Unit Analysis
All of us have been guilty of ignoring the units of a label associated with a number. Unit analysis is a form of organizing information while being conscious of the label. Chemists and physicists often use this strategy. It is particularly used when converting one type of measure to another, such as feet to inches or meters to centimeters. The results of many problems are found by making the units label correct. Sometimes you may not be sure whether to multiply or divide numbers to get the desired results. Place the labels first. The positions for the numbers are then determined by their associated labels. The position, whether numerator or denominator, in turn also determines the math process involved. If both numbers are in the numerator, then multiply; or, if one is in the denominator, then divide by it. Thus the problem tells you what needs to be done. You only need to keep track of the labels by type and whether their position is in the numerator or denominator of a fraction. Of course, it is also advantageous to know some basic conversion facts.
Conversion factors are any quantities which are equivalent. They can be expressed with an equal sign or as a fraction; such as 1 foot = 12 inches or 1 foot/12 inches or 12 inches/1 foot. Notice the fraction can have either value as its numerator. This makes the conversion factor useful for either converting inches to feet or feet to inches. Write some common conversion factors under each of the following categories:
DistanceMass
CapacityTime
Some information is given using the word “per” in a phrase. These can also be written in fraction form but do not imply that the values are equivalent. For example, the expression “miles per hour” means the number of miles traveled per “one” hour. The “one” is understood. So 55 mph. could be written as fraction 55 miles/1 hour. This is understood as a rate of speed. This does not mean that 55 miles is equal to one hour. However, this can mean, in order to travel the distance of 55 miles, it will take 1 hour of time or it can mean, in 1 hour of time, you will have traveled 55 miles.
Example 1: Traveling Abroad
You are traveling around Europe by car. The cost of gasoline in Spain is 189 pesetas (Spanish currency) per liter.
a) How much will it cost you to fill your 15 gallons tank of your vehicle? (Use 1 gallon = 3.79 liters in your conversion.)
b) If the exchange rate is $1 U.S. = 168 pesetas, what is the equivalent cost of gasoline in dollars per gallon?
c) Gasoline in the United States is $1.54 per gallon. If the exchange rate changes to $1 U.S. = 173 pesetas, what is the equivalent cost of gasoline in pesetas per liter?
a) Known: 189 pesetas15 gallon tank
1 liter
You need to know how many pesetas it will cost. You know the volume of gasoline you wish to purchase and the price of gasoline in a cost per a volume. Start with the information you know. Place the given information so that it matches the format of the answer you want. You want cost so the money element must be in the numerator. You do not want the volume label. Since the price is per volume, the volume label is in the denominator. To eliminate the volume, it must cancel with a label in the numerator.
cost = cost x vol. = 189 pesetas x 15 gal. = pesetas-gal.
vol.11 liter1liters
desiredconversionreality
The problem is the volume labels are not the same and therefore do not cancel. Use the conversion factor of gallons to liters. Write it in fraction form with the label you want to purge in the opposite portion of the fraction. In this case, “gallons” was in the numerator and “liters” was in the denominator; therefore, place the conversion factor so that “liters” is in the numerator and “gallons” is in the denominator to eliminate both.
cost = 189 pesetas x 15 gal. x 3.79 liters = 10744.65 pesetas = 10745 pesetas
1 liter11 gal.
desiredgivenconversionanswer (in whole pesetas)
(labels cancel)
b) Known: 189 pesetas $1 U.S___
1 liter168 pesetas
You wish to know a cost per a volume. This matches your given information. Unfortunately, neither the cost label or the volume label is what you desire. Start again with the given information in fraction form aligned with the type of labels you want. Place the conversion factors of liters to gallons and dollars to pesetas so that the labels you do not want are the opposite positions from the given information.
cost = 189 pesetas x $1 x 3.79 liters = $716.31 = $4.26375 $4.26 per gal.
vol.1 liter168 pesetas1 gal.168 gal.1 gal.
desiredgivenconversionanswer (in dollars and cents)
(labels cancel)
c) Known: $1.54 $1 U.S.
