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



Solved Examples - All Measurements and Conversions

Samuel Dominic Chukwuemeka (SamDom For Peace) 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.

Metric to Metric Conversions
Prefix Symbol Multiplication Factor
yocto y $10^{-24}$
zepto z $10^{-21}$
atto a $10^{-18}$
femto f $10^{-15}$
pico p $10^{-12}$
nano n $10^{-9}$
micro $\mu$ $10^{-6}$
milli m $10^{-3}$
centi c $10^{-2}$
deci d $10^{-1}$
deka da $10^1$
hecto h $10^2$
kilo K $10^3$
mega M $10^6$
giga G $10^9$
tera T $10^{12}$
peta P $10^{15}$
exa E $10^{18}$
zetta Z $10^{21}$
yotta Y $10^{24}$


Customary to Customary Conversions
Measurement Customary Customary Unit Conversion Factor
Length inch (in) foot (ft) $12\:inches = 1\:ft$
Length foot (ft) yard (yd) $3\:ft = 1\:yd$
Length yard (yd) mile (mi) $1760\:yd = 1\:mi$
Length foot (ft) mile (mi) $5280\:ft = 1\:mi$
Mass pound (lb) ounce (oz) $1\:lb = 16\:oz$
Mass short ton (ton) pound (lb) $1\:short\:ton = 2000\:lb$
Mass long ton pound (lb) $1\:long\:ton = 2240\:lb$
Mass stone pound (lb) $1\:\:stone = 14\:lb$
Mass long ton stone $1\:long\:ton = 160\:stones$
Area acre (acre) square feet ($ft^2$) $1\:acre = 43560\:ft^2$
Volume quart (qt) pint (pt) $1\:qt = 2\:pt$
Volume pint (pt) cup (cup) $1\:pt = 2\:cups$
Volume quart (qt) cup (cup) $1\:qt = 4\:cups$
Volume quart (qt) fluid ounce (fl. oz) $1\:qt = 32\:fl.\:oz$
Volume pint (pt) fluid ounce (fl. oz) $1\:pt = 16\:fl.\:oz$
Volume cup (cup) fluid ounce (fl. oz) $1\:cup = 8\:fl.\:oz$
Volume gallon (gal) quart (qt) $1\:gal = 4\:qt$
Volume gallon (gal) quart (pt) $1\:gal = 8\:pt$
Volume gallon (gal) cup (cup) $1\:gal = 16\:cups$
Volume gallon (gal) fluid ounce (fl. oz) $1\:gal = 128\:fl.\:oz$
Volume gallon (gal) cubic inches ($in^3$) $1\:gal = 231\:in^3$


Metric to Customary Conversions
Measurement Metric Customary Unit Conversion Factor
Length meter (m) foot (ft) $1\:ft = 0.3048\:m$
Mass gram (g) pound (lb) $1\:lb = 453.59237\:g$
Mass metric ton (tonne) kilogram (kg) $1\:tonne = 1000\:kg$
Volume liter or cubic decimeters (L or $dm^3$) gallons (gal) $1\:L = 0.264\:gal$

Solve all questions.
Use at least two methods as applicable.
State the measurement.
Show all work.


NOTE: Unless specified otherwise:
(1.) Use only the tables provided for you.
(2.) Please do not approximate intermediate calculations.
(3.) Please do not approximate final calculations. Leave your final answer "as is".

(1.) ACT How many minutes would it take a car to travel $120$ miles at a constant speed of $50$ miles per hour?
(Note: There are $60$ minutes in $1$ hour.)

$ F.\:\: 25 \\[3ex] G.\:\: 42 \\[3ex] H.\:\: 70 \\[3ex] J.\:\: 100 \\[3ex] K.\:\: 144 \\[3ex] $

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 $
(2.) CSEC (i) At Bank $A$, $US\$1.00 = BD\$1.96$
Calculate the value of $US\$700$ in $BD\$$

(ii) At Bank $B$, the value of $US\$700$ is $BD\$1386$
Calculate the value of $US\$1.00$ in $BD\$$ at this bank.


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] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] $
$US\$$ $BD\$$
$1.00$ $1.96$
$700$ $x$

$ \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 $
(3.) Convert $7500$ meter per second to kilometer per second


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 $
(4.) ACT There are $16$ ounces in a pound.
If $1.5$ pounds of hamburger costs $\$2.88$, what is the cost per ounce?

