9.3: Simplification of Denominate Numbers
- be able to convert an unsimplified unit of measure to a simplified unit of measure
- be able to add and subtract denominate numbers
- be able to multiply and divide a denominate number by a whole number
Converting to Multiple Units
Numbers that have units of measure associated with them are called denominate numbers . It is often convenient, or even necessary, to simplify a denominate number.
A denominate number is simplified when the number of standard units of measure associated With it does not exceed the next higher type of unit.
The denominate number 55 min is simplified since it is smaller than the next higher type of unit, 1 hr. The denominate number 65 min is not simplified since it is not smaller than the next higher type of unit, 1 hr. The denominate number 65 min can be simplified to 1 hr 5 min. The denominate number 1 hr 5 min is simplified since the next higher type of unit is day, and 1 hr does not exceed 1 day.
Simplify 19 in.
Solution
Since \(\text{12 in. = 1 ft.}\), and \(19 = 12 + 7\).
\(\begin{array} {rcl} {\text{19 in.}} & = & {\text{12 in. + 7 in.}} \\ {} & = & {\text{1 ft + 7 in.}} \\ {} & = & {\text{1 ft 7 in.}} \end{array}\)
Simplify 4 gal 5 qt.
Solution
Since \(\text{4 qt = 1 gal}\), and \(5 = 4 + 1\).
\(\begin{array} {rcl} {\text{4 gal 5 qt}} & = & {\text{4 gal + 4 qt + 1 qt}} \\ {} & = & {\text{4 gal + 1 gal + 1 qt}} \\ {} & = & {\text{5 gal + 1 qt}} \\ {} & = & {\text{5 gal 1 qt}}\end{array}\)
Simplify 2 hr 75 min.
Solution
Since \(\text{60 min = 1 hr}\), and \(75 = 60 + 15\).
\(\begin{array} {rcl} {\text{2 hr 75 min}} & = & {\text{2 hr + 60 min + 15 min}} \\ {} & = & {\text{2 hr + 1 hr + 15 min}} \\ {} & = & {\text{3 hr + 15 min}} \\ {} & = & {\text{3 hr 15 min}}\end{array}\)
Simplify 43 fl oz.
Solution
Since \(\text{8 fl oz = 1 c}\) (1 cup), and \(43 \div 8 = \text{5R3}\).
\(\begin{array} {rcl} {\text{43 fl oz}} & = & {\text{40 fl oz + 3 fl oz}} \\ {} & = & {5 \cdot 8 \text{ fl oz + 3 fl oz}} \\ {} & = & {5 \cdot 1 \text{ c + 3 fl oz}} \\ {} & = & {\text{5 c + 3 fl oz}}\end{array}\)
But, \(\text{2c = 1 pt}\) and \(5 \div 2 = \text{2R1}\). So,
\(\begin{array} {rcl} {\text{5 c + 3 fl oz}} & = & {2 \cdot 2 \text{ c + 1 c + 3 fl oz}} \\ {} & = & {2 \cdot 1 \text{ pt + 1 c + 3 fl oz}} \\ {} & = & {\text{2 pt + 1 c + 3 fl oz}} \end{array}\)
But, \(\text{2 pt = 1 qt}\), so
\(\text{2 pt + 1 c + 3 fl oz = 1 qt 1 c 3 fl oz}\)
Practice Set A
Simplify each denominate number. Refer to the conversion tables given in [link] , if necessary.
18 in.
- Answer
-
1 ft 6 in.
Practice Set A
8 gal 9 qt
- Answer
-
10 gal 1 qt
Practice Set A
5 hr 80 min
- Answer
-
6 hr 20 min
Practice Set A
8 wk 11 da
- Answer
-
9 wk 4 da
Practice Set A
86 da
- Answer
-
12 wk 2 da
Adding and Subtracting Denominate Numbers
Adding and Subtracting Denominate Numbers
Denominate numbers can be added or subtracted by:
- writing the numbers vertically so that the like units appear in the same column.
- adding or subtracting the number parts, carrying along the unit.
- simplifying the sum or difference.
Add 6 ft 8 in. to 2 ft 9 in.
