Solved Examples: All Measurements and Conversions

(a.) $\dfrac{1.057\;quart}{1\;liter}$ because multiplying it by any number of liters would convert the number from liters to quarts by causing the units of liters to cancel without changing number because the factor is equal to 1.
Dividing both sides of the given equation by 1​ liter, shows that the​ factor, $\dfrac{1.057\;quart}{1\;liter}$​, is equivalent to​ 1, allowing it to multiply and cancel the units of liter while leaving the units of quarts and without changing the value of the quantity.​ Therefore, multiplying by the factor only changes the units of what it multiplies.

(b.) Divide the volume by the​ depth, because this will cancel the depth in the volume and give feet square in the numerator which are the units for surface area.
Dividing the volume by the depth cancels a unit of feet from the​ numerator, so the numerator will have square​ feet, and the resulting number will be the area of the surface of the lake since it is the length multiplied by the width of the lake.

(c.) Yes; you need to know how long the light bulb is on because power is the rate at which energy is​ used, and watts are units of power.

(d.) Hot enough to boil​ water, since 110°C is above the boiling point of water.

(e.) Parts per million may be used because it gives the number of molecules of carbon dioxide in a million molecules of air.

(f.) 300 mg per kilogram of body weight per day: 30-kg child every 8​ hours
Convert day to hours: 1 day = 24 hours
So, this can be reworded as:
300 mg per kilogram of body weight per 24-hours: 30-kg child per 8​ hours

$ \underline{Unity\;\;Fraction\;\;Method} \\[3ex] \dfrac{300\;mg}{kg} * \dfrac{1}{24\;hours} * \dfrac{30\;kg} * \dfrac{8\;hours}{1} \\[5ex] $ Multiply 300​ mg/kg/day by 30 kg and divide that result by​ 3, since multiplying will cancel the kg units and dividing by 3 since there are three 8 hour intervals in a day.

(g.) A person with 4 liters of blood has 0.08 * 40 = 3.2 grams of alcohol in his blood because $ \dfrac{0.08g * 40}{100mL * 40} = \dfrac{3.2g}{4L} $

Each gallon means $1$ gallon

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] Convert\:\:5000\:gallons\:\:to\:\:pounds \\[3ex] 5000\:gallons * \dfrac{.....pounds}{.....gallons} \\[5ex] 5000\:gallons * \dfrac{8\:pounds}{1\:gallon} \\[5ex] = \dfrac{5000 * 8}{1} \\[5ex] = 40,000\:pounds \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:the\:\:converted\:\:unit\:\:in\:\:pounds \\[3ex] $

gallons	pounds
$1$	$8$
$5000$	$p$

$ \dfrac{p}{8} = \dfrac{5000}{1} \\[5ex] Cross\:\:Multiply \\[3ex] p * 1 = 8 * 5000 \\[3ex] p = 40,000\:pounds $

Measurement is Mass

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 0.45\;kg \:\:to\:\: lbs \\[3ex] = 0.45\;kg * \dfrac{...lbs}{...kg} \\[5ex] = 0.45\;kg * \dfrac{2.2\;lbs}{1\;kg} \\[5ex] = 0.99\;lbs $

Second Method: Proportional Reasoning Method

$kg$	$lbs$
$1$	$2.2$
$0.45$	$x$

$ \dfrac{2.2}{1} = \dfrac{x}{0.45} \\[5ex] Cross\:\:Multiply \\[3ex] 1(x) = 2.2(0.45) \\[3ex] x = 0.99\;lbs $

cost per ounce means what is the cost for $1$ ounce?
What are the three variables in question?
Ounce, Pounds, and Cost

Third Method: Fast Proportional Reasoning Method

ounce	pounds	cost$(\$)$
$16$	$1$
$x$	$1.5$	$2.88$
$1$		$y$

Let $x$ be the converted unit of $1.5$ pounds in ounce
Let $y$ be the cost of $1$ ounce
We need to find $y$...that is our goal.
But, before we find $y$, we have to find $x$
Use the first two rows to find $x$
Then, use the last two rows to find $y$

$ \dfrac{16}{x} = \dfrac{1}{1.5} \\[5ex] Cross\:\:Multiply \\[3ex] x(1) = 16(1.5) \\[3ex] x = 24\:ounces \\[3ex] Next: \\[3ex] \dfrac{x}{1} = \dfrac{2.88}{y} \\[5ex] \dfrac{24}{1} = \dfrac{2.88}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 24(y) = 1(2.88) \\[3ex] 24y = 2.88 \\[3ex] y = \dfrac{2.88}{24} \\[5ex] y = \$0.12 \\[3ex] $ The cost per ounce is $\$0.12$

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] Given: \\[3ex] 16\:ounces\:\:in\:\:a\:\:pound \\[3ex] 1.5\:pounds\:\:for\:\:\$2.88 \\[3ex] To\:\:find\:\:...cost(\$)\:\:per\:\:ounce \\[3ex] Set\:\:up\:\:first \\[3ex] \dfrac{.....\$}{.....lb} * \dfrac{.....lb}{.....oz} \\[5ex] = \dfrac{\$2.88}{1.5\:lb} * \dfrac{1\:lb}{16\:oz} \\[5ex] = \dfrac{2.88 * 1}{1.5 * 16} \\[5ex] = \dfrac{2.88}{24} \\[5ex] = \$0.12 $

(a.) Weight of 15 bags where 1 bag: 42 pounds (to kilograms)
First Method: Unity Fraction Method

$ 42\;pound * \dfrac{.......gram}{.......pound} * \dfrac{.......kilogram}{.......gram} \\[5ex] = 42\;pound * \dfrac{453.59237\;gram}{1\;pound} * \dfrac{1\;kilogram}{10^3\;gram} \\[5ex] = \dfrac{42 * 453.59237}{1000}\;kilogram \\[5ex] = 19.05087954\;kg \\[3ex] $ Second Method: Proportional Reasoning Method

Let $p$ be the weight of 15 bags in kilogram

bag	kilogram
1	19.05087954
15	$p$

$ \dfrac{p}{15} = \dfrac{19.05087954}{1} \\[5ex] p(1) = 15(19.05087954) \\[3ex] p = 285.7631931\;kilograms \\[3ex] $ (b.) Volume of 15 bags where 1 bag: 2 cubic feet (to cubic yards)
First Method: Unity Fraction Method

$ 2\;ft * ft * ft * \dfrac{.......yd}{.......ft} * \dfrac{.......yd}{.......ft} * \dfrac{.......yd}{.......ft} \\[5ex] = 2\;ft * ft * ft * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;yd}{3\;ft} \\[5ex] = \dfrac{2}{27}\;cubic\;\;yards \\[5ex] \approximately 0.0740740741\;yd^3 \\[3ex] $ Second Method: Proportional Reasoning Method

Let $k$ be the volume of 15 bags in cubic yards

bag	cubic yards
1	0.0740740741
15	$k$

$ \dfrac{k}{15} = \dfrac{0.0740740741}{1} \\[5ex] k(1) = 15(0.0740740741) \\[3ex] k \approximately 1.111111111111\;yd^3 \\[3ex] $

Per hour means 1 hour

Second Method: Proportional Reasoning Method
Let $y$ be the number of miles that you have traveled

$mi$	$hr$
$x$	$1$
$y$	$3$

$ \dfrac{x}{y} = \dfrac{1}{3} \\[5ex] Cross\:\:Multiply \\[3ex] y(1) = x(3) \\[3ex] y = 3x\:miles \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] x\dfrac{miles}{hr} * ....hr \\[5ex] x\dfrac{miles}{1\:hr} * 3\:hrs \\[5ex] = 3x\:miles \\[3ex] $ You've traveled $3x$ miles

Measurement is Speed

Per hour means 1 hour
50 miles per hour means 50 miles in 1 hour
Let x be the number of hours it would take a car to travel 120 miles at a constant speed of 50 miles per hour.
Let y be number of minutes it would take a car to travel 120 miles at a constant speed of 50 miles per hour.

$ \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] $

$miles$	$hour$
$50$	$1$
$120$	$x$

$ \dfrac{50}{120} = \dfrac{1}{x} \\[5ex] Cross\:\:Multiply \\[3ex] 50x = 120(1) \\[3ex] 50x = 120 \\[3ex] x = \dfrac{120}{50} \\[5ex] x = \dfrac{12}{5}\:\:hours \\[5ex] $

$minutes$	$hours$
$60$	$1$
$y$	$\dfrac{12}{5}$

$ \dfrac{60}{y} = 1 \div \dfrac{12}{5} \\[5ex] \dfrac{60}{y} = 1 * \dfrac{5}{12} \\[5ex] \dfrac{60}{y} = \dfrac{5}{12} \\[5ex] Cross\:\:Multiply \\[3ex] 5y = 60(12) \\[3ex] y = \dfrac{60(12)}{5} \\[5ex] y = 12(12) \\[3ex] y = 144\:\:minutes \\[3ex] $ Student: This is a long method.
Is there any faster way to do this question or this kind of questions?
I mean...this is ACT...so we have to solve a question in a minute.
Teacher: Yes!
Good Question
We can use the Fast Proportional Reasoning Method
Or you can just invent your own method.
Yes, you can!!!

Third Method: Fast Proportional Reasoning Method
Let $p$ be the number of minutes it would take a car to travel $120$ miles at a constant speed of $50$ miles per hour.

miles	minutes	hour
$50$	$60$	$1$
$120$	$p$

$ \dfrac{p}{60} = \dfrac{120}{50} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:60 \\[3ex] 60 * \dfrac{p}{60} = 60 * \dfrac{120}{50} \\[5ex] p = 12 * 12 \\[3ex] p = 144\:\:minutes \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] Given: \\[3ex] 50\:miles\:per\:hour \\[3ex] 60\:minutes \\[3ex] 1\:hour \\[3ex] 120\:miles \\[3ex] To\:\:find\:\:...minutes \\[3ex] Set\:\:up\:\:first \\[3ex] 120\:miles * \dfrac{.....hour}{.....miles} * \dfrac{.....minutes}{.....hour} \\[5ex] = 120\:miles * \dfrac{1\:hour}{50\:miles} * \dfrac{60\:minutes}{1\:hour} \\[5ex] = \dfrac{120 * 60}{50} \\[5ex] = \dfrac{7200}{50} \\[5ex] = 144\:\:minutes $

The question asked us to calculate the average in miles per hour

(1.) So, we shall be converting all minutes to hours

The question asked us for the average. So, we shall:

(2.) Calculate the distance, in miles for the first trip.

(3.) Calculate the distance, in miles for the second trip.

(4.) Find the total distance, in miles.

(5.) Divide it by the total time, in hours.

That will give us the average rate, in miles per hour.

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] Per\:hr\:\:means\:\:1\:hr \\[3ex] \underline{First\:\:Trip} \\[3ex] 10\:min\:to\:\:hour \\[3ex] 10\:min * \dfrac{.....hr}{.....min} \\[5ex] = 10\:min * \dfrac{1\:hr}{60\:min} \\[5ex] = \dfrac{1}{6}\:hr \\[5ex] 2\:miles\:per\:hour\:\:for\:\:\dfrac{1}{6}\:hr \\[5ex] = 2 * \dfrac{1}{6} \\[5ex] = \dfrac{1}{3}\:miles \\[5ex] \underline{Second\:\:Trip} \\[3ex] 5\:min\:to\:\:hour \\[3ex] = 5\:min * \dfrac{1\:hr}{60\:min} \\[5ex] = \dfrac{1}{12}\:hr \\[5ex] 3\:miles\:per\:hour\:\:for\:\:\dfrac{1}{12}\:hr \\[5ex] = 3 * \dfrac{1}{12} \\[5ex] = \dfrac{1}{4}\:miles \\[5ex] \underline{Total\:\:Distance} \\[3ex] Total\:\:Distance = \dfrac{1}{3} + \dfrac{1}{4} \\[5ex] = \dfrac{4}{12} + \dfrac{3}{12} \\[5ex] = \dfrac{4 + 3}{12} \\[5ex] = \dfrac{7}{12}\:miles \\[5ex] \underline{Total\:\:Time} \\[5ex] 15\:min\:to\:\:hour \\[3ex] = 15\:min * \dfrac{1\:hr}{60\:min} \\[5ex] = \dfrac{1}{4}\:hr \\[5ex] \underline{Average\:\:Speed\:(mph)} \\[3ex] Average\:\:Speed = \dfrac{Total\:\:Distance}{Total\:\:Time} \\[5ex] = \dfrac{\dfrac{7}{12}}{\dfrac{1}{4}} \\[10ex] = \dfrac{7}{12} \div \dfrac{1}{4} \\[5ex] = \dfrac{7}{12} * \dfrac{4}{1} \\[5ex] = \dfrac{7}{3}\:\:miles\:\:per\:\:hour \\[5ex] $ Third Method: Fast Proportional Reasoning Method

Let:
$d$ = number of hours for $10$ minutes
$c$ = miles per hour for $d$ hours
$f$ = number of hours for $5$ minutes
$e$ = miles per hour for $f$ hours
$h$ = number of hours for $15$ minutes

miles	hour	minutes
$2$	$1$	$60$
$c$	$d$	$10$
$3$	$1$	$60$
$e$	$f$	$5$
	$1$	$60$
	$h$	$15$

