188 + \sqrt{51 + \sqrt{169}}} \\[3ex] \sqrt{169} = 13 \\[3ex] 51 + 13 = 64 \\[3ex] \sqrt{64} = 8 \\[3ex] 188 + 8 = 196 \\[3ex] \sqrt{196} = 14$
(7.) JAMB Simplify $(\sqrt{0.7} + \sqrt{70})^2$

$A.\:\:70.7 \\[3ex] B.\:\: 84.7 \\[3ex] C.\:\: 217.7 \\[3ex] D.\:\: 168.7 \\[3ex]$

$(\sqrt{0.7} + \sqrt{70})^2 \\[3ex] = (\sqrt{0.7} + \sqrt{70})(\sqrt{0.7} + \sqrt{70}) \\[3ex] \sqrt{0.7} * \sqrt{0.7} = (\sqrt{0.7})^2 = 0.7 \\[3ex] \sqrt{0.7} * \sqrt{70} = \sqrt{0.7 * 70} = \sqrt{0.7 * 7 * 10} = \sqrt{7 * 7} = \sqrt{49} = 7 \\[3ex] \sqrt{70} * \sqrt{0.7} = \sqrt{70 * 0.7} = \sqrt{7 * 10 * 0.7} = \sqrt{7 * 7} = \sqrt{49} = 7 \\[3ex] \sqrt{70} * \sqrt{70} = (\sqrt{70})^2 = 70 \\[3ex] = 0.7 + 7 + 7 + 70 \\[3ex] = 84.7$
(8.) ACT $|9(-6) + 5(4)| = ?$

$A.\:\: -34 \\[3ex] B.\:\: 12 \\[3ex] C.\:\: 23 \\[3ex] D.\:\: 34 \\[3ex] E.\:\: 74 \\[3ex]$

$|9(-6) + 5(4)| \\[3ex] = |-54 + 20| \\[3ex] = |-34| \\[3ex] = 34$
(9.) Simplify the expression using the order of operations

$\dfrac{5 * 2 - 3^2}{[2^2 - (-7)]^2} \\[5ex]$

$\dfrac{5 * 2 - 3^2}{[2^2 - (-7)]^2} \\[5ex] \underline{Numerator} \\[3ex] 5 * 2 - 3^2 \\[3ex] = 5 * 2 - 9 \\[3ex] = 10 - 9 \\[3ex] = 1 \\[3ex] \underline{Denominator} \\[3ex] [2^2 - (-7)]^2 \\[3ex] = [4 - (-7)]^2 \\[3ex] = [4 + 7]^2 \\[3ex] = 11^2 \\[3ex] = 121 \\[3ex] \therefore \dfrac{5 * 2 - 3^2}{[2^2 - (-7)]^2} = \dfrac{1}{121}$
(10.) Use the order of operations to simplify the expression

$6^2 - 64 \div 4^2 * 7 - 5 \\[3ex]$

$6^2 - 64 \div 4^2 * 7 - 5 \\[3ex] = 36 - 64 \div 16 * 7 - 5 \\[3ex] = 36 - 4 * 7 - 5 \\[3ex] = 36 - 28 - 5 \\[3ex] = 8 - 5 \\[3ex] = 3$
(11.) Use the order of operations to simplify the expression

$4 - 6[-4(5 - 7) - 7(5 - 4)] \\[3ex]$

$4 - 6[-4(5 - 7) - 7(5 - 4)] \\[3ex] = 4 - 6[-4(-2) - 7(1)] \\[3ex] = 4 - 6[8 - 7] \\[3ex] = 4 - 6(1) \\[3ex] = 4 - 6 \\[3ex] = -2$
(12.) CSEC Using a calculator, or otherwise, calculate the EXACT value of

$(12.8)^2 - (30 \div 0.375) \\[3ex]$

$(12.8)^2 - (30 \div 0.375) \\[3ex] = 163.84 - 80 \\[3ex] = 83.84$
(13.) CMAT $2\:3\:7\:4\:3\:2\:1\:5\:7\:3\:2\:7\:1\:0\:9\:8\:7\:5\:4\:7\:5\:4\:7\:2\:3$
Find the number of $7$ in the given series that are followed by an even number but are not preceded by a prime number?