1 gallon173 pesetas
This is a matter of reworking the previous process. This example starts with the price in dollars per gallon and the different conversion rate. However, instead of eliminating the pesetas and liter labels, we must eradicate the dollars and gallons. Therefore, the conversion factor is reversed; dollars are placed in the denominator and gallons are in the numerator.
cost = $1.54 x 173 pesetas x 1 gal. = 266.42 pesetas = 70.296 pesetas 70 pesetas
vol.1 gal.$13.79 liters3.79 litersliterliter
desiredgivenconversionanswer (in whole pesetas)
(labels cancel)
Example 2: Racing
Helio Castroneves was the Champion Auto Racing Teams winner of the Miller Lite 200 at Mid-Ohio Sports Car Course in 2000. Mid-Ohio is a 2.258 mile road course with 13 turns. Many turns means a lot of shifting gears and changing speeds. Castroneves averaged 119.215 mph.
a) How many laps are required for the Miller Lite 200 (the 200 represents the total race distance of 200 miles)?
b) How much time does it take Castroneves to go around the track one time at his average speed?
c) Time is lost or gained as well as position on the track when cars pit or stop for refueling or new tires. If Castroneves’ first pit-stop was11.7 seconds, what is the distance, in miles, behind where he would have been if he had not needed to pit? How much is this in laps?
d) The cars hold approximately 35 gallons of methanol fuel. If Castroneves first pitted on lap 31, assuming he uses 34.8 gallons before he makes a pit stop, how many miles per gallon does his car get?
e) A lightweight car is faster than a heavier car. Since adding fuel adds weight, what is the minimum amount of fuel Castroneves needed to add during his second pit stop on lap 67 to finish the race? (Use parts a and d to determine your answer.)
Known from the problem: 2.258 miles119.215 miles
1 lap1 hour
a) You are given a total distance and you need to find number of laps. Therefore, you need a relationship between distance and laps. Since your answer is in laps, “laps” needs to be the label in the numerator of the conversion factor. Begin with the given information and transform it to the desired label.
laps = dist x lap = 200 mi. x 1 lap = 200 laps = 88.573959 laps 88.6 laps
1dist.12.258 mi.2.258
desiredgivenconversionanswer
(labels cancel)
b) You want the time per lap. You know the distance per time and distance per lap. Because what is known contains what is wanted, you need to eliminate what you do not want. Fortunately, both contain distance; thus, by manipulating the arrangement of the problem, the distance will cancel out of the problem.
time =time x dist. = 1 hour x 2.258 mi. = 2.258 hours
lapdistlap119.215 mi.1 lap119.215 laps
desiredgivenconversion
(labels cancel)
Yet, time has been given in hours. Logically the time for one lap would not be expressed in hours so the hours need to be converted to minutes. Since 1 hour equals 60 minutes, write this fact as a fraction such that “hours” is placed in the denominator, the opposite position of the fraction than its current position, so that the labels will cancel. Check to be sure the remaining time label, “minutes”, is still in the numerator while “laps” remains in the denominator.
time = 1 hour x 2.258 mi. x 60 min. = 135.48 min. = 1.1364341 min. 1.136 min.
lap119.215 mi.1 lap1 hour119.215 lapslaplap
desiredgivenconversionanswer
(labels cancel)
c) You are given time but you need distance. Again you will need the relationship between the distance and time. The arrangement needs to use the relationship labels to eliminate the time label you have been given. It therefore requires that the time label be
placed in the denominator.
dist. = time x dist. = 11.7 sec. x 119.215 mi. = sec. - mi.
1time11 hourhour
desiredconversionreality
Although the labels are both time, they are not the same and therefore do not cancel. This requires additional conversion factors to eliminate time from the problem. You will need to use the factors that 60 seconds equals 1 minute and 60 minutes equals 1 hour to accomplish the removal of the time element.
dist. = 11.7 sec. x 119.215 mi. x 1 min. x 1 hour = 1394.8155 mi. 0.3874 mi.387 mi.
11 hour60 sec.60 min.3600
desiredgivenconversionanswer
(labels cancel)
d) The amount of fuel used and the laps traveled have been given. You want to know miles per gallon. The label “gallons” is given while the other needed for your solution must be converted. Again the relationship between laps and miles is required.
miles = 31 laps x 1 x 2.258 mi. = 69.998 mi. = 2.0114367 mi. 2.01 mi.
gal.134.8 gal.1 lap34.8 gal.gal.gal.
desiredgivenconversionanswer
(labels cancel)
e) Castroneves has pitted on lap 67. Your task is to find the amount of fuel for the remaining distance. First you need to find that distance.
88.6 laps - 67 laps = 21.6 laps remaining
Once the distance has been determined, the quantity of fuel can be calculated. The distance label must disappear while the volume label appears. Use the relationship you have just found between distance and volume in part d of this problem.
vol. = dist. x vol. = 21.6 laps x 1 gal. = lap-gal.
1dist.12.01 mi.miles
desiredconversionreality
Again the labels do not match and cancel. Use the relationship between the laps and the mileage. Since you want to eliminate both “laps” and “miles”, place “lap” in the denominator opposite its position in the given information. This forces the “miles” into the numerator opposite its position in the miles per gallon factor. Now they cancel and you have the label desired.
vol. = 21.6 laps x 2.258 mi. x 1 gal. = 48.7728 gal.= 24.265 gal. 24.3 gal.
