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

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.

(1.) **ACT** Kaya ran $1\dfrac{2}{5}$ miles on Monday and $2\dfrac{1}{3}$ miles on Tuesday.

What was the total distance, in miles, Kaya ran during those $2$ days?

$ A.\:\: 3\dfrac{2}{15} \\[5ex] B.\:\: 3\dfrac{3}{8} \\[5ex] C.\:\: 3\dfrac{2}{5} \\[5ex] D.\:\: 3\dfrac{7}{15} \\[5ex] E.\:\: 3\dfrac{11}{15} \\[5ex] $

The total distance covered during those $2$ days is the sum of the individual distances.

We can solve this questions in two ways.

Use any method you prefer.

$ Total\:\:distance\:\:for\:\:both\:\:days \\[3ex] = 1\dfrac{2}{5} + 2\dfrac{1}{3} \\[5ex] \underline{First\:\:Method} \\[3ex] Fractions:\:\: \dfrac{2}{5} + \dfrac{1}{3} \\[5ex] = \dfrac{6}{15} + \dfrac{5}{15} \\[5ex] = \dfrac{6 + 5}{15} \\[5ex] = \dfrac{11}{15} \\[5ex] Integers:\:\: 1 + 2 = 3 \\[3ex] Sum = 3 + \dfrac{11}{15} = 3\dfrac{11}{15} \\[5ex] \underline{Second\:\:Method} \\[3ex] 1\dfrac{2}{5} = \dfrac{5 * 1 + 2}{5} = \dfrac{5 + 2}{5} = \dfrac{7}{5} \\[5ex] 2\dfrac{1}{3} = \dfrac{3 * 2 + 1}{3} = \dfrac{6 + 1}{3} = \dfrac{7}{3} \\[5ex] Sum = \dfrac{7}{5} + \dfrac{7}{3} \\[5ex] = \dfrac{21}{15} + \dfrac{35}{15} \\[5ex] = \dfrac{21 + 35}{15} \\[5ex] = \dfrac{56}{15} \\[5ex] = 3\dfrac{11}{15} \\[5ex] $ Kaya ran a total distance of $3\dfrac{11}{15}$ miles during those $2$ days.

What was the total distance, in miles, Kaya ran during those $2$ days?

$ A.\:\: 3\dfrac{2}{15} \\[5ex] B.\:\: 3\dfrac{3}{8} \\[5ex] C.\:\: 3\dfrac{2}{5} \\[5ex] D.\:\: 3\dfrac{7}{15} \\[5ex] E.\:\: 3\dfrac{11}{15} \\[5ex] $

The total distance covered during those $2$ days is the sum of the individual distances.

We can solve this questions in two ways.

Use any method you prefer.

$ Total\:\:distance\:\:for\:\:both\:\:days \\[3ex] = 1\dfrac{2}{5} + 2\dfrac{1}{3} \\[5ex] \underline{First\:\:Method} \\[3ex] Fractions:\:\: \dfrac{2}{5} + \dfrac{1}{3} \\[5ex] = \dfrac{6}{15} + \dfrac{5}{15} \\[5ex] = \dfrac{6 + 5}{15} \\[5ex] = \dfrac{11}{15} \\[5ex] Integers:\:\: 1 + 2 = 3 \\[3ex] Sum = 3 + \dfrac{11}{15} = 3\dfrac{11}{15} \\[5ex] \underline{Second\:\:Method} \\[3ex] 1\dfrac{2}{5} = \dfrac{5 * 1 + 2}{5} = \dfrac{5 + 2}{5} = \dfrac{7}{5} \\[5ex] 2\dfrac{1}{3} = \dfrac{3 * 2 + 1}{3} = \dfrac{6 + 1}{3} = \dfrac{7}{3} \\[5ex] Sum = \dfrac{7}{5} + \dfrac{7}{3} \\[5ex] = \dfrac{21}{15} + \dfrac{35}{15} \\[5ex] = \dfrac{21 + 35}{15} \\[5ex] = \dfrac{56}{15} \\[5ex] = 3\dfrac{11}{15} \\[5ex] $ Kaya ran a total distance of $3\dfrac{11}{15}$ miles during those $2$ days.

(2.) **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

(3.) **ACT** Diana is baking bread, and the original recipe calls for $1\dfrac{1}{2}$ teaspoons
of yeast and $2\dfrac{1}{2}$ cups of flour.

Diana will use the entire contents of a packet that contains $2\dfrac{1}{4}$ teaspoons of yeast and will use the same ratio of ingredients called for in the original recipe.

How many cups of flour will Diana use?

$ F.\:\: 1\dfrac{7}{8} \\[5ex] G.\:\: 3\dfrac{1}{4} \\[5ex] H.\:\: 3\dfrac{1}{2} \\[5ex] J.\:\: 3\dfrac{3}{4} \\[5ex] K.\:\: 4 \\[3ex] $

Original amount of yeast = $1\dfrac{1}{2}$ teaspoons

New amount of yeast = $2\dfrac{1}{4}$ teaspoons

Ratio of New to Original amounts of yeast = $\dfrac{2\dfrac{1}{4}}{1\dfrac{1}{2}}$

Original amount of flour = $2\dfrac{1}{2}$ cups

New amount of yeast = $x$ cups

Ratio of New to Original amounts of yeast = $\dfrac{x}{2\dfrac{1}{2}}$

...same ratio of ingredients means that the ratios are the same

$ \therefore \dfrac{2\dfrac{1}{4}}{1\dfrac{1}{2}} = \dfrac{x}{2\dfrac{1}{2}} \\[7ex] 2\dfrac{1}{4} = \dfrac{4 * 2 + 1}{4} = \dfrac{8 + 1}{4} = \dfrac{9}{4} \\[5ex] 1\dfrac{1}{2} = \dfrac{2 * 1 + 1}{2} = \dfrac{2 + 1}{2} = \dfrac{3}{2} \\[5ex] 2\dfrac{1}{2} = \dfrac{2 * 2 + 1}{2} = \dfrac{4 + 1}{2} = \dfrac{5}{2} \\[5ex] \rightarrow \dfrac{\dfrac{9}{4}}{\dfrac{3}{2}} = \dfrac{x}{\dfrac{5}{2}} \\[7ex] Cross\:\:Multiply \\[3ex] \dfrac{3}{2}x = \dfrac{9}{4} * \dfrac{5}{2} \\[5ex] x = \dfrac{9}{4} * \dfrac{5}{2} * \dfrac{2}{3} \\[5ex] x = \dfrac{15}{4} \\[5ex] x = 3\dfrac{3}{4} \\[5ex] $ Diana will use $3\dfrac{3}{4}$ cups of flour.

Diana will use the entire contents of a packet that contains $2\dfrac{1}{4}$ teaspoons of yeast and will use the same ratio of ingredients called for in the original recipe.

How many cups of flour will Diana use?

$ F.\:\: 1\dfrac{7}{8} \\[5ex] G.\:\: 3\dfrac{1}{4} \\[5ex] H.\:\: 3\dfrac{1}{2} \\[5ex] J.\:\: 3\dfrac{3}{4} \\[5ex] K.\:\: 4 \\[3ex] $

Original amount of yeast = $1\dfrac{1}{2}$ teaspoons

New amount of yeast = $2\dfrac{1}{4}$ teaspoons

Ratio of New to Original amounts of yeast = $\dfrac{2\dfrac{1}{4}}{1\dfrac{1}{2}}$

Original amount of flour = $2\dfrac{1}{2}$ cups

New amount of yeast = $x$ cups

Ratio of New to Original amounts of yeast = $\dfrac{x}{2\dfrac{1}{2}}$

...same ratio of ingredients means that the ratios are the same

$ \therefore \dfrac{2\dfrac{1}{4}}{1\dfrac{1}{2}} = \dfrac{x}{2\dfrac{1}{2}} \\[7ex] 2\dfrac{1}{4} = \dfrac{4 * 2 + 1}{4} = \dfrac{8 + 1}{4} = \dfrac{9}{4} \\[5ex] 1\dfrac{1}{2} = \dfrac{2 * 1 + 1}{2} = \dfrac{2 + 1}{2} = \dfrac{3}{2} \\[5ex] 2\dfrac{1}{2} = \dfrac{2 * 2 + 1}{2} = \dfrac{4 + 1}{2} = \dfrac{5}{2} \\[5ex] \rightarrow \dfrac{\dfrac{9}{4}}{\dfrac{3}{2}} = \dfrac{x}{\dfrac{5}{2}} \\[7ex] Cross\:\:Multiply \\[3ex] \dfrac{3}{2}x = \dfrac{9}{4} * \dfrac{5}{2} \\[5ex] x = \dfrac{9}{4} * \dfrac{5}{2} * \dfrac{2}{3} \\[5ex] x = \dfrac{15}{4} \\[5ex] x = 3\dfrac{3}{4} \\[5ex] $ Diana will use $3\dfrac{3}{4}$ cups of flour.

(4.) **ACT** Of the 804 graduating seniors in a certain high school, approximately $\dfrac{2}{5}$
are going to college and approximately $\dfrac{1}{4}$ of those going to college are going to a state
university.

Which of the following is the closest estimate for how many of the graduating seniors are going to a state university?

$ F.\:\: 80 \\[3ex] G.\:\: 90 \\[3ex] H.\:\: 160 \\[3ex] J.\:\: 200 \\[3ex] K.\:\: 320 \\[3ex] $

"of" means multiplication

We shall be multiplying numbers/fractions

$ Number\:\:of\:\:graduating\:\:seniors = 804 \\[3ex] Going\:\:to\:\:College = \dfrac{2}{5}\:\:of\:\:804 \\[5ex] = \dfrac{2}{5} * 804 \\[5ex] = \dfrac{2 * 804}{5} \\[5ex] Going\:\:to\:\:a\:\:state\:\:university = \dfrac{1}{4}\:\:of\:\:\dfrac{2 * 804}{5} \\[5ex] = \dfrac{1}{4} * \dfrac{2 * 804}{5} \\[5ex] = \dfrac{2 * 201}{5} \\[5ex] = \dfrac{402}{5} \\[5ex] = 80.4 \\[3ex] \approx 81\:\:students...because\:\:these\:\;are\:\:people \\[3ex] We\:\:cannot\:\:have\:\:decimal\:\:number\:\:of\:\:people \\[3ex] Closest\:\:estimate = 80 $

Which of the following is the closest estimate for how many of the graduating seniors are going to a state university?

$ F.\:\: 80 \\[3ex] G.\:\: 90 \\[3ex] H.\:\: 160 \\[3ex] J.\:\: 200 \\[3ex] K.\:\: 320 \\[3ex] $

"of" means multiplication

We shall be multiplying numbers/fractions

$ Number\:\:of\:\:graduating\:\:seniors = 804 \\[3ex] Going\:\:to\:\:College = \dfrac{2}{5}\:\:of\:\:804 \\[5ex] = \dfrac{2}{5} * 804 \\[5ex] = \dfrac{2 * 804}{5} \\[5ex] Going\:\:to\:\:a\:\:state\:\:university = \dfrac{1}{4}\:\:of\:\:\dfrac{2 * 804}{5} \\[5ex] = \dfrac{1}{4} * \dfrac{2 * 804}{5} \\[5ex] = \dfrac{2 * 201}{5} \\[5ex] = \dfrac{402}{5} \\[5ex] = 80.4 \\[3ex] \approx 81\:\:students...because\:\:these\:\;are\:\:people \\[3ex] We\:\:cannot\:\:have\:\:decimal\:\:number\:\:of\:\:people \\[3ex] Closest\:\:estimate = 80 $

(5.) **CSEC** Mark spends $\dfrac{3}{8}$ of his monthly income on housing.

Of the REMAINDER, he spends $\dfrac{1}{3}$ on food and saves what is left.

(i) Calculate the fraction of his monthly income spent on food.

(ii) Calculate the fraction of his monthly income that he saved.