$ \dfrac{d}{1} = \dfrac{10}{60} \\[5ex] d = \dfrac{1}{6}\:hours \\[5ex] \dfrac{c}{2} = \dfrac{d}{1} \\[5ex] \dfrac{c}{2} = \dfrac{\dfrac{1}{6}}{1} \\[7ex] \dfrac{c}{2} = \dfrac{1}{6} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:2 \\[3ex] 2 * \dfrac{c}{2} = 2 * \dfrac{1}{6} \\[5ex] c = \dfrac{1}{3}\:mph \\[5ex] \dfrac{f}{1} = \dfrac{5}{60} \\[5ex] f = \dfrac{1}{12}\:hours \\[5ex] \dfrac{e}{3} = \dfrac{f}{1} \\[5ex] \dfrac{e}{3} = \dfrac{\dfrac{1}{12}}{1} \\[7ex] \dfrac{e}{3} = \dfrac{1}{12} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:3 \\[3ex] 3 * \dfrac{e}{3} = 3 * \dfrac{1}{12} \\[5ex] e = \dfrac{1}{4}\:mph \\[5ex] \dfrac{h}{1} = \dfrac{15}{60} \\[5ex] h = \dfrac{1}{4}\:hours \\[5ex] Average\:\:Speed = \dfrac{Distance}{Time} \\[5ex] = \dfrac{c + e}{h} \\[5ex] = (c + e) \div h \\[3ex] = \left(\dfrac{1}{3} + \dfrac{1}{4}\right) \div \dfrac{1}{4} \\[5ex] = \left(\dfrac{4}{12} + \dfrac{3}{12}\right) \div \dfrac{1}{4} \\[5ex] = \dfrac{4 + 3}{12} \div \dfrac{1}{4} \\[5ex] = \dfrac{7}{12} * \dfrac{4}{1} \\[5ex] = \dfrac{7}{3}\:mph $

The question wants us to convert 900,000 ft² to m²
Measurement: Area
First Method: Unity Fraction Method

$ 900000\;ft * ft * \dfrac{.......m}{.......ft} * \dfrac{.......m}{.......ft} \\[5ex] = 900000\;ft * ft * \dfrac{0.3048\;m}{1\;ft} * \dfrac{0.3048\;m}{1\;ft} \\[5ex] = 83612.736\;m^2 \\[3ex] $ Second Method: Proportional Reasoning Method
Let $p$ represent the area of 900,000 ft² in m²

$ 1\;ft = 0.3048\;m \\[3ex] \implies \\[3ex] (1\;ft)^2 = (0.3048\;m)^2 \\[3ex] 1^2\;ft^2 = (0.3048)^2\;m^2 \\[3ex] 1\;ft^2 = 0.09290304\;m^2 \\[3ex] $

ft²	m²
1	0.09290304
900000	$p$

$ \dfrac{p}{0.09290304} = \dfrac{900000}{1} \\[5ex] p * 1 = 0.09290304 * 900000 \\[3ex] p = 83612.736\;m^2 \\[3ex] $ 900,000 ft² = 83612.736 m²

We are going to use what we were given, rather than using the Tables.

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 3.7\:km/15\:min\:\:to\:\:miles/hr \\[3ex] \dfrac{3.7\:km}{15\:min} * \dfrac{....miles}{....km} * \dfrac{....min}{....hr} \\[5ex] = \dfrac{3.7\:km}{15\:min} * \dfrac{1\:mile}{1.6\:km} * \dfrac{60\:min}{1\:hr} \\[5ex] = \dfrac{3.7 * 1 * 60}{15 * 1.6 * 1} \\[5ex] = \dfrac{222}{24} \\[5ex] = 9.25\:mph \approx 9\:mph \\[3ex] $ Third Method: Fast Proportional Reasoning Method

Let:
$x$ = number of hours for $15$ minutes
$y$ = number of miles for $3.7$ kilometers

$km$	$miles$	$min$	$hr$
$1.6$	$1$	$60$	$1$
$3.7$	$y$	$15$	$x$

$ \dfrac{x}{1} = \dfrac{15}{60} \\[5ex] x = \dfrac{1}{4} \\[5ex] x = 0.25\:hr \\[3ex] \dfrac{y}{1} = \dfrac{3.7}{1.6} \\[5ex] y = 2.3125\:miles \\[3ex] Average\:\:Speed = \dfrac{Distance}{Time} \\[5ex] = \dfrac{2.3125\:miles}{0.25\:hr} \\[5ex] = 9.25\:mph \approx 9\:mph \\[3ex] $ The average speed is approximately $9$ miles per hour.

$ Volume = 26\:ft * 32\:ft * 3\:ft \\[3ex] = 2496\:ft^3 \\[3ex] Convert\:\: 2496\:ft^3 \:\:to\:\: yd^3 \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 2496\:ft^3\:\:to\:\:yd^3 \\[3ex] 2496\:ft * ft * ft * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} \\[5ex] 2496\:ft * ft * ft * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} \\[5ex] = \dfrac{2496 * 1 * 1 * 1}{3 * 3 * 3} \\[5ex] = \dfrac{2496}{27} \\[5ex] = 92.4444444\:yd^3 \\[3ex] $ Second Method: Proportional Reasoning Method

Let $c$ = converted unit from $ft^3$ to $yd^3$

$ 3\:ft = 1\:yd \\[3ex] Cube\:\:both\:\:sides \\[3ex] (3\:ft)^3 = (1\:yd)^3 \\[3ex] 3^3 * ft^3 = 1^3 * yd^3 \\[3ex] 27\:ft^3 = 1\:yd^3 \\[3ex] $

$ft^3$	$yd^3$
$27$	$1$
$2496$	$c$

$ \dfrac{c}{1} = \dfrac{2496}{27} \\[5ex] c = 92.4444444\:yd^3 \\[3ex] \therefore 2496\:ft^3 = 92.4444444\:yd^3 $

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 360\:ft^2\:\:to\:\:yd^2 \\[3ex] 360\:ft * ft * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} \\[5ex] 360\:ft * ft * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} \\[5ex] = \dfrac{360 * 1 * 1}{3 * 3} \\[5ex] = \dfrac{360}{9} \\[5ex] = 40\:yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method

Let $c$ = converted unit from $ft^2$ to $yd^2$

$ 3\:ft = 1\:yd \\[3ex] Square\:\:both\:\:sides \\[3ex] (3\:ft)^2 = (1\:yd)^2 \\[3ex] 3^2 * ft^2 = 1^2 * yd^2 \\[3ex] 9\:ft^2 = 1\:yd^2 \\[3ex] $

$ft^2$	$yd^2$
$9$	$1$
$360$	$c$

$ \dfrac{c}{1} = \dfrac{360}{9} \\[5ex] c = 40\:yd^2 \\[3ex] \therefore 360\:ft^2 = 40\:yd^2 $

Per $kg$ means $1\:kg$
Recommended dosage: $5\:mg$ per $kg$ of body weight.
But: We were given $26$ pounds of body weight, rather than $kg$ of body weight
And asked to find the recommended dosage
This means that we need to find the equivalent of $26\:lb$ in $kg$
So, we need to convert $26\:lb$ to $kg$
Let us use the First Method to convert pounds to kilogram.
But, you can use any method you prefer.

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 26\:lb\:\:to\:\:kg \\[3ex] 26\:lb * \dfrac{.....g}{.....lb} * \dfrac{.....kg}{.....g} \\[5ex] 26\:lb * \dfrac{453.59237\:g}{1\:lb} * \dfrac{1\:kg}{1000\:g} \\[5ex] = \dfrac{26 * 453.59237 * 1}{1 * 1000} \\[5ex] = \dfrac{11793.4016}{1000} \\[5ex] = 11.7934016\:kg \\[3ex] $ We have found the equivalent of $26$ pounds in kilograms
Next, we need to find the recommended dosage for that weight.
Let use the Second Method to find it.

Second Method: Proportional Reasoning Method

Let $d$ = recommended dosage for a weight of $11.7934016$ kilograms

Prescribed ($mg$)	Body Weight ($kg$)
$5$	$1$
$d$	$11.7934016$

$ \dfrac{d}{5} = \dfrac{11.7934016}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] 5 * \dfrac{d}{5} = 5 * \dfrac{11.7934016}{1} \\[5ex] d = 5(11.7934016) \\[3ex] d = 58.967008\:mg \approx 59\:mg \\[3ex] $ The recommended dosage of Ibuprofen for an $18$ month old weighing $26$ pounds is $59\:mg$

Ask students to list at least two differences between Question $13$ and Question $14$

Per $kg$ means $1\:kg$
Recommended dosage: $5\:mg$ per $kg$ of body weight.
But: We were given $19$ pounds of body weight, rather than $kg$ of body weight
And asked to find the recommended dosage
This means that we need to find the equivalent of $19\:lb$ in $kg$
So, we need to convert $19\:lb$ to $kg$
Let us use the First Method to convert pounds to kilogram.
But, you can use any method you prefer.

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 19\:lb\:\:to\:\:kg \\[3ex] 19\:lb * \dfrac{.....kg}{.....lb} \\[5ex] 19\:lb * \dfrac{1\:kg}{2.2\:lb} \\[5ex] = \dfrac{19 * 1}{2.2} \\[5ex] = \dfrac{19}{2.2} \\[5ex] = 8.63636364\:kg \\[3ex] $ We have found the equivalent of $19$ pounds in kilograms
Next, we need to find the recommended dosage for that weight.
Let use the Second Method to find it.

Second Method: Proportional Reasoning Method

Let $d$ = recommended dosage for a weight of $11.7934016$ kilograms

Prescribed ($mg$)	Body Weight ($kg$)
$5$	$1$
$d$	$8.63636364$

$ \dfrac{d}{5} = \dfrac{8.63636364}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] 5 * \dfrac{d}{5} = 5 * \dfrac{8.63636364}{1} \\[5ex] d = 5(8.63636364) \\[3ex] d = 43.1818182\:mg \approx 43\:mg \\[3ex] $ The recommended dosage of Ibuprofen for an $18$ month old weighing $19$ pounds is $43\:mg$

Third Method: Fast Proportional Reasoning Method
Let $x$ be the equivalent amount of water for $100$ grams of sugar

Let $y$ be the equivalent amount of medicine for $100$ grams of sugar

Sugar ($g$)	Water ($L$)	Medicine (package)
$200$	$2$	$1$
$100$	$x$	$y$

$ \underline{Water} \\[3ex] \dfrac{x}{2} = \dfrac{100}{200} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:2 \\[3ex] 2 * \dfrac{x}{2} = 2 * \dfrac{100}{200} \\[5ex] x = \dfrac{2 * 100}{200} \\[5ex] x = \dfrac{200}{200} \\[5ex] x = 1\:L \\[3ex] Similarly: \\[3ex] \underline{Medicine} \\[3ex] \dfrac{y}{1} = \dfrac{100}{200} \\[5ex] y = \dfrac{1}{2}\:package \\[5ex] $ To maintain the same concentration, Rita should use a liter of water and one-half package of the medicine with the hundred grams of sugar.

Second Method: Proportional Reasoning Method

Let $A$ = amount of Medication $A$ in milligrams

Medication $A$($mg$)	Medication $B$($mg$)
$90$	$27$
$A$	$24$

$ \dfrac{A}{90} = \dfrac{24}{27} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:90 \\[3ex] 90 * \dfrac{A}{90} = 90 * \dfrac{24}{27} \\[5ex] A = \dfrac{90 * 24}{27} \\[5ex] A = \dfrac{90 * 8}{9} \\[5ex] A = 10 * 8 \\[3ex] A = 80\:mg \\[3ex] $ Deborah should use $80\:mg$ of Medication $A$

Second Method: Proportional Reasoning Method

Let $p$ = prescription dosage for Timothy

Prescribed ($mL$)	Weight ($pounds$)
$7$	$35$
$p$	$60$

$ \dfrac{p}{7} = \dfrac{60}{35} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:7 \\[3ex] 7 * \dfrac{p}{7} = 7 * \dfrac{60}{35} \\[5ex] p = \dfrac{60}{5} \\[5ex] p = 12\:mL \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 60\:lb * \dfrac{.....mL}{.....lb} \\[5ex] 60\:lb * \dfrac{7\:mL}{35\:lb} \\[5ex] = \dfrac{60}{5} \\[5ex] = 12\:mL \\[3ex] $ The pediatrician will prescribe $12\:mL$ of acetaminophen for Timothy.

Every pound means every $1$ pound

Second Method: Proportional Reasoning Method

Let $d$ = prescription dosage for Smart

Prescribed ($mg$)	Weight ($pounds$)
$7$	$1$
$d$	$70$

$ \dfrac{d}{7} = \dfrac{70}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:7 \\[3ex] 7 * \dfrac{d}{7} = 7 * 70 \\[5ex] d = 490\:mg \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 70\:lb * \dfrac{.....mg}{.....lb} \\[5ex] 70\:lb * \dfrac{7\:mg}{1\:lb} \\[5ex] = 70 * 7 = 490\:mg \\[3ex] $ Smart was given $490\:mg$ of the antibacterial medicine.