$ F.\:\: \$0.12 \\[3ex] G.\:\: \$0.18 \\[3ex] H.\:\: \$0.88 \\[3ex] J.\:\: \$1.92 \\[3ex] K.\:\: \$4.32 \\[3ex] $

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 $
(5.) Convert $7500$ meter per second to kilometer per hour


$ \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 $
(6.) ACT If you've traveled $x$ miles per hours for $3$ hours, how many miles have you traveled?

$ A.\:\: \dfrac{3}{x} \\[5ex] B.\:\: \dfrac{x}{3} \\[5ex] C.\:\: 3x \\[3ex] D.\:\: 60x \\[3ex] E.\:\: 180x \\[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
(7.) Convert $7500$ millimeter per second to kilometer per hour


$ \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 $
(8.) ACT Vanna walked at a rate of $2$ miles per hour for $10$ minutes and then walked at a rate of $3$ miles per hour for $5$ minutes.
Which of the following gives the average rate, in miles per hour, at which she walked over this $15-minute$ period?

$ F.\:\: \dfrac{1}{3} \\[5ex] G.\:\: \dfrac{7}{3} \\[5ex] H.\:\: \dfrac{7}{24} \\[5ex] J.\:\: \dfrac{7}{180} \\[5ex] K.\:\: \dfrac{35}{2} \\[5ex] $

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 $
(9.) Convert $48.9\:mm^2$ into square centimeters


$ \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 $
(10.) ACT Walter recently vacationed in Paris.
While there, he visited the Louvre, a famous art museum.
Afterward, he took a $3.7-kilometer$ cab ride from the Louvre to the Eiffel Tower.
A tour guide named Amélie informed him that $2.5$ million rivets were used to build the tower, which stands $320$ meters tall.

Walter's cab ride lasted $15$ minutes.
Which of the following values is closest to the average speed, in miles per hour, of the cab?
(Note: $1\:mile \approx 1.6\:kilometers)$

$ F.\:\: 9 \\[3ex] G.\:\: 15 \\[3ex] H.\:\: 21 \\[3ex] J.\:\: 24 \\[3ex] K.\:\: 35 \\[3ex] $

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.
(11.) A basement for a $26$ feet by $32$ feet house is to be dug at a depth of $3$ feet.
How many cubic yards of earth need to be hauled away?
Assume the surface of the ground is level.


$ 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 $
(12.) ACT Tameka calculates that she needs $360$ square feet of new carpet.
But the type of carpet that she wants is priced by the square yard.
How many square yards of carpet does she need?

$ F.\:\: 15 \\[3ex] G.\:\: 40 \\[3ex] H.\:\: 60 \\[3ex] J.\:\: 90 \\[3ex] K.\:\: 120 \\[3ex] $

$ \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 $
(13.) When Ibuprofen is given for fever to children $6$ months of age up to $2$ years, the usual does is $5$ milligrams ($mg$) per kilogram ($kg$) of body weight when the fever is under $102.5$ degrees Fahrenheit.
How much medicine would be usual dose for an $18$ month old weighing $26$ pounds?


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 $59\:mg$
(14.) When Ibuprofen is given for fever to children $6$ months of age up to $2$ years, the usual does is $5$ milligrams ($mg$) per kilogram ($kg$) of body weight when the fever is under $102.5$ degrees Fahrenheit.
How much medicine would be usual dose for an $18$ month old weighing $19$ pounds?
$1\:kg = 2.2\:pounds$


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 $43\:mg$
(15.) A medication (drink) calls for $200$ grams of sugar, $2$ liters of water, and $1$ package of a prepared medicine.
While gathering the items, Rita realizes that there is only $100$ grams of sugar available.
How can Rita still make the drink so that it has the correct concentration?


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.
(16.) As a nurse, part of the daily duties of Deborah is to mix medications in the proper proportions for patients.
For one of her regular patients, she always mixes Medication $A$ with Medication $B$ in the same proportion.
Last week, her patient's doctor indicated that she should mix $90$ milligrams of Medication $A$ with $27$ milligrams of Medication $B$.
However, this week, the doctor said she should only use $24$ milligrams of Medication $B$.
How many milligrams of Medication $A$ should be mixed this week?


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 $A$
(17.) Pediatricians prescribe $7$ milliliters ($mL$) of acetaminophen for $35$ pounds of a child's weight.
How many milliliters of acetaminophen will a pediatrician prescribe for Timothy, who weighs $60$ pounds?


