Fractions, Ratios, and Proportions

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 a wrong answer.

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

Solve all questions.
Use at least two (two or more) methods whenever applicable.
Show all work.
For all my applicable students: Calculators ARE NOT allowed for all questions.

(1.) CSEC Using a calculator, or otherwise, calculate the EXACT value of:

$ \dfrac{3\dfrac{1}{2} * 1\dfrac{2}{3}}{4\dfrac{1}{5}} \\[5ex] $

$ \dfrac{3\dfrac{1}{2} * 1\dfrac{2}{3}}{4\dfrac{1}{5}} \\[10ex] \underline{Numerator} \\[3ex] 3\dfrac{1}{2} = \dfrac{2 * 3 + 1}{2} = \dfrac{6 + 1}{2} = \dfrac{7}{2} \\[5ex] 1\dfrac{2}{3} = \dfrac{3 * 1 + 2}{3} = \dfrac{3 + 2}{3} = \dfrac{5}{3} \\[5ex] 3\dfrac{1}{2} * 1\dfrac{2}{3} \\[5ex] = \dfrac{7}{2} * \dfrac{5}{3} \\[5ex] = \dfrac{7 * 5}{2 * 3} \\[5ex] = \dfrac{35}{6} \\[5ex] \underline{Denominator} \\[3ex] 4\dfrac{1}{5} = \dfrac{5 * 4 + 1}{5} = \dfrac{20 + 1}{5} = \dfrac{21}{5} \\[5ex] \underline{Entire\:\:Question} \\[3ex] Quotient = \dfrac{35}{6} \div \dfrac{21}{5} \\[5ex] = \dfrac{35}{6} * \dfrac{5}{21} \\[5ex] = \dfrac{35 * 5}{6 * 21} \\[5ex] = \dfrac{5 * 5}{6 * 3} \\[5ex] = \dfrac{25}{18} \\[5ex] = 1\dfrac{7}{18} $

(2.) ACT What is the least common denominator of the fractions

$ \dfrac{4}{21}, \dfrac{1}{12},\:\:and\:\:\dfrac{3}{8} \\[5ex] F.\:\: 56 \\[3ex] G.\:\: 168 \\[3ex] H.\:\: 252 \\[3ex] J.\:\: 672 \\[3ex] K.\:\: 2,016 \\[3ex] $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCD.
Then, include the rest.

$ Denominators = 21, 12, 8 \\[3ex] 8 = \color{darkblue}{2} * \color{purple}{2} * 2 \\[3ex] 12 = \color{darkblue}{2} * \color{purple}{2} * \color{black}{3} \\[3ex] 21 = \color{black}{3} * 7 \\[5ex] LCD = \color{darkblue}{2} * \color{purple}{2} * \color{black}{3} * 2 * 7 \\[3ex] LCD = 168 $

(3.) CSEC Using a calculator, or otherwise, calculate

$ 5\dfrac{1}{2} \div 3\dfrac{2}{3} + 1\dfrac{4}{5} \\[5ex] $ giving your answer as a fraction in its lowest terms

$ PEMDAS \\[3ex] 5\dfrac{1}{2} \div 3\dfrac{2}{3} + 1\dfrac{4}{5} \\[7ex] 5\dfrac{1}{2} = \dfrac{2 * 5 + 1}{2} = \dfrac{10 + 1}{2} = \dfrac{11}{2} \\[5ex] 3\dfrac{2}{3} = \dfrac{3 * 3 + 2}{3} = \dfrac{9 + 2}{3} = \dfrac{11}{3} \\[5ex] 1\dfrac{4}{5} = \dfrac{5 * 1 + 4}{5} = \dfrac{5 + 4}{5} = \dfrac{9}{5} \\[5ex] 5\dfrac{1}{2} \div 3\dfrac{2}{3} \\[5ex] = \dfrac{11}{2} \div \dfrac{11}{3} \\[5ex] = \dfrac{11}{2} * \dfrac{3}{11} \\[5ex] = \dfrac{3}{2} \\[5ex] = 1\dfrac{1}{2} \\[5ex] 1\dfrac{1}{2} + 1\dfrac{4}{5} \\[5ex] We\:\:can\:\:solve\:\:in\:\:two\:\:ways \\[3ex] Use\:\:any\:\:method\:\:you\:\:prefer \\[3ex] \underline{First\:\:Method} \\[3ex] Fractions:\:\: \dfrac{1}{2} + \dfrac{4}{5} \\[5ex] LCD = 10 \\[3ex] = \dfrac{5}{10} + \dfrac{8}{10} \\[5ex] = \dfrac{5 + 8}{10} \\[5ex] = \dfrac{13}{10} \\[5ex] = 1\dfrac{3}{10} \\[5ex] Integers:\:\: 1 + 1 = 2 \\[3ex] Answer:\:\: = 2 + 1\dfrac{3}{10} = 3\dfrac{3}{10} \\[5ex] \therefore 5\dfrac{1}{2} \div 3\dfrac{2}{3} + 1\dfrac{4}{5} = 3\dfrac{3}{10} \\[5ex] \underline{Second\:\:Method} \\[3ex] 1\dfrac{1}{2} + 1\dfrac{4}{5} \\[5ex] = \dfrac{3}{2} + \dfrac{9}{5} \\[5ex] Denominators = 2, 5 \\[3ex] LCD = 10 \\[3ex] = \dfrac{15}{10} + \dfrac{18}{10} \\[5ex] = \dfrac{15 + 18}{10} \\[5ex] = \dfrac{33}{10} \\[5ex] = 3\dfrac{3}{10} \\[5ex] \therefore 5\dfrac{1}{2} \div 3\dfrac{2}{3} + 1\dfrac{4}{5} = 3\dfrac{3}{10} $