$ mph\:\:means\:\:miles\:\:per\:\:hour \\[3ex] mi/hr\:\:also\:\:means\:\:miles\:\:per\:\:hour \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 1056\:ft/s\:\:to\:\:mi/hr \\[3ex] 1056\dfrac{ft}{s} * \dfrac{.....miles}{.....ft} * \dfrac{.....s}{.....min} * \dfrac{.....min}{.....hr} \\[5ex] 1056\dfrac{ft}{s} * \dfrac{1\:mile}{5280\:ft} * \dfrac{60\:s}{1\:min} * \dfrac{60\:min}{1\:hr} \\[5ex] = \dfrac{1056 * 1 * 60 * 60}{5280 * 1 * 1} \\[5ex] = \dfrac{3801600}{5280} \\[5ex] = 720\:mph \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Per\:\:second\:\:means\:\:1\:\:second \\[3ex] Per\:\:hour\:\:means\:\:1\:\:hour \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:miles \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hour \\[3ex] Let\:\:k\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hr\:\:for\:\:60\:s \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hr\:\:for\:\:1\:s \\[3ex] \dfrac{1056\:ft}{1\:s} = \dfrac{x\:mile}{y\:hr} \\[5ex] $

Numerator

$ft$	$mile$
$5280$	$1$
$1056$	$x$

$ \dfrac{x}{1} = \dfrac{1056}{5280} \\[5ex] x = 0.2\:mile \\[3ex] $

Denominator

$s$	$min$	$hr$
$60$	$1$	$k$
	$60$	$1$
$1$		$y$

$ \dfrac{1}{60} = \dfrac{k}{1} \\[5ex] \dfrac{1}{60} = k \\[5ex] k = \dfrac{1}{60} \\[5ex] Next: \\[3ex] \dfrac{60}{1} = \dfrac{k}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 60y = k(1) \\[3ex] 60y = k \\[3ex] y = \dfrac{k}{60} \\[5ex] y = k \div 60 \\[3ex] y = \dfrac{1}{60} \div \dfrac{60}{1} \\[5ex] y = \dfrac{1}{60} * \dfrac{1}{60} \\[5ex] y = \dfrac{1 * 1}{60 * 60} \\[5ex] y = \dfrac{1}{3600} \\[5ex] \implies 1\:s = \dfrac{1}{3600}\:hr \\[5ex] \dfrac{x\:mi}{y\:hr} = x \div y \\[5ex] = 0.2 \div \dfrac{1}{3600} \\[5ex] = 0.2 * \dfrac{3600}{1} \\[5ex] = 0.2(3600) \\[3ex] = 720\:mi/hr \\[3ex] \therefore 1056\:ft/s = 720\:mph $

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 3168\:ft/min\:\:to\:\:mi/s \\[3ex] 3168\dfrac{ft}{min} * \dfrac{.....miles}{.....ft} * \dfrac{.....min}{.....s} \\[5ex] 3168\dfrac{ft}{min} * \dfrac{1\:mile}{5280\:ft} * \dfrac{1\:min}{60\:s} \\[5ex] = \dfrac{3168 * 1 * 1}{5280 * 60} \\[5ex] = \dfrac{3168}{316800} \\[5ex] = 0.01\:mile/second \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Per\:\:second\:\:means\:\:1\:\:second \\[3ex] Per\:\:minute\:\:means\:\:1\:\:hour \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:miles \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:second \\[3ex] \dfrac{3168\:ft}{1\:min} = \dfrac{x\:mile}{y\:s} \\[5ex] $

Numerator

$ft$	$mile$
$5280$	$1$
$3168$	$x$

$ \dfrac{x}{1} = \dfrac{3168}{5280} \\[5ex] x = 0.6\:mile \\[3ex] \underline{Denominator} \\[3ex] 1\:min = 60\:s \\[3ex] \implies y = 60\:s \\[3ex] \dfrac{x\:mi}{y\:s} = x \div y \\[5ex] = \dfrac{0.6}{60} \\[5ex] = 0.01\:mile/second \\[3ex] \therefore 3186\:ft/m = 0.01\:mi/s $

$ \underline{One\:\:Basket} \\[3ex] Total\:\:Length\:\:of\:\:5\:\:ribbons = 5(8) = 40\:inches \\[3ex] Total\:\:Length\:\:of\:\:3\:\:ribbons = 3(16) = 48\:inches \\[3ex] Total\:\:Length\:\:of\:\:2\:\:ribbons = 2(10) = 20\:inches \\[3ex] Total\:\:Length = 40 + 48 + 20 = 108\:inches \\[3ex] \underline{Ten\:\:Baskets} \\[3ex] Total\:\:Length = 108(10) = 1080\:inches \\[3ex] Convert\:\:1080\:inches\:\:to\:\:yards \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 1080\:inches * \dfrac{.....yd}{.....in} \\[5ex] = 1080\:inches * \dfrac{1\:yd}{36\:inches} \\[5ex] = 30\:yards \\[3ex] Per\:\:yard\:\:means\:\:1\:\:yard \\[3ex] Cost\:\:of\:\:1\:\:yard = \$0.98 \\[3ex] Cost\:\:of\:\:30\:yards\:\:@\:\:\$0.98\:\:per\:\:yard \\[3ex] = 30 * 0.98 \\[3ex] = \$29.4 \approx \$30 \\[3ex] $ In this case, it is in order to approximate to $\$30$ rather than $\$29$ because who will pay that $40$ cents?

Per hour means $1$ hour

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] \underline{Car\:\:A} \\[3ex] 60\:\dfrac{miles}{1\:hr} * ....hr \\[5ex] \dfrac{60\:miles}{1\:hr} * 1\dfrac{1}{2}\:hrs \\[5ex] = 60\:\dfrac{miles}{1\:hr} * \dfrac{3}{2}\:hrs \\[5ex] = 30 * 3 \\[3ex] = 90\:miles \\[3ex] \underline{Car\:\:B} \\[3ex] 40\:\dfrac{miles}{1\:hr} * ....hr \\[5ex] \dfrac{40\:miles}{1\:hr} * 2\:hrs \\[5ex] = 40 * 2 \\[3ex] = 80\:miles \\[3ex] \underline{Difference\:\:in\:\:miles\:\:between\:\:Car\:\:A\:\:and\:\:Car\:\:B} \\[3ex] Difference = 90 - 80 = 10\:miles $

Second Method: Proportional Reasoning Method

Let $c$ = number of miles represented by $1.6\:inches$

inches	miles
$1.5$	$90$
$1.6$	$c$

$ \dfrac{c}{1.6} = \dfrac{90}{1.5} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:1.6 \\[3ex] 1.6 * \dfrac{c}{1.6} = 1.6 * \dfrac{90}{1.5} \\[5ex] c = \dfrac{1.6 * 90}{1.5} \\[5ex] c = \dfrac{144}{1.5} \\[5ex] c = 96\:miles \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 1.6\:inches * \dfrac{.....miles}{.....inches} \\[5ex] 1.6\:inches * \dfrac{90\:miles}{1.5\:inches} \\[5ex] = \dfrac{1.6 * 90}{1.5} \\[5ex] = \dfrac{144}{1.5} \\[5ex] = 96\:miles $

Unit of weight means $1$ weight
This means that $1\:stone = 14\:lbs$

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] Convert\:\:177\:lbs\:\:to\:\:stone \\[3ex] 177\:lbs * \dfrac{.....stone}{.....lbs} \\[5ex] 177\:lbs * \dfrac{1\:stone}{14\:lbs} \\[5ex] = \dfrac{177 * 1}{14} \\[5ex] = 12.6428571\:stone \approx 12.6\:stone \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:the\:\:converted\:\:unit\:\:in\:\:stone \\[3ex] $

stone	pounds
$1$	$14$
$p$	$177$

$ \dfrac{p}{1} = \dfrac{177}{14} \\[5ex] p = 12.6428571\:stone \approx 12.6\:stone \\[3ex] \therefore 177\:lbs = 12.6428571\:stone $

Per minute means $1$ minute
Per hour means $1$ hour

Second Method: Proportional Reasoning Method

The radiator is leaking fluid at $4$ ounces per minute
First: We need to find how many minutes it will take for $480$ ounces to leak

The truck is traveling at $35$ miles per hour

Second: We need to convert the minutes in the First step to hours

Third:Then, we find how many miles the truck will travel for those number of hours

Let $c$ = number of minutes it will take for $480$ ounces to leak
Let $d$ = number of miles the truck will travel before the radiator is empty

ounces	minute
$40$	$1$
$480$	$c$

$ \dfrac{c}{1} = \dfrac{480}{4} \\[5ex] c = 120\:minutes \\[3ex] 60\:minutes = 1\:hour \\[3ex] 120\:minutes = 2\:hours \\[3ex] \rightarrow c = 2\:hours \\[3ex] $

miles	hour
$35$	$1$
$d$	$2$

$ \dfrac{d}{2} = \dfrac{35}{1} \\[5ex] Cross\:\:Multiply \\[3ex] d * 1 = 35 * 2 \\[3ex] d = 70\:miles \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 480\:ounces * \dfrac{.....minutes}{.....ounces} * \dfrac{.....hour}{.....minutes} * \dfrac{.....miles}{.....hour} \\[5ex] 480\:ounces * \dfrac{1\:minute}{4\:ounces} * \dfrac{1\:hour}{60\:minutes} * \dfrac{35\:miles}{1\:hour} \\[5ex] = \dfrac{480 * 1 * 1 * 35}{4 * 60} \\[5ex] = 2 * 35 \\[3ex] = 70\:miles \\[3ex] $ The truck will drive for $70$ miles before the radiator fluid is empty.

Measurement is Money/Currency

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] (i)\:\:Bank\:\:A \\[3ex] US\$700 \:\:to\:\: BD\$ \\[3ex] = US\$700 * \dfrac{1.96\:BD\$}{1.00\:US\$} \\[5ex] US\$700 = BD\$1372 \\[3ex] (ii)\:\:Bank\:\:B \\[3ex] US\$1.00\:\:to\:\:BD\$ \\[3ex] = US\$1.00 * \dfrac{1386\:BD\$}{700\:US\$} \\[5ex] US\$1.00 = BD\$1.98 \\[3ex] $

Second Method: Proportional Reasoning Method

$US\$$	$BD\$$
$1.00$	$1.96$
$700$	$x$

$ \dfrac{1.00}{700} = \dfrac{1.96}{x} \\[5ex] Cross\:\:Multiply \\[3ex] 1.00(x) = 700(1.96) \\[3ex] x = BD\$1372 \\[3ex] $

$US\$$	$BD\$$
$700$	$1386$
$1$	$y$

$ \dfrac{700}{1} = \dfrac{1386}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 700(y) = 1(1386) \\[3ex] 700y = 1386 \\[3ex] y = \dfrac{1386}{700} \\[5ex] y = BD\$1.98 $

Per second means $1$ second

Second Method: Proportional Reasoning Method

Let $p$ = number of miles the beam travels in $2$ hours

$ 60\:seconds = 1\:minute \\[3ex] 60\:minutes = 1\:hour \\[3ex] \rightarrow (60 * 60)\:seconds = 1\:hour \\[3ex] 3600\:seconds = 1\:hour \\[3ex] \rightarrow 2 * 3600\:seconds = 2 * 1\:hour \\[3ex] \therefore 7200\:seconds = 2\:hours \\[3ex] 2\:hours = 7200\:seconds \\[3ex] $

miles	seconds
$186000$	$1$
$p$	$7200$

$ \dfrac{p}{186000} = \dfrac{7200}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:18600 \\[3ex] 18600 * \dfrac{p}{186000} = 18600 * \dfrac{7200}{1} \\[5ex] p = 186000(7200) \\[3ex] p = 1339200000\:miles \\[3ex] p = 1.3392 * 10^9\:miles \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 186000\:\dfrac{miles}{seconds} * \dfrac{.....seconds}{.....hours} * 2\:hours \\[5ex] \dfrac{186000\:miles}{1\:second} * \dfrac{3600\:seconds}{1\:hour} * 2\:hours \\[5ex] = \dfrac{186000 * 3600 * 2}{1 * 1} \\[5ex] = 186000 * 3600 * 2 \\[5ex] = 1339200000\:miles \\[3ex] = 1.3392 * 10^9\:miles $

Driving $200$ miles in $5$ hours means how many miles per hour?

$ 200\:miles\:\:in\:\:5\:hours\:\:means \\[3ex] \dfrac{200\:miles}{5\:hours} \\[5ex] = 40\:miles\:per\:hour \\[3ex] OR \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:the\:\:number\:\:of\:\:miles\:\:in\:\:1\:hour \\[3ex] $

miles	hours
$200$	$5$
$p$	$1$

$ \dfrac{p}{1} = \dfrac{200}{5} \\[5ex] p = 40\:miles \\[3ex] This\:\:means\:\:40\:\:miles\:\:in\:\:1\:hour \\[3ex] $ But if Kaya was driving at an average of $10\:mph$ faster

This means that she was driving at $40 + 10 = 50\:mph$ faster

So, how many hours will she need to drive $200\:miles$ if she was driving at $50$ miles per hour?

$ Speed = \dfrac{Distance}{Time} \\[5ex] Time = \dfrac{Distance}{Speed} \\[5ex] = \dfrac{200\:miles}{50\:mph} \\[5ex] = 4\:hours \\[3ex] Actual\:\:Driving\:\:Time\:\:@\:\:40mph = 5\:hours \\[3ex] Supposed\:\:Driving\:\:Time\:\:@\:\:50mph = 4\:hours \\[3ex] Savings\:\:in\:\:time = 5\:hours - 4\:hours = 1\:hour $

Per gallon means $1$ gallon
Per cubic centimeter means $1$ cubic centimeter
We are restricted to using only that information.
As you can see, we do not have any direct conversion from $gal$ to $cm^3$
So, we need to establish that conversion.
How?
We have the conversion from $inch$ to $cm$
We have the conversion from $gal$ to $inch^3$
We have to figure a way to do it. ☺

$ 1\:inch = 2.54\:cm \\[3ex] Cube\:\:both\:\:sides \\[3ex] (1\:inch)^3 = (2.54\:cm)^3 \\[3ex] 1^3\:inch^3 = 2.54^3\:cm^3 \\[3ex] 1\:inch^3 = 16.387064\:cm^3 \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 10\:lb/gal\:\:to\:\:g/cm^3 \\[3ex] \dfrac{10\:lb}{1\:gal} * \dfrac{.....g}{.....lb} * \dfrac{.....gal}{.....inch^3} * \dfrac{.....inch^3}{.....cm^3} \\[5ex] \dfrac{10\:lb}{1\:gal} * \dfrac{453.6\:g}{1\:lb} * \dfrac{1\:gal}{231\:inch^3} * \dfrac{1\:inch^3}{16.387064\:cm^3} \\[5ex] = \dfrac{10 * 453.6}{231 * 16.387064} \\[5ex] = \dfrac{4536}{3785.41178} \\[5ex] = 1.19828443\:g/cm^3 \\[3ex] \approx 1.20\:g/cm^3 \\[3ex] $

---

$ \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:g\:\:for\:\:10\:lb \\[3ex] $

Numerator

$lb$	$g$
$1$	$453.6$
$10$	$x$

$ \dfrac{x}{10} = \dfrac{453.6}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:10 \\[3ex] 10 * \dfrac{x}{10} = 10 * \dfrac{453.6}{1} \\[5ex] x = 10(453.6) \\[3ex] x = 4536\:g \\[3ex] \underline{Third\:\:Method:\:\:Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:cm^3\:\:for\:\:1\:gal(231\:inch^3) \\[3ex] $

Denominator

$inch^3$	$gal$	$cm^3$
$1$		$16.387064$
$231$	$1$	$y$

$ \dfrac{y}{16.387064} = \dfrac{231}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:16.387064 \\[3ex] 16 * \dfrac{y}{16.387064} = 16 * \dfrac{231}{1} \\[5ex] y = 16(231) \\[3ex] y = 3696\:cm^3 \\[3ex] \underline{Entire\:\:Question} \\[3ex] 10\:lb/gal \\[3ex] = \dfrac{x}{y} \\[5ex] = \dfrac{4536}{3696} \\[5ex] = 1.22727273\:g/cm^3...oops... \\[3ex] $ Teacher: There is a problem here...
It is not the same exact value that we got using the Unity Fraction Method.
*Student: Why is that?...speaking like an American lol
*Teacher: Can you figure out why?...asking questions with questions...typical of a Nigerian lol
Student: Well, I think it is because we are working with rounded values (approximate values) rather than exact values.
Teacher: That is correct.
So, what do we do to get the same answer as the First Method?
Student: Work with exact values. Do not work with approximate values.
Teacher: Okay.
But...specifically to this question, what should we use?
Student: I think we should convert $1\:inch^3$ to gallons first...
Then, we use the $gal$ to $cm^3$ measurement and find $y$
Teacher: That is correct!