$ \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] Total\:\:Percent = 100\% \\[3ex] \therefore Total\:\:Fraction = 100\% = \dfrac{100}{100} = 1 \\[5ex] \rightarrow Monthly\:\:income = 1 \\[3ex] Spent\:\:on\:\:housing = \dfrac{3}{8}\:\:of\:\:1 = \dfrac{3}{8} * 1 = \dfrac{3}{8} \\[5ex] Remainder = 1 - \dfrac{3}{8} = \dfrac{8}{8} - \dfrac{3}{8} = \dfrac{8 - 3}{8} = \dfrac{5}{8} \\[5ex] (i) \\[3ex] Spent\:\:on\:\:food = \dfrac{1}{3}\:\:of\:\:Remainder \\[3ex] = \dfrac{1}{3} * \dfrac{5}{8} \\[5ex] = \dfrac{1 * 5}{3 * 8} \\[5ex] = \dfrac{5}{24} \\[5ex] (ii) \\[3ex] Saves = rest\:\:of\:\:Remainder \\[3ex] Two\:\:ways\:\:we\:\:can\:\:do\:\:it \\[3ex] First\:\:Way:\:\: Saves = 1 - Spent \\[3ex] Spent = \dfrac{3}{8} + \dfrac{5}{24} \\[5ex] = \dfrac{9}{24} + \dfrac{5}{24} \\[5ex] = \dfrac{14}{24} \\[5ex] Saves = 1 - \dfrac{14}{24} \\[5ex] = \dfrac{24}{24} - \dfrac{14}{24} \\[5ex] = \dfrac{24 - 14}{24} \\[5ex] = \dfrac{10}{24} \\[5ex] = \dfrac{5}{12} \\[5ex] OR \\[3ex] Second\:\:Way:\:\: Saves = \dfrac{2}{3}\:\:of\:\:Remainder \\[5ex] Why\:\:\dfrac{2}{3}? \\[5ex] If\:\:he\:\:spent\:\:\dfrac{1}{3}\:\:of\:\:Remainder\:\:on\:\:food, \\[5ex] This\:\:means\:\:he\:\:saved\:\:\dfrac{2}{3}\:\:of\:\:Remainder \\[5ex] Because\:\:1 - \dfrac{2}{3} = \dfrac{3}{3} - \dfrac{1}{3} = \dfrac{2}{3} \\[5ex] Saves = \dfrac{2}{3} * \dfrac{5}{8} \\[5ex] = \dfrac{1 * 5}{3 * 4} \\[5ex] = \dfrac{5}{12} \\[5ex] \underline{Second\:\:Method:\:\:Algebraically} \\[3ex] Let\:\:the\:\:monthly\:\:income = x \\[3ex] Housing:\:\: \dfrac{3}{8} * x = \dfrac{3x}{8} \\[5ex] Remainder:\:\: x - \dfrac{3x}{8} \\[5ex] = \dfrac{8x}{8} - \dfrac{3x}{8} \\[5ex] = \dfrac{8x - 3x}{8} \\[5ex] = \dfrac{5x}{8} \\[5ex] (i)\:\: \\[3ex] Food:\:\: \dfrac{1}{3} * \dfrac{5x}{8} \\[5ex] = \dfrac{5x}{24} \\[5ex] Fraction\:\:spent\:\:on\:\:food = \dfrac{5}{24} \\[5ex] (ii)\:\: \\[3ex] Save:\:\: \dfrac{5x}{8} - \dfrac{5x}{24} \\[5ex] = \dfrac{15x}{24} - \dfrac{5x}{24} \\[5ex] = \dfrac{10x}{24} \\[5ex] = \dfrac{5x}{12} \\[5ex] Fraction\:\:saved = \dfrac{5}{12} $

Of the REMAINDER, he spends $\dfrac{1}{3}$ on food and saves what is left.

(i) Calculate the fraction of his monthly income spent on food.

(ii) Calculate the fraction of his monthly income that he saved.

$ \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] Total\:\:Percent = 100\% \\[3ex] \therefore Total\:\:Fraction = 100\% = \dfrac{100}{100} = 1 \\[5ex] \rightarrow Monthly\:\:income = 1 \\[3ex] Spent\:\:on\:\:housing = \dfrac{3}{8}\:\:of\:\:1 = \dfrac{3}{8} * 1 = \dfrac{3}{8} \\[5ex] Remainder = 1 - \dfrac{3}{8} = \dfrac{8}{8} - \dfrac{3}{8} = \dfrac{8 - 3}{8} = \dfrac{5}{8} \\[5ex] (i) \\[3ex] Spent\:\:on\:\:food = \dfrac{1}{3}\:\:of\:\:Remainder \\[3ex] = \dfrac{1}{3} * \dfrac{5}{8} \\[5ex] = \dfrac{1 * 5}{3 * 8} \\[5ex] = \dfrac{5}{24} \\[5ex] (ii) \\[3ex] Saves = rest\:\:of\:\:Remainder \\[3ex] Two\:\:ways\:\:we\:\:can\:\:do\:\:it \\[3ex] First\:\:Way:\:\: Saves = 1 - Spent \\[3ex] Spent = \dfrac{3}{8} + \dfrac{5}{24} \\[5ex] = \dfrac{9}{24} + \dfrac{5}{24} \\[5ex] = \dfrac{14}{24} \\[5ex] Saves = 1 - \dfrac{14}{24} \\[5ex] = \dfrac{24}{24} - \dfrac{14}{24} \\[5ex] = \dfrac{24 - 14}{24} \\[5ex] = \dfrac{10}{24} \\[5ex] = \dfrac{5}{12} \\[5ex] OR \\[3ex] Second\:\:Way:\:\: Saves = \dfrac{2}{3}\:\:of\:\:Remainder \\[5ex] Why\:\:\dfrac{2}{3}? \\[5ex] If\:\:he\:\:spent\:\:\dfrac{1}{3}\:\:of\:\:Remainder\:\:on\:\:food, \\[5ex] This\:\:means\:\:he\:\:saved\:\:\dfrac{2}{3}\:\:of\:\:Remainder \\[5ex] Because\:\:1 - \dfrac{2}{3} = \dfrac{3}{3} - \dfrac{1}{3} = \dfrac{2}{3} \\[5ex] Saves = \dfrac{2}{3} * \dfrac{5}{8} \\[5ex] = \dfrac{1 * 5}{3 * 4} \\[5ex] = \dfrac{5}{12} \\[5ex] \underline{Second\:\:Method:\:\:Algebraically} \\[3ex] Let\:\:the\:\:monthly\:\:income = x \\[3ex] Housing:\:\: \dfrac{3}{8} * x = \dfrac{3x}{8} \\[5ex] Remainder:\:\: x - \dfrac{3x}{8} \\[5ex] = \dfrac{8x}{8} - \dfrac{3x}{8} \\[5ex] = \dfrac{8x - 3x}{8} \\[5ex] = \dfrac{5x}{8} \\[5ex] (i)\:\: \\[3ex] Food:\:\: \dfrac{1}{3} * \dfrac{5x}{8} \\[5ex] = \dfrac{5x}{24} \\[5ex] Fraction\:\:spent\:\:on\:\:food = \dfrac{5}{24} \\[5ex] (ii)\:\: \\[3ex] Save:\:\: \dfrac{5x}{8} - \dfrac{5x}{24} \\[5ex] = \dfrac{15x}{24} - \dfrac{5x}{24} \\[5ex] = \dfrac{10x}{24} \\[5ex] = \dfrac{5x}{12} \\[5ex] Fraction\:\:saved = \dfrac{5}{12} $

(6.) **ACT** A formula for the volume $V$ of a sphere with radius $r$ is $V = \dfrac{4}{3}\pi r^3$.

If the radius of a spherical rubber ball is $1\dfrac{1}{4}$ inches, what is its volume to the nearest cubic inch?

$ A.\:\: 5 \\[3ex] B.\:\: 7 \\[3ex] C.\:\: 8 \\[3ex] D.\:\: 16 \\[3ex] E.\:\: 65 \\[3ex] $

$ V = \dfrac{4}{3}\pi r^3 \\[5ex] V = \dfrac{4}{3} * \pi * r * r * r \\[5ex] r = 1\dfrac{1}{4} = \dfrac{4 * 1 + 1}{4} = \dfrac{4 + 1}{4} = \dfrac{5}{4} \\[5ex] \pi = \dfrac{22}{7} \\[5ex] V = \dfrac{4}{3} * \dfrac{22}{7} * \dfrac{5}{4} * \dfrac{5}{4} * \dfrac{5}{4} \\[5ex] = \dfrac{1}{3} * \dfrac{11}{7} * \dfrac{5}{1} * \dfrac{5}{2} * \dfrac{5}{4} \\[5ex] = \dfrac{1 * 11 * 5 * 5 * 5}{3 * 7 * 1 * 2 * 4} \\[5ex] = \dfrac{1375}{168}\:\:cubic\:\:inches....answer\:\:in\:\:fraction \\[5ex] = 8.18452381 \\[3ex] \approx 8\:\:cubic\:\:inches $

If the radius of a spherical rubber ball is $1\dfrac{1}{4}$ inches, what is its volume to the nearest cubic inch?

$ A.\:\: 5 \\[3ex] B.\:\: 7 \\[3ex] C.\:\: 8 \\[3ex] D.\:\: 16 \\[3ex] E.\:\: 65 \\[3ex] $

$ V = \dfrac{4}{3}\pi r^3 \\[5ex] V = \dfrac{4}{3} * \pi * r * r * r \\[5ex] r = 1\dfrac{1}{4} = \dfrac{4 * 1 + 1}{4} = \dfrac{4 + 1}{4} = \dfrac{5}{4} \\[5ex] \pi = \dfrac{22}{7} \\[5ex] V = \dfrac{4}{3} * \dfrac{22}{7} * \dfrac{5}{4} * \dfrac{5}{4} * \dfrac{5}{4} \\[5ex] = \dfrac{1}{3} * \dfrac{11}{7} * \dfrac{5}{1} * \dfrac{5}{2} * \dfrac{5}{4} \\[5ex] = \dfrac{1 * 11 * 5 * 5 * 5}{3 * 7 * 1 * 2 * 4} \\[5ex] = \dfrac{1375}{168}\:\:cubic\:\:inches....answer\:\:in\:\:fraction \\[5ex] = 8.18452381 \\[3ex] \approx 8\:\:cubic\:\:inches $

(7.) **ACT** Patrick and Ayako are painting a room in the city hall.

They started with $6$ gallons of paint.

On the first day, Patrick used $\dfrac{1}{2}$ gallon of paint and Ayako used $1\dfrac{3}{4}$ gallons of paint.

How many gallons of paint were left when they completed their first day of painting?

$ A.\:\: 2\dfrac{1}{4} \\[5ex] B.\:\: 3\dfrac{3}{4} \\[5ex] C.\:\: 4\dfrac{1}{4} \\[5ex] D.\:\: 4\dfrac{3}{4} \\[5ex] E.\:\: 5\dfrac{1}{2} \\[5ex] $

$ Total\:\:gallons\:\:of\:\:paint = 6 \\[3ex] \underline{Used\:\:on\:\:First\:\:Day} \\[3ex] Patrick:\:\: \dfrac{1}{2} \\[5ex] Ayako:\:\: 1\dfrac{3}{4} = \dfrac{4 * 1 + 3}{4} = \dfrac{4 + 3}{4} = \dfrac{7}{4} \\[5ex] \dfrac{1}{2} + \dfrac{7}{4} \\[5ex] = \dfrac{2}{4} + \dfrac{7}{4} \\[5ex] = \dfrac{2 + 7}{4} \\[5ex] = \dfrac{9}{4} \\[5ex] Remaining:\:\: 6 - \dfrac{9}{4} \\[5ex] = \dfrac{6}{1} - \dfrac{9}{4} \\[5ex] = \dfrac{24}{4} - \dfrac{9}{4} \\[5ex] = \dfrac{24 - 9}{4} \\[5ex] = \dfrac{15}{4} \\[5ex] = 3\dfrac{3}{4}\:\:gallons $

They started with $6$ gallons of paint.

On the first day, Patrick used $\dfrac{1}{2}$ gallon of paint and Ayako used $1\dfrac{3}{4}$ gallons of paint.

How many gallons of paint were left when they completed their first day of painting?

$ A.\:\: 2\dfrac{1}{4} \\[5ex] B.\:\: 3\dfrac{3}{4} \\[5ex] C.\:\: 4\dfrac{1}{4} \\[5ex] D.\:\: 4\dfrac{3}{4} \\[5ex] E.\:\: 5\dfrac{1}{2} \\[5ex] $

$ Total\:\:gallons\:\:of\:\:paint = 6 \\[3ex] \underline{Used\:\:on\:\:First\:\:Day} \\[3ex] Patrick:\:\: \dfrac{1}{2} \\[5ex] Ayako:\:\: 1\dfrac{3}{4} = \dfrac{4 * 1 + 3}{4} = \dfrac{4 + 3}{4} = \dfrac{7}{4} \\[5ex] \dfrac{1}{2} + \dfrac{7}{4} \\[5ex] = \dfrac{2}{4} + \dfrac{7}{4} \\[5ex] = \dfrac{2 + 7}{4} \\[5ex] = \dfrac{9}{4} \\[5ex] Remaining:\:\: 6 - \dfrac{9}{4} \\[5ex] = \dfrac{6}{1} - \dfrac{9}{4} \\[5ex] = \dfrac{24}{4} - \dfrac{9}{4} \\[5ex] = \dfrac{24 - 9}{4} \\[5ex] = \dfrac{15}{4} \\[5ex] = 3\dfrac{3}{4}\:\:gallons $

(8.) **WASSCE** A petrol tanker is $\dfrac{2}{5}$ full.

When $35,000$ litres of petrol are added, the tanker will be $\dfrac{3}{4}$ full.

What is the capacity of the tanker in litres?

$ A.\:\: 70,000 \\[3ex] B.\:\: 75,000 \\[3ex] C.\:\: 90,000 \\[3ex] D.\:\: 100,000 \\[3ex] $

We can solve this question in two ways.

Use any method you prefer.

One method will be explained here.

The other method is explained here (Question $76$)

Litres or Liters...no worries

United State: Liters

Nigeria, Britain: Litres

WASSCE is West African question...so I am using litres

__First:__ Let us find the fraction that accounts for that $35,000$ litres

__Second:__ We shall use Proportional Reasoning to find the capacity of the tanker

$ Total\:\:Percent = 100\% \\[3ex] \therefore Total\:\:Fraction = 100\% = \dfrac{100}{100} = 1 \\[5ex] Initial\:\:volume/fraction = \dfrac{2}{5} \\[5ex] 35000\:\:litres\:\:are\:\:added \\[3ex] New\:\:volume/fraction = \dfrac{3}{4} \\[5ex] Difference = \dfrac{3}{4} - \dfrac{2}{5} \\[5ex] = \dfrac{15}{20} - \dfrac{8}{20} \\[5ex] = \dfrac{15 - 8}{20} \\[5ex] = \dfrac{7}{20} \\[5ex] $ $35000\:\:litres$ accounts for this fraction (difference)

So, how many litres will account for the capacity (the entire new volume/fraction)?