Solution
\(\begin{array} {r} {\text{6 ft 8 in.}} \\ {\underline{\text{+ 2 ft 9 in.}}} \\ {\text{8 ft 17 in.}} \end{array}\) Simplify this denominate number.
Since \(\text{12 in. = 1 ft.}\)
\(\begin{array} {rcl} {\text{8 ft + 12 in. + 5 in.}} & = & {\text{8 ft + 1 ft + 5 in.}} \\ {} & = & {\text{9 ft + 5 in.}} \\ {} & = & {\text{9 ft 5 in.}} \end{array}\)
Subtract 5 da 3 hr from 8 da 11 hr.
Solution
\(\begin{array} {r} {\text{8 da 11 hr}} \\ {\underline{\text{- 5 da 3 hr}}} \\ {\text{3 da 8 hr}} \end{array}\)
Subtract 3 lb 14 oz from 5 lb 3 oz.
Solution
\(\begin{array} {r} {\text{5 lb 3 oz}} \\ {\underline{\text{- 3 lb 14 oz}}} \end{array}\)
We cannot directly subtract 14 oz from 3 oz, so we must borrow 16 oz from the pounds.
\(\begin{array} {rcl} {\text{5 lb 3 oz}} & = & {\text{5 lb + 3 oz}} \\ {} & = & {\text{4 lb + 1 lb + 3 oz}} \\ {} & = & {\text{4 lb + 16 oz + 3 oz (Since 1 lb = 16 oz.)}} \\ {} & = & {\text{4 lb + 19 oz}} \\ {} & = & {\text{4 lb 19 oz}} \end{array}\)
\(\begin{array} {r} {\text{4 lb 19 oz}} \\ {\underline{\text{- 3 lb 14 oz}}} \\ {\text{1 lb 5 oz}} \end{array}\)
Subtract 4 da 9 hr 21 min from 7 da 10 min.
Solution
\(\begin{array} {r} {\text{7 da 0 hr 10 min}} \\ {\underline{\text{- 4 da 9 hr 21 min}}} \end{array}\) Borrow 1 da from the 7 da.
\(\begin{array} {r} {\text{6 da 24 hr 10 min}} \\ {\underline{\text{- 4 da 9 hr 21 min}}} \end{array}\) Borrow 1 hr from the 24 hr.
\(\begin{array} {r} {\text{6 da 23 hr 70 min}} \\ {\underline{\text{- 4 da 9 hr 21 min}}} \\ {\text{2 da 14 hr 49 min}} \end{array}\)
Practice Set B
Perform each operation. Simplify when possible.
Add 4 gal 3 qt to 1 gal 2 qt.
- Answer
-
6 gal 1 qt
Practice Set B
Add 9 hr 48 min to 4 hr 26 min.
- Answer
-
14 hr 14 min
Practice Set B
Subtract 2 ft 5 in. from 8 ft 7 in.
- Answer
-
6 ft 2in.
Practice Set B
Subtract 15 km 460 m from 27 km 800 m.
- Answer
-
12 km 340 m
Practice Set B
Subtract 8 min 35 sec from 12 min 10 sec.
- Answer
-
3 min 35 sec
Practice Set B
Add 4 yd 2 ft 7 in. to 9 yd 2 ft 8 in.
- Answer
-
14 yd 2 ft 3 in
Practice Set B
Subtract 11 min 55 sec from 25 min 8 sec.
- Answer
-
13 min 13 sec
Multiplying a Denominate Number by a Whole Number
Let's examine the repeated sum
\(\underbrace{\text{4 ft 9 in. + 4 ft 9 in. + 4 ft 9 in.}}_{\text{3 times}} = \text{12 ft 27 in.}\)
Recalling that multiplication is a description of repeated addition, by the distributive property we have
\(\begin{array} {rcl} {\text{3(4 ft 9 in.)}} & = & {\text{3 (4ft + 9 in.)}} \\ {} & = & {3 \cdot 4 \text{ ft } + 3 \cdot 9 \text{ in.}} \\ {} & = & {\text{12 ft + 27 in. Now, 27 in. = 2 ft 3 in.}} \\ {} & = & {\text{12 ft + 2 ft + 3 in.}} \\ {} & = & {\text{14 ft + 3 in.}} \\ {} & = & {\text{14 ft 3 in.}} \end{array}\)
From these observations, we can suggest the following rule.