$1.\:\: 1 \\[3ex] 2.\:\: 2 \\[3ex] 3.\:\: 3\\[3ex] 4.\:\: 4 \\[3ex]$

$1st\:\:7 \\[3ex] Preceed\:\:3 = prime\:\:number \\[3ex] Succeed\:\:4 = even\:\:number \\[3ex] NO \\[3ex] 2nd\:\:7 \\[3ex] Preceed\:\:5 = prime\:\:number \\[3ex] Succeed\:\:3 = odd\:\:number \\[3ex] NO \\[3ex] 3rd\:\:7 \\[3ex] Preceed\:\:2 = prime\:\:number \\[3ex] Succeed\:\:1 = odd\:\:number \\[3ex] NO \\[3ex] 4th\:\:7 \\[3ex] Preceed\:\:8 = not\:\:prime\:\:number \\[3ex] Succeed\:\:5 = odd\:\:number \\[3ex] NO \\[3ex] 5th\:\:7 \\[3ex] Preceed\:\:4 = not\:\:prime\:\:number \\[3ex] Succeed\:\:5 = odd\:\:number \\[3ex] NO \\[3ex] 6th\:\:7 \\[3ex] Preceed\:\:4 = not\:\:prime\:\:number \\[3ex] Succeed\:\:2 = even\:\:number \\[3ex] YES \\[3ex] There\:\:is\:\:only\:\:one\:\:7$
(14.) ACT The square root of a certain number is approximately $9.2371$
The certain number is between what $2$ integers?

$F.\:\: 3\:\:and\:\:4 \\[3ex] G.\:\: 4\:\:and\:\:5 \\[3ex] H.\:\: 9\:\:and\:\:10 \\[3ex] J.\:\: 18\:\:and\:\:19 \\[3ex] K.\:\: 81\:\:and\:\:99 \\[3ex]$

$Let\:\:the\:\:number = x \\[3ex] \sqrt{x} = 9.2371 \\[3ex] Square\:\:both\:\:sides \\[3ex] (\sqrt{x})^2 = 9.2371^2 \\[3ex] x \gt 9^2 \\[3ex] x \lt 10^2 \\[3ex] 9^2 \lt x \lt 10^2 \\[3ex] 81 \lt x \lt 100$
(15.) ACT For all nonzero values of $a$ and $b$, the value of which of the following expressions is always negative?

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

Let us analyze each of the options.
$a$ is any nonzero value
This means that $a$ could be positive or negative

$b$ is any nonzero value
This means that $b$ could be positive or negative

Neither $a$ nor $b$ is zero

$Option\:\:F \\[3ex] a - b \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] 5 - 2 = 3 \\[3ex] 3\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:G \\[3ex] -a - b \\[3ex] Assume\:\: a = -5, b = 2 \\[3ex] -(-5) - 2 = 5 - 2 = 3 \\[3ex] 3\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:H \\[3ex] |a| + |b| \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] |5| + |2| = 5 + 2 = 7 \\[3ex] 7\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:J \\[3ex] |a| - |b| \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] |5| - |2| = 5 - 2 = 3 \\[3ex] 3\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:K \\[3ex] -|a| - |b| \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] -|5| - |2| = -5 - 2 = -7...okay \\[3ex] Assume\:\: a = -5, b = 2 \\[3ex] -|-5| - |2| = -5 - 2 = -7...okay \\[3ex] Assume\:\: a = 5, b = -2 \\[3ex] -|5| - |-2| = -5 - 2 = -7...okay \\[3ex] Assume\:\: a = -5, b = -2 \\[3ex] -|-5| - |-2| = -5 - 2 = -7...okay \\[3ex] Correct!$
(16.) ACT What is the smallest integer greater than $\sqrt{85}$

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

$\sqrt{85} \approx 9.21954 \\[3ex]$ Next integer $\gt 9.21954$ is $10$
(17.) JAMB Without using tables, evaluate $(343)^{\dfrac{1}{3}} * (0.14)^{-1} * (25)^{-\dfrac{1}{2}}$

$A.\:\: 12 \\[3ex] B.\:\: 10 \\[3ex] C.\:\: 8 \\[3ex] D.\:\: 7 \\[3ex]$

$(343)^{\dfrac{1}{3}} * (0.14)^{-1} * (25)^{-\dfrac{1}{2}} \\[7ex] (343)^{\dfrac{1}{3}} = \sqrt{343} = 7 \\[7ex] (0.14)^{-1} = \dfrac{1}{(0.14)^1} = \dfrac{1}{0.14} \\[5ex] (25)^{-\dfrac{1}{2}} = \dfrac{1}{(25)^{\dfrac{1}{2}}} \\[7ex] (25)^{\dfrac{1}{2}} = \sqrt{25} = 5 \\[3ex] \rightarrow (25)^{-\dfrac{1}{2}} = \dfrac{1}{5} \\[5ex] = 7 * \dfrac{1}{0.14} * \dfrac{1}{5} \\[5ex] = \dfrac{7 * 1 * 1}{0.14 * 5} \\[5ex] = \dfrac{7 * 100}{0.14 * 100 * 5} \\[5ex] = \dfrac{7 * 100}{14 * 5} \\[5ex] = \dfrac{1 * 20}{2 * 1} \\[5ex] = \dfrac{20}{2} \\[5ex] = 10$
(18.) ACT What is $|5 - x|$ when $x = 9?$