$ Re-do \\[3ex] \underline{Third\:\:Method:\:\:Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:c\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:gal\:\:for\:\:1\:inch^3 \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:cm^3\:\:for\:\:1\:gal(231\:inch^3) \\[3ex] $

Denominator

$inch^3$	$gal$	$cm^3$
$1$	$c$	$16.387064$
$231$	$1$	$y$

$ \dfrac{c}{1} = \dfrac{1}{231} \\[5ex] c = \dfrac{1}{231}\:gal \\[5ex] Next \\[3ex] \dfrac{y}{1} = \dfrac{16.387064}{c} \\[5ex] y = 16.387064 \div c \\[3ex] y = 16.387064 \div \dfrac{1}{231} \\[5ex] y = 16.387064 * \dfrac{231}{1} \\[5ex] y = 3785.41178\:cm^3 \\[3ex] \underline{Entire\:\:Question} \\[3ex] 10\:lb/gal \\[3ex] = \dfrac{x}{y} \\[5ex] = \dfrac{4536}{3785.41178} \\[5ex] = 1.19828443\:g/cm^3 \\[3ex] \approx 1.20\:g/cm^3 $

Second Method: Proportional Reasoning Method

Let $c$ the number of liters of sauce required for $80$ servings

servings	liters
$150$	$4.5$
$80$	$c$

$ \dfrac{c}{4.5} = \dfrac{80}{150} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: 4.5 \\[3ex] 4.5 * \dfrac{c}{4.5} = 4.5 * \dfrac{80}{150} \\[5ex] c = \dfrac{4.5 * 80}{150} \\[5ex] c = \dfrac{360}{150} \\[5ex] c = 2.4\:liters \\[3ex] $ $2.4$ liters of sauce is required for $80$ servings

Per hour means $1$ hour
Per second means $1$ second

$ 60\:seconds = 1\:minute \\[3ex] 60\:minutes = 1\:hour \\[3ex] \rightarrow (60 * 60)\:seconds = 1\:hour \\[3ex] 3600\:seconds = 1\:hour \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 1245\:ft/hr\:\:to\:\:cm/s \\[3ex] 1245\dfrac{ft}{hr} * \dfrac{.....m}{.....ft} * \dfrac{.....cm}{.....m} * \dfrac{.....hr}{.....s} \\[5ex] 1245\dfrac{ft}{hr} * \dfrac{0.3048\:m}{1\:ft} * \dfrac{1\:cm}{10^{-2}\:m} * \dfrac{1\:hr}{3600\:s} \\[5ex] = 1245\dfrac{ft}{hr} * \dfrac{0.3048\:m}{1\:ft} * \dfrac{1\:cm}{0.01\:m} * \dfrac{1\:hr}{3600\:s} \\[5ex] = \dfrac{1245 * 0.3048}{0.01 * 3600} \\[5ex] = \dfrac{379.476}{36} \\[5ex] = 10.541\:cm/s \\[3ex] \underline{Third\:\:Method:\:\:Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] 1\:cm = 10^{-2}\:m \\[3ex] 1\:cm = 0.01\:m \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:cm\:\:for\:\:0.3048\:m \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:cm\:\:for\:\:1245\:ft \\[3ex] $

Numerator

$ft$	$m$	$cm$
	$0.01$	$1$
$1$	$0.3048$	$x$
$1245$		$y$

$ \dfrac{x}{1} = \dfrac{0.3048}{0.01} \\[5ex] x = 30.48\:cm \\[3ex] Next: \\[3ex] \dfrac{y}{x} = \dfrac{1245}{1} \\[5ex] \dfrac{y}{30.48} = 1245 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:30.48 \\[3ex] 30.48 * \dfrac{y}{30.48} = 30.48(1245) \\[5ex] y = 37947.6\:cm \\[3ex] \underline{Denominator} \\[3ex] 1\:hr = 3600\:s \\[3ex] \underline{Entire\:\:Question} \\[3ex] \implies \dfrac{1245\:ft}{1\:hr} \\[5ex] = \dfrac{37947.6\:cm}{3600\:s} \\[5ex] = 10.541\:cm/s $

Per hour means $1$ hour
Per minute means $1$ minute

$ 60\:minutes = 1\:hour \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 1245\:ft/hr\:\:to\:\:inch/min \\[3ex] 1245\dfrac{ft}{hr} * \dfrac{.....inch}{.....ft} * \dfrac{.....hr}{.....min} \\[5ex] 1245\dfrac{ft}{hr} * \dfrac{12\:inch}{1\:ft} * \dfrac{1\:hr}{60\:min} \\[5ex] = \dfrac{1245 * 12}{60} \\[5ex] = \dfrac{1245}{5} \\[5ex] = 249\:inches/min \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:inch\:\:for\:\:1245\:ft \\[3ex] $

Numerator

$ft$	$inch$
$1$	$12$
$1245$	$p$

$ \dfrac{p}{12} = \dfrac{1245}{1} \\[5ex] Cross\:\:Multiply \\[3ex] p(1) = 12(1245) \\[3ex] p = 14940\:inches \\[3ex] \underline{Denominator} \\[3ex] 1\:hr = 60\:min \\[3ex] \underline{Entire\:\:Question} \\[3ex] \implies \dfrac{1245\:ft}{1\:hr} \\[5ex] = \dfrac{14940\:inch}{60\:min} \\[5ex] = 249\:inches/min $

Per hour means $1$ hour
Per second means $1$ second

$ 60\:seconds = 1\:minute \\[3ex] 60\:minutes = 1\:hour \\[3ex] \rightarrow (60 * 60)\:seconds = 1\:hour \\[3ex] 3600\:seconds = 1\:hour \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 240\:miles/hr\:\:to\:\:ft/s \\[3ex] 240\dfrac{miles}{hr} * \dfrac{.....ft}{.....miles} * \dfrac{.....hr}{.....s} \\[5ex] 240\dfrac{miles}{hr} * \dfrac{5280\:ft}{1\:mile} * \dfrac{1\:hr}{3600\:s} \\[5ex] = \dfrac{240 * 5280}{3600}...Option\:C \\[5ex] = \dfrac{1267200}{3600} \\[5ex] = 352\:ft/s \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:feet \\[3ex] $

Numerator

$miles$	$ft$
$1$	$5280$
$240$	$p$

$ \dfrac{p}{240} = \dfrac{5280}{1} \\[5ex] Cross\:\:Multiply \\[3ex] p(1) = 240(5280) \\[3ex] p = \dfrac{5280}{240}\:ft \\[5ex] \underline{Denominator} \\[3ex] 1\:hr = 3600\:s \\[3ex] \underline{Entire\:\:Question} \\[3ex] \implies \dfrac{240\:miles}{1\:hr} \\[5ex] = \dfrac{240(5280)\:ft}{3600\:s} \\[5ex] = 352\:ft/s $

First Method: Unity Fraction Method

$ 4\dfrac{1}{2}\;inches * \dfrac{......miles}{.........inches} \\[5ex] - \dfrac{9}{2}\;inches * \dfrac{10\;miles}{\dfrac{1}{2}\;inches} \\[7ex] = \dfrac{9}{2} * 10 \div \dfrac{1}{2} \\[5ex] = \dfrac{9}{2} * \dfrac{10}{1} * \dfrac{2}{1} \\[5ex] = 90\;miles \\[3ex] $ Second Method: Proportional Reasoning Method
Let $p$ be the number of actual miles that is represented by $4\dfrac{1}{2}$ inches

inch	miles
$\dfrac{1}{2}$	$10$
$4\dfrac{1}{2} = \dfrac{9}{2}$	$p$

$ 4\dfrac{1}{2} = \dfrac{2 * 4 + 1}{2} = \dfrac{8 + 1}{2} = \dfrac{9}{2} \\[5ex] \dfrac{p}{10} = \dfrac{\dfrac{9}{2}}{\dfrac{1}{2}} \\[7ex] Multiply\:\:both\:\:sides\:\:by\:\: 10 \\[3ex] 10 * \dfrac{p}{10} = 10 * \dfrac{\dfrac{9}{2}}{\dfrac{1}{2}} \\[7ex] p = \dfrac{\dfrac{10 * 9}{2}}{\dfrac{1}{2}} \\[7ex] p = \dfrac{5 * 9}{\dfrac{1}{2}} \\[7ex] p = \dfrac{45}{\dfrac{1}{2}} \\[7ex] p = 45 \div \dfrac{1}{2} \\[5ex] p = 45 * \dfrac{2}{1} \\[5ex] p = 45 * 2 \\[3ex] p = 90\:miles \\[3ex] $ $4\dfrac{1}{2}$ inches represents $90$ miles

Per hour means $1$ hour
Per hour squared means $1^2\:hr^2 = 1\:hr^2$
Per minute means $1$ minute
Per minute squared means $1^2\:min^2 = 1\:min^2$

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 0.00005\:km/min^2\:\:to\:\:cm/hr^2 \\[3ex] \dfrac{0.00005\:km}{min * min} * \dfrac{.....min}{.....hr} * \dfrac{.....min}{.....hr} * \dfrac{.....m}{.....km} * \dfrac{.....cm}{.....m} \\[5ex] \dfrac{0.00005\:km}{min * min} * \dfrac{60\:min}{1\:hr} * \dfrac{60\:min}{1\:hr} * \dfrac{10^3\:m}{1\:km} * \dfrac{1\:cm}{10^{-2}\:m} \\[5ex] = \dfrac{0.00005\:km}{min * min} * \dfrac{60\:min}{1\:hr} * \dfrac{60\:min}{1\:hr} * \dfrac{1000\:m}{1\:km} * \dfrac{1\:cm}{0.01\:m} \\[5ex] = \dfrac{0.00005 * 60 * 60 * 1000}{0.01} \\[5ex] = \dfrac{180}{0.01} \\[5ex] = 18000\:cm/hr^2 \\[3ex] $

---

$ \underline{Third\:\:Method:\:\:Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:cm\:\:for\:\:0.00005\:km \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:km\:\:for\:\:1\:cm \\[3ex] 10^{-2} = \dfrac{1}{10^2} = \dfrac{1}{100} = 0.01 \\[5ex] 10^3 = 1000 \\[3ex] $

Numerator

$km$	$m$	$cm$
$y$	$0.01$	$1$
$1$	$1000$
$0.00005$		$x$

$ \dfrac{y}{1} = \dfrac{0.01}{1000} \\[5ex] y = 0.00001 \\[3ex] Next \\[3ex] \dfrac{x}{1} = \dfrac{0.00005}{y} \\[5ex] x = \dfrac{0.00005}{0.00001} \\[5ex] x = 5\:cm \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hr^2\:\:for\:\:1\:min^2 \\[3ex] 60\:minutes = 1\:hr \\[3ex] Square\:\:both\:\:sides \\[3ex] (60\:minutes)^2 = (1\:hr)^2 \\[3ex] 60^2\:minute^2 = 1^2\:hr^2 \\[3ex] 3600\:minute^2 = 1\:hr^2 \\[3ex] $

Denominator

$min^2$	$hr^2$
$3600$	$1$
$1$	$p$

$ \dfrac{p}{1} = \dfrac{1}{3600} \\[5ex] p = \dfrac{1}{3600}\:hr^2 \\[5ex] \underline{Entire\:\:Question} \\[3ex] 0.00005\:km/min^2 \\[3ex] = \dfrac{x}{p} \\[5ex] = x \div p \\[3ex] = 5 \div \dfrac{1}{3600} \\[5ex] = 5 * 3600 \\[3ex] = 18000\:cm/hr^2 $

Per cm means $1$ cm
Per square cm means $1^2\:cm^2 = 1\:cm^2$
Per inch means $1$ inch
Per square inch means $1^2\:inch^2 = 1\:inch^2$

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 3450\:N/cm^2\:\:to\:\:lb/inch^2 \\[3ex] \dfrac{3450\:N}{cm * cm} * \dfrac{.....cm}{.....inch} * \dfrac{.....cm}{.....inch} * \dfrac{.....lb}{.....N} \\[5ex] \dfrac{3450\:N}{cm * cm} * \dfrac{2.54\:cm}{1\:inch} * \dfrac{2.54\:cm}{1\:inch} * \dfrac{1\:lb}{4.448\:N} \\[5ex] = \dfrac{3450 * 2.54 * 2.54}{4.448} \\[5ex] = \dfrac{22258.02}{4.448} \\[5ex] = 5004.05126 \\[3ex] \approx 5004.05\:lb/inch^2 \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:lb\:\:for\:\:3450\:N \\[3ex] $

