Solved Examples: Applications of Measurements and Units

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.

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.) Assume an average human heart beats 71 times per minute and an average human lifetime is 78 years.
How many times does the average human heart beat in an average lifetime?
Assume that each year is a non-leap year.

$ 78\;years * \dfrac{.......times}{.......minute} * \dfrac{.......minute}{.......hour} * \dfrac{.......hour}{.......day} * \dfrac{.......day}{.......year} \\[5ex] 78\;years * \dfrac{71\;times}{1\;hour} * \dfrac{24\;hours}{1\;day} * \dfrac{36\;days}{1\;year} \\[5ex] 2910772800\;times \\[3ex] $ The average human heart beat in an average lifetime is 2910772800 heart beats.

(2.) You plan to take a 2055-mile trip in your car, which averages 30 miles per gallon.
(a.) How many gallons of gasoline should you expect to use?
(b.) Would a car that has only half the gas mileage require twice as much gasoline for the same trip?
Explain.

$ (a.) \\[3ex] 2055\;mile * \dfrac{1\;gallon}{30\;miles} = 68.5\;gallons \\[5ex] $ You should expect to use about 68.5 gallons of gasoline.

(b.) only half the gas mileage ⇒ 30 ÷ 2 = 15 miles

$ 2055\;mile * \dfrac{1\;gallon}{15\;miles} = 137\;gallons \\[5ex] $ 2 * 68.5 = 137
Yes; the car would use 137 gallons, which is twice as many gallons.

(3.) Gas mileage actually varies slightly with the driving speed of a car (as well as with highway vs. city driving).
Suppose your car averages 35 miles per gallon on the highway if your average speed is 53 miles per hour, and it averages 20 miles per gallon on the highway if your average speed is 67 miles per hour.

(a.) What is the driving time for a 1800-mile trip if you drive at an average speed of 53 miles per hour?
(b.) What is the driving time for a 1800-mile trip if you drive at an average speed of 67 miles per hour?
Assume a gasoline price of $4.22 per gallon.
(c.) What is the gasoline cost for a 1800-mile trip if you drive at an average speed of 53 miles per hour?
(d.) What is the gasoline cost at 67 miles per hour?

$ (a.) \\[3ex] 1800\;mile * \dfrac{1\;hour}{53\;miles} = 33.96226415\;hours \\[5ex] (b.) \\[3ex] 1800\;mile * \dfrac{1\;hour}{67\;miles} = 26.86567164\;hours \\[5ex] (c.) \\[3ex] 53\;miles\;\;per\;\;hour \equiv 35\;miles\;\;per\;\;gallon \\[3ex] \implies \\[3ex] 1800\;mile * \dfrac{1\;gallon}{35\;miles} * \dfrac{\$4.22}{1\;gallon} = \$217.0285714 \approx \$217.03 \\[5ex] (d.) \\[3ex] 67\;miles\;\;per\;\;hour \equiv 20\;miles\;\;per\;\;gallon \\[3ex] \implies \\[3ex] 1800\;mile * \dfrac{1\;gallon}{20\;miles} * \dfrac{\$4.22}{1\;gallon} = \$379.80 $

(4.) (a.) A 10% dextrose solution (D10W) contains 10 mg of dextrose per 100 mL of solution.
How many milligrams of dextrose is in 1750mL of a 10% solution?

(b.) How many milliliters of D10W solution should be given to a patient needing 160 mg of dextrose?

(c.) Normal saline solution (NS) has a concentration of 0.9% sodium chloride, or 0.9 mg per 100 mL.
How many milligrams of sodium chloride is in 0.7 L of NS?

(d.) How many milliliters of NS should be given to a patient needing 185 mg of sodium chloride?

$ (a.) \\[3ex] 1750\;mL * \dfrac{.......mg}{.......mL} \\[5ex] 1750\;mL * \dfrac{10\;mg}{100\;mL} \\[5ex] 175\;mg \\[3ex] (b.) \\[3ex] 160\;mg * \dfrac{.......mL}{.......mg} \\[5ex] 160\;mg * \dfrac{100\;mL}{10\;mg} \\[5ex] 1600mL \\[3ex] (c.) \\[3ex] 0.7L * \dfrac{.......mL}{.......L} * \dfrac{.......mg}{.......mL} \\[5ex] 0.7L * \dfrac{1\;mL}{10^{-3}\;L} * \dfrac{0.9\;mg}{100\;mL} \\[5ex] 6.3\;mg \\[3ex] (d.) \\[3ex] 185\;mg * \dfrac{.......mL}{.......mg} \\[5ex] 185\;mg * \dfrac{100\;mL}{0.9\;mg} \\[5ex] 20555.5556 \approx 20556\;mL $

(5.) Suppose that a certain basketball athlete signed 4-year contract worth $184.96 million.
Assume he plays an average of 80 games per season (including playoffs).
(a.) How much does the athlete earn in a single season?
(b.) How much does the athlete earn per game?
(c.) Suppose that, over the course of each year, the athlete spends an average of 20 hours training for and playing in each of the 80 games.
Including the training time, what is his hourly salary?

