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 Chukwuemeka

# Fractions, Ratios, and Proportions

Calculators for Fractions

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.

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.

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

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

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

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

$Numbers = 50, 30, 70 \\[3ex] 50 = \color{black}{2} * \color{darkblue}{5} * 5 \\[3ex] 30 = \color{black}{2} * 3 * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[3ex] LCM = \color{black}{2} * \color{darkblue}{5} * 5 * 3 * 7 \\[3ex] LCM = 1,050$
(17.) JAMB Simplify $3\dfrac{1}{2} - \left(2\dfrac{1}{3} * 1\dfrac{1}{4}\right) + \dfrac{3}{5}$

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

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

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

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