Numerator

$N$	$lb$
$4.448$	$1$
$3450$	$p$

$ \dfrac{p}{1} = \dfrac{3450}{4.448} \\[5ex] p = 775.629496\:lb \\[3ex] 1\:inch = 2.54\:cm \\[3ex] Square\:\:both\:\:sides \\[3ex] (1\:inch)^2 = (2.54\:cm)^2 \\[3ex] 1^2\:inch^2 = 2.54^2\:cm^2 \\[3ex] 1\:inch^2 = 6.4516\:cm^2 \\[3ex] $

Denominator

$cm^2$	$inch^2$
$6.4516$	$1$
$1$	$c$

$ \dfrac{c}{1} = \dfrac{1}{6.4516} \\[5ex] c = 0.15500031\:inch^2 \\[3ex] \underline{Entire\:\:Question} \\[3ex] 3450\:N/cm^2 \\[3ex] = \dfrac{p}{c} \\[5ex] = \dfrac{775.629496}{0.15500031} \\[5ex] = 5004.05126 \\[3ex] \approx 5004.05\:lb/inch^2 $

Per meter means $1$ meter
Per cubic meter means $1^3\:m^3 = 1\:m^3$
Per foot means $1$ foot
Per cubic feet means $1^3\:ft^3 = 1\:ft^3$

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 84520\:g/m^3\:\:to\:\:lb/ft^3 \\[3ex] \dfrac{84520\:g}{m * m * m} * \dfrac{.....m}{.....ft} * \dfrac{.....m}{.....ft} * \dfrac{.....m}{.....ft} * \dfrac{.....lb}{.....g} \\[5ex] \dfrac{84520\:g}{m * m * m} * \dfrac{0.3048\:m}{1\:ft} * \dfrac{0.3048\:m}{1\:ft} * \dfrac{0.3048\:m}{1\:ft} * \dfrac{1\:lb}{453.59237\:g} \\[5ex] = \dfrac{84520 * 0.3048 * 0.3048 * 0.3048}{453.59237} \\[5ex] = \dfrac{2393.33987}{453.59237} \\[5ex] = 5.27641122\:lb/ft^3 \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:lb\:\:for\:\:84520\:g \\[3ex] $

Numerator

$g$	$lb$
$453.59237$	$1$
$84520$	$p$

$ \dfrac{p}{1} = \dfrac{84520}{453.59237} \\[5ex] p = 186.334704\:lb \\[3ex] \underline{Denominator} \\[3ex] 1\:ft = 0.3048\:m \\[3ex] Cube\:\:both\:\:sides \\[3ex] (1\:ft)^3 = (0.3048\:m)^3 \\[3ex] 1^3\:ft^3 = 0.3048^3\:m^3 \\[3ex] 1\:ft^3 = 0.0283168466\:m^3 \\[3ex] $

Denominator

$m^3$	$ft^3$
$0.0283168466$	$1$
$1$	$c$

$ \dfrac{c}{1} = \dfrac{1}{0.0283168466} \\[5ex] c = 35.3146667\:ft^3 \\[3ex] \underline{Entire\:\:Question} \\[3ex] 84520\:g/m^3 \\[3ex] = \dfrac{p}{c} \\[5ex] = \dfrac{186.334704}{35.3146667} \\[5ex] = 5.27641123\:lb/ft^3 $

Per hour means $1$ hour
Per hour squared means $1^2\:hr^2 = 1\:hr^2$
Per minute means $1$ minute
Per minute squared means $1^2\:min^2 = 1\:min^2$

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 12450\:ft/hr^2\:\:to\:\:inch/min^2 \\[3ex] \dfrac{12450\:ft}{hr * hr} * \dfrac{.....hr}{.....min} * \dfrac{.....hr}{.....min} * \dfrac{....inch}{.....ft} \\[5ex] = \dfrac{12450\:ft}{hr * hr} * \dfrac{1\:hr}{60\:min} * \dfrac{1\:hr}{60\:min} * \dfrac{12\:inch}{1\:ft} \\[5ex] = \dfrac{12450 * 12}{60 * 60} \\[5ex] = \dfrac{149400}{3600} \\[5ex] = 41.5\:inch/min^2 \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:inch\:\:for\:\:12450\:ft \\[3ex] $

Numerator

$feet$	$inch$
$1$	$12$
$12450$	$p$

$ \dfrac{p}{12} = \dfrac{12450}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: 12 \\[3ex] 12 * \dfrac{p}{12} = 12 * \dfrac{12450}{1} \\[5ex] p = 12(12450) \\[3ex] p = 149400\:inch \\[3ex] \underline{Denominator} \\[3ex] 60\:minutes = 1\:hr \\[3ex] Square\:\:both\:\:sides \\[3ex] (60\:minutes)^2 = (1\:hr)^2 \\[3ex] 60^2\:minute^2 = 1^2\:hr^2 \\[3ex] 3600\:minute^2 = 1\:hr^2 \\[3ex] 1\:hr^2 = 3600\:minute^2 \\[3ex] \underline{Entire\:\:Question} \\[3ex] 12450\:ft/hr^2 \\[3ex] = \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{149400}{3600} \\[5ex] = 41.5\:inch/min^2 $

Third Method: Fast Proportional Reasoning Method
Let $p$ be the total cost for fuel over the $350$ miles of highway driving

Highway

miles	gallon	cost($\$)
$32$	$1$	$4$
$350$		$p$

$ \dfrac{p}{4} = \dfrac{350}{32} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:4 \\[3ex] 4 * \dfrac{p}{4} = 4 * \dfrac{350}{32} \\[5ex] p = \dfrac{4 * 350}{32} \\[5ex] p = \dfrac{350}{8} \\[5ex] p = \$43.75 \\[3ex] $ The total cost for fuel over the $350$ miles of highway driving is $\$43.75$

Third Method: Fast Proportional Reasoning Method
Let $k$ be the certain number of miles driven per year on which the cost estimates are based

City/Highway

miles	gallon	cost($\$$)
$25$	$1$	$4$
$k$		$2400$

$ \dfrac{k}{25} = \dfrac{2400}{4} \\[5ex] \dfrac{k}{25} = 600 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:25 \\[3ex] 25 * \dfrac{k}{25} = 25(600) \\[5ex] k = 15000\:miles $ The certain number of miles driven per year on which the cost estimates are based is $15,000\:miles$

$ \underline{Carl's\:\:New\:\:Car\:\:versus\:\:Average\:\:New\:\:Vehicle} \\[3ex] Savings\:\:in\:\:fuel\:\:costs\:\:over\:\:5\:\:years = \$3500 \\[3ex] Savings\:\:in\:\:fuel\:\:costs\:\:over\:\:1\:\:year = \dfrac{3500}{5} = \$700 \\[5ex] Annual\:\:fuel\:\:cost = \$2400 \\[3ex] \underline{Average\:\:New\:\:Vehicle} \\[3ex] Expected\:\:Annual\:\:Fuel\:\:Cost \\[3ex] = \$700 + \$2400 \\[3ex] = \$3100 \\[3ex] $ The expected annual fuel cost of an average new vehicle is thirty one hundred dollars (American way) OR three thousand, one hundred dollars (Nigerian way)

Second Method: Proportional Reasoning Method

Let $p$ the number of tons of sand required to produce $3000$ barrels

tons	barrels
$100000$	$60000$
$p$	$3000$

$ \dfrac{p}{3000} = \dfrac{100000}{60000} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: 3000 \\[3ex] 3000 * \dfrac{p}{3000} = 3000 * \dfrac{100000}{60000} \\[5ex] p = \dfrac{3000 * 100000}{60000} \\[5ex] p = 500 * 10 \\[3ex] p = 5000\:tons \\[3ex] $ $5,000$ tons of sand will be required to produce $3,000$ barrels of the tarry material

$ Area\:\;in\:\:square\:\:feet = 90\:feet * 30\:feet = 2700\:square\:feet \\[3ex] Convert\:\: 2700\:feet^2 \:\:to\:\: yard^2 \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 2700\:ft^2\:\:to\:\:yd^2 \\[3ex] 2700\:ft * ft * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} \\[5ex] 2700\:ft * ft * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} \\[5ex] = \dfrac{2700 * 1 * 1}{3 * 3} \\[5ex] = \dfrac{2700}{9} \\[5ex] = 300\:yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method
Let $c$ = converted unit from $ft^2$ to $yd^2$

$ 3\:ft = 1\:yd \\[3ex] Square\:\:both\:\:sides \\[3ex] (3\:ft)^2 = (1\:yd)^2 \\[3ex] 3^2 * ft^2 = 1^2 * yd^2 \\[3ex] 9\:ft^2 = 1\:yd^2 \\[3ex] $

$ft^2$	$yd^2$
$9$	$1$
$2700$	$c$

$ \dfrac{c}{1} = \dfrac{2700}{9} \\[5ex] c = 300\:yd^2 \\[3ex] \therefore 2700\:ft^2 = 300\:yd^2 $

$70$ beats per minute implies $70$ beats in one minute

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] First:\:\:Convert\:\: 70\:\:beats\:\:per\:\:minute\:\:to\:\:beats\:\:per\:\:day \\[3ex] \dfrac{70\:beats}{1\:minute} * \dfrac{...minute}{...hour} * \dfrac{...hour}{...day} \\[5ex] \dfrac{70\:beats}{1\:minute} * \dfrac{60\:minutes}{1\:hour} * \dfrac{24\:hours}{1\:day} \\[5ex] = \dfrac{70 * 60 * 24}{1 * 1 * 1} \\[5ex] = 100800\:beats/day \\[3ex] Second:\:\: Calculate\:\:the\:\:number\:\:of\:\:beats\:\:in\:\:7\:\:days \\[3ex] $ Second Method: Proportional Reasoning Method

Let $c$ = number of beats in $7$ days

beats	days
$100800$	$1$
$c$	$7$

$ \dfrac{c}{100800} = \dfrac{7}{1} \\[5ex] \dfrac{c}{100800} = 7 \\[5ex] 100800 * \dfrac{c}{100800} = 100800(7) \\[5ex] c = 705600\:beats \\[3ex] To\:\:the\:\:nearest\:\:thousand,\:\: c \approx 706,000\:beats \\[3ex] $ Student: Mr. C, could you not just multiply by $7$?
Much faster than setting it like a table.
This is what I mean.
In one day, she has $100800$ beats
So, in seven days, it will be $7 * 100800 = 705600$ beats
Teacher: You are correct.
I did it that way to accommodate everyone...including some people who may not understand it the way you did it.

(I.)
(a.) $\dfrac{m}{hr}$
(b.) meters per hour.

(II.)
(a.) $\dfrac{\$}{mi}$
(b.) dollars per mile.

(III.)
(a.) $\dfrac{L}{min}$
(b.) liters per minute.

(IV.)
multiplying the production rate in begels per baker per hour by the number of hours and by the number of bakers.

$ (a.) \\[3ex] \dfrac{bagels}{\dfrac{baker}{hour}} * hours * bakers \\[7ex] \dfrac{bagels}{baker} \div hour * hours * bakers \\[5ex] \dfrac{bagels}{baker} \div \dfrac{hour}{1} * hours * bakers \\[5ex] \dfrac{bagels}{baker} * \dfrac{1}{hour} * hours * bakers \\[5ex] bagels \\[3ex] $ (b.) bagels.

(V.)
cost of gas divided by the gas mileage and and multiplying the result by the length of the trip
cost of the gas in: $ per gallon
gas mileage in: miles per gallon
length of the trip in: miles

$ (a.) \\[3ex] \dfrac{\$}{gallon} \div \dfrac{miles}{gallon} * miles \\[5ex] \dfrac{\$}{gallon} * \dfrac{gallon}{miles} * miles \\[5ex] \$ $
(b.) dollars.

Let us solve the question.
Then, we can select the correct answer option.
per kilogram means 1 kg
per day means 1 day

First, let us deal with the conversion one at a time.

First Method: Unity Fraction Method

$ 30\;kg \;\;to\;\;how\;\;many\;\;mg\;\;per\;\;day \\[3ex] 30\;kg * \dfrac{.......mg}{.......kg} \;\;per\;\;day \\[5ex] = 30\;kg * \dfrac{300\;mg}{1\;kg} \;\;per\;\;day \\[5ex] = 9000\;mg \;\;per\;\;day \\[3ex] 9000\;\;mg\;\;per\;\;day \;\;to\;\;how\;\;mg\;\;per\;\;every\;\;8\;\;hours \\[3ex] 24\;hours = 1 \;day \\[3ex] 3\;\;every\;\;8\;\;hours = 1\;\;day \\[3ex] \implies \\[3ex] \dfrac{9000\;mg}{1\;day} * \dfrac{.......day}{3\;\;every\;\;8-hours} \\[5ex] = \dfrac{9000\;mg}{1\;day} * \dfrac{1\;day}{3\;\;every\;\;8-hours} \\[5ex] = 3000\;mg\;\;per\;\;every\;\;8\;\;hours \\[3ex] $ Based on the steps in our solution, the correct option is B.

(I.) 1st: 15​-gallon tank of gas for ​$48.35
2nd: ​$3.10​/gallon
Let us determine the cost per gallon (cost for 1 gallon of gas)

$ \underline{cost\;\;per\;\;gallon} \\[3ex] \underline{1st:} \\[3ex] 1\;gallon * \dfrac{\$48.35}{15\;gallon} \\[5ex] = \$3.223333333 \\[3ex] \approx \$3.22 \\[3ex] \underline{2nd:} \\[3ex] \$3.10 \\[3ex] \$3.22 \gt \$3.10 \\[3ex] $ The better deal is to buy the gas for ​$3.10​/gallon because the price to fill a 15​-gallon tank of gas for ​$48.35 is ​$3.122 per gallon

(II.) 1st: ​6-ounce bottle for ​$2.69
2nd: ​15-ounce bottle for ​​$9.59
Let us determine the cost per ounce

$ \underline{cost\;\;per\;\;ounce} \\[3ex] \underline{1st:} \\[3ex] 1\;ounce * \dfrac{\$2.69}{6\;ounce} \\[5ex] = \$0.4483333333 \\[3ex] \approx \$0.45 \\[3ex] \underline{2nd:} \\[3ex] 1\;ounce * \dfrac{\$9.59}{15\;ounce} \\[5ex] = \$0.6393333333 \\[3ex] \approx \$0.64 \\[3ex] $ The 6​-ounce bottle is the better deal because the cost per ounce is $0.45 per ounce while the 15​-ounce bottle is ​$0.64 per ounce

(III.) We have at least two options to solve this question.
We can:
(1.) Convert the rent for Storage Locker A to dollars per square feet per week
or
(2.) Convert the rent for Storage Locker B to dollars per square yard per month.
Let us do both approaches. However, use any approach you profer.