$A.\:\: -14 \\[3ex] B.\:\: -4 \\[3ex] C.\:\: 4 \\[3ex] D.\:\: 9 \\[3ex] E.\:\: 14 \\[3ex]$

$|5 - x| \\[3ex] x = 9 \\[3ex] = |5 - 9| \\[3ex] = |-4| \\[3ex] = 4$
(19.) Evaluate the expression: $3(-9)(2 - 10 - 2(10))$

$3(-9)(2 - 10 - 2(10)) \\[3ex] = -27(2 - 10 - 20) \\[3ex] = -27(-8 - 20) \\[3ex] = -27(-28) \\[3ex] = 756$
(20.) Simplify $5[2 + 2(3 * 9 - 19)]$

$5[2 + 2(3 * 9 - 19)] \\[3ex] = 5[2 + 2(27 - 19)] \\[3ex] = 5[2 + 2(8)] \\[3ex] = 5[2 + 16] \\[3ex] = 5 \\[3ex] = 90$

(21.) 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?

$A.\:\: 0 \\[3ex] B.\:\: 1 \\[3ex] C.\:\: 2 \\[3ex] D.\:\: 3 \\[3ex] E.\:\: 4 \\[3ex]$

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.
The answer is Option $B$
(22.) ACT Which of the following numbers has the greatest value?

$A.\:\: 0.\bar{3} \\[3ex] B.\:\: 0.3 \\[3ex] C.\:\: 0.33 \\[3ex] D.\:\: 0.333 \\[3ex] E.\:\: 0.3333 \\[3ex]$

$0.\bar{3} = 0.33333\bar{3} \\[3ex] 0.33333\bar{3} \gt 0.3333 \gt 0.333 \gt 0.33 \gt 0.3 \\[3ex] \therefore Greatest\:\:value = 0.\bar{3}$
(23.) ACT What is the value of the expression

$\dfrac{|-3 - 2|^2 + (-1)^3}{16 \div 4 * 2 - 5}? \\[5ex] F.\:\: -8 \\[3ex] G.\:\: -\dfrac{2}{3} \\[5ex] H.\:\: \dfrac{2}{3} \\[5ex] J.\:\: \dfrac{26}{3} \\[5ex] K.\:\: 8 \\[3ex]$

$\dfrac{|-3 - 2|^2 + (-1)^3}{16 \div 4 * 2 - 5} \\[5ex] \underline{Numerator} \\[3ex] |-3 - 2|^2 \\[3ex] = |-5|^2 \\[3ex] = 5^2 \\[3ex] = 25 \\[3ex] (-1)^3 \\[3ex] = (-1)(-1)(-1) \\[3ex] = -1 \\[3ex] \therefore |-3 - 2|^2 + (-1)^3 \\[3ex] = 25 + -1 \\[3ex] = 25 - 1 \\[3ex] = 24 \\[3ex] \underline{Denominator} \\[3ex] 16 \div 4 * 2 - 5 \\[3ex] = 4 * 2 - 5 \\[3ex] = 8 - 5 \\[3ex] = 3 \\[3ex] \underline{Entire\:\:Question} \\[3ex] = \dfrac{24}{3} \\[5ex] = 8$
(24.) ACT The difference $\dfrac{3}{5} - \dfrac{-1}{3}$ lies in which of the following intervals graphed on the real number line? $\dfrac{3}{5} - \dfrac{-1}{3} \\[5ex] = \dfrac{3}{5} - -\dfrac{1}{3} \\[5ex] = \dfrac{3}{5} + \dfrac{1}{3} \\[5ex] LCD = 15 \\[3ex] = \dfrac{9}{15} + \dfrac{5}{15} \\[5ex] = \dfrac{9 + 5}{15} \\[5ex] = \dfrac{14}{15} \\[5ex] = 0.933333333 \\[3ex] \dfrac{4}{5} = 0.8 \\[3ex]$ Looking at the options, $\dfrac{14}{15}$ is between $\dfrac{4}{5}$ and $1$
(25.) ACT How many integers, but not including, $20$ and $30$ have a prime factorization with exactly $3$ factors that are NOT necessarily unique?
(Note: $1$ is NOT a prime number.)