11 lap2.01 mi.2.01
desiredgivenconversionanswer
(labels cancel)
Problems
1.Basic Conversions: Use the common conversion factors for Standard to Standard and for Metric to Metric.
a.13.6 days to hoursf.608 fluid ounces to quarts
b.253 meters to centimetersg.378 feet to yards
c.1783 ounces to poundsh.33792 feet to miles
d.45824 millimeters to metersi.279 hours to days
e.1848 inches to feetj.2.9 miles to feet
2.More Conversions: Use 1 meter = 3.281 feet, 1 ounce = 28.35 grams, and 1 gallon = 3.785 liters to convert between Standard and Metric quantities.
a.47 meters to feetf.130 miles to meters
b.18 liters to quartsg.93 centimeters to inches
c.3.5 years to hoursh.317 feet to centimeters
d.157 pounds to kilogramsi.326 kilometers to miles
e.63 inches to millimetersj.112 pints to liters
3.Traveling: The U.S. dollar is currently worth approximately 9.2 Swedish crowns.
a.If a Swedish traveler bought a $23.40 meal in the U.S., how much would he have spent in his native currency?
b.Suppose flying from Sweden one-way to the U.S. cost 4800 Swedish crowns. What would the return flight cost assuming the flights are equivalent?
c.Suppose a Swedish student came to the U.S. with a 100,000 Swedish crown student loan. If the exchange rate changed so that one U.S. dollar equaled 7.4 Swedish crowns, how much would the student gain or lose due to the change? Would the change have benefited the student or not?
d.If gasoline costs 10.476 Swedish crowns per liter, how much will it cost in dollars to fill your 12-gallon tank?
4.Use the facts that one kilobyte (1K) of computer memory is equal to 1024 bytes, one megabyte (1 MB) is equal to (1 x 1) K, and each letter the computer saves in memory requires 1 byte.
a.If 2875 bytes of memory are required to save a one-page resume and the average number of letters per word is 5.5, how many words are in this resume?
b.How many bytes of memory are required to save a 614-word report if the average word has eight letters?
c.Some 3 ½-inch floppy disks have 1.4 MB of memory. How many words averaging 9 letters could be saved on the disk?
d.Some Zip disks have 250 MB of memory. If you transferred the information from 3 ½-inch floppy disks with 1.4 MB of memory, how many 3 ½-inch disks will it take to fill one Zip disk?
e.A 600 MB CD-ROM has approximately how many times the memory of a 3 ½-inch floppy with 1.4 MB of memory?
f.Suppose 4.3 MB of memory is needed to store the words of an encyclopedia. How many such books can be saved on a CD-ROM with 600 MB of memory?
5.The people of a small township raised money for their community recreation area by donating pennies. One-foot-long cardboard strips were designed to hold a row of 16 pennies.
a.If the cardboard strips were placed end to end for a distance of one mile, how much money is in the strips?
b.If $1317.76 is obtained in one foot strips, what is the length of the strips if they are placed end to end?
c.How many dollars are 13850 full cardboard strips worth?
d.It is 3.8 miles along Main Street from one end of town to the other. If the town raised $3210.24 by the cardboard strips, would the strips reach the entire length of Main Street?
6.Jimmy Vasser was the Champion Auto Racing Teams winner of the Texaco-Havoline Grand Prix in Houston, Texas for the year 2000. Houston is a 1.52 mile temporary street course run through downtown Houston. It has 10 turns. Because it is a street course, the turns are true street turns unlike the sweeping turns of many road courses. The turns mean a lot of shifting gears and changing speeds. Vasser averaged 89.206 mph.
a) If the Texaco-Havoline Grand Prix is 100 laps, what is the total race distance ?
b) How much time does it take Vasser to go around the track one time at his average speed?
c) Time is lost or gained as well as position on the track when cars pit or stop for refueling or new tires. If Vasser’s pit-stop was12.4 seconds, what is the distance, in miles, behind where he would have been if he had not needed to pit? How much is this in laps?
d) The cars hold approximately 35 gallons of methanol fuel. If Vasser first pitted on lap 51, assuming he uses 34.8 gallons before he makes a pit stop, how many miles per gallon does his car get?
e) Driver Greg Moore holds the Houston course record of completing one lap in 59.508 seconds during the race in 1998. What is the course record in miles per hour?
7.Smoking has been estimated to shorten a person’s life by 10 to 15 minutes for each cigarette he or she has smoked. If a person smoked three packs a day for 25 years, assuming the worst that each cigarette shortened their life by 15 minutes, how much shorter would their life be? Give the answer in entire years, days, hours, minutes. For example: 41 years, 128 days, 14 hours, and 23 minutes. (Note: 20 cigarettes = 1 pack).
Unit Analysis1