$ \underline{Option\;1} \\[3ex] \dfrac{\$30}{yd * yd} * \dfrac{1}{.......month} * \dfrac{.......yd}{.......ft} * \dfrac{.......yd}{.......ft} * \dfrac{.......month}{.......weeks} \\[5ex] \dfrac{\$30}{yd * yd} * \dfrac{1}{1\;month} * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;month}{4\;weeks} \\[5ex] = \$0.833333333 \\[3ex] \approx \$0.83 \\[3ex] $ Rent for Storage Locker A = $\dfrac{\$0.83}{ft^2}$ per week.

Rent for Storage Locker B = $\dfrac{\$1.36}{ft^2}$ per week.

Storage Locker A is the better deal.

$ \underline{Option\;2} \\[3ex] \dfrac{\$1.36}{ft * ft} * \dfrac{1}{.......week} * \dfrac{.......ft}{.......yd} * \dfrac{.......ft}{.......yd} * \dfrac{.......week}{.......month} \\[5ex] \dfrac{\$1.36}{ft * ft} * \dfrac{1}{1\;week} * \dfrac{3\;ft}{1\;yd} * \dfrac{3\;ft}{1\;yd} * \dfrac{4\;weeks}{1\;month} \\[5ex] = \$48.96 \\[3ex] $ Rent for Storage Locker A = $\dfrac{\$30}{yd^2}$ per month.

Rent for Storage Locker B = $\dfrac{\$48.96}{yd^2}$ per month.

Storage Locker A is the better deal.

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 7500\:mm/s\:\:to\:\:km/hr \\[3ex] 7500\dfrac{mm}{s} * \dfrac{.....m}{.....mm} * \dfrac{.....km}{.....m} * \dfrac{.....s}{.....min} * \dfrac{.....min}{.....hr} \\[5ex] 7500\dfrac{mm}{s} * \dfrac{10^{-3}\:m}{1\:mm} * \dfrac{1\:km}{10^3\:m} * \dfrac{60\:s}{1\:min} * \dfrac{60\:min}{1\:hr} \\[5ex] 7500\dfrac{mm}{s} * \dfrac{0.001\:m}{1\:mm} * \dfrac{1\:km}{1000\:m} * \dfrac{60\:s}{1\:min} * \dfrac{60\:min}{1\:hr} \\[5ex] = \dfrac{7500 * 0.001 * 60 * 60}{1000 * 1 * 1} \\[5ex] = \dfrac{27000}{1000} \\[5ex] = 27\:km/hr \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Per\:\:second\:\:means\:\:1\:\:second \\[3ex] Per\:\:hour\:\:means\:\:1\:\:hour \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:km\:\:for\:\:1\:mm \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:km\:\:for\:\:7500\:mm \\[3ex] Let\:\:k\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hr\:\:for\:\:60\:s \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hr\:\:for\:\:1\:s \\[3ex] \dfrac{7500\:mm}{1\:s} = \dfrac{x\:km}{y\:hr} \\[5ex] $

Numerator

$mm$	$m$	$km$
$1$	$10^{-3} = 0.001$	$p$
	$10^3 = 1000$	$1$
$7500$		$x$

$ \dfrac{0.001}{1000} = \dfrac{p}{1} \\[5ex] 0.000001 = p \\[3ex] p = 0.000001\:km \\[3ex] Next: \\[3ex] \dfrac{1}{7500} = {p}{y} \\[5ex] Cross\:\:Multiply \\[3ex] x(1) = 7500(p) \\[3ex] x = 7500(0.000001) \\[3ex] x = 0.0075\:km \\[3ex] $

Denominator

$s$	$min$	$hr$
$60$	$1$	$k$
	$60$	$1$
$1$		$y$

$ \dfrac{1}{60} = \dfrac{k}{1} \\[5ex] \dfrac{1}{60} = k \\[5ex] k = \dfrac{1}{60} \\[5ex] Next: \\[3ex] \dfrac{60}{1} = \dfrac{k}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 60y = k(1) \\[3ex] 60y = k \\[3ex] y = \dfrac{k}{60} \\[5ex] y = k \div 60 \\[3ex] y = \dfrac{1}{60} \div \dfrac{60}{1} \\[5ex] y = \dfrac{1}{60} * \dfrac{1}{60} \\[5ex] y = \dfrac{1 * 1}{60 * 60} \\[5ex] y = \dfrac{1}{3600} \\[5ex] \implies 1\:s = \dfrac{1}{3600}\:hr \\[5ex] \dfrac{x\:km}{y\:hr} = x \div y \\[5ex] = 0.0075 \div \dfrac{1}{3600} \\[5ex] = 0.0075 * \dfrac{3600}{1} \\[5ex] = 0.0075(3600) \\[3ex] = 27\:km/hr \\[3ex] \therefore 7500\:mm/s = 27\:km/hr $

Conversion between​ Celsius, Fahrenheit, and Kelvin can be done using conversion formulas.​
To convert 60°F to Celsius and Kelvin use the formulas $C = \dfrac{F - 32}{1.8}$ to convert to Celsius and then use $K = C + 273.15$ to convert to Kelvin to get $60^\circ F = 15.56^\circ C = 288.71K$.

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 48.9\:mm^2\:\:to\:\:cm^2 \\[3ex] 48.9\:mm * mm * \dfrac{.....m}{.....mm} * \dfrac{.....m}{.....mm} * \dfrac{.....cm}{.....m} * \dfrac{.....cm}{.....m} \\[5ex] 48.9\:mm * mm * \dfrac{10^{-3}\:m}{1\:mm} * \dfrac{10^{-3}\:m}{1\:mm} * \dfrac{1\:cm}{10^{-2}\:m} * \dfrac{1\:cm}{10^{-2}\:m} \\[7ex] = \dfrac{48.9 * 10^{-3 + -3} * 1 * 1}{1 * 1 * 10^{-2 + -2}} \\[7ex] = \dfrac{48.9 * 10^{-3 - 3}}{10^{-2 - 2}} \\[7ex] = \dfrac{48.9 * 10^{-6}}{10^{-4}} \\[7ex] = \dfrac{48.9}{10^{-4 - (-6)}} \\[7ex] = \dfrac{48.9}{10^{-4 + 6}} \\[7ex] = \dfrac{48.9}{10^2} \\[7ex] = \dfrac{48.9}{100} \\[5ex] = 0.489\:\:cm^2 \\[3ex] $ Second Method: Proportional Reasoning Method

Let:
$p$ = converted unit from $mm^2$ to $m^2$
$d$ = converted unit from $m^2$ to $cm^2$

$ 1\:mm = 10^{-3}\:m \\[3ex] 1\:mm = 0.001\:m \\[3ex] Square\:\:both\:\:sides \\[3ex] (1\:mm)^2 = (0.001\:m)^2 \\[3ex] 1^2 * mm^2 = 0.001^2 * m^2 \\[3ex] 1\:mm^2 = 0.000001\:m^2 \\[3ex] $

$mm^2$	$m^2$
$1$	$0.000001$
$48.9$	$p$

$ \dfrac{p}{48.9} = \dfrac{0.000001}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:48.9 \\[3ex] 48.9 * \dfrac{p}{48.9} = 48.9 * \dfrac{0.000001}{1} \\[5ex] p = 48.9(0.000001) \\[3ex] p = 0.0000489\:m^2 \\[3ex] 1\:cm = 10^{-2}\:m \\[3ex] 1\:cm = 0.01\:m \\[3ex] Square\:\:both\:\:sides \\[3ex] (1\:cm)^2 = (0.01\:m)^2 \\[3ex] 1^2 * cm^2 = 0.0001^2 * m^2 \\[3ex] 1\:cm^2 = 0.0001\:m^2 \\[3ex] $

$m^2$	$cm^2$
$0.0001$	$1$
$0.0000489$	$d$

$ \dfrac{d}{1} = \dfrac{0.0000489}{0.0001} \\[5ex] d = 0.489\:cm^2 $

First Method: Unity Fraction Method

$ (a.)\;\;cubic\;feet\;\;to\;\;pounds \\[3ex] cubic\;feet * \dfrac{.......gallon}{.......cubic\;feet} * \dfrac{.......pound}{.......gallon} \\[5ex] 7.2\;cubic\;feet * \dfrac{7.48\;gallon}{1\;cubic\;feet} * \dfrac{8.33\;pound}{1\;gallon} \\[5ex] = 448.62048\;pounds \\[3ex] (b.)\;\;cubic\;feet\;\;to\;\;tons \\[3ex] cubic\;feet * \dfrac{.......gallon}{.......cubic\;feet} * \dfrac{.......pound}{.......gallon} * \dfrac{.......ton}{.......pound} \\[5ex] 7.2\;cubic\;feet * \dfrac{7.48\;gallon}{1\;cubic\;feet} * \dfrac{8.33\;pound}{1\;gallon} * \dfrac{1\;ton}{2000\;pound} \\[5ex] = 0.22431024\;tons $

First Method: Unity Fraction Method

$ To\;\;Find:\;\;calories\;\;per\;\;tablespoon \\[3ex] \dfrac{66\;calories}{15\;grams} * \dfrac{........grams}{..........tablespoon} \\[5ex] = \dfrac{66\;calories}{15\;grams} * \dfrac{5\;grams}{1\;tablespoon} \\[5ex] = 22\;calories \\[3ex] Calories\;\;from\;\;fat \\[3ex] = 59\%\;\;of\;\; 22\;calories \\[3ex] = 0.59(22) \\[3ex] = 12.98 \\[3ex] \approx 13\;calories $

Second Method: Proportional Reasoning Method

Compound Y

ounces	cost ($)
16	2.40
1	p

$ \dfrac{p}{1} = \dfrac{2.40}{16} \\[5ex] p = \$0.15 \\[3ex] $

Compounds Y and X

cost ($)	% of X
0.15	10%
k	100%

$ k * 10\% = 0.15 * 100\% \\[3ex] k * 0.1 = 0.15 * 1 \\[3ex] 0.1k = 0.15 \\[3ex] k = \dfrac{0.15}{0.1} \\[5ex] k = \$ 1.5 \\[3ex] $ The chemist must pay $1.50 in order to ensure that she receives 1 ounce of element X

First Method: Unity Fraction Method

$ (a.)\;\; furlong\;\;to\;\;yards \\[3ex] 10\;furlong * \dfrac{.......rod}{.......furlong} * \dfrac{.......yd}{.......rd} \\[5ex] = 10\;furlong * \dfrac{40\;rod}{1\;furlong} * \dfrac{5.5\;yd}{1\;rod} \\[5ex] = 2200\;yards \\[3ex] (b.)\;\;furlong\;\;to\;\;miles \\[3ex] 10\;furlong * \dfrac{.......rod}{.......furlong} * \dfrac{.......yd}{.......rd} * \dfrac{.......miles}{.......yd} \\[5ex] = 10\;furlong * \dfrac{40\;rod}{1\;furlong} * \dfrac{5.5\;yd}{1\;rod} * \dfrac{1\;mi}{1760\;yd} \\[5ex] = \dfrac{2200}{1760}\;miles \\[5ex] = 1.25\;miles $

Unity Fraction method is used as applicable.

$ (a.) \\[3ex] L\;\;to\;\;gal \\[3ex] 1\;L = 0.26417205\;gal \\[3ex] 1.3L * \dfrac{0.26417205gal}{1L} \\[5ex] = 0.343423665gal \\[3ex] $ This statement makes sense because 1.3 liters ≈ ​gallon(s), which is a reasonable amount of water to drink in one day.

$ (b.) \\[3ex] kg\;\;to\;\;lb \\[3ex] 1\;lb = 453.59237\;g \\[3ex] 250kg * \dfrac{10^3g}{1kg} * \dfrac{1lb}{453.59237g} \\[5ex] = 551.15565555lb \\[3ex] $ This statement does not make sense because 250 kilograms ≈ 551.3 ​pound(s), which is an unreasonable weight for a professional bicyclist.

(c.) This statement does not make sense because a meter is not a unit of volume.

(d.) This statement makes sense because one calorie equals​ 4,184 joules.
A typical adult consumes about​ 2,500 calories of energy each​ day, which equals about 10 million joules per day.

(e.) This statement does not make sense because 10 grams per cubic centimeter is far greater than the density of​ water, which is 1 gram per cubic centimeter.
Not only would this beach ball be​ heavy, it would sink in water.

(f.) The statement does not make sense because the population density given is small so the city would be a small city rather than a big city.

