# 1.13: Rates as "Hidden" Conversion Factors

While most conversion factors are derived from measured equalities and prefix modifier equalities, rates found within problems can also be used to change one unit to another.  These rates are sometimes called "hidden" conversion factors, as they are not explicitly stated to be conversion factors within the context of a word problem.  Instead, students must recognize key words, such as "per," in order to identify a rate-based conversion factor.  Therefore, the word "per" is often called an indicator word, which is a word or a phrase that has a deeper meaning or application.  In this case, the word "per" means "use this information as a conversion factor."  Finally, while the word "per" is generally written out explicitly, as in the phrase "words per minute," it can also be even further hidden in an abbreviation, such as "mph" (which means "miles per hour") or "ppm" (which stands for "parts per million").

Note that "hidden" conversion factors are not exact quantities.  While there are always 12 inches in 1 foot, the speed limit is not always 45 miles per hour.  Therefore, the number of significant figures that are present in a "hidden" conversion factor must be considered when determining the number of significant figures that should be applied to a calculated answer.

Consider the following word problem:  How many hours are required for a train to travel 275 miles, if it is traveling at a constant speed of 55 mph?

Much like in the previous sections, the six-step dimensional analysis process must be applied to correctly solve this problem.  Step 1, in particular, will need to be considered more thoughtfully, in order to arrive at the correct answer.

1. Determine the "given" unit in the problem.  This problem contains two numbers, "275" and "55".  Determining which of these is considered the "given" number is the most important aspect of a word problem.

The unit associated with the number "55" is "mph," which should be understood to mean "miles per hour."  The "per" in this unit indicates that this quantity actually represents a "hidden" conversion factor and should be expressed as such.  In particular, the word "per" implies a ratio, which is linked to the mathematical operation of division.  Therefore, the word "per" is written as the fraction bar in a conversion factor.  Any number or unit read before the word "per" should be written in the numerator of the conversion, and any number or unit found after the word "per" should be written in the denominator.  The phrase "55 miles per hour" should then be written as

$$\dfrac{ \text{55 mi}}{\text{h}}$$

Remember that every conversion factor has a second form, in which the quantities in the numerator and the denominator are interchanged.  Therefore, the phrase "55 miles per hour" could also be represented as

$$\dfrac{ \text{h}}{\text{55 mi}}$$

Because the value "55 mph" should be expressed as a conversion factor, it should not be selected as the "given" quantity in a problem.  Therefore, "275 miles" must be the given quantity, and "miles" should be selected as the "given" unit.

2. Determine the "desired" unit in the problem.  The process of selecting the "desired" unit also changes for a word problem.  Instead of looking for the word "to" or "into," identify a phrase like "how many" or "how much" to determine the desired unit.  In this case, the problem is asking about "[h]ow many hours" are required for the train to travel a particular distance.  Therefore, "hours" is the desired unit.

3. Determine which equality or equalities relate the given and desired units.  Section 1.6 does not reference an equality that directly relates miles, a unit of distance, to hours, which is a unit for time.  In fact, this section does not relate any distance-based unit to any time-based unit!  However, the "hidden" conversion factor written above (in Step 1), does provide this relationship, and so is applied as the conversion factor in this problem.

4. Use the appropriate conversion factors to achieve unit cancelation.  Using the methodology developed in the previous sections, the second form of this conversion factor, in which the unit "mi" is written in the denominator, must be used.

$${275 \; \cancel{\rm{mi}}} \times \dfrac{\rm{h}}{55 \; \cancel{\rm{mi}}}$$

5. Perform the calculation that remains once the units have been canceled.  In this case,

$${275} \times {\text {((1) h ÷ 55)}} = {\text {5 h}}$$

6. Apply the correct number of significant figures to the calculated quantity.  Since the conversion used is based on a "hidden" conversion factor, the "55," which has 2 significant figures, must be considered when determining the number of significant figures that the answer should have.  Note that the "1" in the "hidden" conversion factor is not explicitly-provided, and so is not considered.  The given number, "275," has 3 significant figures.