Each season is one year.
Per season means per year.
Per season means 1 season.
Per game means 1 game.
1 season = 1 year.

$ (a.) \\[3ex] 1\;year * \dfrac{.......\$}{.......year} \\[5ex] 1\;year * \dfrac{\$184.96\;millions}{4\;years} \\[5ex] 46.24\;millions \\[3ex] 46.24 * 10^6 \\[3ex] \$46240000\;\;per\;\;year \\[5ex] (b.) \\[3ex] 1\;game * \dfrac{.......\$}{.......year} * \dfrac{.......year}{.......game} \\[5ex] 1\;game * \dfrac{\$46240000}{1\;year} * \dfrac{1\;year}{80\;games} \\[5ex] \$578000\;\;per\;\;game \\[5ex] (c.) \\[3ex] 1\;hour * \dfrac{.......game}{.......hour} * \dfrac{.......\$}{.......game} \\[5ex] 1\;hour * \dfrac{1\;game}{20\;hours} * \dfrac{\$578000}{1\;game} \\[5ex] \$28900\;\;per\;\;hour $

(6.) A certain antihistamine is often prescribed for allergies.
A typical dose for a 100-pound person is 18 mg every six hours.

(a.) Following this dosage, how many 12.2 mg chewable tablets would be taken in a week?
(b.) His antihistamine also comes in a liquid form with a concentration of 12.2 mg/10 mL.
Following the prescribed dosage, how much liquid antihistamine should a 100-pound person take in a week?

(a.) Let us first find how many mg per week

$ \dfrac{18\;mg}{6\;hours} * \dfrac{24\;hours}{1\;day} * \dfrac{7\;days}{1\;week} \\[5ex] = \dfrac{504\;mg}{1\;week} \\[5ex] $ How many 12.2 mg are in 504 mg?

$ \dfrac{504\;mg}{1\;week} * \dfrac{1}{12.2\;mg} \\[5ex] = 41.31147541 \\[3ex] \approx 41\;\;tablets\;\;of\;\;12.2\;mg\;\;per\;\;week \\[5ex] (b.) \\[3ex] \dfrac{504\;mg}{1\;week} * \dfrac{10\;mL}{12.2\;mg} \\[5ex] = 413.1147541 \\[3ex] \approx 413\;mL\;\;per\;\;week $

(7.) Assume that when you take a bath, you fill a tub to the halfway point.
The tub measures 6 feet by 3 feet by 2.6 feet.
When you take a shower, you use a shower head with a flow rate of 1.28 gallons per minute, and you typically spend 10 minutes in the shower.
There are 7.5 gallons in one cubic foot.

(a.) Do you use more water taking a shower or taking a bath?
(b.) By how much is the water used?
(c.) How long would you need to shower in order to use as much water as you use taking a bath?
(d.) Assuming your shower is in a bath tub, propose a non-mathematical way to compare, in one experiment, the amounts of water you use taking a shower and a bath.

$ (a.) \\[3ex] \underline{Tub} \\[3ex] Volume = 6 * 3 * 2.6 = 46.8\;ft^3 \\[5ex] \underline{Bath} \\[3ex] Halfway\;\;of\;\;tub \\[3ex] Volume = \dfrac{1}{2} * 46.8 = 23.4\;ft^3 \\[5ex] \underline{Shower} \\[3ex] Volume = \dfrac{1.28\;gallons}{minute} * 10\;minutes * \dfrac{1\;ft^3}{7.5\;gallons} \\[5ex] Volume = 1.706666667\;ft^3 \\[5ex] (b.) \\[3ex] \underline{More\;\;Water\;\;and\;\;by\;\;how\;\;much} \\[3ex] Bath\;\;uses\;\;more\;\;water \\[3ex] By\;\;how\;\;much \\[3ex] = Volume\;\;of\;\;Bath - Volume\;\;of\;\;Tub \\[3ex] = 23.4 - 1.706666667 \\[3ex] = 21.69333333\;ft^3 \\[3ex] $ (c.) How long would you need to shower in order to use as much water as you use taking a bath?
Paraphrasing: For what amount of time in the shower will the volume of the shower be equal to the volume of the bath?
Rather than 10 minutes, we do not know how long.
Let us just denote the amount of time in the shower as Time

$ (c.) \\[3ex] \underline{Shower} \\[3ex] Volume = \dfrac{1.28\;gallons}{minute} * \color{darkblue}{Time}\;minutes * \dfrac{1\;ft^3}{7.5\;gallons} \\[5ex] Volume = 0.1706666667 * \color{darkblue}{Time} \;ft^3 \\[3ex] \underline{Bath} \\[3ex] Volume = 23.4\;ft^3 \\[3ex] Equate\;\;both\;\;volumes\;\;to\;\;calculate\;\;\color{darkblue}{Time} \\[3ex] \implies \\[3ex] 0.1706666667 * \color{darkblue}{Time} = 23.4 \\[3ex] \color{darkblue}{Time} = \dfrac{23.4}{0.1706666667} \\[5ex] \color{darkblue}{Time} = 137.109375 \\[3ex] \color{darkblue}{Time} \approx 137\;minutes \\[3ex] $ (d.) Put the plug in the drain when taking a shower.