Multiplying a Denominate Number by a Whole Number
To multiply a denominate number by a whole number, multiply the number part of each unit by the whole number and affix the unit to this product.
Perform the following multiplications. Simplify if necessary.
\(\begin{array} {rcl} {6 \cdot \text{(2 ft 4 in.)}} & = & {6 \cdot 2 \text{ ft + 6} \cdot 4 \text{in.}} \\ {} & = & {\text{12 ft + 24 in.}} \end{array}\)
Since \(\text{3 ft = 1 yd}\) and \(\text{12 in. = 1 ft.}\)
\(\begin{array} {rcl} {\text{12 ft + 24 in.}} & = & {\text{4 yd + 2 ft}} \\ {} & = & {\text{4 yd 2 ft}} \end{array}\)
\(\begin{array} {rcl} {8 \cdot \text{(5 hr 21 min 55 sec)}} & = & {8 \cdot 5 \text{ hr} + 8 \cdot 21 \text{ min} + 8 \cdot 55 \text{ sec}} \\ {} & = & {\text{40 hr + 168 min + 440 sec}} \\ {} & = & {\text{40 hr + 168 min + 7 min + 20 sec}} \\ {} & = & {\text{40 hr + 175 min + 20 sec}} \\ {} & = & {\text{40 hr + 2 hr + 55 min + 20 sec}} \\ {} & = & {\text{42 hr + 55 min + 20 sec}} \\ {} & = & {\text{24 hr + 18 hr + 55 min + 20 sec}} \\ {} & = & {\text{1 da + 18 hr + 55 min + 20 sec}} \\ {} & = & {\text{1 da 18 hr 55 min 20 sec}} \end{array}\)
Practice Set C
Perform the following multiplications. Simplify.
\(2 \cdot \text{(10 min)}\)
- Answer
-
20 min
Practice Set C
\(5 \cdot \text{(3 qt)}\)
- Answer
-
\(\text{15 qt = 3 gal 3 qt}\)
Practice Set C
\(4 \cdot \text{(5 ft 8 in.)}\)
- Answer
-
\(\text{20 ft 32 in. = 7 yd 1 ft 8 in.}\)
Practice Set C
\(10 \cdot \text{(2 hr 15 min 40 sec)}\)
- Answer
-
\(\text{20 hr 150 min 400 sec = 22 hr 36 min 40 sec}\)
Dividing a Denominate Number by a Whole Number
Dividing a Denominate Number by a Whole Number
To divide a denominate number by a whole number, divide the number part of each unit by the whole number beginning with the largest unit. Affix the unit to this quotient. Carry any remainder to the next unit.
Perform the following divisions. Simplify if necessary.
\(\text{(12 min 40 sec)} \div 4\)
Solution
Thus \(\text{(12 min 40 sec)} \div 4 = \text{3 min 10 sec}\)
\(\text{(5 yd 2 ft 9 in.)} \div 3\)
Solution
\(\begin{array} {c} {\text{Convert to feet: 2 yd 2 ft = 8 ft}} \\ {\text{Convert to inches: 2 ft 9 in. = 33 in.}} \end{array}\)
Thus \(\text{(5 yd 2 ft 9 in.)} \div 3 = \text{1 yd 2 ft 11 in.}\)
Practice Set D
Perform the following divisions. Simplify if necessary.
\(\text{(18 hr 36 min)} \div 9\)
- Answer
-
2 hr 4 min
Practice Set D
\(\text{(36 hr 8 min)} \div 8\)
- Answer
-
4 hr 18 min
Practice Set D
\(\text{(13 yd 7 in.)} \div 5\)
- Answer
-
2 yd 1 ft 11 in
Practice Set D
\(\text{(47 gal 2 qt 1 pt)} \div 3\)
- Answer
-
15 gal 3 qt 1 pt
Exercises
For the following 15 problems, simplify the denominate numbers.
Exercise \(\PageIndex{1}\)
16 in.