First Method: Unity Fraction Method

$ 2\dfrac{3}{4}\;inches * \dfrac{......miles}{.........inches} \\[5ex] - \dfrac{11}{4}\;inches * \dfrac{16\;miles}{\dfrac{1}{4}\;inches} \\[7ex] = \dfrac{11}{4} * 16 \div \dfrac{1}{4} \\[5ex] = \dfrac{11}{4} * \dfrac{16}{1} * \dfrac{4}{1} \\[5ex] = 176\;miles \\[3ex] $ Second Method: Proportional Reasoning Method
Let $p$ be the number of actual miles that is represented by $2\dfrac{3}{4}$ inches

inch	miles
$\dfrac{1}{4}$	$16$
$2\dfrac{3}{4} = \dfrac{11}{4}$	$p$

$ 2\dfrac{3}{4} = \dfrac{4 * 2 + 3}{4} = \dfrac{8 + 3}{4} = \dfrac{11}{4} \\[5ex] \dfrac{p}{16} = \dfrac{\dfrac{11}{4}}{\dfrac{1}{4}} \\[7ex] Multiply\:\:both\:\:sides\:\:by\:\: 16 \\[3ex] 16 * \dfrac{p}{16} = 16 * \dfrac{\dfrac{11}{4}}{\dfrac{1}{4}} \\[7ex] p = \dfrac{\dfrac{16 * 11}{4}}{\dfrac{1}{4}} \\[7ex] p = \dfrac{\dfrac{176}{4}}{\dfrac{1}{4}} \\[7ex] p = \dfrac{176}{4} \div \dfrac{1}{4} \\[5ex] p = \dfrac{176}{4} * \dfrac{4}{1} \\[5ex] p = 176\:miles \\[3ex] $ $2\dfrac{3}{4}$ inches represents $176$ miles

First Method: Unity Fraction Method

$ (a.)\;\; feet\;\;to\;\;fathom \\[3ex] 36054\;feet * \dfrac{.......fathom}{.......feet} \\[5ex] = 36054 * \dfrac{1\;fathom}{6\;ft} \\[5ex] = 6009\;ft \\[3ex] (b.)\;\; feet\;\;to\;\;league \\[3ex] 36054\;feet * \dfrac{.......meter}{.......feet} * \dfrac{.......kilometer}{.......meter} * \dfrac{.......nautical\;\;mile}{.......kilometer} * \dfrac{.......league}{.......nautical\;\;mile} \\[5ex] = 36054\;feet * \dfrac{0.3048\;meter}{1\;feet} * \dfrac{1\;kilometer}{10^3\;meter} * \dfrac{1\;nautical\;\;mile}{1.852\;kilometer} * \dfrac{1\;league}{3\;nautical\;\;mile} \\[5ex] = \dfrac{36054 * 0.3048}{1000 * 1.852 * 3}\;league \\[5ex] = \dfrac{10989.2592}{5556}\;league \\[5ex] = 1.977908423\;league $

Unity Fraction method is used as applicable

(a.) 28 knots to miles per hour
1 knot = 1 nautical mile per​ hour
⇒ 28 nautical miles per hour to miles per hour
1 nautical mile = 1.852 kilometers
1 kilometer = $10^3$ meters = 1000 meters
1 feet = 0.3048 meter
1 mile = 5280 feet

$ \dfrac{28\;nautical\;\;miles}{hour} * \dfrac{.......km}{.......nautical\;\;miles} * \dfrac{.......m}{.......km} * \dfrac{.......ft}{.......m} * \dfrac{.......mi}{.......ft} \\[5ex] \dfrac{28\;nautical\;\;miles}{hour} * \dfrac{1.852km}{1\;nautical\;\;miles} * \dfrac{10^3m}{1km} * \dfrac{1ft}{0.3048m} * \dfrac{1mi}{5280ft} \\[5ex] \dfrac{51856\;mi}{1609.344\;hour} \\[5ex] 32.22182454\;miles\;\;per\;\;hour \\[3ex] $ (b.) 94,000 tons to metric tonnes
94,000 short tons to metric tonnes
1 short ton = 2000 pounds
1 pound = 453.59237 grams
1 kg = $10^3$ grams = 1000grams
1 metric tonne = 1000 ​kilograms

$ 94000\;short\;\;ton * \dfrac{.......pound}{.......short\;\;ton} * \dfrac{.......gram}{.......pound} * \dfrac{.......kilogram}{.......gram} * \dfrac{.......metric\;\;tonne}{.......kilogram} \\[5ex] 94000\;short\;\;ton * \dfrac{2000\;pound}{1\;short\;\;ton} * \dfrac{453.59237\;gram}{1\;pound} * \dfrac{1\;kilogram}{1000\;gram} * \dfrac{1\;metric\;\;tonne}{1000\;kilogram} \\[5ex] 85275.36556\;metric\;\;tonne $

First Method: Unity Fraction Method

$ 12\;minutes * \dfrac{........hour}{........minutes} * \dfrac{.........miles}{........hour} \\[5ex] = 12\;minutes * \dfrac{1\;hour}{60\;minutes} * \dfrac{8\;miles}{1\;hour} \\[5ex] = \dfrac{8\;miles}{5} \\[5ex] = 1\dfrac{3}{5}\;miles \\[5ex] $ Third Method: Fast Proportional Reasoning Method
Let $p$ be the number of miles she will run in 12 minutes

miles	hour	minutes
8	1	60
$p$		12

$ \dfrac{p}{8} = \dfrac{12}{60} \\[5ex] p = \dfrac{8 * 12}{60} \\[5ex] p = \dfrac{8 * 1}{5} \\[5ex] p = 1\dfrac{3}{5}\;miles \\[5ex] $ She will run $1\dfrac{3}{5}$ miles in 12 minutes.

(I.) Understand: How much does the water in the pool​ weigh?
The unknown in the problem is: The weight of the water in the pool

Solve: 1 cubic foot of water weighs 64.2 pounds.
So, 20 cubic feet of water will weigh: 20(64.2) = 1284 pounds.

Explain:
First Method: Unity Fraction Method

$ 20\;ft^3 * \dfrac{........pounds}{........ft^3} \\[5ex] = 20\;ft^3 * \dfrac{64.2\;pounds}{1\;ft^3} \\[5ex] = 1284\;pounds \\[3ex] $ OR
Second Method: Proportional Reasoning Method

Volume (ft³)	Pound (lbs)
1	64.2
20	what

$ 1 * what = 20 * 64.2 \\[3ex] what = 1284\;pounds \\[3ex] $ 20 cubic feet of water will weigh: 1284 pounds.

(II.) Understand: How fast does Earth rotate?
The unknown in the problem is: The speed

Solve: and Explain: travels 25,000 miles in 24 hours.
So, in unit of miles per hour:

$ speed = \dfrac{distance}{time} \\[5ex] distance = 25000\;miles \\[3ex] time = 24\;hours \\[3ex] speed = \dfrac{25000}{24} = 1041.666667 \approx 1041.7\;miles\;\;per\;\;hour \\[5ex] $ The earth rotates at a speed of approximately 1041.7 miles per hour.

(I.) Per pound means per 1 pound
$ 3.1\;pounds * \dfrac{........\$}{........pound} \\[5ex] = 3.1\;pounds * \dfrac{0.73\;\$}{1\;pound} \\[5ex] = \$2.263 \\[3ex] \approx \$2.26 \\[3ex] $ Units in question are: pound and dollars

Proportional Reasoning

pound	dollars, $
1	0.73
3.1	what

$ \dfrac{what}{3.1} = \dfrac{0.73}{1} \\[5ex] what = 3.1 * 0.73 \\[3ex] = \$2.263 \\[3ex] $ The price the person will pay is 2.26 dollars

(II.) The person's hourly wage is the amount the person is paid per hour

$ 1\;hour * \dfrac{........\$}{........hour} \\[5ex] = 1\;hour * \dfrac{490\;\$}{27\;hours} \\[5ex] = \$18.14814815 \\[3ex] \approx \$18.15 \\[3ex] $ Units in question are: hours and dollars

Proportional Reasoning

hour	dollars, $
1	0.73
3.1	what

$ \dfrac{what}{3.1} = \dfrac{0.73}{1} \\[5ex] what = 3.1 * 0.73 \\[3ex] = \$2.263 \\[3ex] $ The person's wage is 18.15 dollars per hour.

(III.) Each basketball weighs 25 ounces

$ 12\;basketballs * \dfrac{........ounces}{........basketball} \\[5ex] = 12\;basketballs * \dfrac{25\;ounces}{1\;basketball} \\[5ex] = 300\;ounces \\[3ex] $ The total weight of the basketballs is 300 ounces.

(IV.) Each bus holds 25 people

$ 300\;people * \dfrac{........bus}{........people} \\[5ex] = 300\;people * \dfrac{1\;bus}{25\;people} \\[5ex] = 12\;buses \\[3ex] $ To transport 300 people, 12 buses are needed.

First Method: Unity Fraction Method

$ 5\;eggs * \dfrac{........servings}{........eggs} \\[5ex] = 5\;eggs * \dfrac{6\;servings}{3\;eggs} \\[5ex] = 10\;servings \\[3ex] $ Second Method: Proportional Reasoning Method
Let $p$ be the total number of servings the modified recipe will make

eggs	servings
3	6
5	$p$

$ \dfrac{p}{6} = \dfrac{5}{3} \\[5ex] 3p = 5(6) \\[3ex] p = \dfrac{5(6)}{3} \\[5ex] p = 10\;servings \\[3ex] $ The total number of servings the modified recipe will make is 10 servings.

Measurement is Speed

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 7500\:m/s\:\:to\:\:km/s \\[3ex] 7500\dfrac{m}{s} * \dfrac{.....km}{.....m} \\[5ex] 7500\dfrac{m}{s} * \dfrac{1\:km}{10^3\:m} \\[5ex] 7500\dfrac{m}{s} * \dfrac{1\:km}{1000\:m} \\[5ex] = 7.5\:km/s \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:km\:\:for\:\:7500\:m \\[3ex] \dfrac{7500\:m}{1\:s} = \dfrac{x\:km}{1\:s} \\[5ex] $

$m$	$km$
$10^3 = 1000$	$1$
$7500$	$x$

$ \dfrac{1000}{7500} = \dfrac{1}{x} \\[5ex] OR \\[3ex] \dfrac{7500}{1000} = \dfrac{x}{1} \\[5ex] 7.5 = x \\[3ex] x = 7.5\:km \\[3ex] \therefore \dfrac{7500\:m}{1\:s} = \dfrac{7.5\:km}{1\:s} \\[5ex] 7500\:m/s = 7.5\:km/s $

$ (a.) \\[3ex] 1\;liter = 0.26417205\;gallon \\[3ex] 1\;gallon = 4\;quart \\[3ex] 25\;liters * \dfrac{.......gallons}{.......liters} * \dfrac{.......quarts}{.......gallons} \\[5ex] 25\;liters * \dfrac{0.26417205\;gallon}{1\;liter} * \dfrac{4\;quarts}{1\;gallon} \\[5ex] 26.417205\;quarts \\[5ex] (b.) \\[3ex] 1\;foot = 0.3048\;meter \\[3ex] 4.6\;meters * \dfrac{.......feet}{.......meters} \\[5ex] 4.6\;meters * \dfrac{1\;feet}{0.3048\;meters} \\[5ex] 15.0918635\;feet \\[5ex] (c.) \\[3ex] 1\;pound = 453.59237\;grams \\[3ex] 6\;pounds * \dfrac{.......grams}{.......pounds} * \dfrac{.......kilograms}{.......grams} \\[5ex] 6\;pounds * \dfrac{453.59237\;grams}{1\;pounds} * \dfrac{1\;kilograms}{10^3\;grams} \\[5ex] \dfrac{6 * 453.59237}{1000} kilograms \\[5ex] 2.72155422\;kilograms \\[5ex] $

$ Area\:\;in\:\:square\:\:feet = 15\:feet * 21\:feet = 315\:square\:feet \\[3ex] Convert\:\: 315\:feet^2 \:\:to\:\: yard^2 \\[3ex] $ First Method: Unity Fraction Method

$ 315\:ft^2\:\:to\:\:yd^2 \\[3ex] 315\:ft * ft * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} \\[5ex] 315\:ft * ft * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} \\[5ex] = \dfrac{315 * 1 * 1}{3 * 3} \\[5ex] = \dfrac{315}{9} \\[5ex] = 35\:yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method
Let $c$ = converted unit from $ft^2$ to $yd^2$

$ 3\:ft = 1\:yd \\[3ex] Square\:\:both\:\:sides \\[3ex] (3\:ft)^2 = (1\:yd)^2 \\[3ex] 3^2 * ft^2 = 1^2 * yd^2 \\[3ex] 9\:ft^2 = 1\:yd^2 \\[3ex] $

$ft^2$	$yd^2$
$9$	$1$
$315$	$c$

$ \dfrac{c}{1} = \dfrac{315}{9} \\[5ex] c = 35\:yd^2 \\[3ex] \therefore 2700\:ft^2 = 35\:yd^2 $

$ \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 7500\:m/s\:\:to\:\:km/hr \\[3ex] 7500\dfrac{m}{s} * \dfrac{.....km}{.....m} * \dfrac{.....s}{.....min} * \dfrac{.....min}{.....hr} \\[5ex] 7500\dfrac{m}{s} * \dfrac{1\:km}{10^3\:m} * \dfrac{60\:s}{1\:min} * \dfrac{60\:min}{1\:hr} \\[5ex] 7500\dfrac{m}{s} * \dfrac{1\:km}{1000\:m} * \dfrac{60\:s}{1\:min} * \dfrac{60\:min}{1\:hr} \\[5ex] = \dfrac{7500 * 60 * 60}{1000 * 1 * 1} \\[5ex] = \dfrac{27000000}{1000} \\[5ex] = 27000\:km/hr \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] Per\:\:second\:\:means\:\:1\:\:second \\[3ex] Per\:\:hour\:\:means\:\:1\:\:hour \\[3ex] Let\:\:x\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:km\:\:for\:\:7500\:m \\[3ex] Let\:\:p\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hr\:\:for\:\:60\:s \\[3ex] Let\:\:y\:\:be\:\:the\:\:converted\:\:unit\:\:in\:\:hr\:\:for\:\:1\:s \\[3ex] \dfrac{7500\:m}{1\:s} = \dfrac{x\:km}{y\:hr} \\[5ex] $

Numerator

$m$	$km$
$10^3 = 1000$	$1$
$7500$	$x$

$ \dfrac{7500}{1000} = \dfrac{x}{1} \\[5ex] 7.5 = x \\[3ex] x = 7.5\:km \\[3ex] $

Denominator

$s$	$min$	$hr$
$60$	$1$	$p$
	$60$	$1$
$1$		$y$

$ \dfrac{1}{60} = \dfrac{p}{1} \\[5ex] \dfrac{1}{60} = p \\[5ex] p = \dfrac{1}{60} \\[5ex] Next: \\[3ex] \dfrac{60}{1} = \dfrac{x}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 60y = p(1) \\[3ex] 60y = p \\[3ex] y = \dfrac{p}{60} \\[5ex] y = p \div 60 \\[3ex] y = \dfrac{1}{60} \div \dfrac{60}{1} \\[5ex] y = \dfrac{1}{60} * \dfrac{1}{60} \\[5ex] y = \dfrac{1 * 1}{60 * 60} \\[5ex] y = \dfrac{1}{3600} \\[5ex] \implies 1\:s = \dfrac{1}{3600}\:hr \\[5ex] \dfrac{x\:km}{y\:hr} = x \div y \\[5ex] = 7.5 \div \dfrac{1}{3600} \\[5ex] = 7.5 * \dfrac{3600}{1} \\[5ex] = 7.5(3600) \\[3ex] = 27000\:km/hr \\[3ex] \therefore 7500\:m/s = 27000\:km/hr $

(a.)
Length is the distance.
In other words, 1494 meters is ............ percent of 2 miles.
Let us convert 1494 meters to miles

$ 1494\;meters * \dfrac{.......feet}{.......meters} * \dfrac{.......miles}{.......feet} \\[5ex] 1494\;meters * \dfrac{3.28084\;feet}{1\;meters} * \dfrac{1\;miles}{5280\;feet} \\[5ex] 0.9283285909\;miles \\[3ex] $ In other words, 0.9283285909 miles is ............ percent of 2 miles.