(4.) CSEC Evaluate $\dfrac{\dfrac{1}{2} * \dfrac{3}{5}}{3\dfrac{1}{2}}$, giving your answer as fraction in its lowest terms.

$ \dfrac{\dfrac{1}{2} * \dfrac{3}{5}}{3\dfrac{1}{2}} \\[10ex] \underline{Numerator} \\[3ex] \dfrac{1}{2} * \dfrac{3}{5} \\[5ex] = \dfrac{1 * 3}{2 * 5} \\[5ex] = \dfrac{3}{10} \\[5ex] \underline{Denominator} \\[3ex] 3\dfrac{1}{2} = \dfrac{2 * 3 + 1}{2} = \dfrac{6 + 1}{2} = \dfrac{7}{2} \\[5ex] \underline{Entire\:\:Question} \\[3ex] Quotient = \dfrac{3}{10} \div \dfrac{7}{2} \\[5ex] = \dfrac{3}{10} * \dfrac{2}{7} \\[5ex] = \dfrac{3}{5} * \dfrac{1}{7} \\[5ex] = \dfrac{3 * 1}{5 * 7} \\[5ex] = \dfrac{3}{35} $

(5.) ACT

$ \dfrac{1}{4} * \dfrac{2}{5} * \dfrac{3}{6} * \dfrac{4}{7} * \dfrac{5}{8} * \dfrac{6}{9} * \dfrac{7}{n} = ? \\[5ex] F.\:\: 1 \\[3ex] G.\:\: \dfrac{1}{n} \\[5ex] H.\:\: \dfrac{1}{12n} \\[5ex] J.\:\: \dfrac{2}{9n} \\[5ex] K.\:\: \dfrac{6}{17n} \\[5ex] $

$ \dfrac{1}{4} * \dfrac{2}{5} * \dfrac{3}{6} * \dfrac{4}{7} * \dfrac{5}{8} * \dfrac{6}{9} * \dfrac{7}{n} \\[5ex] = \dfrac{1}{2} * \dfrac{1}{1} * \dfrac{1}{2} * \dfrac{1}{1} * \dfrac{1}{1} * \dfrac{1}{3} * \dfrac{1}{n} \\[5ex] = \dfrac{1 * 1 * 1 * 1 * 1 * 1 * 1}{2 * 1 * 2 * 1 * 1 * 3 * n} \\[5ex] = \dfrac{1}{12n} $

(6.) ACT What is the least common denominator of the fractions

$ \dfrac{4}{21}, \dfrac{1}{24},\:\:and\:\:\dfrac{3}{16} \\[5ex] F.\:\: 112 \\[3ex] G.\:\: 336 \\[3ex] H.\:\: 504 \\[3ex] J.\:\: 2,688 \\[3ex] K.\:\: 8,064 \\[3ex] $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCD.
Then, include the rest.

$ Denominators = 21, 24, 16 \\[3ex] 21 = \color{black}{3} * 7 \\[3ex] 24 = \color{darkblue}{2} * \color{purple}{2} * \color{grey}{2} * \color{black}{3} \\[3ex] 16 = \color{darkblue}{2} * \color{purple}{2} * \color{grey}{2} * 2 \\[5ex] LCD = \color{black}{3} * \color{darkblue}{2} * \color{purple}{2} * \color{grey}{2} * 7 * 2 \\[3ex] LCD = 336 $