Let the capacity of the tanker in litres be $p$

$ \dfrac{p}{1} = \dfrac{35000}{\dfrac{7}{20}} \\[7ex] p = 35000 \div \dfrac{7}{20} \\[5ex] p = 35000 * \dfrac{20}{7} \\[5ex] p = 5000 * 20 \\[3ex] p = 10000 \\[3ex] $ The capacity of the tank is $100,000$ litres

When $35,000$ litres of petrol are added, the tanker will be $\dfrac{3}{4}$ full.

What is the capacity of the tanker in litres?

$ A.\:\: 70,000 \\[3ex] B.\:\: 75,000 \\[3ex] C.\:\: 90,000 \\[3ex] D.\:\: 100,000 \\[3ex] $

We can solve this question in two ways.

Use any method you prefer.

One method will be explained here.

The other method is explained here (Question $76$)

Litres or Liters...no worries

United State: Liters

Nigeria, Britain: Litres

WASSCE is West African question...so I am using litres

$ Total\:\:Percent = 100\% \\[3ex] \therefore Total\:\:Fraction = 100\% = \dfrac{100}{100} = 1 \\[5ex] Initial\:\:volume/fraction = \dfrac{2}{5} \\[5ex] 35000\:\:litres\:\:are\:\:added \\[3ex] New\:\:volume/fraction = \dfrac{3}{4} \\[5ex] Difference = \dfrac{3}{4} - \dfrac{2}{5} \\[5ex] = \dfrac{15}{20} - \dfrac{8}{20} \\[5ex] = \dfrac{15 - 8}{20} \\[5ex] = \dfrac{7}{20} \\[5ex] $ $35000\:\:litres$ accounts for this fraction (difference)

So, how many litres will account for the capacity (the entire new volume/fraction)?

Let the capacity of the tanker in litres be $p$

volume (in fractions) | volume (in litres) |
---|---|

$\dfrac{7}{20}$ | $35000$ |

$1$ | $p$ |

$ \dfrac{p}{1} = \dfrac{35000}{\dfrac{7}{20}} \\[7ex] p = 35000 \div \dfrac{7}{20} \\[5ex] p = 35000 * \dfrac{20}{7} \\[5ex] p = 5000 * 20 \\[3ex] p = 10000 \\[3ex] $ The capacity of the tank is $100,000$ litres

(9.) **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?

$ F.\:\: 28\dfrac{8}{9} \\[5ex] G.\:\: 30 \\[3ex] H.\:\: 43\dfrac{1}{3} \\[5ex] J.\:\: 48 \\[3ex] K.\:\: 54 \\[3ex] $

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 \\[5ex] 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 $

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?

$ F.\:\: 28\dfrac{8}{9} \\[5ex] G.\:\: 30 \\[3ex] H.\:\: 43\dfrac{1}{3} \\[5ex] J.\:\: 48 \\[3ex] K.\:\: 54 \\[3ex] $

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 \\[5ex] 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 $

(10.) **CSEC** A sum of money is shared between Aaron and Betty in the ratio $2:5$.

Aaron received $\$60$.

How much money was shared**altogether**?

$ \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] \dfrac{Aaron's\:\:ratio}{Total\:\:ratio} = \dfrac{Aaron's\:\:share}{Total\:\:share} \\[5ex] Total\:\:ratio = 2 + 5 = 7 \\[3ex] \rightarrow \dfrac{2}{7} = \dfrac{60}{Total\:\:share} \\[5ex] \dfrac{60}{2} = 30 \\[3ex] 30 * 7 = 210 \\[3ex] \therefore Total\:\:share = \$210 \\[3ex] \underline{Second\:\:Method:\:\:Algebraically} \\[3ex] Let\:\:the\:\:money\:\:shared\:\:altogether = p \\[3ex] Ratio\:\:shared\:\:between\:\:Aaron\:\:and\:\:Betty = 2:5 \\[3ex] Sum\:\:of\:\:ratios = 2 + 5 = 7 \\[3ex] Aaron's\:\:share = \dfrac{2}{5}\:\:of\:\:p = \dfrac{2}{7} * p \\[5ex] Aaron's\:\:share = 60 \\[3ex] \rightarrow \dfrac{2}{7} * p = 60 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:\dfrac{7}{2} \\[5ex] \dfrac{7}{2} * \dfrac{2}{7} * p = \dfrac{7}{2} * 60 \\[5ex] p = 7 * 30 \\[3ex] p = 210 \\[3ex] $ They shared a sum of $\$210$

$ \underline{Check} \\[3ex] Money\:\:shared = 210 \\[3ex] Ratio\:\:shared\:\:between\:\:Aaron\:\:and\:\:Betty = 2:5 \\[3ex] Sum\:\:of\:\:ratios = 2 + 5 = 7 \\[3ex] Aaron's\:\:share \\[3ex] = \dfrac{2}{7} * 210 \\[5ex] = 2 * 30 \\[3ex] = \$60 $

Aaron received $\$60$.

How much money was shared

$ \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] \dfrac{Aaron's\:\:ratio}{Total\:\:ratio} = \dfrac{Aaron's\:\:share}{Total\:\:share} \\[5ex] Total\:\:ratio = 2 + 5 = 7 \\[3ex] \rightarrow \dfrac{2}{7} = \dfrac{60}{Total\:\:share} \\[5ex] \dfrac{60}{2} = 30 \\[3ex] 30 * 7 = 210 \\[3ex] \therefore Total\:\:share = \$210 \\[3ex] \underline{Second\:\:Method:\:\:Algebraically} \\[3ex] Let\:\:the\:\:money\:\:shared\:\:altogether = p \\[3ex] Ratio\:\:shared\:\:between\:\:Aaron\:\:and\:\:Betty = 2:5 \\[3ex] Sum\:\:of\:\:ratios = 2 + 5 = 7 \\[3ex] Aaron's\:\:share = \dfrac{2}{5}\:\:of\:\:p = \dfrac{2}{7} * p \\[5ex] Aaron's\:\:share = 60 \\[3ex] \rightarrow \dfrac{2}{7} * p = 60 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:\dfrac{7}{2} \\[5ex] \dfrac{7}{2} * \dfrac{2}{7} * p = \dfrac{7}{2} * 60 \\[5ex] p = 7 * 30 \\[3ex] p = 210 \\[3ex] $ They shared a sum of $\$210$

$ \underline{Check} \\[3ex] Money\:\:shared = 210 \\[3ex] Ratio\:\:shared\:\:between\:\:Aaron\:\:and\:\:Betty = 2:5 \\[3ex] Sum\:\:of\:\:ratios = 2 + 5 = 7 \\[3ex] Aaron's\:\:share \\[3ex] = \dfrac{2}{7} * 210 \\[5ex] = 2 * 30 \\[3ex] = \$60 $

(11.) **ACT** The *specific gravity* of a substance is the ratio of the weight of the substance to
the weight of an equal volume of water.

If $1$ cubic foot of water weighs $62.5$ pounds, what is the specific gravity of a liquid that weighs $125$ pounds per cubic foot?

$ A.\:\: 1 \\[3ex] B.\:\: 1.25 \\[3ex] C.\:\: 2 \\[3ex] D.\:\: 6.25 \\[3ex] E.\:\: 125 \\[3ex] $

$ Specific\:\:Gravity\:\:of\:\:the\:\:Liquid = \dfrac{Weight\:\:of\:\:Liquid}{Weight\:\:of\:\:Equal\:\:Volume\:\:of\:\:Water} \\[5ex] = \dfrac{125}{62.5} \\[5ex] = 2 $

If $1$ cubic foot of water weighs $62.5$ pounds, what is the specific gravity of a liquid that weighs $125$ pounds per cubic foot?

$ A.\:\: 1 \\[3ex] B.\:\: 1.25 \\[3ex] C.\:\: 2 \\[3ex] D.\:\: 6.25 \\[3ex] E.\:\: 125 \\[3ex] $

$ Specific\:\:Gravity\:\:of\:\:the\:\:Liquid = \dfrac{Weight\:\:of\:\:Liquid}{Weight\:\:of\:\:Equal\:\:Volume\:\:of\:\:Water} \\[5ex] = \dfrac{125}{62.5} \\[5ex] = 2 $

(12.) **ACT** In a poll of $500$ registered voters, $337$ voters favored a proposal to increase funding
for local schools.

Suppose the poll is indicative of how the $22,000$ registered voters will vote on the proposal.

Which of the following values is closest to how many of the $22,000$ registered voters will be expected to vote in favor of the proposal?

$ A.\:\: 13,200 \\[3ex] B.\:\: 14,830 \\[3ex] C.\:\: 21,840 \\[3ex] D.\:\: 22,000 \\[3ex] E.\:\: 32,640 \\[3ex] $

Let the number of the voters out of the $22,000$ registered voters that is expected to vote in favor of the proposal be $p$

$ \dfrac{p}{22000} = \dfrac{337}{500} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:22000 \\[3ex] 22000 * \dfrac{p}{22000} = 22000 * \dfrac{337}{500} \\[5ex] p = \dfrac{220 * 337}{5} \\[5ex] p = 44 * 337 \\[3ex] p = 14,828\:\:voters \\[3ex] $ About $14,830$ (option that is closest to $14,828$) voters out of the $22,000$ registered voters will vote in favor of the proposal

Suppose the poll is indicative of how the $22,000$ registered voters will vote on the proposal.

Which of the following values is closest to how many of the $22,000$ registered voters will be expected to vote in favor of the proposal?

$ A.\:\: 13,200 \\[3ex] B.\:\: 14,830 \\[3ex] C.\:\: 21,840 \\[3ex] D.\:\: 22,000 \\[3ex] E.\:\: 32,640 \\[3ex] $

Let the number of the voters out of the $22,000$ registered voters that is expected to vote in favor of the proposal be $p$

Registered Voters | Vote in Favor |
---|---|

$500$ | $337$ |

$22000$ | $p$ |

$ \dfrac{p}{22000} = \dfrac{337}{500} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:22000 \\[3ex] 22000 * \dfrac{p}{22000} = 22000 * \dfrac{337}{500} \\[5ex] p = \dfrac{220 * 337}{5} \\[5ex] p = 44 * 337 \\[3ex] p = 14,828\:\:voters \\[3ex] $ About $14,830$ (option that is closest to $14,828$) voters out of the $22,000$ registered voters will vote in favor of the proposal

(13.) **ACT** A survey about 3 issues affecting Bluff City Park was given to 60 residents.

The results of the survey are shown below.

Assume that the results in the table accurately predict the response ratios for the town's $1200$ residents.

How many of the $1200$ residents would respond No on the curfew issue?

$ F.\:\: 240 \\[3ex] G.\:\: 300 \\[3ex] H.\:\: 600 \\[3ex] J.\:\: 680 \\[3ex] K.\:\: 960 \\[3ex] $

$ Sample = 60\:\:residents \\[3ex] Responded\:\:No\:\:on\:\:Curfew = 12 \\[3ex] 1st\:\:Ratio = \dfrac{12}{60} \\[5ex] Population = 1200\:\:residents \\[3ex] Would\:\:Respond\:\:No\:\:on\:\:Curfew = x \\[3ex] 2nd\:\:Ratio = \dfrac{x}{1200} \\[5ex] 1st\:\:Ratio\:\:accurately\:\:predicts\:\:2nd\:\:Ratio \\[3ex] \dfrac{12}{60} = \dfrac{x}{1200} \\[3ex] \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] \dfrac{1200}{60} = 20 \\[3ex] 20 * 12 = x \\[3ex] 240 = x \\[3ex] x = 240\:\:residents \\[3ex] \underline{Second\:\:Method:\:\:Algebraically} \\[3ex] \dfrac{12}{60} = \dfrac{x}{1200} \\[5ex] \dfrac{x}{1200} = \dfrac{12}{60} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:1200 \\[3ex] 1200 * \dfrac{x}{1200} = 1200 * \dfrac{12}{60} \\[5ex] x = 20(12) \\[3ex] x = 240\:\:residents \\[3ex] $ $240$ residents of the population of $1200$ residents would respond No on the curfew issue.

The results of the survey are shown below.

Issue | Yes | No |
---|---|---|

Curfew Skateboard use Children under $14$ accompanied by a person at least $14$ years old |
$48$ $26$ $38$ |
$12$ $34$ $22$ |

Assume that the results in the table accurately predict the response ratios for the town's $1200$ residents.

How many of the $1200$ residents would respond No on the curfew issue?