$F.\:\: 1 \\[3ex] G.\:\: 2 \\[3ex] H.\:\: 3 \\[3ex] J.\:\: 4 \\[3ex] K.\:\: 5 \\[3ex]$

$21 = 3 * 7 \\[3ex] 22 = 2 * 11 \\[3ex] 23 = 23 \\[3ex] 24 = 2 * 2 * 2 * 3 \\[3ex] 25 = 5 * 5 \\[3ex] 26 = 2 * 13 \\[3ex] 27 = 3 * 3 * 3...three\:\:factors \\[3ex] 28 = 2 * 2 * 7 ...three\:\:factors \\[3ex] 29 = 29 \\[3ex]$ Two numbers, $27$ and $28$ have exactly three factors.
(26.) ACT What is the smallest positive integer having exactly $5$ different positive integer divisors?

$A.\:\: 5 \\[3ex] B.\:\: 6 \\[3ex] C.\:\: 12 \\[3ex] D.\:\: 16 \\[3ex] E.\:\: 18 \\[3ex]$

Positive integer divisors of $5 = 1, 5$
Number of positive integer divisors = $2$

Positive integer divisors of $6 = 1, 2, 3, 6$
Number of positive integer divisors = $4$

Positive integer divisors of $12 = 1, 2, 3, 4, 6, 12$
Number of positive integer divisors = $6$

Positive integer divisors of $16 = 1, 2, 4, 8, 16$
Number of positive integer divisors = $5$

Positive integer divisors of $18 = 1, 2, 3, 6, 9, 18$
Number of positive integer divisors = $6$

The correct option is $D.$
(27.) ACT Which of the following arranges the numbers

$\dfrac{9}{5}, 1.\overline{8}, 1.08$, and $1.\overline{08}$ into ascending order? (Note: The overbar notation shows that the digits under the bar will repeat. For example, $1.\overline{73} = 1.737373...$)

$F.\:\: \dfrac{9}{5} \lt 1.\overline{08} \lt 1.08 \lt 1.\overline{8} \\[5ex] G.\:\: \dfrac{9}{5} \lt 1.08 \lt 1.\overline{08} \lt 1.\overline{8} \\[5ex] H.\:\: 1.\overline{08} \lt 1.08 \lt \dfrac{9}{5} \lt 1.\overline{8} \\[5ex] J.\:\: 1.08 \lt 1.\overline{08} \lt 1.\overline{8} \lt \dfrac{9}{5} \\[5ex] K.\:\: 1.08 \lt 1.\overline{08} \lt \dfrac{9}{5} \lt 1.\overline{8} \\[5ex]$

Ascending order means from least to greatest

$\dfrac{9}{5}, 1.\overline{8}, 1.08, 1.\overline{08} \\[5ex] \dfrac{9}{5} = 1.8 \\[5ex] 1.\overline{8} = 1.888888888888... \\[3ex] 1.08 \\[3ex] 1.\overline{08} = 1.0808080808080808... \\[3ex] 1.88 \gt 1.8 \\[3ex] 1.8 \gt 1.0808 \\[3ex] 1.0808 \gt 1.08 \\[3ex] 1.08 \lt 1.0808 \lt 1.8 \lt 1.88 \\[3ex] \therefore 1.08 \lt 1.\overline{08} \lt \dfrac{9}{5} \lt 1.\overline{8}$
(28.) ACT Walter recently vacationed in Paris.
While there, he visited the Louvre, a famous art museum.
Afterward, he took a $3.7-kilometer$ cab ride from the Louvre to the Eiffel Tower.
A tour guide named Amélie informed him that $2.5$ million rivets were used to build the tower, which stands $320$ meters tall.

When written in scientific notation, the number of rivets used to build the Eiffel Tower is equal to which of the following expressions?

$A.\:\: 2.5 * 10^6 \\[3ex] B.\:\: 2.5 * 10^7 \\[3ex] C.\:\: 2.5 * 10^8 \\[3ex] D.\:\: 25 * 10^6 \\[3ex] E.\:\: 25 * 10^7 \\[3ex]$

$2.5$ million is two million, five hundred thousand

$2.5\:\:million \\[3ex] = 2,500,000 \\[3ex] = 2.5 * 10^6$
(29.) ACT Let $a$ and $b$ be real numbers.
If $(a + b)^2 = a^2 + b^2$, it must be true that:

$A.$ either $a$ or $b$ is zero
$B.$ both $a$ and $b$ are zero.
$C.$ both $a$ and $b$ are positive.
$D.$ $a$ is positive and $b$ is negative.
$E.$ $a$ is negative and $b$ is positive.