(8.) The Greenland ice sheet contains about 3 million cubic kilometers of ice.
If completely melted, this ice would release about 1.9 million cubic kilometers of water, which would spread out over Earth's approximate 336 million square kilometers of ocean surface.
How much would the sea level rise?
Round your answer to three decimal places.

If ice sheet melts, water will be released because the melting of ice gives water.
The water released would make the sea level to rise.

How much would the sea level rise means the height of the sea level due to the release of the water, which was caused by the melting of the ice sheet.

cubic kilometers is a unit of Volume
square kilometers is a unit of Area
kilometers is a unit of Length (Height)

km³ = km² * km
Volume = Area * Length
Therefore:
Volume of water = Area of ocean surface * Height of sea Level

$ 1.9 * 10^6\;\;km^3 = 336 * 10^6\;km^2 * height \\[3ex] height = \dfrac{1.9 * 10^6\;km^3}{336 * 10^6\;km^2} \\[5ex] height = 0.0056547619\;km \\[3ex] To\;\;3\;\;decimal\;\;places,\;\;height \approx 0.006 \\[3ex] $ The sea level would rise approximately 0.006 km

(11.) Experts estimate that when levees break in the aftermath of a hurricane, water can flow into a city at a peak rate of 10 billion gallons per day.
There are 7.5 gallons in 1 cubic foot.

(a.) Determine the flow rate in units of cubic feet per second (cfs).

(b.) Compare this flow rate to a river with an average flow rate of 25000 cfs.
The flow rate is approximately ............ of the flow rate of the river.

(c.) Assume that the flooded part of the city had an area of 7 square miles.
Estimate how much (in feet) the water level rose in one day at the given flow rate.

(a.) Paraphrasing: Convert 10 billion gallons per day to cubic feet per second.

$ \dfrac{10 * 10^9\;gallons}{1\;day} * \dfrac{.......ft^3}{.......gallons} * \dfrac{.......day}{.......hour} * \dfrac{.......hour}{.......minute} * \dfrac{.......minute}{.......second} \\[5ex] \dfrac{10^{10}\;gallons}{1\;day} * \dfrac{1\;ft^3}{7.5\;gallons} * \dfrac{1\;day}{24\;hour} * \dfrac{1\;hour}{60\;minute} * \dfrac{1\;minute}{60\;second} \\[5ex] = \dfrac{10^{10}\;ft^3}{7.5(24)(60)(60)\;seconds} \\[5ex] = 15432.09877 \\[3ex] \approx 15432\;ft^3/second \\[3ex] \approx 15432\;cfs \\[3ex] $ (b.) 15432.09877 cfs is approximately ............ of 25000 cfs?

$ \dfrac{is}{of} = \dfrac{\%}{100}...Percent-Proportion \\[5ex] \dfrac{15432.09877}{25000} = \dfrac{what\%}{100} \\[5ex] what\% = \dfrac{15432.09877(100)}{25000} \\[5ex] what\% = 61.72839506\% \approx 62\% \\[3ex] $ (c.) Given:
Area in square miles per day
Volume in cubic feet per second (cfs)
To find:
height in feet
First: We need to convert the area in square miles per day to square feet per second so we can have the same unit of feet per second
Second: We find the height by multiplying it by the area and equating it to the volume

$ Convert\;\; 7\;\;square\;\;miles\;\;per\;\;day\;\;to\;\;square\;\;feet\;\;per\;\;second \\[3ex] \dfrac{7\;mile * mile}{1\;day} * \dfrac{.......feet}{.......mile} * \dfrac{.......feet}{.......mile} * \dfrac{.......day}{.......hour} * \dfrac{.......hour}{.......minute} * \dfrac{.......minute}{.......second} \\[5ex] \dfrac{7\;mile * mile}{1\;day} * \dfrac{5280\;feet}{1\;mile} * \dfrac{5280\;feet}{1\;mile} * \dfrac{1\;day}{24\;hours} * \dfrac{1\;hour}{60\;minutes} * \dfrac{1\;minute}{60\;seconds} \\[5ex] = 2258.666667\;feet^2/second \\[3ex] Area * height = Volume \\[3ex] 2258.666667\;\dfrac{feet^2}{second} * height\;feet = 15432.09877\;\dfrac{feet^3}{second} \\[5ex] height = \dfrac{15432.09877}{2258.666667}\;feet \\[5ex] height = 6.832393198\;feet \\[3ex] height \approx 6.8\;feet $

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