$ F.\:\: 240 \\[3ex] G.\:\: 300 \\[3ex] H.\:\: 600 \\[3ex] J.\:\: 680 \\[3ex] K.\:\: 960 \\[3ex] $

$ Sample = 60\:\:residents \\[3ex] Responded\:\:No\:\:on\:\:Curfew = 12 \\[3ex] 1st\:\:Ratio = \dfrac{12}{60} \\[5ex] Population = 1200\:\:residents \\[3ex] Would\:\:Respond\:\:No\:\:on\:\:Curfew = x \\[3ex] 2nd\:\:Ratio = \dfrac{x}{1200} \\[5ex] 1st\:\:Ratio\:\:accurately\:\:predicts\:\:2nd\:\:Ratio \\[3ex] \dfrac{12}{60} = \dfrac{x}{1200} \\[3ex] \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] \dfrac{1200}{60} = 20 \\[3ex] 20 * 12 = x \\[3ex] 240 = x \\[3ex] x = 240\:\:residents \\[3ex] \underline{Second\:\:Method:\:\:Algebraically} \\[3ex] \dfrac{12}{60} = \dfrac{x}{1200} \\[5ex] \dfrac{x}{1200} = \dfrac{12}{60} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:1200 \\[3ex] 1200 * \dfrac{x}{1200} = 1200 * \dfrac{12}{60} \\[5ex] x = 20(12) \\[3ex] x = 240\:\:residents \\[3ex] $ $240$ residents of the population of $1200$ residents would respond No on the curfew issue.

(14.) **ACT** For the next school year, a college will use $\dfrac{1}{9}$ of the money in its operating
budget for library books and $\dfrac{1}{6}$ of the money in its operating budget for scholarships.

What fraction of the operating budget remains for other uses?

$ F.\:\: \dfrac{1}{18} \\[5ex] G.\:\: \dfrac{5}{18} \\[5ex] H.\:\: \dfrac{13}{18} \\[5ex] J.\:\: \dfrac{20}{27} \\[5ex] K.\:\: \dfrac{8}{9} \\[5ex] $

Due to the nature of this question, because we were not given the amount of the operating budget; we shall use the total fraction as $1$

$ Total\:\:Percentage = 100\% = 1 \\[3ex] Likewise\:\:Total\:\:Fraction = 1 \\[3ex] Library\:\:Boooks = \dfrac{1}{9} \\[5ex] Scholarships = \dfrac{1}{6} \\[5ex] Other\:\:uses = 1 - \left(\dfrac{1}{9} + \dfrac{1}{6}\right) \\[5ex] PEMDAS \\[3ex] \dfrac{1}{9} + \dfrac{1}{6} \\[5ex] = \dfrac{2}{18} + \dfrac{3}{18} \\[5ex] = \dfrac{2 + 3}{18} \\[5ex] = \dfrac{5}{18} \\[5ex] Other\:\:uses \\[3ex] = 1 - \dfrac{5}{18} \\[5ex] = \dfrac{18}{18} - \dfrac{5}{18} \\[5ex] = \dfrac{18 - 5}{18} \\[5ex] = \dfrac{13}{18} \\[5ex] $ The fraction of the operating budget remaining for other uses is $\dfrac{13]{18}$

What fraction of the operating budget remains for other uses?

$ F.\:\: \dfrac{1}{18} \\[5ex] G.\:\: \dfrac{5}{18} \\[5ex] H.\:\: \dfrac{13}{18} \\[5ex] J.\:\: \dfrac{20}{27} \\[5ex] K.\:\: \dfrac{8}{9} \\[5ex] $

Due to the nature of this question, because we were not given the amount of the operating budget; we shall use the total fraction as $1$

$ Total\:\:Percentage = 100\% = 1 \\[3ex] Likewise\:\:Total\:\:Fraction = 1 \\[3ex] Library\:\:Boooks = \dfrac{1}{9} \\[5ex] Scholarships = \dfrac{1}{6} \\[5ex] Other\:\:uses = 1 - \left(\dfrac{1}{9} + \dfrac{1}{6}\right) \\[5ex] PEMDAS \\[3ex] \dfrac{1}{9} + \dfrac{1}{6} \\[5ex] = \dfrac{2}{18} + \dfrac{3}{18} \\[5ex] = \dfrac{2 + 3}{18} \\[5ex] = \dfrac{5}{18} \\[5ex] Other\:\:uses \\[3ex] = 1 - \dfrac{5}{18} \\[5ex] = \dfrac{18}{18} - \dfrac{5}{18} \\[5ex] = \dfrac{18 - 5}{18} \\[5ex] = \dfrac{13}{18} \\[5ex] $ The fraction of the operating budget remaining for other uses is $\dfrac{13]{18}$

(15.) **ACT** The Harrisburg Recreation Center recently changed its hours to open $1$ hour later and
close $3$ hours later than it had previously.

Residents of Harrisburg age $16$ or older were given a survey, and $560$ residents replied.

The survey asked each resident his or her student status (high school, college, or nonstudent) and what he or she thought about the change in hours (approve, disapprove, or no opinion).

The results are summarized in the table below.

What fraction of these nonstudent residents replied that they disapproved of the change in hours?

$ F.\:\: \dfrac{1}{3} \\[5ex] G.\:\: \dfrac{4}{45} \\[5ex] H.\:\: \dfrac{14}{75} \\[5ex] J.\:\: \dfrac{353}{367} \\[5ex] K.\:\: \dfrac{353}{485} \\[5ex] $

$ Total\:\:number\:\:of\:\:nonstudent\:\:residents = 85 + 353 + 47 = 485 \\[3ex] Number\:\:of\:\:nonstudent\:\:residents\:\:that\:\:disapproved = 353 \\[3ex] Fraction\:\:of\:\:nonstudent\:\:residents\:\:that\:\:disapproved = \dfrac{353}{485} $

Residents of Harrisburg age $16$ or older were given a survey, and $560$ residents replied.

The survey asked each resident his or her student status (high school, college, or nonstudent) and what he or she thought about the change in hours (approve, disapprove, or no opinion).

The results are summarized in the table below.

Student status | Approve | Disapprove | No opinion |

High school College Nonstudent |
$30$ $14$ $85$ |
$4$ $10$ $353$ |
$11$ $6$ $47$ |

Total | $129$ | $367$ | $64$ |

What fraction of these nonstudent residents replied that they disapproved of the change in hours?

$ F.\:\: \dfrac{1}{3} \\[5ex] G.\:\: \dfrac{4}{45} \\[5ex] H.\:\: \dfrac{14}{75} \\[5ex] J.\:\: \dfrac{353}{367} \\[5ex] K.\:\: \dfrac{353}{485} \\[5ex] $

$ Total\:\:number\:\:of\:\:nonstudent\:\:residents = 85 + 353 + 47 = 485 \\[3ex] Number\:\:of\:\:nonstudent\:\:residents\:\:that\:\:disapproved = 353 \\[3ex] Fraction\:\:of\:\:nonstudent\:\:residents\:\:that\:\:disapproved = \dfrac{353}{485} $

(16.) **ACT** A group of $60$ students and $4$ sponsors took a field trip to a local museum.

For their first guided tour, students were given a choice of $1$ of $3$ art exhibits.

Of the $60$ students, $\dfrac{1}{2}$ chose Modern, $\dfrac{1}{4}$ chose American Folk, and $\dfrac{1}{6}$ chose Western.

Each student that expressed a choice chose exactly $1$ exhibit.

The remaining students expressed no choice.

How many of the students expressed no choice?

$ A.\:\: 5 \\[3ex] B.\:\: 6 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 15 \\[3ex] E.\:\: 30 \\[3ex] $

$ \underline{First\:\:Method} \\[3ex] Total\:\:number\:\:of\:\:students = 60 \\[3ex] Modern:\:\: \dfrac{1}{2}\:\:of\:\:60 = \dfrac{1}{2} * 60 = 30\:\:students \\[5ex] American\:\:Folk\:\: \dfrac{1}{4}\:\:of\:\:60 = \dfrac{1}{4} * 60 = 15\:\:students \\[5ex] Western:\:\: \dfrac{1}{6}\:\:of\:\:60 = \dfrac{1}{6} * 60 = 10\:\:students \\[5ex] No\:\:choice \\[3ex] = 60 - (30 + 15 + 10) \\[3ex] = 60 - 55 \\[3ex] = 5\:\:students \\[3ex] \underline{Second\:\:Method} \\[3ex] Total\:\:Percentage = 100\% = 1 \\[3ex] Likewise\:\:Total\:\:Fraction...for\:\:this\:\:question = 1 \\[3ex] Modern = \dfrac{1}{2} \\[5ex] American\:\:Folk = \dfrac{1}{4} \\[5ex] Western = \dfrac{1}{6} \\[5ex] No\:\:choice = 1 - \left(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6}\right) \\[5ex] PEMDAS \\[3ex] \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} \\[5ex] = \dfrac{6}{12} + \dfrac{3}{12} + \dfrac{2}{12} \\[5ex] = \dfrac{6 + 3 + 2}{12} \\[5ex] = \dfrac{11}{12} \\[5ex] \rightarrow No\:\:choice = 1 - \dfrac{11}{12} \\[5ex] = \dfrac{12}{12} - \dfrac{11}{12} \\[5ex] = \dfrac{12 - 11}{12} \\[5ex] = \dfrac{1}{12} \\[5ex] No\:\:choice\:\:students = \dfrac{1}{12}\:\:of\:\:60 \\[5ex] = \dfrac{1}{12} * 60 \\[5ex] = 1(5) \\[3ex] = 5\:\:students \\[3ex] $ $5$ students expressed no choice.

For their first guided tour, students were given a choice of $1$ of $3$ art exhibits.

Of the $60$ students, $\dfrac{1}{2}$ chose Modern, $\dfrac{1}{4}$ chose American Folk, and $\dfrac{1}{6}$ chose Western.

Each student that expressed a choice chose exactly $1$ exhibit.

The remaining students expressed no choice.

How many of the students expressed no choice?

$ A.\:\: 5 \\[3ex] B.\:\: 6 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 15 \\[3ex] E.\:\: 30 \\[3ex] $

$ \underline{First\:\:Method} \\[3ex] Total\:\:number\:\:of\:\:students = 60 \\[3ex] Modern:\:\: \dfrac{1}{2}\:\:of\:\:60 = \dfrac{1}{2} * 60 = 30\:\:students \\[5ex] American\:\:Folk\:\: \dfrac{1}{4}\:\:of\:\:60 = \dfrac{1}{4} * 60 = 15\:\:students \\[5ex] Western:\:\: \dfrac{1}{6}\:\:of\:\:60 = \dfrac{1}{6} * 60 = 10\:\:students \\[5ex] No\:\:choice \\[3ex] = 60 - (30 + 15 + 10) \\[3ex] = 60 - 55 \\[3ex] = 5\:\:students \\[3ex] \underline{Second\:\:Method} \\[3ex] Total\:\:Percentage = 100\% = 1 \\[3ex] Likewise\:\:Total\:\:Fraction...for\:\:this\:\:question = 1 \\[3ex] Modern = \dfrac{1}{2} \\[5ex] American\:\:Folk = \dfrac{1}{4} \\[5ex] Western = \dfrac{1}{6} \\[5ex] No\:\:choice = 1 - \left(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6}\right) \\[5ex] PEMDAS \\[3ex] \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} \\[5ex] = \dfrac{6}{12} + \dfrac{3}{12} + \dfrac{2}{12} \\[5ex] = \dfrac{6 + 3 + 2}{12} \\[5ex] = \dfrac{11}{12} \\[5ex] \rightarrow No\:\:choice = 1 - \dfrac{11}{12} \\[5ex] = \dfrac{12}{12} - \dfrac{11}{12} \\[5ex] = \dfrac{12 - 11}{12} \\[5ex] = \dfrac{1}{12} \\[5ex] No\:\:choice\:\:students = \dfrac{1}{12}\:\:of\:\:60 \\[5ex] = \dfrac{1}{12} * 60 \\[5ex] = 1(5) \\[3ex] = 5\:\:students \\[3ex] $ $5$ students expressed no choice.

(17.) **ACT** A school admissions office accepts $2$ out of every $7$ applicants.

Given that the school accepted $630$ students, how many applicants were NOT accepted?

$ F.\:\: 140 \\[3ex] G.\:\: 180 \\[3ex] H.\:\: 490 \\[3ex] J.\:\: 1,260 \\[3ex] K.\:\: 1,575 \\[3ex] $

$ Sample = 7\:\:applicants \\[3ex] Accepted = 2 \\[3ex] 1st\:\:Ratio = \dfrac{2}{7} \\[5ex] Population = p\:\:applicants \\[3ex] Accepted\:\:for\:\:Population = 630 \\[3ex] 2nd\:\:Ratio = \dfrac{630}{p} \\[5ex] 1st\:\:Ratio\:\:accurately\:\:predicts\:\:2nd\:\:Ratio \\[3ex] \dfrac{2}{7} = \dfrac{630}{p} \\[3ex] \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] \dfrac{630}{2} = 315 \\[3ex] 315 * 7 = p \\[3ex] 2205 = p \\[3ex] Population = 2205\:\:applicants \\[3ex] NOT\:\:accepted = 2205 - 630 = 1,575\:\:applicants \\[3ex] \underline{Second\:\:Method:\:\:Algebraically - Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Out\:\:of\:\:7\:\:applicants: \\[3ex] Accepted = 2 \\[3ex] Not\:\:accepted = 7 - 2 = 5 \\[3ex] Let\:\:k\:\:be\:\:the\:\:number\:\:of\:\:applicants\:\:out\:\:of\:\:630\:\:NOT\:\:accepted \\[3ex] $

$ \dfrac{k}{5} = \dfrac{630}{2} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] 5 * \dfrac{k}{5} = 5 * \dfrac{630}{2} \\[5ex] k = {5 * 630}{2} \\[5ex] k = 5(315) \\[3ex] k = 1,575\:\:applicants \\[3ex] $ $1,575$ applicants out of $2,205$ applicants were not accepted.