$Real\:\:Numbers\:\:includes\:\:positive\:\:numbers,\:\:negative\:\:numbers,\:\:and\:\:zero \\[3ex] \underline{Arithmetic\:\:Method:\:\:Test\:\:Numbers} \\[3ex] If\:\:(a + b)^2 = a^2 + b^2 \\[3ex] Option\:A:\:\:either\:\:a\:\:or\:\:b\:\: = 0 \\[3ex] This\:\:means\:\:that\:\:a\:\:could\:\:be\:\:0\:\:OR\:\:b\:\:could\:\:be\:\:0\:\:OR\:\:both\:\:a\:\:and\:\:b\:are\:\:zeros \\[3ex] Test\:\:a = 0,\:\: b = -3 \\[3ex] LHS:\:\:(a + b)^2 \implies (0 + -3)^2 = (0 - 3)^2 = (-3)^2 = 9 \\[3ex] RHS:\:\:a^2 + b^2 \implies 0^2 + (-3)^2 = 0 + 9 = 9 \\[3ex] Test\:\:a = 3,\:\: b = 0 \\[3ex] LHS:\:\:(a + b)^2 \implies (3 + 0)^2 = (3)^2 = 9 \\[3ex] RHS:\:\:a^2 + b^2 \implies 3^2 + 0^2 = 9 + 0 = 9 \\[3ex] Test\:\:a = 0,\:\: b = 0 \\[3ex] LHS:\:\:(a + b)^2 \implies (0 + 0)^2 = 0^2 = 0 \\[3ex] RHS:\:\:a^2 + b^2 \implies 0^2 + 0^2 = 0 + 0 = 0 \\[3ex] Works...but\:\:let\:\:us\:\:analyze\:\:the\:\:other\:\:options \\[3ex] Option\:B:\:\:both\:\:a\:\:and\:\:b\:\: = 0 \\[3ex] This\:\:option\:\:is\:\:included\:\:in\:\:Option\:\:A \\[3ex] So,\:\:it\:\:cannot\:\:be\:\:the\:\:answer \\[3ex] Option\:C:\:\:both\:\:a\:\:and\:\:b\:\:are\:\:positive \\[3ex] Test\:\:a = 1,\:\: b = 3 \\[3ex] LHS:\:\:(a + b)^2 \implies (1 + 3)^2 = 4^2 = 16 \\[3ex] RHS:\:\:a^2 + b^2 \implies 1^2 + 3^2 = 1 + 9 = 10 \\[3ex] 16 \ne 10 \\[3ex] This\:\:does\:\:not\:\:work...NO \\[3ex] Option\:D:\:\:a\:\:is\:\:positive\:\:and\:\:b\:\:is\:\:negative \\[3ex] Test\:\:a = 1,\:\: b = -3 \\[3ex] LHS:\:\:(a + b)^2 \implies (1 + -3)^2 = (1 - 3)^2 = (-2)^2 = 4 \\[3ex] RHS:\:\:a^2 + b^2 \implies 1^2 + (-3)^2 = 1 + 9 = 10 \\[3ex] 4 \ne 10 \\[3ex] This\:\:does\:\:not\:\:work...NO \\[3ex] Option\:E:\:\:a\:\:is\:\:negative\:\:and\:\:b\:\:is\:\:positive \\[3ex] Test\:\:a = -1,\:\: b = 3 \\[3ex] LHS:\:\:(a + b)^2 \implies (-1 + 3)^2 = 2^2 = 4 \\[3ex] RHS:\:\:a^2 + b^2 \implies (-1)^2 + 3^2 = 1 + 9 = 10 \\[3ex] 4 \ne 10 \\[3ex] This\:\:does\:\:not\:\:work...NO \\[3ex] \therefore the\:\:correct\:\:option\:\:is\:\:A$
(30.) ACT What is the value of $\left(9^{\dfrac{1}{2}} + 16^{\dfrac{1}{2}}\right)^2 ?$

$A.\:\: 7 \\[3ex] B.\:\: 25 \\[3ex] C.\:\: 49 \\[3ex] D.\:\: 337 \\[3ex] E.\:\: 625 \\[3ex]$

$\left(9^{\dfrac{1}{2}} + 16^{\dfrac{1}{2}}\right)^2 \\[5ex] = \left(\sqrt{9} + \sqrt{16}\right)^2 \\[3ex] = (3 + 4)^2 \\[3ex] = 7^2 \\[3ex] = 49$
(31.) ACT What is the greatest common factor of $45$, $50$, and $84$?