(7.) CSEC Using a calculator, or otherwise, determine the exact value of

$ \dfrac{3\dfrac{1}{3} - 2\dfrac{3}{5}}{2\dfrac{1}{5}} \\[7ex] $

$ \dfrac{3\dfrac{1}{3} - 2\dfrac{3}{5}}{2\dfrac{1}{5}} \\[10ex] \underline{Numerator} \\[3ex] 3\dfrac{1}{3} - 2\dfrac{3}{5} \\[5ex] \underline{First\:\:Method} \\[3ex] Fractions:\:\: \dfrac{1}{3} - \dfrac{3}{5} \\[3ex] LCD = 15 \\[3ex] = \dfrac{5}{15} - \dfrac{9}{15} \\[5ex] = \dfrac{5 - 9}{15} \\[5ex] = -\dfrac{4}{15}...Not\:\:what\:\:we\:\:want \\[5ex] Borrow\:\:1\:\:from\:\:the\:\:Integer, 3 \\[3ex] Remaining:\:\: 3 - 1 = 2 \\[3ex] Add\:\:that\:\:1\:\:to\:\:\dfrac{1}{3} \\[5ex] 1 + \dfrac{1}{3} \\[5ex] = \dfrac{3}{3} + \dfrac{1}{3} \\[5ex] = \dfrac{3 + 1}{3} \\[5ex] = \dfrac{4}{3} \\[5ex] Fractions\:\:again:\:\: \dfrac{4}{3} - \dfrac{3}{5} \\[5ex] LCD = 15 \\[3ex] = \dfrac{20}{15} - \dfrac{9}{15} \\[5ex] = \dfrac{20 - 9}{15} \\[5ex] = \dfrac{11}{15} \\[5ex] Integers:\:\: 2 - 2 = 0 \\[3ex] Difference = 0 + \dfrac{11}{15} = \dfrac{11}{15} \\[5ex] \underline{Second\:\:Method} \\[3ex] 3\dfrac{1}{3} = \dfrac{3 * 3 + 1}{3} = \dfrac{9 + 1}{3} = \dfrac{10}{3} \\[5ex] 2\dfrac{3}{5} = \dfrac{5 * 2 + 3}{5} = \dfrac{10 + 3}{5} = \dfrac{13}{5} \\[5ex] \dfrac{10}{3} - \dfrac{13}{5} \\[5ex] LCD = 15 \\[3ex] = \dfrac{50}{15} - \dfrac{39}{15} \\[5ex] = \dfrac{50 - 39}{15} \\[5ex] = \dfrac{11}{15} \\[5ex] \underline{Denominator} \\[3ex] 2\dfrac{1}{5} = \dfrac{5 * 2 + 1}{5} = \dfrac{10 + 1}{5} = \dfrac{11}{5} \\[5ex] \underline{Entire\:\:Question} \\[3ex] Quotient = \dfrac{11}{15} \div \dfrac{11}{5} \\[5ex] = \dfrac{11}{15} * \dfrac{5}{11} \\[5ex] = \dfrac{1}{3} $

(8.) CSEC Using a calculator, or otherwise, calculate the exact value of

$ \dfrac{2\dfrac{1}{4} * \dfrac{4}{5}}{\dfrac{3}{5} - \dfrac{1}{2}} \\[5ex] $

$ \dfrac{2\dfrac{1}{4} * \dfrac{4}{5}}{\dfrac{3}{5} - \dfrac{1}{2}} \\[10ex] \underline{Numerator} \\[3ex] 2\dfrac{1}{4} = \dfrac{4 * 2 + 1}{4} = \dfrac{8 + 1}{4} = \dfrac{9}{4} \\[5ex] \dfrac{9}{4} * \dfrac{4}{5} = \dfrac{9}{5} \\[5ex] \underline{Denominator} \\[3ex] \dfrac{3}{5} - \dfrac{1}{2} \\[5ex] = \dfrac{6}{10} - \dfrac{5}{10} \\[5ex] = \dfrac{6 - 5}{10} \\[5ex] = \dfrac{1}{10} \\[5ex] \underline{Entire\:\:Question} \\[3ex] \dfrac{9}{5} \div \dfrac{1}{10} \\[5ex] = \dfrac{9}{5} * \dfrac{10}{1} \\[5ex] = \dfrac{9}{1} * \dfrac{2}{1} \\[5ex] = 9 * 2 \\[3ex] = 18 $

(9.) WASSCE Simplify:

$ \dfrac{1\dfrac{7}{8} * 2\dfrac{2}{5}}{6\dfrac{3}{4} \div \dfrac{3}{4}} \\[7ex] A.\:\: 9 \\[3ex] B.\:\: 4\dfrac{1}{2} \\[5ex] C.\:\: 2 \\[3ex] D.\:\: \dfrac{1}{2} \\[5ex] $