Given that the school accepted $630$ students, how many applicants were NOT accepted?

$ F.\:\: 140 \\[3ex] G.\:\: 180 \\[3ex] H.\:\: 490 \\[3ex] J.\:\: 1,260 \\[3ex] K.\:\: 1,575 \\[3ex] $

$ Sample = 7\:\:applicants \\[3ex] Accepted = 2 \\[3ex] 1st\:\:Ratio = \dfrac{2}{7} \\[5ex] Population = p\:\:applicants \\[3ex] Accepted\:\:for\:\:Population = 630 \\[3ex] 2nd\:\:Ratio = \dfrac{630}{p} \\[5ex] 1st\:\:Ratio\:\:accurately\:\:predicts\:\:2nd\:\:Ratio \\[3ex] \dfrac{2}{7} = \dfrac{630}{p} \\[3ex] \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] \dfrac{630}{2} = 315 \\[3ex] 315 * 7 = p \\[3ex] 2205 = p \\[3ex] Population = 2205\:\:applicants \\[3ex] NOT\:\:accepted = 2205 - 630 = 1,575\:\:applicants \\[3ex] \underline{Second\:\:Method:\:\:Algebraically - Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Out\:\:of\:\:7\:\:applicants: \\[3ex] Accepted = 2 \\[3ex] Not\:\:accepted = 7 - 2 = 5 \\[3ex] Let\:\:k\:\:be\:\:the\:\:number\:\:of\:\:applicants\:\:out\:\:of\:\:630\:\:NOT\:\:accepted \\[3ex] $

Accept | Not Accepted | Applicants |
---|---|---|

$2$ | $5$ | $7$ |

$630$ | $k$ |

$ \dfrac{k}{5} = \dfrac{630}{2} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] 5 * \dfrac{k}{5} = 5 * \dfrac{630}{2} \\[5ex] k = {5 * 630}{2} \\[5ex] k = 5(315) \\[3ex] k = 1,575\:\:applicants \\[3ex] $ $1,575$ applicants out of $2,205$ applicants were not accepted.

(18.) **ACT** The ratio of Jane's age to her daughter's age is $9:2$.

The sum of their ages is $44$.

How old is Jane?

$ A.\:\: 22 \\[3ex] B.\:\: 33 \\[3ex] C.\:\: 35 \\[3ex] D.\:\: 36 \\[3ex] E.\:\: 40 \\[3ex] $

$ Jane:Daughter = 9:2 \\[3ex] Sum\:\:of\:\:ratios = 9 + 2 = 11 \\[3ex] Sum\:\:of\:\:their\:\:ages = 44 \\[3ex] Jane's\:\:age \\[3ex] = \dfrac{9}{11} * 44 \\[5ex] = 9(4) \\[3ex] = 36 \\[3ex] $ Jane is $36$ years old

The sum of their ages is $44$.

How old is Jane?

$ A.\:\: 22 \\[3ex] B.\:\: 33 \\[3ex] C.\:\: 35 \\[3ex] D.\:\: 36 \\[3ex] E.\:\: 40 \\[3ex] $

$ Jane:Daughter = 9:2 \\[3ex] Sum\:\:of\:\:ratios = 9 + 2 = 11 \\[3ex] Sum\:\:of\:\:their\:\:ages = 44 \\[3ex] Jane's\:\:age \\[3ex] = \dfrac{9}{11} * 44 \\[5ex] = 9(4) \\[3ex] = 36 \\[3ex] $ Jane is $36$ years old

(19.) **CSEC** Concrete tiles are made using buckets of cement, sand, and gravel mixed in the ratio
$1:4:6$

(i) How many buckets of gravel are needed for $4$ buckets of cement?

(ii) If $20$ buckets of sand are used, how many buckets of EACH of the following will be needed?

(a.) Cement

(b.) Gravel

We can solve this questions in__at least two ways__

Use any method you prefer

$ Cement:Sand:Gravel \\[3ex] 1:4:6 \\[3ex] For\:\:each\:(1)\:\:bucket\:\:of\:\:Cement \\[3ex] You\:\:need\:\:4\:\:buckets\:\:of\:\:Sand \\[3ex] You\:\:need\:\:6\:\:buckets\:\:of\:\:Gravel \\[3ex] \underline{First\:\:Method:\:\:Arithmetically - Proportion} \\[3ex] (i) \\[3ex] Given:\:\:4\:\:buckets\:\:of\:\:Cement \\[3ex] \dfrac{Cement}{Gravel} = \dfrac{1}{6} = \dfrac{4}{???} \\[5ex] \dfrac{4}{1} = 4 \\[5ex] 6 * 4 = 24 \\[3ex] 24\:\:buckets\:\:of\:\:Gravel\:\:is\:\:needed \\[3ex] (ii) \\[3ex] Used:\:\:20\:\:buckets\:\:of\:\:Sand \\[3ex] (a.) \\[3ex] \dfrac{Cement}{Sand} = \dfrac{1}{4} = \dfrac{???}{20} \\[5ex] \dfrac{20}{4} = 5 \\[3ex] 1 * 5 = 5 \\[3ex] 5\:\:buckets\:\:of\:\:Cement\:\:will\:\:be\:\:needed \\[3ex] (b.) \\[3ex] \dfrac{Sand}{Gravel} = \dfrac{4}{6} = \dfrac{20}{???} \\[5ex] \dfrac{20}{4} = 5 \\[3ex] 6 * 5 = 30 \\[3ex] 30\:\:buckets\:\:of\:\:Gravel\:\:will\:\:be\:\:needed \\[3ex] \underline{Second\:\:Method:\:\:Algebraically - Proportional\:\:Reasoning\:\:Method} \\[3ex] (i) \\[3ex] Let\:\:p\:\:be\:\:the\:\:number\:\:of\:\:buckets\:\:of\:\:Gravel\:\:needed\:\:for\:\:6\:\:buckets\:\:of\:\:Cement \\[3ex] $

$ \dfrac{p}{4} = \dfrac{6}{1} \\[5ex] \dfrac{p}{4} = 6 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:4 \\[3ex] 4 * \dfrac{p}{4} = 4 * 6 \\[5ex] p = 24 \\[3ex] 24\:\:buckets\:\:of\:\:Gravel\:\:is\:\:needed \\[3ex] (ii) \\[3ex] \underline{Third\:\:Method:\:\:Algebraically - Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:x\:\:be\:\:the\:\:number\:\:of\:\:buckets\:\:of\:\:Cement\:\:needed\:\:for\:\:20\:\:buckets\:\:of\:\:Sand \\[3ex] Let\:\:y\:\:be\:\:the\:\:number\:\:of\:\:buckets\:\:of\:\:Gravel\:\:needed\:\:for\:\:20\:\:buckets\:\:of\:\:Sand \\[3ex] $

$ \dfrac{x}{1} = \dfrac{20}{4} \\[5ex] x = 5 \\[3ex] 5\:\:buckets\:\:of\:\:Cement\:\:will\:\:be\:\:needed \\[3ex] Next \\[3ex] \dfrac{y}{6} = \dfrac{20}{4} \\[5ex] \dfrac{y}{6} = 5 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:6 \\[3ex] 6 * \dfrac{y}{6} = 6 * 5 \\[5ex] y = 30 \\[3ex] 30\:\:buckets\:\:of\:\:Gravel\:\:will\:\:be\:\:needed $

(i) How many buckets of gravel are needed for $4$ buckets of cement?

(ii) If $20$ buckets of sand are used, how many buckets of EACH of the following will be needed?

(a.) Cement

(b.) Gravel

We can solve this questions in

Use any method you prefer

$ Cement:Sand:Gravel \\[3ex] 1:4:6 \\[3ex] For\:\:each\:(1)\:\:bucket\:\:of\:\:Cement \\[3ex] You\:\:need\:\:4\:\:buckets\:\:of\:\:Sand \\[3ex] You\:\:need\:\:6\:\:buckets\:\:of\:\:Gravel \\[3ex] \underline{First\:\:Method:\:\:Arithmetically - Proportion} \\[3ex] (i) \\[3ex] Given:\:\:4\:\:buckets\:\:of\:\:Cement \\[3ex] \dfrac{Cement}{Gravel} = \dfrac{1}{6} = \dfrac{4}{???} \\[5ex] \dfrac{4}{1} = 4 \\[5ex] 6 * 4 = 24 \\[3ex] 24\:\:buckets\:\:of\:\:Gravel\:\:is\:\:needed \\[3ex] (ii) \\[3ex] Used:\:\:20\:\:buckets\:\:of\:\:Sand \\[3ex] (a.) \\[3ex] \dfrac{Cement}{Sand} = \dfrac{1}{4} = \dfrac{???}{20} \\[5ex] \dfrac{20}{4} = 5 \\[3ex] 1 * 5 = 5 \\[3ex] 5\:\:buckets\:\:of\:\:Cement\:\:will\:\:be\:\:needed \\[3ex] (b.) \\[3ex] \dfrac{Sand}{Gravel} = \dfrac{4}{6} = \dfrac{20}{???} \\[5ex] \dfrac{20}{4} = 5 \\[3ex] 6 * 5 = 30 \\[3ex] 30\:\:buckets\:\:of\:\:Gravel\:\:will\:\:be\:\:needed \\[3ex] \underline{Second\:\:Method:\:\:Algebraically - Proportional\:\:Reasoning\:\:Method} \\[3ex] (i) \\[3ex] Let\:\:p\:\:be\:\:the\:\:number\:\:of\:\:buckets\:\:of\:\:Gravel\:\:needed\:\:for\:\:6\:\:buckets\:\:of\:\:Cement \\[3ex] $

Cement | Gravel |
---|---|

$1$ | $4$ |

$6$ | $p$ |

$ \dfrac{p}{4} = \dfrac{6}{1} \\[5ex] \dfrac{p}{4} = 6 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:4 \\[3ex] 4 * \dfrac{p}{4} = 4 * 6 \\[5ex] p = 24 \\[3ex] 24\:\:buckets\:\:of\:\:Gravel\:\:is\:\:needed \\[3ex] (ii) \\[3ex] \underline{Third\:\:Method:\:\:Algebraically - Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:x\:\:be\:\:the\:\:number\:\:of\:\:buckets\:\:of\:\:Cement\:\:needed\:\:for\:\:20\:\:buckets\:\:of\:\:Sand \\[3ex] Let\:\:y\:\:be\:\:the\:\:number\:\:of\:\:buckets\:\:of\:\:Gravel\:\:needed\:\:for\:\:20\:\:buckets\:\:of\:\:Sand \\[3ex] $

Cement | Sand | Gravel |
---|---|---|

$1$ | $4$ | $6$ |

$x$ | $20$ | $y$ |

$ \dfrac{x}{1} = \dfrac{20}{4} \\[5ex] x = 5 \\[3ex] 5\:\:buckets\:\:of\:\:Cement\:\:will\:\:be\:\:needed \\[3ex] Next \\[3ex] \dfrac{y}{6} = \dfrac{20}{4} \\[5ex] \dfrac{y}{6} = 5 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:6 \\[3ex] 6 * \dfrac{y}{6} = 6 * 5 \\[5ex] y = 30 \\[3ex] 30\:\:buckets\:\:of\:\:Gravel\:\:will\:\:be\:\:needed $

(20.) **ACT** Lian has $6\dfrac{1}{2}$ yards of ribbon she will use to make bows.

She will use $\dfrac{3}{4}$ yard of ribbon to make each bow.

After Lian has made all the bows possible with the ribbon, what length of ribbon, in yards, will NOT have been used to make bows?

$ A.\:\: 0 \\[3ex] B.\:\: \dfrac{1}{2} \\[5ex] C.\:\: \dfrac{21}{32} \\[5ex] D.\:\: \dfrac{2}{3} \\[5ex] E.\:\: \dfrac{7}{8} \\[5ex] $

*
Make up real-world scenarios with integers *

__First Scenario:__ You are given $\$20$ to buy books.

Each book costs $\$3$

Neglect taxes.

How many books can you buy?

How much money is the balance?