$A.\;\; 0 \\[3ex] B.\;\; 1 \\[3ex] C.\;\; 2 \\[3ex] D.\;\; 3 \\[3ex] E.\;\; 5 \\[3ex]$

$\underline{Prime\;\;Factorization\;\;Method} \\[3ex] 45 = 3 * 3 * 5 \\[3ex] 50 = 2 * 5 * 5 \\[3ex] 84 = 2 * 2 * 3 * 7 \\[3ex] GCF = 1 \\[3ex]$ THis is because there is no other commom factor of the three numbers.
$1$ is a factor of everything because $1$ * any thing is that thing.
$1$ was not listed in that method because it is not a prime number.
However, $1$ is a factor of those three numbers.
(32.) ACT What is the $358th$ digit after the decimal point in the repeating decimal $0.\overline{3178}?$

$F.\:\: 0 \\[3ex] G.\:\: 3 \\[3ex] H.\:\: 1 \\[3ex] J.\:\: 7 \\[3ex] K.\:\: 8 \\[3ex]$

$0.\overline{3178} \\[3ex] means\:\: 0.\overline{3178317831783178....} \\[3ex] The\:\:digits\:\: 3178 \:\:repeats \\[3ex] Every\:\:4th\:\:digit = 8 \\[3ex] Every\:\:8th\:\:digit = 8 \\[3ex] Every\:\:12th\:\:,16th\:\:,20th\:\:...digits = 8 \\[3ex] Every\:\:multiple\:\:of\:\:4\:\:-th\:\:digit = 8 \\[3ex] For\:\:358th\:\:digit, \\[3ex] \dfrac{358}{4} = 89.5 \\[3ex] Integer\:\:part = 89 \\[3ex] 89 * 4 = 356 \\[3ex] 356th\:\:digit = 8 \\[3ex] Back-to-repeating \\[3ex] 357th\:\:digit = 3 \\[3ex] 358th\:\:digit = 1 \\[3ex] OR \\[3ex] \dfrac{358}{4} = 89.5 \\[3ex] Decimal\:\:part = 0.5 \\[3ex] 0.5 * 4 = 2 \\[3ex] Count\:\:the\:\:2nd\:\:digit\:\:in\:\:the\:\:sequence \\[3ex] The\:\:2nd\:\:digit = 358th\:\:digit \\[3ex] 2nd\:\:digit = 1 \\[3ex] \therefore the\:\: 358th\:\:digit = 1$
(33.) ACT For real numbers $a$, $b$, and $c$ such that $a \gt b \gt c$ and $b \gt 0$, which of the statements below is(are) always true?

$I.\:\: |a| \gt |b| \\[3ex] II.\:\: |a| \gt |c| \\[3ex] III.\:\: |b| \gt |c| \\[3ex] A.\:\: I \:\:only \\[3ex] B.\:\: II \:\:only \\[3ex] C.\:\: I \:\:and\:\: II \:\:only \\[3ex] D.\:\: II \:\:and\:\: III \:\:only \\[3ex] E.\:\: I, \:\:II,\:\:and\:\: III \\[3ex]$

$\underline{First\:\:Method:\:\:Arithmetically} \\[3ex] Try\:\:several\:\:numbers \\[3ex] Real\:\:Numbers\:\:includes\:\:positive\:\:numbers,\:\:negative\:\:numbers,\:\:and\:\:zero \\[3ex] Use\:\:both\:\:positive\:\:and\:\:negative\:\:real\:\:numbers\:\:as\:\:applicable \\[3ex] b \gt 0 \\[3ex] Let\:\: b = 1 \\[3ex] a \gt b \\[3ex] Let\:\: a = 2 \\[3ex] b \gt c \\[3ex] Let\:\: c = -1 \\[3ex] a \gt b \gt c \implies 2 \gt 1 \gt -1 \\[3ex] I.\:\: |a| \gt |b| \\[3ex] |2| \gt |1| ? \\[3ex] 2 \gt 1 ...correct...YES \\[3ex] I\:\:works \\[3ex] II.\:\: |a| \gt |c| \\[3ex] |2| \gt |-1| ? \\[3ex] 2 \gt 1 ...correct \\[3ex] But\:\:what\:\:if\:\:c = -2 \\[3ex] c\:\:can\:\:be\:\:-2\:\:because\:\: b \gt c \implies 1 \gt -2 \\[3ex] Try\:\: c = -2 \\[3ex] |2| \gt |-2| ? \\[3ex] 2 \gt 2 ...NO \\[3ex] II\:\:will\:\:not\:\:work \\[3ex] III.\:\: |b| \gt |c| \\[3ex] |1| \gt |-1| ? \\[3ex] 1 \gt -1...NO \\[3ex] III\:\:will\:\:not\:\:work \\[3ex] Correct\:\:Option = I\:\:only = A$
(34.) ACT Which of the following inequalities orders the numbers $0.2$, $0.03$, and $\dfrac{1}{4}$ from least to greatest?