$ \dfrac{is}{of} = \dfrac{\%}{100}...Percent-Proportion \\[5ex] \dfrac{0.9283285909}{2} = \dfrac{what}{100} \\[5ex] what = \dfrac{0.9283285909 * 100}{2} \\[5ex] what = 46.41642955 \\[3ex] what \approx 46\% \\[5ex] $ For each race, convert the time to hours.
Then divide the respective miles by the time (in hours) to convert to miles per hour.

Student
$ (b.) \\[3ex] \underline{Hometown} \\[3ex] 13:15:34 \\[3ex] 13\;minutes \\[3ex] 15\;seconds \\[3ex] 34\;fractional\;seconds \\[3ex] \implies \\[3ex] 13\;minutes * \dfrac{.......hours}{.......minutes} \\[5ex] + 15\;seconds * \dfrac{.......minutes}{.......seconds} * \dfrac{.......hours}{.......minutes} \\[5ex] + 34\;fractional\;seconds * \dfrac{.......seconds}{.......fractional\;seconds} * \dfrac{.......minutes}{.......seconds} * \dfrac{.......hours}{.......minutes} \\[5ex] = 13\;minutes * \dfrac{1\;hours}{60\;minutes} \\[5ex] + 15\;seconds * \dfrac{1\;minutes}{60\;seconds} * \dfrac{1\;hours}{60\;minutes} \\[5ex] + 34\;fractional\;seconds * \dfrac{1\;seconds}{60\;fractional\;seconds} * \dfrac{1\;minutes}{60\;seconds} * \dfrac{1\;hours}{60\;minutes} \\[5ex] = \dfrac{13}{60}hours + \dfrac{15}{60 * 60}hours + \dfrac{34}{60 * 60 * 60}hours \\[5ex] = 0.2209907407\;hours \\[3ex] time = 0.2209907407\;hours \\[3ex] distance = 2\;miles \\[3ex] Average\;\;speed = \dfrac{distance}{time} \\[5ex] = \dfrac{2}{0.2209907407} \\[5ex] = 9.050152931 \\[3ex] \approx 9.05\;mph...to\;\;2...decimal\;\;places \\[5ex] (c.) \\[3ex] \underline{College\;\;Town} \\[3ex] 12:33:00 \\[3ex] 12\;minutes \\[3ex] 33\;seconds \\[3ex] 00\;\;fractional\;seconds \\[3ex] \implies \\[3ex] 12\;minutes * \dfrac{.......hours}{.......minutes} \\[5ex] + 33\;seconds * \dfrac{.......minutes}{.......seconds} * \dfrac{.......hours}{.......minutes} \\[5ex] 12\;minutes * \dfrac{1\;hours}{60\;minutes} \\[5ex] + 33\;seconds * \dfrac{1\;minutes}{60\;seconds} * \dfrac{1\;hours}{60\;minutes} \\[5ex] = \dfrac{12}{60}hours + \dfrac{33}{60 * 60}hours \\[5ex] = \dfrac{251}{1200}hours \\[5ex] time = \dfrac{251}{1200}\;hours \\[5ex] distance = 14942\;meters = 0.9283285909\;miles \\[3ex] Average\;\;speed = \dfrac{distance}{time} \\[5ex] = distance \div time \\[3ex] = 0.9283285909 \div \dfrac{251}{1200} \\[5ex] = \dfrac{0.9283285909 * 1200}{251} \\[3ex] = 4.438224339 \\[3ex] \approx 4.44\;mph...to\;\;2...decimal\;\;places \\[3ex] $ Other Student
$ (d.) \\[3ex] \underline{Hometown} \\[3ex] 14:19:41 \\[3ex] 14\;minutes \\[3ex] 19\;seconds \\[3ex] 41\;fractional\;seconds \\[3ex] \implies \\[3ex] 14\;minutes * \dfrac{.......hours}{.......minutes} \\[5ex] + 19\;seconds * \dfrac{.......minutes}{.......seconds} * \dfrac{.......hours}{.......minutes} \\[5ex] + 41\;fractional\;seconds * \dfrac{.......seconds}{.......fractional\;seconds} * \dfrac{.......minutes}{.......seconds} * \dfrac{.......hours}{.......minutes} \\[5ex] = 14\;minutes * \dfrac{1\;hours}{60\;minutes} \\[5ex] + 19\;seconds * \dfrac{1\;minutes}{60\;seconds} * \dfrac{1\;hours}{60\;minutes} \\[5ex] + 41\;fractional\;seconds * \dfrac{1\;seconds}{60\;fractional\;seconds} * \dfrac{1\;minutes}{60\;seconds} * \dfrac{1\;hours}{60\;minutes} \\[5ex] = \dfrac{14}{60}hours + \dfrac{19}{60 * 60}hours + \dfrac{41}{60 * 60 * 60}hours \\[5ex] = 0.2388009259\;hours \\[3ex] time = 0.2388009259\;hours \\[3ex] distance = 2\;miles \\[3ex] Average\;\;speed = \dfrac{distance}{time} \\[5ex] = \dfrac{2}{0.2388009259} \\[5ex] = 8.375176906 \\[3ex] \approx 8.38\;mph...to\;\;2...decimal\;\;places \\[5ex] (e.) \\[3ex] \underline{College\;\;Town} \\[3ex] 13:41:27 \\[3ex] 13\;minutes \\[3ex] 41\;seconds \\[3ex] 27\;fractional\;seconds \\[3ex] \implies \\[3ex] 13\;minutes * \dfrac{.......hours}{.......minutes} \\[5ex] + 41\;seconds * \dfrac{.......minutes}{.......seconds} * \dfrac{.......hours}{.......minutes} \\[5ex] + 27\;fractional\;seconds * \dfrac{.......seconds}{.......fractional\;seconds} * \dfrac{.......minutes}{.......seconds} * \dfrac{.......hours}{.......minutes} \\[5ex] = 13\;minutes * \dfrac{1\;hours}{60\;minutes} \\[5ex] + 41\;seconds * \dfrac{1\;minutes}{60\;seconds} * \dfrac{1\;hours}{60\;minutes} \\[5ex] + 27\;fractional\;seconds * \dfrac{1\;seconds}{60\;fractional\;seconds} * \dfrac{1\;minutes}{60\;seconds} * \dfrac{1\;hours}{60\;minutes} \\[5ex] = \dfrac{13}{60}hours + \dfrac{41}{60 * 60}hours + \dfrac{27}{60 * 60 * 60}hours \\[5ex] = 0.2281805556\;hours \\[3ex] time = 0.2281805556\;hours \\[3ex] distance = 14942\;meters = 0.9283285909\;miles \\[3ex] Average\;\;speed = \dfrac{distance}{time} \\[5ex] = \dfrac{0.9283285909}{0.2281805556} \\[5ex] = 4.068394823 \\[3ex] \approx 4.07\;mph...to\;\;2...decimal\;\;places \\[3ex] $ (f.) If the student runs the hometown race at the same average speed as in the college town​ race, he will beat his personal record.
If other students run the hometown race at the same average speed as in the college town​ race, they will beat their record time.

First Method: Unity Fraction Method

$ 7600\;in * in * \dfrac{.......ft}{.......in} * \dfrac{.......ft}{.......in} * \dfrac{.......yd}{.......ft} * \dfrac{.......yd}{.......ft} \\[5ex] = 7600\;in * in * \dfrac{1\;ft}{12\;in} * \dfrac{1\;ft}{12\;in} * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;yd}{3\;ft} \\[5ex] = \dfrac{7600}{1296}\;yd^2 \\[5ex] = 5.864197531\;yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method

$ 12\;in = 1\;ft \\[3ex] (12\;in)^2 = (1\;ft)^2 \\[3ex] 144\;in^2 = 1\;ft^2 \\[3ex] 7600\;in^2 = what\;ft^2 \\[3ex] $

in²	ft²
144	1
7600	what

$ \dfrac{what}{1} = \dfrac{7600}{144} \\[5ex] what = 52.77777778\;ft^2 \\[3ex] $ 3\;ft = 1\;yd \\[3ex] (3\;ft)^2 = (1\;yd)^2 \\[3ex] 9\;ft^2 = 1\;yd^2 \\[3ex] 52.77777778\;ft^2 = what\;yd^2 \\[3ex] $

ft²	yd²
9	1
52.77777778	what

$ \dfrac{what}{1} = \dfrac{52.77777778}{9} \\[5ex] what = 5.864197531\;yd^2 $

First Method: Unity Fraction Method

$ 59400\;ft * ft * \dfrac{.......yd}{.......ft} * \dfrac{.......yd}{.......ft} \\[5ex] = 59400\;ft * ft * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;yd}{3\;ft} \\[5ex] = \dfrac{59400}{9}\;yd^2 \\[5ex] = 6600\;yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method

$ 3\;ft = 1\;yd \\[3ex] (3\;ft)^2 = (1\;yd)^2 \\[3ex] 9\;ft^2 = 1\;yd^2 \\[3ex] 59400\;ft^2 = what\;yd^2 \\[3ex] $

ft²	yd²
9	1
59400	what

$ \dfrac{what}{1} = \dfrac{59400}{9} \\[5ex] what = 6600\;yd^2 $

First Method: Unity Fraction Method

$ 460\;ft * ft * ft * \dfrac{1}{s} * \dfrac{.......yd}{.......ft} * \dfrac{.......yd}{.......ft} * \dfrac{.......yd}{.......ft} * \dfrac{.......s}{.......min} \\[5ex] = 460\;ft * ft * ft * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;yd}{3\;ft} * \dfrac{1\;yd}{3\;ft} * \dfrac{60\;s}{1\;min} \\[5ex] = \dfrac{27600}{27}\dfrac{yd^3}{min} \\[5ex] = 1022.222222\;\dfrac{yd^3}{min} \\[5ex] $ Second Method: Proportional Reasoning Method

$ \underline{Numerator} \\[3ex] 3\;ft = 1\;yd \\[3ex] (3\;ft)^3 = (1\;yd)^3 \\[3ex] 27\;ft^3 = 1\;yd^3 \\[3ex] 460\;ft^3 = what\;yd^3 \\[3ex] $

Numerator

ft³	yd³
27	1
460	what

$ \dfrac{what}{1} = \dfrac{460}{27} \\[5ex] what = \dfrac{460}{27}\;yd^3 \\[5ex] Also: \\[3ex] 60\;s = 1\;min \\[3ex] $

Denominator

s	min
60	1
1	what

$ \dfrac{what}{1} = \dfrac{1}{60} \\[5ex] what = \dfrac{1}{60}\;min \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] Numerator \div Denominator \\[3ex] \dfrac{460}{27}\;yd^3 \div \dfrac{1}{60}\;min \\[5ex] \dfrac{460}{27} * \dfrac{60}{27} \\[5ex] \dfrac{27600}{27}\dfrac{yd^3}{min} \\[5ex] 1022.222222\;\dfrac{yd^3}{min} $

First Method: Unity Fraction Method

$ \$ 156\; * \dfrac{1}{ft} * \dfrac{1}{ft} * \dfrac{.......euro}{.......\$} * \dfrac{.......ft}{.......m} * \dfrac{.......ft}{.......m} \\[5ex] = \$ 156\; * \dfrac{1}{ft} * \dfrac{1}{ft} * \dfrac{1\;euro}{\$\;1.225} * \dfrac{3.28\;ft}{1\;m} * \dfrac{3.28\;ft}{1\;m} \\[5ex] = \dfrac{1678.3104}{1.225}\dfrac{euro}{m^2} \\[5ex] = 1370.049306\;euro/m^2 \\[5ex] $ Second Method: Proportional Reasoning Method
$156/ft² = $156 ÷ 1 ft²

$ \underline{Numerator} \\[3ex] \$1.225 = 1\;euro \\[3ex] \$156 = what\;euro \\[3ex] $

Numerator

$	euro
1.225	1
156	what

$ \dfrac{what}{1} = \dfrac{156}{1.225} \\[5ex] what = 127.3469388\;euro \\[5ex] \underline{Denominator} \\[3ex] 3.28\;ft = 1\;m \\[3ex] (3.28\;ft)^2 = (1\;m)^2 \\[3ex] 10.7584\;ft^2 = 1\;m^2 \\[3ex] 1\;ft^2 = what\;m^2 \\[3ex] $

Denominator

ft²	m²
10.7584	1
1	what

$ \dfrac{what}{1} = \dfrac{1}{10.7584} \\[5ex] what = 0.0929506246\;m^2 \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] Numerator \div Denominator \\[3ex] 127.3469388\;euro \div 0.0929506246\;m^2 \\[3ex] = 1370.049306\;euro/m^2 $