Ask students how they came up with their answers...both answers

The first answer is got by a__division__

The second answer involves__multiplication__, then __subtraction__

__Second Scenario:__ You are given $30$ yards of ribbon to decorate hats

Each hat is decorated using exactly $5$ ribbons

How many hats can be decorated? ...found by division

How many ribbons will be left?...found by multiplication, then subtraction

__Third Scenario:__ You are given $30$ yards of ribbon to decorate hats

Each hat is decorated using exactly $7$ ribbons

How many hats can be decorated? ...found by division

How many ribbons will be left?...found by multiplication, then subtraction

Bring to to Fractions

Take note of any *tricks*

$ 6\dfrac{1}{2} = \dfrac{2 * 6 + 1}{2} = \dfrac{12 + 1}{2} = \dfrac{13}{2} \\[5ex] Yards\:\:of\:\:ribbons\:\:to\:\:make\:\:bows = 6\dfrac{1}{2}\:yards \\[5ex] Yard\:\:of\:\:ribbon\:\:used\:\:to\:\:make\:\:each\:\:bow(1\:\:bow) = \dfrac{3}{4}\:yards \\[5ex] How\:\:many\:\:bows\:\:can\:\:be\:\:made\:\:with\:\:6\dfrac{1}{2}\:yards \\[5ex] = 6\dfrac{1}{2} \div \dfrac{3}{4} \\[5ex] = \dfrac{13}{2} \div \dfrac{3}{4} \\[5ex] = \dfrac{13}{2} * \dfrac{4}{3} \\[5ex] = \dfrac{13}{1} * \dfrac{2}{3} \\[5ex] = \dfrac{13 * 2}{1 * 3} \\[5ex] = \dfrac{26}{3} \\[5ex] = 8\dfrac{2}{3} \\[5ex] 6\dfrac{1}{2}\:yards\:\:is\:\:used\:\:to\:\:make\:\:8\dfrac{2}{3}\:\:bows \\[5ex] But: \\[3ex] \dfrac{2}{3}\:bow\:\:is\:\:not\:\:a\:\:bow...recycle\:\:it \\[3ex] Keep\:\:8\:\:bows \\[3ex] So\:\:how\:\:many\:\:yards\:\:are\:\:used\:\:to\:\:make\:\:8\:\:bows \\[3ex] dfrac{3}{4}\:yard\:\:is\:\:used\:\:to\:\:make\:\:1\:\:bow \\[5ex] \therefore \dfrac{3}{4} * 8\:\:yards\:\:will\:\:be\:\:used\:\:to\:\:make\:\:8\:\:bows \\[5ex] \dfrac{3}{4} * 8 = 3(2) = 6 \\[3ex] 6\:yards\:\:is\:\:used\:\:to\:\:make\:\:8\:\:bows \\[3ex] Total\:\:yards\:\:Given = 6\dfrac{1}{2}\:yards \\[5ex] Used = 6\:yards \\[3ex] Remaining = 6\dfrac{1}{2} - 6 = \dfrac{1}{2} \\[5ex] $ $\dfrac{1}{2}\:yard$ will NOT have been used to make bows.

She will use $\dfrac{3}{4}$ yard of ribbon to make each bow.

After Lian has made all the bows possible with the ribbon, what length of ribbon, in yards, will NOT have been used to make bows?

$ A.\:\: 0 \\[3ex] B.\:\: \dfrac{1}{2} \\[5ex] C.\:\: \dfrac{21}{32} \\[5ex] D.\:\: \dfrac{2}{3} \\[5ex] E.\:\: \dfrac{7}{8} \\[5ex] $

Each book costs $\$3$

Neglect taxes.

How many books can you buy?

How much money is the balance?

Ask students how they came up with their answers...both answers

The first answer is got by a

The second answer involves

Each hat is decorated using exactly $5$ ribbons

How many hats can be decorated? ...found by division

How many ribbons will be left?...found by multiplication, then subtraction

Each hat is decorated using exactly $7$ ribbons

How many hats can be decorated? ...found by division

How many ribbons will be left?...found by multiplication, then subtraction

Bring to to Fractions

Take note of any *tricks*

$ 6\dfrac{1}{2} = \dfrac{2 * 6 + 1}{2} = \dfrac{12 + 1}{2} = \dfrac{13}{2} \\[5ex] Yards\:\:of\:\:ribbons\:\:to\:\:make\:\:bows = 6\dfrac{1}{2}\:yards \\[5ex] Yard\:\:of\:\:ribbon\:\:used\:\:to\:\:make\:\:each\:\:bow(1\:\:bow) = \dfrac{3}{4}\:yards \\[5ex] How\:\:many\:\:bows\:\:can\:\:be\:\:made\:\:with\:\:6\dfrac{1}{2}\:yards \\[5ex] = 6\dfrac{1}{2} \div \dfrac{3}{4} \\[5ex] = \dfrac{13}{2} \div \dfrac{3}{4} \\[5ex] = \dfrac{13}{2} * \dfrac{4}{3} \\[5ex] = \dfrac{13}{1} * \dfrac{2}{3} \\[5ex] = \dfrac{13 * 2}{1 * 3} \\[5ex] = \dfrac{26}{3} \\[5ex] = 8\dfrac{2}{3} \\[5ex] 6\dfrac{1}{2}\:yards\:\:is\:\:used\:\:to\:\:make\:\:8\dfrac{2}{3}\:\:bows \\[5ex] But: \\[3ex] \dfrac{2}{3}\:bow\:\:is\:\:not\:\:a\:\:bow...recycle\:\:it \\[3ex] Keep\:\:8\:\:bows \\[3ex] So\:\:how\:\:many\:\:yards\:\:are\:\:used\:\:to\:\:make\:\:8\:\:bows \\[3ex] dfrac{3}{4}\:yard\:\:is\:\:used\:\:to\:\:make\:\:1\:\:bow \\[5ex] \therefore \dfrac{3}{4} * 8\:\:yards\:\:will\:\:be\:\:used\:\:to\:\:make\:\:8\:\:bows \\[5ex] \dfrac{3}{4} * 8 = 3(2) = 6 \\[3ex] 6\:yards\:\:is\:\:used\:\:to\:\:make\:\:8\:\:bows \\[3ex] Total\:\:yards\:\:Given = 6\dfrac{1}{2}\:yards \\[5ex] Used = 6\:yards \\[3ex] Remaining = 6\dfrac{1}{2} - 6 = \dfrac{1}{2} \\[5ex] $ $\dfrac{1}{2}\:yard$ will NOT have been used to make bows.

(21.) **ACT** At a refinery, $100,000$ tons of sand are required to produce each $60,000$ barrels
of a tarry material.

How many tons of sand are required to produce $3,000$ barrels of this tarry material?

$ A.\:\: 5,000 \\[3ex] B.\:\: 18,000 \\[3ex] C.\:\: 20,000 \\[3ex] D.\:\: 40,000 \\[3ex] E.\:\: 50,000 \\[3ex] $

Let the tons of sand required to produce $3,000$ barrels of this tarry material = $d$

$ \dfrac{d}{100000} = \dfrac{3000}{60000} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:100000 \\[5ex] 100000 * \dfrac{d}{100000} = 100000 * \dfrac{3000}{60000} \\[5ex] d = \dfrac{10000 * 3000}{60000} \\[5ex] d = \dfrac{5000 * 1}{1} \\[5ex] d = 5,000 \\[3ex] $ $5,000$ tons of sand are required to produce $3,000$ barrels of the tarry material

How many tons of sand are required to produce $3,000$ barrels of this tarry material?

$ A.\:\: 5,000 \\[3ex] B.\:\: 18,000 \\[3ex] C.\:\: 20,000 \\[3ex] D.\:\: 40,000 \\[3ex] E.\:\: 50,000 \\[3ex] $

Let the tons of sand required to produce $3,000$ barrels of this tarry material = $d$

$tons$ | $barrels$ |
---|---|

$100000$ | $60000$ |

$d$ | $3000$ |

$ \dfrac{d}{100000} = \dfrac{3000}{60000} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:100000 \\[5ex] 100000 * \dfrac{d}{100000} = 100000 * \dfrac{3000}{60000} \\[5ex] d = \dfrac{10000 * 3000}{60000} \\[5ex] d = \dfrac{5000 * 1}{1} \\[5ex] d = 5,000 \\[3ex] $ $5,000$ tons of sand are required to produce $3,000$ barrels of the tarry material

(22.) **ACT** The combined length of $3$ pieces of a board is $60$ inches.

The lengths of the pieces are in the ratio $3:5:7$

What is the length, in inches, of the longest piece?

$ A.\:\: 4 \\[3ex] B.\:\: 12 \\[3ex] C.\:\: 15 \\[3ex] D.\:\: 20 \\[3ex] E.\:\: 28 \\[3ex] $

The length of the longest piece is the piece with the greatest ratio value.

$ Length\:\:of\:\:the\:\:board = 60\:inches \\[3ex] Greatest\:\:ratio\:\:value = 7 \\[3ex] Sum\:\:of\:\:ratios = 3 + 5 + 7 = 15 \\[3ex] Length\:\:of\:\:longest\:\:piece = \dfrac{7}{15} * 60 \\[5ex] = 7(4) \\[3ex] = 28\:inches \\[3ex] $ The length of the longest piece is $28\:inches$

The lengths of the pieces are in the ratio $3:5:7$

What is the length, in inches, of the longest piece?

$ A.\:\: 4 \\[3ex] B.\:\: 12 \\[3ex] C.\:\: 15 \\[3ex] D.\:\: 20 \\[3ex] E.\:\: 28 \\[3ex] $

The length of the longest piece is the piece with the greatest ratio value.

$ Length\:\:of\:\:the\:\:board = 60\:inches \\[3ex] Greatest\:\:ratio\:\:value = 7 \\[3ex] Sum\:\:of\:\:ratios = 3 + 5 + 7 = 15 \\[3ex] Length\:\:of\:\:longest\:\:piece = \dfrac{7}{15} * 60 \\[5ex] = 7(4) \\[3ex] = 28\:inches \\[3ex] $ The length of the longest piece is $28\:inches$

(23.) **ACT** Maria ordered a pizza.

She ate only $\dfrac{2}{9}$ of it and gave the remaining pizza to her $3$ brothers.

What fraction of the whole pizza will each of Maria's brothers receive, if they share the remaining pizza equally?

$ F.\:\: \dfrac{7}{9} \\[5ex] G.\:\: \dfrac{3}{7} \\[5ex] H.\:\: \dfrac{1}{3} \\[5ex] J.\:\: \dfrac{7}{27} \\[5ex] K.\:\: \dfrac{2}{27} \\[5ex] $

$ Let\:\:the\:\:whole\:\:pizza = 1 \\[3ex] Because\:\:Total\:\:Fraction = 1 \\[3ex] Ate:\:\: \dfrac{2}{9} \\[5ex] Remaining:\:\: 1 - \dfrac{2}{9} = \dfrac{9}{9} - \dfrac{2}{9} = \dfrac{9 - 2}{9} = \dfrac{7}{9} \\[5ex] Shared\:\:the\:\:remaining\:\:among\:\:three\:\:brothers \\[3ex] Each\:\:share = \dfrac{7}{9} \div 3 \\[5ex] = \dfrac{7}{9} \div \dfrac{3}{1} \\[5ex] = \dfrac{7}{9} * \dfrac{1}{3} \\[5ex] = \dfrac{7 * 1}{9 * 3} \\[5ex] = \dfrac{7}{27} $

She ate only $\dfrac{2}{9}$ of it and gave the remaining pizza to her $3$ brothers.

What fraction of the whole pizza will each of Maria's brothers receive, if they share the remaining pizza equally?

$ F.\:\: \dfrac{7}{9} \\[5ex] G.\:\: \dfrac{3}{7} \\[5ex] H.\:\: \dfrac{1}{3} \\[5ex] J.\:\: \dfrac{7}{27} \\[5ex] K.\:\: \dfrac{2}{27} \\[5ex] $

$ Let\:\:the\:\:whole\:\:pizza = 1 \\[3ex] Because\:\:Total\:\:Fraction = 1 \\[3ex] Ate:\:\: \dfrac{2}{9} \\[5ex] Remaining:\:\: 1 - \dfrac{2}{9} = \dfrac{9}{9} - \dfrac{2}{9} = \dfrac{9 - 2}{9} = \dfrac{7}{9} \\[5ex] Shared\:\:the\:\:remaining\:\:among\:\:three\:\:brothers \\[3ex] Each\:\:share = \dfrac{7}{9} \div 3 \\[5ex] = \dfrac{7}{9} \div \dfrac{3}{1} \\[5ex] = \dfrac{7}{9} * \dfrac{1}{3} \\[5ex] = \dfrac{7 * 1}{9 * 3} \\[5ex] = \dfrac{7}{27} $

(24.) **ACT** Marcus's favorite casserole recipe requires $3$ eggs and makes $6$ servings.

Marcus will modify the recipe by using $5$ eggs and increasing all other ingredients in the recipe proportionally.

What is the total number of servings the modified recipe will make?

$ A.\:\: 6 \\[3ex] B.\:\: 8 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 15 \\[3ex] $

Let the number of servings to be made by the modified recipe = $n$

$ \dfrac{n}{5} = \dfrac{6}{3} \\[5ex] \dfrac{n}{5} = 2 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[5ex] 5 * \dfrac{n}{5} = 5(2) \\[5ex] n = 10 \\[3ex] $ $10$ servings will be made from $5$ eggs

Marcus will modify the recipe by using $5$ eggs and increasing all other ingredients in the recipe proportionally.

What is the total number of servings the modified recipe will make?

$ A.\:\: 6 \\[3ex] B.\:\: 8 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 15 \\[3ex] $

Let the number of servings to be made by the modified recipe = $n$

$eggs$ | $servings$ |
---|---|

$3$ | $6$ |

$5$ | $n$ |

$ \dfrac{n}{5} = \dfrac{6}{3} \\[5ex] \dfrac{n}{5} = 2 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[5ex] 5 * \dfrac{n}{5} = 5(2) \\[5ex] n = 10 \\[3ex] $ $10$ servings will be made from $5$ eggs

(25.) **JAMB** Three boys shared some oranges.

The first received $\dfrac{1}{3}$ of the oranges.

The second received $\dfrac{2}{3}$ of the remainder.

If the third boy received the remaining $12$ oranges, how many oranges did they share?