$ \dfrac{1\dfrac{7}{8} * 2\dfrac{2}{5}}{6\dfrac{3}{4} \div \dfrac{3}{4}} \\[10ex] \underline{Numerator} \\[3ex] 1\dfrac{7}{8} = \dfrac{8 * 1 + 7}{8} = \dfrac{8 + 7}{8} = \dfrac{15}{8} \\[5ex] 2\dfrac{2}{5} = \dfrac{5 * 2 + 2}{5} = \dfrac{10 + 2}{5} = \dfrac{12}{5} \\[5ex] \dfrac{15}{8} * \dfrac{12}{5} \\[5ex] = \dfrac{3}{2} * \dfrac{3}{1} \\[5ex] = \dfrac{3 * 3}{2 * 1} \\[5ex] = \dfrac{9}{2} \\[5ex] \underline{Denominator} \\[3ex] 6\dfrac{3}{4} = \dfrac{4 * 6 + 3}{4} = \dfrac{24 + 3}{4} = \dfrac{27}{4} \\[5ex] \dfrac{27}{4} \div \dfrac{3}{4} \\[5ex] = \dfrac{27}{4} * \dfrac{4}{3} \\[5ex] = \dfrac{9}{1} * \dfrac{1}{1} \\[5ex] = 9 * 1 \\[3ex] = 9 \\[3ex] \underline{Entire\:\:Question} \\[3ex] \dfrac{9}{2} \div 9 \\[5ex] = \dfrac{9}{2} \div \dfrac{9}{1} \\[5ex] =\dfrac{9}{2} * \dfrac{1}{9} \\[5ex] = \dfrac{1}{2} * \dfrac{1}{1} \\[5ex] = \dfrac{1}{2} $

(10.) ACT For all real numbers $x$ such that $x \ne 0$, $\dfrac{4}{5} + \dfrac{7}{x} = ?$

$ A.\:\: \dfrac{11}{5x} \\[5ex] B.\:\: \dfrac{28}{5x} \\[5ex] C.\:\: \dfrac{11}{5 + x} \\[5ex] D.\:\: \dfrac{7x + 20}{5 + x} \\[5ex] E.\:\: \dfrac{4x + 35}{5x} \\[5ex] $

$ \dfrac{4}{5} + \dfrac{7}{x} \\[5ex] LCD\:\:of\:\:5\:\:and\:\:x = 5 * x = 5x \\[3ex] = \dfrac{4x}{5x} + \dfrac{35}{5x} \\[5ex] = \dfrac{4x + 35}{5x} $

(11.) CSEC Using a calculator, or otherwise, determine the EXACT value of:

$ 2\dfrac{2}{5} - 1\dfrac{1}{3} + 3\dfrac{1}{2} \\[5ex] $

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

$ 2\dfrac{2}{5} - 1\dfrac{1}{3} + 3\dfrac{1}{2} \\[7ex] PEMDAS \\[3ex] \underline{First\:\:Method} \\[3ex] Fractions:\:\: \dfrac{2}{5} - \dfrac{1}{3} + \dfrac{1}{2} \\[5ex] LCD = 30 \\[3ex] = \dfrac{12}{30} - \dfrac{10}{30} + \dfrac{15}{30} \\[5ex] = \dfrac{12 - 10 + 15}{30} \\[5ex] = \dfrac{17}{30} \\[5ex] Integers:\:\: 2 - 1 + 3 = 4 \\[3ex] Answer:\:\: = 4 + \dfrac{17}{30} = 4\dfrac{17}{30} \\[5ex] \therefore 2\dfrac{2}{5} - 1\dfrac{1}{3} + 3\dfrac{1}{2} = 4\dfrac{17}{30} \\[5ex] \underline{Second\:\:Method} \\[3ex] 2\dfrac{2}{5} = \dfrac{5 * 2 + 2}{5} = \dfrac{10 + 2}{5} = \dfrac{12}{5} \\[5ex] 1\dfrac{1}{3} = \dfrac{3 * 1 + 1}{3} = \dfrac{3 + 1}{3} = \dfrac{4}{3} \\[5ex] 3\dfrac{1}{2} = \dfrac{2 * 3 + 1}{2} = \dfrac{6 + 1}{2} = \dfrac{7}{2} \\[5ex] \dfrac{12}{5} - \dfrac{4}{3} + \dfrac{7}{2} \\[5ex] LCD = 30 \\[3ex] = \dfrac{72}{30} - \dfrac{40}{30} + \dfrac{105}{30} \\[5ex] = \dfrac{72 - 40 + 105}{30} \\[5ex] = \dfrac{137}{30} \\[5ex] = 4\dfrac{17}{30} \\[5ex] \therefore 2\dfrac{2}{5} - 1\dfrac{1}{3} + 3\dfrac{1}{2} = 4\dfrac{17}{30} $

(12.) CSEC Evaluate $\dfrac{\dfrac{1}{2} * \dfrac{3}{5}}{3\dfrac{1}{2}}$, giving your answer as fraction in its lowest terms.