$F.\:\: 0.2 \lt 0.03 \lt \dfrac{1}{4} \\[5ex] G.\:\: 0.03 \lt 0.2 \lt \dfrac{1}{4} \\[5ex] H.\:\: 0.03 \lt \dfrac{1}{4} \lt 0.2 \\[5ex] J.\:\: \dfrac{1}{4} \lt 0.03 \lt 0.2 \\[5ex] K.\:\: \dfrac{1}{4} \lt 0.2 \lt 0.03 \\[5ex]$

$\dfrac{1}{4} = 0.25 \\[5ex] 0.03 \lt 0.2 \lt 0.25 \\[3ex] \therefore 0.03 \lt 0.2 \lt \dfrac{1}{4}$
(35.)

$\underline{Prime\;\;Factorization\;\;Method} \\[3ex] 45 = 3 * 3 * 5 \\[3ex] 50 = 2 * 5 * 5 \\[3ex] 84 = 2 * 2 * 3 * 7 \\[3ex] GCF = 1 \\[3ex]$ THis is because there is no other commom factor of the three numbers.
$1$ is a factor of everything because $1$ * any thing is that thing.
$1$ was not listed in that method because it is not a prime number.
However, $1$ is a factor of those three numbers.
(36.) ACT If both $x$ and $\left(\dfrac{x}{3} + \dfrac{x}{7} + \dfrac{x}{9}\right)$ are positive integers, what is the least possible value of $x$?

$F.\:\: 21 \\[3ex] G.\:\: 27 \\[3ex] H.\:\: 36 \\[3ex] J.\:\: 63 \\[3ex] K.\:\: 189 \\[3ex]$

We need a positive integer that can be divided by $3$, $7$, and $9$ without a remainder.
The least possible value of that positive integer is the least common multiple of $3$, $7$, and $9$

$\underline{Prime\;\;Factorization\;\;Method} \\[3ex] 3 = \color{black}{3} \\[3ex] 7 = 7 \\[3ex] 9 = \color{black}{3} * 3 \\[3ex] LCM = \color{black}{3} * 7 * 3 = 63 \\[3ex] x = 63$
(37.) ACT Given consecutive positive integers $a$, $b$, $c$, and $d$ such that $a \lt b \lt c \lt d$, which of the following expressions has the greatest value?

$F.\:\: \dfrac{a}{b} \\[5ex] G.\:\: \dfrac{b}{c} \\[5ex] H.\:\: \dfrac{c}{d} \\[5ex] J.\:\: \dfrac{a + b}{b + c} \\[5ex] K.\:\: \dfrac{b + c}{c + d} \\[5ex]$

We shall solve this question by testing with numbers
I am going to solve by fractions because I would like for you to be comfortable with fractions.
However, you may choose to convert to decimals if you want to.

$a, b, c, d \;\;are\;\; positive \;\; integers \\[3ex] a \lt b \lt c \lt d \\[3ex] Let: \\[3ex] a = 1 \\[3ex] b = 2 \\[3ex] c = 3 \\[3ex] d = 4 \\[3ex] Option\;F \\[3ex] \dfrac{a}{b} = \dfrac{1}{2} \\[5ex] Option\;G \\[3ex] \dfrac{b}{c} = \dfrac{2}{3} \\[5ex] Option\;H \\[3ex] \dfrac{c}{d} = \dfrac{3}{4} \\[5ex] Option\;J \\[3ex] \dfrac{a + b}{b + c} = \dfrac{1 + 2}{2 + 3} = \dfrac{3}{5} \\[5ex] Option\;K \\[3ex] \dfrac{b + c}{c + d} = \dfrac{2 + 3}{3 + 4} = \dfrac{5}{7} \\[5ex] LCM \;\; 2, 3, 4, 5, 7 = 4 * 3 * 5 * 7 = 420 \\[3ex] Option\;F:\;\; \dfrac{1}{2} = \dfrac{210}{420} \\[5ex] Option\;G:\;\; \dfrac{2}{3} = \dfrac{280}{420} \\[5ex] Option\;H:\;\; \dfrac{3}{4} = \dfrac{315}{420} \\[5ex] Option\;J:\;\; \dfrac{3}{5} = \dfrac{252}{420} \\[5ex] Option\;K:\;\; \dfrac{5}{7} = \dfrac{300}{420} \\[5ex] Same\;\;Denominator = 420 \\[3ex] Greatest\;\;Numerator = 315 \\[3ex] \therefore Greatest\;\;Value = Option\;H:\;\; \dfrac{c}{d}$
(38.) ACT If both $x$ and $\left(\dfrac{x}{3} + \dfrac{x}{7} + \dfrac{x}{9}\right)$ are positive integers, what is the least possible value of $x$?