While the given number normally dictates the number of significant figures that should be applied to a calculated answer, the "hidden" conversion factor is the quantity that contains the lesser count of significant figures for this particular problem.  Therefore, the final answer must have 2 significant figures.  Since the number of digits that the answer actually has is less than the number of digits that it should have, significant zeroes must be added to the calculated answer.  Therefore, the final answer should be reported as 5.0 h.
Example $$\PageIndex{1}$$

A student types at an average rate of 45 words per minute.  How many words can the student type in 1.10 hours?

Solution

The "per" in the phrase "45 words per minute" indicates that this quantity represents a "hidden" conversion factor and should be expressed as such.  Therefore, the only remaining quantity, "1.10 hours," must be the given quantity, and "hours" should be selected as the "given" unit.  Since the problem is asking about "[h]ow many words" can be typed in a particular time, "words" is the desired unit.  The "hidden" conversion factor relates "words" (the desired unit) to "minutes."  While "minutes" is a time-based unit, it is not the given unit.  Therefore, the measured equality $${ \text{1 h}} = { \text{60 min}}$$ must also be used to solve this problem, as shown below.

$${1.10 \; \cancel{\rm{h}}} \times \dfrac{60 \; \bcancel{ \rm{min}}}{1 \; \cancel{\rm{h}}}\ \times \dfrac{45 \; \rm{words}}{\bcancel{\rm{min}}} = {\text {2,970 words}}$$

The given number, "1.10," has 3 significant figures.  Since the first conversion used is based on a prefix modifier equality, neither the "1" nor the "60" are considered when determining the number of significant figures that the answer should have.  However, the second conversion used is based on a "hidden" conversion factor, so its quantity ("45") must be considered.

Since the answer must be limited to the lesser count of significant figures, the "hidden" conversion factor is the limiting quantity, and the final answer must have 2 significant figures.  Since the number of digits that the answer actually has is more than the number of digits that it should have, the calculated answer must be rounded to 3.0 x 103 words.  Note that this quantity must be expressed using proper scientific notation, as its two possible decimal forms, "3,000 words" and "3,000. words", would have too few (1) and too many (4) significant figures, respectively.

Exercise $$\PageIndex{1}$$

Due to the process by which it is harvested, saffron (a spice) costs $114.35 per ounce, making it more expensive than gold! How many kilograms of saffron can a chef purchase for$2,500?

The "per" in the phrase "$114.35 per ounce" indicates that this quantity represents a "hidden" conversion factor and should be expressed as such. Therefore, the only remaining quantity, "$2,500," must be the given quantity, and "$(dollars)" should be selected as the "given" unit. Since the problem is asking about "[h]ow many kilograms" can be purchased for a particular amount of money, "kilograms" is the desired unit. The "hidden" conversion factor relates "dollars" (the given unit) to "ounces." While "ounces" is a mass-based unit, it is not the desired unit. Therefore, the measured equalities $${ \text{1 lb}} = { \text{16 oz}}$$ and $${ \text{1 kg}} = { \text{2.2 lb}}$$ must also be used to solve this problem, as shown below. $${2,500 \; \cancel{\rm{dollars}}} \times \dfrac{\bcancel{ \rm{oz}}}{114.35 \; \cancel{\rm{dollars}}}\ \times \dfrac{1 \; \cancel{ \rm{lb}}}{16 \; \bcancel{\rm{oz}}}\ \times \dfrac{1 \; \rm{kg}}{2.2 \; \cancel{\rm{lb}}} = {\text {0.621099495... kg}}$$ The given number, "$2,500," has 2 significant figures.  Since the second and third conversions used are based on prefix modifier equalities, none of their quantities are considered when determining the number of significant figures that the answer should have.  However, the first conversion used is based on a "hidden" conversion factor, so its quantity ("114.35") must be considered.