(13.) CSEC Calculate the EXACT value of

$ \left(1\dfrac{3}{4} - \dfrac{1}{8}\right) + \left(\dfrac{5}{6} \div \dfrac{2}{3}\right) \\[5ex] $

$ PEMDAS \\[3ex] \underline{First\:\:Part} \\[3ex] \left(1\dfrac{3}{4} - \dfrac{1}{8}\right) \\[5ex] 1\dfrac{3}{4} = \dfrac{4 * 1 + 3}{4} = \dfrac{4 + 3}{4} = \dfrac{7}{4} \\[5ex] = \dfrac{7}{4} - \dfrac{1}{8} \\[5ex] LCD = 8 \\[3ex] = \dfrac{14}{8} - \dfrac{1}{8} \\[5ex] = \dfrac{14 - 1}{8} \\[5ex] = \dfrac{13}{8} \\[5ex] \underline{Second\:\:Part} \\[3ex] \dfrac{5}{6} \div \dfrac{2}{3} \\[5ex] = \dfrac{5}{6} * \dfrac{3}{2} \\[5ex] = \dfrac{5}{2} * \dfrac{1}{2} \\[5ex] = \dfrac{5 * 1}{2 * 2} \\[5ex] = \dfrac{5}{4} \\[5ex] \underline{First\:\:Part + Second\:\:Part} \\[3ex] \dfrac{13}{8} + \dfrac{5}{4} \\[5ex] LCD = 8 \\[3ex] = \dfrac{13}{8} + \dfrac{10}{8} \\[5ex] = \dfrac{13 + 10}{8} \\[5ex] = \dfrac{23}{8} \\[5ex] = 2\dfrac{7}{8} $

(14.) ACT $8\%$ of $60$ is $\dfrac{1}{5}$ of what number?

$ A.\:\: 0.96 \\[3ex] B.\:\: 12 \\[3ex] C.\:\: 24 \\[3ex] D.\:\: 240 \\[3ex] E.\:\: 3,750 \\[3ex] $

$ 8\%\:\:of\:\:60 \\[3ex] = \dfrac{8}{100} * 60 \\[5ex] = \dfrac{4}{5} * 6 \\[5ex] = \dfrac{24}{5} \\[5ex] \rightarrow \dfrac{24}{5}\:\:is\:\:\dfrac{1}{5}\:\:of\:\:what\:\:number \\[5ex] Let\:\:the\:\:number = n \\[3ex] \dfrac{24}{5} = \dfrac{1}{5} * n \\[5ex] Denominators\:\:are\:\:the\:\:same \\[3ex] Equate\:\:the\:\:numerators \\[3ex] 24 = 1n \\[3ex] n = 24 $

(15.) ACT What is the least common denominator of the fractions $\dfrac{4}{35}$, $\dfrac{1}{77}$, and $\dfrac{3}{22}$

$ A.\:\: 110 \\[3ex] B.\:\: 770 \\[3ex] C.\:\: 2,695 \\[3ex] D.\:\: 8,470 \\[3ex] E.\:\: 59,290 \\[3ex] $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCD.
Then, include the rest.

$ Denominators = 35, 77, 22 \\[3ex] 35 = 5 * \color{black}{7} \\[3ex] 77 = \color{black}{7} * \color{darkblue}{11} \\[3ex] 22 = 2 * \color{darkblue}{11} \\[5ex] LCD = \color{black}{7} * \color{darkblue}{11} * 5 * 2 \\[3ex] LCD = 770 $

(16.) ACT What is the least common multiple of 50, 30, and 70?

$ F.\:\: 50 \\[3ex] G.\:\: 105 \\[3ex] H.\:\: 150 \\[3ex] J.\:\: 1,050 \\[3ex] K.\:\: 105,000 \\[3ex] $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCD.
Then, include the rest.

$ Numbers = 50, 30, 70 \\[3ex] 50 = \color{black}{2} * \color{darkblue}{5} * 5 \\[3ex] 30 = \color{black}{2} * 3 * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 5 * 3 * 7 \\[3ex] LCM = 1,050 $

(17.) ACT $\dfrac{2}{3} - \dfrac{5}{6}\left(\dfrac{2}{5} + \dfrac{1}{10}\right)$ = ?

$ F.\;\; -\dfrac{1}{3} \\[5ex] G.\;\; -\dfrac{1}{9} \\[5ex] H.\;\; -\dfrac{1}{12} \\[5ex] J.\;\; \dfrac{1}{4} \\[5ex] K.\;\; \dfrac{13}{30} \\[5ex] $

$ PEMDAS \\[3ex] \dfrac{2}{3} - \dfrac{5}{6}\left(\dfrac{2}{5} + \dfrac{1}{10}\right) \\[5ex] \underline{Parenthesis} \\[3ex] \dfrac{2}{5} + \dfrac{1}{10} \\[5ex] LCD = 10 \\[3ex] = \dfrac{4}{10} + \dfrac{1}{10} \\[5ex] = \dfrac{4 + 1}{10} \\[5ex] = \dfrac{5}{10} \\[5ex] = \dfrac{1}{2} \\[5ex] \underline{Multiplication} \\[3ex] \dfrac{5}{6} * \dfrac{1}{2} \\[5ex] = \dfrac{5}{12} \\[5ex] \underline{Subtraction} \\[3ex] \dfrac{2}{3} - \dfrac{5}{12} \\[5ex] LCD = 12 \\[3ex] = \dfrac{8}{12} - \dfrac{5}{12} \\[5ex] = \dfrac{8 - 5}{12} \\[5ex] = \dfrac{3}{12} \\[5ex] = \dfrac{1}{4} $