$ A.\:\: 60 \\[3ex] B.\:\: 54 \\[3ex] C.\:\: 48 \\[3ex] D.\:\: 42 \\[3ex] $

We can solve this in at least two ways.

One way is to solve it Algebraically...Question $68$ of Word Problems on Linear Equations

Another method is to use the*Proportional Reasoning Method* (as shown below)

$ Total\:\:Fraction = 1 \\[3ex] First\:\:boy's\:\:share = \dfrac{1}{3} \\[5ex] Remainder = 1 - \dfrac{1}{3} = \dfrac{3}{3} - \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3} \\[5ex] Second\:\:boy's\:\:share = \dfrac{2}{3}\:\:of\:\:Remainder = \dfrac{2}{3} * \dfrac{2}{3} = \dfrac{2 * 2}{3 * 3} = \dfrac{4}{9} \\[5ex] First\:\:boy's\:\:and\:\:Second\:\:boys's\:\:shares = \dfrac{1}{3} + \dfrac{4}{9} = \dfrac{3}{9} + \dfrac{4}{9} = \dfrac{3 + 4}{9} = \dfrac{7}{9} \\[5ex] Third\:\:boy's\:\:share = Remainder = 1 - \dfrac{7}{9} = \dfrac{9}{9} - \dfrac{7}{9} = \dfrac{9 - 7}{9} = \dfrac{2}{9} \\[5ex] Also:\:\:Third\:\:boy\:\:received\:\:12\:\:oranges \\[3ex] $ Let the number of oranges shared be $p$

$ \dfrac{p}{1} = \dfrac{12}{\dfrac{2}{9}} \\[7ex] p = 12 \div \dfrac{2}{9} \\[5ex] p = 12 * \dfrac{9}{2} \\[5ex] p = 6 * 9 \\[3ex] p = 54 \\[3ex] $ $54$ oranges were shared among the three boys.

$ \underline{Check} \\[3ex] First\:\:boy's\:\:share = \dfrac{1}{3}\:\:of\:\:54 = \dfrac{1}{3} * 54 = 18\:\:oranges \\[5ex] Second\:\:boy's\:\:share = \dfrac{4}{9}\:\:of\:\:54 = \dfrac{4}{9} * 54 = 4 * 6 = 24\:\:oranges \\[5ex] Third\:\:boy's\:\:share = 12\:\:oranges \\[3ex] Total = 18 + 24 + 12 = 54\:\:oranges $

The first received $\dfrac{1}{3}$ of the oranges.

The second received $\dfrac{2}{3}$ of the remainder.

If the third boy received the remaining $12$ oranges, how many oranges did they share?

$ A.\:\: 60 \\[3ex] B.\:\: 54 \\[3ex] C.\:\: 48 \\[3ex] D.\:\: 42 \\[3ex] $

We can solve this in at least two ways.

One way is to solve it Algebraically...Question $68$ of Word Problems on Linear Equations

Another method is to use the

$ Total\:\:Fraction = 1 \\[3ex] First\:\:boy's\:\:share = \dfrac{1}{3} \\[5ex] Remainder = 1 - \dfrac{1}{3} = \dfrac{3}{3} - \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3} \\[5ex] Second\:\:boy's\:\:share = \dfrac{2}{3}\:\:of\:\:Remainder = \dfrac{2}{3} * \dfrac{2}{3} = \dfrac{2 * 2}{3 * 3} = \dfrac{4}{9} \\[5ex] First\:\:boy's\:\:and\:\:Second\:\:boys's\:\:shares = \dfrac{1}{3} + \dfrac{4}{9} = \dfrac{3}{9} + \dfrac{4}{9} = \dfrac{3 + 4}{9} = \dfrac{7}{9} \\[5ex] Third\:\:boy's\:\:share = Remainder = 1 - \dfrac{7}{9} = \dfrac{9}{9} - \dfrac{7}{9} = \dfrac{9 - 7}{9} = \dfrac{2}{9} \\[5ex] Also:\:\:Third\:\:boy\:\:received\:\:12\:\:oranges \\[3ex] $ Let the number of oranges shared be $p$

amount (in fractions) | actual amount |
---|---|

$\dfrac{2}{9}$ | $12$ |

$1$ | $p$ |

$ \dfrac{p}{1} = \dfrac{12}{\dfrac{2}{9}} \\[7ex] p = 12 \div \dfrac{2}{9} \\[5ex] p = 12 * \dfrac{9}{2} \\[5ex] p = 6 * 9 \\[3ex] p = 54 \\[3ex] $ $54$ oranges were shared among the three boys.

$ \underline{Check} \\[3ex] First\:\:boy's\:\:share = \dfrac{1}{3}\:\:of\:\:54 = \dfrac{1}{3} * 54 = 18\:\:oranges \\[5ex] Second\:\:boy's\:\:share = \dfrac{4}{9}\:\:of\:\:54 = \dfrac{4}{9} * 54 = 4 * 6 = 24\:\:oranges \\[5ex] Third\:\:boy's\:\:share = 12\:\:oranges \\[3ex] Total = 18 + 24 + 12 = 54\:\:oranges $

(26.) **ACT** Every camera lens has a measurement called the focal length, $f$, such that when
an object is in focus, the distance from the object to the center of the lens, $D_o$, and the
distance from the center of the lens to the film, $D_i$, satisfy the equation
$\dfrac{1}{D_o} + \dfrac{1}{D_i} = \dfrac{1}{f}$.

If the object is in focus, $D_o = 36$ centimeters, and $D_i = 12$ centimeters, what is the focal length of the lens, in centimeters?

$ A\:\: 3 \\[3ex] B.\:\: 6 \\[3ex] C.\:\: 9 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 24 \\[3ex] $

$ \dfrac{1}{D_o} + \dfrac{1}{D_i} = \dfrac{1}{f} \\[5ex] D_o = 36\:cm \\[3ex] D_i = 12\:cm \\[3ex] f = ? \\[3ex] \dfrac{1}{36} + \dfrac{1}{12} = \dfrac{1}{f} \\[5ex] \dfrac{1}{f} = \dfrac{1}{36} + \dfrac{1}{12} \\[5ex] \dfrac{1}{f} = \dfrac{1}{36} + \dfrac{3}{36} = \dfrac{1 + 3}{36} \\[5ex] \dfrac{1}{f} = \dfrac{4}{36} \\[5ex] f = \dfrac{36}{4} \\[5ex] f = 9\:cm $

If the object is in focus, $D_o = 36$ centimeters, and $D_i = 12$ centimeters, what is the focal length of the lens, in centimeters?

$ A\:\: 3 \\[3ex] B.\:\: 6 \\[3ex] C.\:\: 9 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 24 \\[3ex] $

$ \dfrac{1}{D_o} + \dfrac{1}{D_i} = \dfrac{1}{f} \\[5ex] D_o = 36\:cm \\[3ex] D_i = 12\:cm \\[3ex] f = ? \\[3ex] \dfrac{1}{36} + \dfrac{1}{12} = \dfrac{1}{f} \\[5ex] \dfrac{1}{f} = \dfrac{1}{36} + \dfrac{1}{12} \\[5ex] \dfrac{1}{f} = \dfrac{1}{36} + \dfrac{3}{36} = \dfrac{1 + 3}{36} \\[5ex] \dfrac{1}{f} = \dfrac{4}{36} \\[5ex] f = \dfrac{36}{4} \\[5ex] f = 9\:cm $

(27.) **ACT** Siblings Peter, Paul, and Mary earned a total of $\$200$ shoveling snow.

If Peter earned $37\%$ of the total and Paul earned $\$16$ what fraction of the $\$200$ did Mary earn?

$ A.\:\: \dfrac{1}{3} \\[5ex] B.\:\: \dfrac{1}{2} \\[5ex] C.\:\: \dfrac{11}{20} \\[5ex] D.\:\: \dfrac{131}{200} \\[5ex] E.\:\: \dfrac{147}{200} \\[5ex] $

$ Total = \$200 \\[3ex] Peter's\:\:share = 37\%\:\:of\:\:200 = \dfrac{37}{100} * 200 = 37(2) = \$74 \\[5ex] Paul's\:\:share = \$16 \\[3ex] Peter's\:\:and\:\:Paul's\:\:shares = 74 + 16 = \$90 \\[3ex] Mary's\:\:share = 200 - 90 = \$110 \\[3ex] What\:\:fraction\:\:of\:\:200\:\:is\:\:110? \\[3ex] = \dfrac{is}{of} \\[5ex] = \dfrac{110}{200} \\[5ex] = \dfrac{11}{20} \\[5ex] \underline{Check} \\[3ex] \dfrac{11}{20}\:\:of\:\:200 = \dfrac{11}{20} * 200 = 11(10) = 110 $

If Peter earned $37\%$ of the total and Paul earned $\$16$ what fraction of the $\$200$ did Mary earn?

$ A.\:\: \dfrac{1}{3} \\[5ex] B.\:\: \dfrac{1}{2} \\[5ex] C.\:\: \dfrac{11}{20} \\[5ex] D.\:\: \dfrac{131}{200} \\[5ex] E.\:\: \dfrac{147}{200} \\[5ex] $

$ Total = \$200 \\[3ex] Peter's\:\:share = 37\%\:\:of\:\:200 = \dfrac{37}{100} * 200 = 37(2) = \$74 \\[5ex] Paul's\:\:share = \$16 \\[3ex] Peter's\:\:and\:\:Paul's\:\:shares = 74 + 16 = \$90 \\[3ex] Mary's\:\:share = 200 - 90 = \$110 \\[3ex] What\:\:fraction\:\:of\:\:200\:\:is\:\:110? \\[3ex] = \dfrac{is}{of} \\[5ex] = \dfrac{110}{200} \\[5ex] = \dfrac{11}{20} \\[5ex] \underline{Check} \\[3ex] \dfrac{11}{20}\:\:of\:\:200 = \dfrac{11}{20} * 200 = 11(10) = 110 $

(28.) **ACT** A roof rises $4$ inches for each $12$ inches of horizontal run.

This roof rises $30\dfrac{1}{2}\:inches$ in how many inches of horizontal run?

$ F.\:\: 10\dfrac{1}{6} \\[5ex] G.\:\: 22\dfrac{1}{2} \\[5ex] H.\:\: 38\dfrac{1}{2} \\[5ex] J.\:\: 91\dfrac{1}{2} \\[5ex] K.\:\: 122 \\[3ex] $

Let the number of inches of horizontal run for $30\dfrac{1}{2}\:inches$ of rise = $n$

$ \dfrac{n}{12} = \dfrac{30\dfrac{1}{2}}{4} \\[7ex] \dfrac{n}{12} = 30\dfrac{1}{2} \div 4 \\[5ex] 30\dfrac{1}{2} = \dfrac{2 * 30 + 1}{2} = \dfrac{60 + 1}{2} = \dfrac{61}{2} \\[5ex] \dfrac{n}{12} = \dfrac{61}{2} \div 4 \\[5ex] \dfrac{n}{12} = \dfrac{61}{2} * \dfrac{1}{4} \\[5ex] \dfrac{n}{12} = \dfrac{61}{8} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[5ex] 12 * \dfrac{n}{12} = 12 * \dfrac{61}{8} \\[5ex] n = \dfrac{3 * 61}{2} = \dfrac{183}{2} \\[5ex] n = 91\dfrac{1}{2}\:inches \\[5ex] $ This roof rises $30\dfrac{1}{2}\:inches$ in how many $91\dfrac{1}{2}\:inches$ of horizontal run

This roof rises $30\dfrac{1}{2}\:inches$ in how many inches of horizontal run?

$ F.\:\: 10\dfrac{1}{6} \\[5ex] G.\:\: 22\dfrac{1}{2} \\[5ex] H.\:\: 38\dfrac{1}{2} \\[5ex] J.\:\: 91\dfrac{1}{2} \\[5ex] K.\:\: 122 \\[3ex] $

Let the number of inches of horizontal run for $30\dfrac{1}{2}\:inches$ of rise = $n$

$rise$ | $run$ |
---|---|

$4$ | $12$ |

$30\dfrac{1}{2}$ | $n$ |

$ \dfrac{n}{12} = \dfrac{30\dfrac{1}{2}}{4} \\[7ex] \dfrac{n}{12} = 30\dfrac{1}{2} \div 4 \\[5ex] 30\dfrac{1}{2} = \dfrac{2 * 30 + 1}{2} = \dfrac{60 + 1}{2} = \dfrac{61}{2} \\[5ex] \dfrac{n}{12} = \dfrac{61}{2} \div 4 \\[5ex] \dfrac{n}{12} = \dfrac{61}{2} * \dfrac{1}{4} \\[5ex] \dfrac{n}{12} = \dfrac{61}{8} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[5ex] 12 * \dfrac{n}{12} = 12 * \dfrac{61}{8} \\[5ex] n = \dfrac{3 * 61}{2} = \dfrac{183}{2} \\[5ex] n = 91\dfrac{1}{2}\:inches \\[5ex] $ This roof rises $30\dfrac{1}{2}\:inches$ in how many $91\dfrac{1}{2}\:inches$ of horizontal run

(29.) **WASSCE** Thirty five coloured balls were shared among four teams such that one team takes all
the red balls.

If the remainder is shared to the other teams in the ratio $4:3:2$ and the smallest share was $6$ balls, how many red balls were there?