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.
(18.) A veterinarian prescribed to Smart, a $70-pound$ dog, an antibacterial medicine in case an infection emerges after her teeth were cleaned.
If the dosage is $7\:mg$ for every pound, how much medicine was Smart given?


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.
(19.) Convert $1056$ feet per second to mile per hour


$ 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 $
(20.) Convert $3168$ feet per minute to mile per second


$ \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 $




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(21.) ACT In decorating baskets for a retirement party, Rudy needs the following amounts of ribbons for each basket:
number of ribbons length of "each" ribbon (inches)
$5$ $8$
$3$ $16$
$2$ $10$

If the ribbon costs $\$0.98$ per yard, which of the following would be the approximate cost of ribbon for $10$ baskets?
(Note: $1$ yard = $36$ inches)

$ F.\:\: \$3 \\[3ex] G.\:\: \$9 \\[3ex] H.\:\: \$30 \\[3ex] J.\:\: \$35 \\[3ex] K.\:\: \$90 \\[3ex] $

$ \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?
(22.) ACT Car $A$ travels $60$ miles per hour for $1\dfrac{1}{2}$ hours; Car $B$ travels $40$ miles per hour for $2$ hours.
What is the difference between the number of miles traveled by Car $A$ and the number of miles traveled by Car $B$?

$ F.\:\: 0 \\[3ex] G.\:\: 10 \\[3ex] H.\:\: 80 \\[3ex] J.\:\: 90 \\[3ex] K.\:\: 170 \\[3ex] $

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 $
(23.) ACT A road map is drawn to scale so that $1.5$ inches represents $90$ miles.
How many miles does $1.6$ inches represent?

$ F.\:\: 91 \\[3ex] G.\:\: 96 \\[3ex] H.\:\: 99 \\[3ex] J.\:\: 100 \\[3ex] K.\:\: 106 \\[3ex] $

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 $
(24.) ACT A stone is a unit of weight equivalent to $14$ pounds.
If a person weighs $177$ pounds, how many stone, to the nearest tenth, does this person weigh?

$ A.\:\: 247.8 \\[3ex] B.\:\: 126.4 \\[3ex] C.\:\: 79.1 \\[3ex] D.\:\: 12.6 \\[3ex] E.\:\: 7.9 \\[3ex] $

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 $
(25.) ACT A truck sprang a leak at the bottom of its radiator, which held $480$ ounces of fluid when it started to leak, and started losing radiator fluid at a constant rate of $4$ ounces per minute.
Suppose that the radiator continued to leak at this constant rate and that the truck, traveling at $35$ miles per hour, could continue traveling at this rate until its radiator was completely empty.
In how many miles would the radiator be empty?

$ F.\:\: 13.7 \\[3ex] G.\:\: 17.5 \\[3ex] H.\:\: 35.0 \\[3ex] J.\:\: 70.0 \\[3ex] K.\:\: 120.0 \\[3ex] $

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.
(26.) ACT The end-on view of a cylindrical milk tank on its support is shown in the figure below.
The interior radius of the tank's circular end is $4$ feet.
The interior length of the tank is $25$ feet.
cylindrical tank

The tank currently holds $5,000$ gallons of milk.
Each gallon of milk weighs about $8$ pounds.
About how many pounds does this milk weigh?

$ F.\:\: 625 \\[3ex] G.\:\: 4,000 \\[3ex] H.\:\: 4,992 \\[3ex] J.\:\: 5,008 \\[3ex] K.\:\: 40,000 \\[3ex] $

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 $
(27.) ACT Traveling at approximately $186,000$ miles per second, about how many miles does a beam of light travel in $2$ hours?

$ F.\:\: 3.72 * 10^5 \\[3ex] G.\:\: 2.23 * 10^6 \\[3ex] H.\:\: 2.68 * 10^7 \\[3ex] J.\:\: 6.70 * 10^8 \\[3ex] K.\:\: 1.34 * 10^9 \\[3ex] $

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 $
(28.) ACT Kaya drove $200$ miles in $5$ hours of actual driving time.
By driving an average of $10$ miles per hour faster, Kaya could have saved how many hours of actual driving time?

$ A.\:\: \dfrac{1}{6} \\[5ex] B.\:\: \dfrac{2}{3} \\[5ex] C.\:\: \dfrac{7}{10} \\[5ex] D.\:\: 1 \\[3ex] E.\:\: 4 \\[3ex] $

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 $
(29.) A liquid substance has a density of $10$ pounds per gallon.
Convert this into grams per cubic centimeter using only the information:

$ 1\:gal \approx 231\:inch^3 \\[3ex] 1\:lb \approx 453.6\:g \\[3ex] 1\:inch = 2.54\:cm \\[3ex] $ Round your answer to the nearest hundredth as needed.