(18.) ACT If the positive integers $x$ and $y$ are relatively prime (their greatest common factor is $1$) and $\dfrac{1}{2} + \dfrac{1}{3} * \dfrac{1}{4} \div \dfrac{1}{5} = \dfrac{x}{y}$, then $x + y = ?$

$ A.\:\: 23 \\[3ex] B.\:\: 25 \\[3ex] C.\:\: 49 \\[3ex] D.\:\: 91 \\[3ex] E.\:\: 132 \\[3ex] $

$ \dfrac{1}{2} + \dfrac{1}{3} * \dfrac{1}{4} \div \dfrac{1}{5} \\[5ex] PEMDAS \\[3ex] \dfrac{1}{2} + \dfrac{1 * 1}{3 * 4} \div \dfrac{1}{5} \\[5ex] = \dfrac{1}{2} + \dfrac{1}{12} \div \dfrac{1}{5} \\[5ex] = \dfrac{1}{2} + \dfrac{1}{12} * \dfrac{5}{1} \\[5ex] = \dfrac{1}{2} + \dfrac{1 * 5}{12 * 1} \\[5ex] = \dfrac{1}{2} + \dfrac{5}{12} \\[5ex] LCD = 12 \\[3ex] = \dfrac{6}{12} + \dfrac{5}{12} \\[5ex] = \dfrac{6 + 5}{12} \\[5ex] = \dfrac{11}{12} = \dfrac{x}{y} \\[5ex] \implies x = 11,\:\: y = 12 \\[3ex] x + y = 11 + 12 = 23 $

(19.) JAMB Simplify $3\dfrac{1}{2} - \left(2\dfrac{1}{3} * 1\dfrac{1}{4}\right) + \dfrac{3}{5}$

$ A.\:\: 2\dfrac{11}{60} \\[5ex] B.\:\: 2\dfrac{1}{60} \\[5ex] C.\:\: 1\dfrac{11}{60} \\[5ex] D.\:\: 1\dfrac{1}{60} \\[5ex] $

$ PEMDAS \\[3ex] 2\dfrac{1}{3} = \dfrac{3 * 2 + 1}{3} = \dfrac{6 + 1}{3} = \dfrac{7}{3} \\[5ex] 1\dfrac{1}{4} = \dfrac{4 * 1 + 1}{4} = \dfrac{4 + 1}{4} = \dfrac{5}{4} \\[5ex] 2\dfrac{1}{3} * 1\dfrac{1}{4} \\[5ex] = \dfrac{7}{3} * \dfrac{5}{4} \\[5ex] = \dfrac{35}{12} \\[5ex] 3\dfrac{1}{2} = \dfrac{2 * 3 + 1}{2} = \dfrac{6 + 1}{2} = \dfrac{7}{2} \\[5ex] 3\dfrac{1}{2} - \left(2\dfrac{1}{3} * 1\dfrac{1}{4}\right) + \dfrac{3}{5} \\[5ex] = \dfrac{7}{2} - \dfrac{35}{12} + \dfrac{3}{5} \\[5ex] LCD\:\:of\:\:2,12,5 = 60 \\[3ex] = \dfrac{210}{60} - \dfrac{175}{60} + \dfrac{36}{60} \\[5ex] = \dfrac{210 - 175 + 36}{60} \\[5ex] = \dfrac{71}{60} \\[5ex] = 1\dfrac{11}{60} $

(20.) ACT What is the least comon multiple of 3, 5x, 4y, and 6xy?

$ A.\;\; 30xy \\[3ex] B.\;\; 60xy \\[3ex] C.\;\; 60x^2y \\[3ex] D.\;\; 120xy \\[3ex] E.\;\; 120x^2y \\[3ex] $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCD.
Then, include the rest.

$ Numbers = 3, 5x, 4y, 6xy \\[3ex] 3 = \color{black}{3} \\[3ex] 5x = 5 * \color{darkblue}{x} \\[3ex] 4y = \color{green}{2} * 2 * \color{purple}{y} \\[3ex] 6xy = \color{green}{2} * \color{black}{3} * \color{darkblue}{x} * \color{purple}{y} \\[5ex] LCM = \color{black}{3} * \color{green}{2} * \color{darkblue}{x} * \color{purple}{y} * 5 * 2 \\[3ex] LCM = 60xy $

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(22.) ACT What is the least common multiple of 70, 60, and 50?