- Answer
-
1 foot 4 inches
Exercise \(\PageIndex{2}\)
19 ft
Exercise \(\PageIndex{3}\)
85 min
- Answer
-
1 hour 25 minutes
Exercise \(\PageIndex{4}\)
90 min
Exercise \(\PageIndex{5}\)
17 da
- Answer
-
2 weeks 3 days
Exercise \(\PageIndex{6}\)
25 oz
Exercise \(\PageIndex{7}\)
240 oz
- Answer
-
15 pounds
Exercise \(\PageIndex{8}\)
3,500 lb
Exercise \(\PageIndex{9}\)
26 qt
- Answer
-
6 gallons 2 quarts
Exercise \(\PageIndex{10}\)
300 sec
Exercise \(\PageIndex{11}\)
135 oz
- Answer
-
8 pounds 7 ounces
Exercise \(\PageIndex{12}\)
14 tsp
Exercise \(\PageIndex{13}\)
18 pt
- Answer
-
2 gallons 1 quart
Exercise \(\PageIndex{14}\)
3,500 m
Exercise \(\PageIndex{15}\)
16,300 mL
- Answer
-
16 liters 300 milliliters (or 1daL 6 L 3dL)
For the following 15 problems, perform the indicated operations and simplify the answers if possible.
Exercise \(\PageIndex{16}\)
Add 6 min 12 sec to 5 min 15 sec.
Exercise \(\PageIndex{17}\)
Add 14 da 6 hr to 1 da 5 hr.
- Answer
-
15 days 11 hours
Exercise \(\PageIndex{18}\)
Add 9 gal 3 qt to 2 gal 3 qt.
Exercise \(\PageIndex{19}\)
Add 16 lb 10 oz to 42 lb 15 oz.
- Answer
-
59 pounds 9 ounces
Exercise \(\PageIndex{20}\)
Subtract 3 gal 1 qt from 8 gal 3 qt.
Exercise \(\PageIndex{21}\)
Subtract 3 ft 10 in. from 5 ft 8 in.
- Answer
-
1 foot 10 inches
Exercise \(\PageIndex{22}\)
Subtract 5 lb 9 oz from 12 lb 5 oz.
Exercise \(\PageIndex{23}\)
Subtract 10 hr 10 min from 11 hr 28 min.
- Answer
-
1 hour 18 minutes
Exercise \(\PageIndex{24}\)
Add 3 fl oz 1 tbsp 2 tsp to 5 fl oz 1 tbsp 2 tsp.
Exercise \(\PageIndex{25}\)
Add 4 da 7 hr 12 min to 1 da 8 hr 53 min.
- Answer
-
5 days 16 hours 5 minutes
Exercise \(\PageIndex{26}\)
Subtract 5 hr 21 sec from 11 hr 2 min 14 sec.
Exercise \(\PageIndex{27}\)
Subtract 6 T 1,300 lb 10 oz from 8 T 400 lb 10 oz.
- Answer
-
1 ton 1,100 pounds (or 1T 1,100 lb)
Exercise \(\PageIndex{28}\)
Subtract 15 mi 10 in. from 27 mi 800 ft 7 in.
Exercise \(\PageIndex{29}\)
Subtract 3 wk 5 da 50 min 12 sec from 5 wk 6 da 20 min 5 sec.
- Answer
-
2 weeks 23 hours 29 minutes 53 seconds
Exercise \(\PageIndex{30}\)
Subtract 3 gal 3 qt 1 pt 1 oz from 10 gal 2 qt 2 oz.
Exercises for Review
Exercise \(\PageIndex{31}\)
Find the value: \((\dfrac{5}{8})^2 + \dfrac{39}{64}\).
- Answer
-
1
Exercise \(\PageIndex{32}\)
Find the sum: \(8 + 6 \dfrac{3}{5}\).
Exercise \(\PageIndex{33}\)
Convert \(2.05 \dfrac{1}{11}\) to a fraction.
- Answer
-
\(2 \dfrac{14}{275}\)
Exercise \(\PageIndex{34}\)
An acid solution is composed of 3 parts acid to 7 parts water. How many parts of acid are there in a solution that contains 126 parts water?
Exercise \(\PageIndex{35}\)
Convert 126 kg to grams.
- Answer
-
126,000 g