$ 35\:\:coloured\:\:balls\:\:including\:\:red\:\:balls \\[3ex] Four\:\:teams \\[3ex] Let\:\:the\:\:number\:\:of\:\:red\:\:balls = R \\[3ex] Remaining\:\:colored\:\:balls = 35 - R \\[3ex] One\:\:team\:\:took\:\:R\:\:balls \\[3ex] Three\:\:teams\:\:took\:\:(35 - R)\:\:balls \\[3ex] (35 - R)\:\:balls\:\:shared\:\:among\:\:three\:\:teams\:\:in\:\:the\:\:ratio\:\:of\:\:4:3:2 \\[3ex] Ratio = 4:3:2 \\[3ex] Sum\:\:of\:\:ratios = 4 + 3 + 2 = 9 \\[3ex] Smallest\:\:ratio = 2 \\[3ex] Smallest\:\:share = \dfrac{2}{9} * (35 - R) \\[5ex] Smallest\:\:share = 6\:\:balls \\[3ex] \implies \dfrac{2}{9} * (35 - R) = 6 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{9}{2} \\[5ex] \dfrac{9}{2} * \dfrac{2}{9} * (35 - R) = \dfrac{9}{2} * 6 \\[5ex] 35 - R = 9(3) \\[3ex] 35 - R = 27 \\[3ex] 35 - 27 = R \\[3ex] 8 = R \\[3ex] R = 8 \\[3ex] $ There were $8$ red balls.

If the remainder is shared to the other teams in the ratio $4:3:2$ and the smallest share was $6$ balls, how many red balls were there?

$ 35\:\:coloured\:\:balls\:\:including\:\:red\:\:balls \\[3ex] Four\:\:teams \\[3ex] Let\:\:the\:\:number\:\:of\:\:red\:\:balls = R \\[3ex] Remaining\:\:colored\:\:balls = 35 - R \\[3ex] One\:\:team\:\:took\:\:R\:\:balls \\[3ex] Three\:\:teams\:\:took\:\:(35 - R)\:\:balls \\[3ex] (35 - R)\:\:balls\:\:shared\:\:among\:\:three\:\:teams\:\:in\:\:the\:\:ratio\:\:of\:\:4:3:2 \\[3ex] Ratio = 4:3:2 \\[3ex] Sum\:\:of\:\:ratios = 4 + 3 + 2 = 9 \\[3ex] Smallest\:\:ratio = 2 \\[3ex] Smallest\:\:share = \dfrac{2}{9} * (35 - R) \\[5ex] Smallest\:\:share = 6\:\:balls \\[3ex] \implies \dfrac{2}{9} * (35 - R) = 6 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{9}{2} \\[5ex] \dfrac{9}{2} * \dfrac{2}{9} * (35 - R) = \dfrac{9}{2} * 6 \\[5ex] 35 - R = 9(3) \\[3ex] 35 - R = 27 \\[3ex] 35 - 27 = R \\[3ex] 8 = R \\[3ex] R = 8 \\[3ex] $ There were $8$ red balls.

(30.) **WASSCE** There are $5$ more girls than boys in a class.

If $2$ boys join the class, the ratio of girls to boys will be $5:4$.

Find the:

(i) number of girls in the class;

(ii) total number of pupils in the class.

$ \underline{Initial\:\:Count} \\[3ex] Let: \\[3ex] number\:\:of\:\:boys = p \\[3ex] number\:\:of\:\:girls = p + 5...(5\:\:more\:\:girls\:\:than\:\:boys) \\[3ex] \underline{Later\:\:Count...based\:\:on\:\:the\:\:condition,\:\: "if"} \\[3ex] number\:\:of\:\:boys = p + 2...(2\:\:more\:\:boys\:\:join) \\[3ex] number\:\:of\:\:girls = p + 5...(no\:\:change) \\[3ex] Ratio\:\:of\:\:girls:boys = 5:4 \\[3ex] \implies \dfrac{p + 5}{p + 2} = \dfrac{5}{4} \\[5ex] 4(p + 5) = 5(p + 2) \\[3ex] 4p + 20 = 5p + 10 \\[3ex] 20 - 10 = 5p - 4p \\[3ex] 10 = p \\[3ex] p = 10 \\[3ex] Number\:\:of\:\:boys = p = 10 \\[3ex] 10\:\:boys \\[3ex] $*
*__Student:__ Excuse me Ma'am/Sir

I thought the boys would be $10 + 2 = 12$

Rather than $10$

__Teacher:__ Yes, you have a point.

The initial number of boys is $p = 10$

However, the later count is based on a conditional statement, "if"

"If" $2$ boys join the class...then the ratio is ...

This does not imply that $2$ boys "actually" joined them

This was to assist us in determining the number of girls and boys in the class

So, we have to go by the Initial Count, rather than the "Conditional" Later Count.

$ (i) \\[3ex] Number\:\:of\:\:girls \\[3ex] = p + 5 \\[3ex] = 10 + 5 = 15 \\[3ex] 15\:\:girls \\[3ex] (ii) \\[3ex] total\:\:number\:\:of\:\:pupils\:\:in\:\:the\:\:class \\[3ex] 10 + 15 = 25 \\[3ex] 15\:\:pupils $

If $2$ boys join the class, the ratio of girls to boys will be $5:4$.

Find the:

(i) number of girls in the class;

(ii) total number of pupils in the class.

$ \underline{Initial\:\:Count} \\[3ex] Let: \\[3ex] number\:\:of\:\:boys = p \\[3ex] number\:\:of\:\:girls = p + 5...(5\:\:more\:\:girls\:\:than\:\:boys) \\[3ex] \underline{Later\:\:Count...based\:\:on\:\:the\:\:condition,\:\: "if"} \\[3ex] number\:\:of\:\:boys = p + 2...(2\:\:more\:\:boys\:\:join) \\[3ex] number\:\:of\:\:girls = p + 5...(no\:\:change) \\[3ex] Ratio\:\:of\:\:girls:boys = 5:4 \\[3ex] \implies \dfrac{p + 5}{p + 2} = \dfrac{5}{4} \\[5ex] 4(p + 5) = 5(p + 2) \\[3ex] 4p + 20 = 5p + 10 \\[3ex] 20 - 10 = 5p - 4p \\[3ex] 10 = p \\[3ex] p = 10 \\[3ex] Number\:\:of\:\:boys = p = 10 \\[3ex] 10\:\:boys \\[3ex] $

I thought the boys would be $10 + 2 = 12$

Rather than $10$

The initial number of boys is $p = 10$

However, the later count is based on a conditional statement, "if"

"If" $2$ boys join the class...then the ratio is ...

This does not imply that $2$ boys "actually" joined them

This was to assist us in determining the number of girls and boys in the class

So, we have to go by the Initial Count, rather than the "Conditional" Later Count.

$ (i) \\[3ex] Number\:\:of\:\:girls \\[3ex] = p + 5 \\[3ex] = 10 + 5 = 15 \\[3ex] 15\:\:girls \\[3ex] (ii) \\[3ex] total\:\:number\:\:of\:\:pupils\:\:in\:\:the\:\:class \\[3ex] 10 + 15 = 25 \\[3ex] 15\:\:pupils $

(31.)

(32.) **ACT** The ratio of Alani's height to Baahir's height is $5:7$

The ratio of Baahir's height to Connor's height is $4:3$

What is the ratio of Alani's height to Connor's height?

$ A.\:\: 2:3 \\[3ex] B.\:\: 8:11 \\[3ex] C.\:\: 15:28 \\[3ex] D.\:\: 20:21 \\[3ex] E.\:\: 28:15 \\[3ex] $

Let Alani's height = $A$

Let Baahir's height = $B$

Let Connor's height = $C$

$ \dfrac{A}{B} = \dfrac{5}{7} \\[5ex] \dfrac{B}{C} = \dfrac{4}{3} \\[5ex] \dfrac{A}{C} = \dfrac{A}{B} * \dfrac{B}{C} \\[5ex] = \dfrac{5}{7} * \dfrac{4}{3} \\[5ex] = \dfrac{20}{21} \\[5ex] $*
Student: Why did you multiply? *

Teacher: The question asked for the ratio of Alani's height to Connor's height

That is $\dfrac{A}{C}$

$ \dfrac{A}{C} = \dfrac{A}{B} * \dfrac{B}{C} \\[5ex] $ The $B$ cancels out.

The ratio of Baahir's height to Connor's height is $4:3$

What is the ratio of Alani's height to Connor's height?

$ A.\:\: 2:3 \\[3ex] B.\:\: 8:11 \\[3ex] C.\:\: 15:28 \\[3ex] D.\:\: 20:21 \\[3ex] E.\:\: 28:15 \\[3ex] $

Let Alani's height = $A$

Let Baahir's height = $B$

Let Connor's height = $C$

$ \dfrac{A}{B} = \dfrac{5}{7} \\[5ex] \dfrac{B}{C} = \dfrac{4}{3} \\[5ex] \dfrac{A}{C} = \dfrac{A}{B} * \dfrac{B}{C} \\[5ex] = \dfrac{5}{7} * \dfrac{4}{3} \\[5ex] = \dfrac{20}{21} \\[5ex] $

Teacher: The question asked for the ratio of Alani's height to Connor's height

That is $\dfrac{A}{C}$

$ \dfrac{A}{C} = \dfrac{A}{B} * \dfrac{B}{C} \\[5ex] $ The $B$ cancels out.

(33.)

(34.) **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 (2.); 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 (2.); 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

(35.)

(36.)

(37.)

(38.) **ACT** According to the United State Department of Commerce, the approximate area of Rhode Island is
$1.5 * 10^3$ square miles and the approximate area of Alaska is $6.6 * 10^5$ square miles.

Using these measurements, which of the following is closest to the ratio of the area of Alaska to the area of Rhode Island?

$ A.\;\; 22:5 \\[3ex] B.\;\; 440:1 \\[3ex] C.\;\; 484:25 \\[3ex] D.\;\; 510:1 \\[3ex] E.\;\; 810:1 \\[3ex] $

Both areas are the same unit (square miles)

The ratio is unitless

$ Ratio = \dfrac{area\;\;of\;\;Alaska}{area\;\;of\;\;Rhode\;\;Island} \\[5ex] = \dfrac{6.6 * 10^5}{1.5 * 10^3} \\[5ex] = \dfrac{660000}{1500} \\[5ex] = \dfrac{440}{1} \\[5ex] = 440:1 $

Using these measurements, which of the following is closest to the ratio of the area of Alaska to the area of Rhode Island?

$ A.\;\; 22:5 \\[3ex] B.\;\; 440:1 \\[3ex] C.\;\; 484:25 \\[3ex] D.\;\; 510:1 \\[3ex] E.\;\; 810:1 \\[3ex] $

Both areas are the same unit (square miles)

The ratio is unitless

$ Ratio = \dfrac{area\;\;of\;\;Alaska}{area\;\;of\;\;Rhode\;\;Island} \\[5ex] = \dfrac{6.6 * 10^5}{1.5 * 10^3} \\[5ex] = \dfrac{660000}{1500} \\[5ex] = \dfrac{440}{1} \\[5ex] = 440:1 $

(39.)

(40.)

(41.)

(42.)

(43.)

(44.) **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] $

Similar to Questions (2.) and (34.); 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] $

Similar to Questions (2.) and (34.); 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

(45.)

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(49.)

(50.) **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

(51.)

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(53.)

(54.) **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.

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(60.) **ACT** A survey in a study skills class asked the 20 students enrolled in the class how many hours
(rounded to the nearest hour) they has spent studying on the previous evening.

The 20 responses are summarized by the histogram below.

What fraction of the students responded that they had spent less than 3 hours studying?

$ A.\;\; \dfrac{13}{100} \\[5ex] B.\;\; \dfrac{1}{5} \\[5ex] C.\;\; \dfrac{3}{10} \\[5ex] D.\;\; \dfrac{13}{20} \\[5ex] E.\;\; \dfrac{17}{20} \\[5ex] $

$ Total\;\;number\;\;of\;\;students = whole = 20 \\[3ex] \underline{Less\;\;than\;\;3\;\;hours} \\[3ex] 0\;hour \rightarrow 2\;\;students \\[3ex] 1\;\;hour \rightarrow 5\;\;students \\[3ex] 2\;hours \rightarrow 6\;\;students \\[3ex] Number\;\;of\;\;students = part = 2 + 5 + 6 = 13 \\[3ex] Fraction\;\;of\;\;students = \dfrac{part}{whole} = \dfrac{13}{20} $

The 20 responses are summarized by the histogram below.

What fraction of the students responded that they had spent less than 3 hours studying?

$ A.\;\; \dfrac{13}{100} \\[5ex] B.\;\; \dfrac{1}{5} \\[5ex] C.\;\; \dfrac{3}{10} \\[5ex] D.\;\; \dfrac{13}{20} \\[5ex] E.\;\; \dfrac{17}{20} \\[5ex] $

$ Total\;\;number\;\;of\;\;students = whole = 20 \\[3ex] \underline{Less\;\;than\;\;3\;\;hours} \\[3ex] 0\;hour \rightarrow 2\;\;students \\[3ex] 1\;\;hour \rightarrow 5\;\;students \\[3ex] 2\;hours \rightarrow 6\;\;students \\[3ex] Number\;\;of\;\;students = part = 2 + 5 + 6 = 13 \\[3ex] Fraction\;\;of\;\;students = \dfrac{part}{whole} = \dfrac{13}{20} $

(61.)

(62.)