$ F.\:\: 60 \\[3ex] G.\:\: 180 \\[3ex] H.\:\: 210 \\[3ex] J.\:\: 2,100 \\[3ex] K.\:\: 210,000 \\[3ex] $

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 = 70, 60, 50 \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[3ex] 60 = \color{black}{2} * 2 * 3 * \color{darkblue}{5} \\[3ex] 50 = \color{black}{2} * \color{darkblue}{5} * 5 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 7 * 2 * 3 * 5 \\[3ex] LCM = 2,100 $

(24.) ACT What is the least common denominator of the fractions $\dfrac{4}{15}$, $\dfrac{1}{20}$, and $\dfrac{3}{8}$

$ A.\:\: 24 \\[3ex] B.\:\: 120 \\[3ex] C.\:\: 300 \\[3ex] D.\:\: 480 \\[3ex] E.\:\: 2,400 \\[3ex] $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCD.
Then, include the rest.

$ Denominators = 15, 20, 8 \\[3ex] 15 = 3 * \color{black}{5} \\[3ex] 20 = \color{darkblue}{2} * \color{purple}{2} * \color{black}{5} \\[3ex] 8 = \color{darkblue}{2} * \color{purple}{2} * 2 \\[5ex] LCD = \color{darkblue}{2} * \color{purple}{2} * \color{black}{5} * 3 * 2 \\[3ex] LCD = 120 $

(26.) ACT What is the least common multiple of 60, 70, and 90?

$ F.\:\: 60 \\[3ex] G.\:\: 220 \\[3ex] H.\:\: 630 \\[3ex] J.\:\: 1,260 \\[3ex] K.\:\: 378,000 \\[3ex] $

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 = 60, 70, 90 \\[3ex] 60 = \color{black}{2} * 2 * \color{purple}{3} * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[3ex] 90 = \color{black}{2} * 3 * \color{purple}{3} * \color{darkblue}{5} \\[5ex] LCM = \color{black}{2} * \color{purple}{3} * \color{darkblue}{5} * 2 * 7 * 3 \\[3ex] LCM = 1,260 $

(27.) ACT There are 100 fractions in the following set.

$ \left\{ \dfrac{1}{4}, \dfrac{4}{7}, \dfrac{7}{10}, \dfrac{10}{13}, ... , \dfrac{292}{295}, \dfrac{295}{298}, \dfrac{298}{301} \right\} $

Each fraction after the first is found by adding 3 to the preceding fraction's numerator and denominator.
What is the product of these 100 fractions?

$ A.\;\; 1 \\[3ex] B.\;\; \dfrac{1}{3} \\[5ex] C.\;\; \dfrac{1}{4} \\[5ex] D.\;\; \dfrac{1}{100} \\[5ex] E.\;\; \dfrac{1}{301} \\[5ex] $

Product means multiplication
If you observe the sequence:
the denominator of the first fraction completely divides the numerator of the second fraction (because they are the same numbers)
the denominator of the second fraction completely divides the numerator of the third fraction (because they are the same numbers)
...and so on and so forth ...up until
the denominator of the ninety-ninth fraction completely divides the numerator of the one-hundredth fraction (because they are the same numbers)
So, we are left with:
the numerator of the first fraction and the denominator of the one-hundredth fraction

$ \dfrac{1}{\cancel{4}} * \dfrac{\cancel{4}}{\cancel{7}} * ... * \dfrac{\cancel{295}}{\cancel{298}} * \dfrac{\cancel{298}}{301} \\[5ex] = \dfrac{1}{301} $

(32.) ACT The least common multiple (LCM) of 2 numbers is 216.
The larger of the 2 numbers is 108.
What is the greatest value the other number can have?

$ A.\;\; 2 \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 36 \\[3ex] D.\;\; 54 \\[3ex] E.\;\; 72 \\[3ex] $

Because the ACT is a timed test, my advice in solving this question is to begin with the highest number
Why? Because the question is asking for the greatest value.
Begin with the highest number and if it does not work, try the next higher number, and keep going that way through the options.
The highest number = 72
So, let us find the LCM of 108 and 72

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 = 108, 72 \\[3ex] 108 = \color{black}{2} * \color{darkblue}{2} * 3 * \color{purple}{3} * \color{green}{3} \\[3ex] 72 = \color{black}{2} * \color{darkblue}{2} * 2 * \color{purple}{3} * \color{green}{3} \\[5ex] LCM = \color{black}{2} * \color{darkblue}{2} * \color{purple}{3} * \color{green}{3} * 3 * 2 \\[3ex] LCM = 216 \\[3ex] $ This works.
The other number is 72

(38.) ACT If $x = \dfrac{3}{4} + \dfrac{4}{3}$, $y = \dfrac{2}{3} + \dfrac{3}{2}$, and $z = 1 + 1$, which of the following orders x, y, and z from least to greatest?