$F.\:\: 21 \\[3ex] G.\:\: 27 \\[3ex] H.\:\: 36 \\[3ex] J.\:\: 63 \\[3ex] K.\:\: 189 \\[3ex]$

We need a positive integer that can be divided by $3$, $7$, and $9$ without a remainder.
The least possible value of that positive integer is the least common multiple of $3$, $7$, and $9$

$\underline{Prime\;\;Factorization\;\;Method} \\[3ex] 3 = \color{black}{3} \\[3ex] 7 = 7 \\[3ex] 9 = \color{black}{3} * 3 \\[3ex] LCM = \color{black}{3} * 7 * 3 = 63 \\[3ex] x = 63$
(39.) ACT

We shall solve this question by testing with numbers
I am going to solve by fractions because I would like for you to be comfortable with fractions.
However, you may choose to convert to decimals if you want to.

$a, b, c, d \;\;are\;\; positive \;\; integers \\[3ex] a \lt b \lt c \lt d \\[3ex] Let: \\[3ex] a = 1 \\[3ex] b = 2 \\[3ex] c = 3 \\[3ex] d = 4 \\[3ex] Option\;F \\[3ex] \dfrac{a}{b} = \dfrac{1}{2} \\[5ex] Option\;G \\[3ex] \dfrac{b}{c} = \dfrac{2}{3} \\[5ex] Option\;H \\[3ex] \dfrac{c}{d} = \dfrac{3}{4} \\[5ex] Option\;J \\[3ex] \dfrac{a + b}{b + c} = \dfrac{1 + 2}{2 + 3} = \dfrac{3}{5} \\[5ex] Option\;K \\[3ex] \dfrac{b + c}{c + d} = \dfrac{2 + 3}{3 + 4} = \dfrac{5}{7} \\[5ex] LCM \;\; 2, 3, 4, 5, 7 = 4 * 3 * 5 * 7 = 420 \\[3ex] Option\;F:\;\; \dfrac{1}{2} = \dfrac{210}{420} \\[5ex] Option\;G:\;\; \dfrac{2}{3} = \dfrac{280}{420} \\[5ex] Option\;H:\;\; \dfrac{3}{4} = \dfrac{315}{420} \\[5ex] Option\;J:\;\; \dfrac{3}{5} = \dfrac{252}{420} \\[5ex] Option\;K:\;\; \dfrac{5}{7} = \dfrac{300}{420} \\[5ex] Same\;\;Denominator = 420 \\[3ex] Greatest\;\;Numerator = 315 \\[3ex] \therefore Greatest\;\;Value = Option\;H:\;\; \dfrac{c}{d}$
(40.) ACT Which of the following must be true for each set of $4$ consecutive positive integers?
$I.$ At least $1$ of the $4$ integers is prime.
$II.$ At least $2$ of the $4$ integers have a common prime factor.
$III.$ At least $1$ of the $4$ integers is a factor of at least $1$ of the $3$ other integers.

$F.$ $I$ only
$G.$ $II$ only
$H.$ $I$ and $III$ only
$J.$ $II$ and $III$ only
$K.$ $I$, $II$, and $III$

For this question, we have to find a contradiction to each statement.
In other words, we have to try to prove each statement incorrect.
If we cannot, then that is the correct option.
Try to write several examples of $4$ consecutive positive integers
Let us begin with $I.$
$I.$ At least $1$ of the $4$ integers is prime.
Test Case: $32, 33, 34, 35$...these are $4$ consecutive positive integers
As we can see, all $4$ integers are not prime numbers.
They are all composite numbers.
Hence $I.$ is incorrect.
This eliminates three options: $F.$, $H.$ and $K.$

Next is $II.$
$II.$ At least $2$ of the $4$ integers have a common prime factor.
This is a true statement because $4$ consecutive positive integers must include $2$ even integers
$2$ is a prime factor of every even integer
Therefore, two of the four integers must have a common prime factor, and that common prime factor is $2$
At least two means two or more
So, this statement is true

Then, for $III.$
Using the same example we used for $I.$
Test Case: $32, 33, 34, 35$...these are $4$ consecutive positive integers
$32$ is not a factor of $33$, $34$, or $35$
Neither is any of those integers a factor of any of the other integers.
Hence, this is a false statement.

The only true statement is $II.$
The correct option is $G.$