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 $
(30.) ACT A recipe for $150$ servings requires $4.5$ liters of sauce.
About how many liters of sauce are required for $80$ servings?

$ A.\:\: 2.1 \\[3ex] B.\:\: 2.4 \\[3ex] C.\:\: 3.0 \\[3ex] D.\:\: 8.4 \\[3ex] E.\:\: 15.5 \\[3ex] $

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
(31.) Convert $1245$ feet per hour to centimeter per second


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 $
(32.) Convert $1245$ feet per hour to inch per minute


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 $
(33.) ACT A certain race car has a maximum speed of $240$ miles per hour.
Which of the following is an expression for this maximum speed in feet per second?
(Note: $1$ mile = $5,280$ feet)

$ A.\:\: \dfrac{240(5,280)}{36,000} \\[5ex] B.\:\: \dfrac{240(3,600)}{5,280} \\[5ex] C.\:\: \dfrac{240(5,280)}{3,600} \\[5ex] D.\:\: \dfrac{60(5,280)}{240} \\[5ex] E.\:\: \dfrac{240(5,280)}{360} \\[5ex] $

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 $
(34.) ACT On a map, $\dfrac{1}{2}\:inch$ represents $10$ actual miles.

Two miles that are $4\dfrac{1}{2}\:inches$ apart on this map are how many actual miles apart?

$ A.\:\: 10 \\[3ex] B.\:\: 20 \\[3ex] C.\:\: 22\dfrac{1}{2} \\[5ex] D.\:\: 45 \\[3ex] E.\:\: 90 \\[3ex] $

Second Method: Proportional Reasoning Method

Let $p$ 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
(35.) Convert $0.00005$ kilometer per minute squared to centimeter per hour squared


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 $
(36.) Convert $3450$ Newtons per square centimeter to pounds per square inch.
Use the conversion:$1\:inch = 2.54\:cm$ and $1\:lb = 4.448\:N$
Round your answer to the nearest hundredth as needed.


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 $
(37.) Convert $84520$ grams per cubic meter to pounds per cubic feet.


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 $
(38.) Convert $12450$ feet per hour squared to inch per minute squared


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 $


Use the following information to answer Questions $39$, $40$, and $41$

ACT Carl purchased a new car.
The fuel economy window sticker on the new car contained the information shown below.
In the figure, MPG is miles per gallon.

conversions


(39.) ACT Carl is planning a trip in his new car that will include $350$ miles of highway driving.
Using the average fuel cost per gallon given in the fuel economy window sticker, which of the following dollar amounts is closest to his total cost for fuel over the $350$ miles of highway driving?

$ F.\:\: \$43.75 \\[3ex] G.\:\: \$51.85 \\[3ex] H.\:\: \$56.00 \\[3ex] J.\:\: \$63.64 \\[3ex] K.\:\: \$87.50 \\[3ex] $

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 $\$43.75$
(40.) ACT The cost estimates are based on a certain number of miles driven per year.
To the nearest $1,000$ miles, what is this number?

$ A.\:\: 13,000 \\[3ex] B.\:\: 15,000 \\[3ex] C.\:\: 16,000 \\[3ex] D.\:\: 19,000 \\[3ex] E.\:\: 22,000 \\[3ex] $

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 $15,000\:miles$
(41.) ACT Based on the annual fuel cost estimate for this car and the estimate for how much Carl will save in fuel costs over the next $5$ years, what would be the expected annual fuel cost of an average new vehicle?

$ F.\:\: \$1,180 \\[3ex] G.\:\: \$2,950 \\[3ex] H.\:\: \$3,100 \\[3ex] J.\:\: \$3,500 \\[3ex] K.\:\: \$3,980 \\[3ex] $

$ \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)
(42.) 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] $

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
(43.) ACT A rectangular stage is $90$ feet long and $30$ feet wide.
What is the area, in square yards, of this stage?

$ A.\:\: 30\sqrt{3} \\[3ex] B.\:\: 300 \\[3ex] C.\:\: 675 \\[3ex] D.\:\: 900 \\[3ex] E.\:\: 2,700 \\[3ex] $

$ 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 $
(44.)