$ A.\;\; x \lt y \lt z \\[3ex] B.\;\; y \lt x \lt z \\[3ex] C.\;\; y \lt z \lt x \\[3ex] D.\;\; z \lt x \lt y \\[3ex] E.\;\; z \lt y \lt x \\[3ex] $

Write all the sums as fractions with the same LCD

$ x = \dfrac{3}{4} + \dfrac{4}{3} \\[5ex] LCD = 12 \\[3ex] x = \dfrac{9}{12} + \dfrac{16}{12} = \dfrac{9 + 16}{12} \\[5ex] x = \dfrac{25}{12} \\[5ex] y = \dfrac{2}{3} + \dfrac{3}{2} \\[5ex] LCD = 6 \\[3ex] y = \dfrac{4}{6} + \dfrac{9}{6} = \dfrac{13}{6} = \dfrac{26}{12} \\[5ex] y = \dfrac{26}{12} \\[5ex] z = 1 + 1 \\[3ex] z = 2 = \dfrac{24}{12} \\[5ex] z = \dfrac{24}{12} \\[5ex] \implies \\[3ex] \dfrac{24}{12} \lt \dfrac{25}{12} \lt \dfrac{26}{12} \\[5ex] \therefore z \lt x \lt y $

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(42.) ACT The largest square in the figure below is partitioned into 36 congruent smaller squares.
What fraction of the interior of the largest square is black?

$ F.\;\; \dfrac{5}{12} \\[5ex] G.\;\; \dfrac{1}{2} \\[5ex] H.\;\; \dfrac{7}{12} \\[5ex] J.\;\; \dfrac{5}{8} \\[5ex] K.\;\; \dfrac{5}{6} \\[5ex] $

6 by 6 = 36 congruent squares
1st row = all white
2nd row = 5 white, 1 black
3rd row = 4 white, 2 black
4th row = 3 white, 3 black
5th row = 2 white, 4 black
6th row = 1 white, 5 black

$ number\;\;of\;\;black\;\;squares = 1 + 2 + 3 + 4 + 5 = 15 \\[3ex] fraction\;\;that\;\;is\;\;black = \dfrac{15}{36} = \dfrac{5}{12} $

(46.) ACT What is the least common multiple of the numbers 1, 2, 3, 4, 5, and 6?

$ A.\;\; 720 \\[3ex] B.\;\; 180 \\[3ex] C.\;\; 60 \\[3ex] D.\;\; 30 \\[3ex] E.\;\; 1 \\[3ex] $

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 = 1, 2, 3, 4, 5, 6 \\[3ex] 1 = 1 \\[3ex] 2 = \color{black}{2} \\[3ex] 3 = \color{darkblue}{3} \\[3ex] 4 = \color{black}{2} * 2 \\[3ex] 5 = 5 \\[3ex] 6 = \color{black}{2} * \color{darkblue}{3} \\[5ex] LCM = \color{black}{2} * \color{darkblue}{3} * 1 * 2 * 5 \\[3ex] LCM = 60 $

(50.) ACT What is the least common multiple of 30, 20, and 70?

$ F.\:\: 40 \\[3ex] G.\:\: 42 \\[3ex] H.\:\: 120 \\[3ex] J.\:\: 420 \\[3ex] K.\:\: 42,000 \\[3ex] $

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 = 30, 20, 70 \\[3ex] 30 = \color{black}{2} * 3 * \color{darkblue}{5} \\[3ex] 20 = \color{black}{2} * 2 * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 3 * 2 * 7 \\[3ex] LCM = 420 $

(54.) ACT What is the least common multiple of 80, 70, and 30?

$ F.\:\: 60 \\[3ex] G.\:\: 168 \\[3ex] H.\:\: 180 \\[3ex] J.\:\: 1,680 \\[3ex] K.\:\: 168,000 \\[3ex] $

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 = 80, 70, 30 \\[3ex] 80 = \color{black}{2} * 2 * 2 * 2 * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[3ex] 30 = \color{black}{2} * 3 * \color{darkblue}{5} \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 2 * 2 * 2 * 7 * 3 \\[3ex] LCM = 1680 $

(60.) ACT For all values of x where the expression is defined, $ \dfrac{\dfrac{3}{x - 5}}{1 - \dfrac{2}{x - 5}} = ? \\[9ex] A.\;\; -3 \\[3ex ] B.\;\; -\dfrac{3}{2} \\[5ex] C.\;\; -\dfrac{3}{x^2 - 25} \\[5ex] D.\;\; \dfrac{3}{x - 7} \\[5ex] E.\;\; \dfrac{3}{x - 3} \\[5ex] $

$ \underline{Denominator} \\[3ex] 1 - \dfrac{2}{x - 5} \\[5ex] \dfrac{x - 5}{x - 5} - \dfrac{2}{x - 5} \\[5ex] \dfrac{(x - 5) - 2}{x - 5} \\[5ex] \dfrac{x - 5 - 2}{x - 5} \\[5ex] \dfrac{x - 7}{x - 5} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = Numerator \div Denominator \\[3ex] = \dfrac{3}{x - 5} \div \dfrac{x - 7}{x - 5} \\[5ex] = \dfrac{3}{x - 5} * \dfrac{x - 5}{x - 7} \\[5ex] = \dfrac{3}{x - 7} $

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