9.3: Simplification of Denominate Numbers

Learning Objectives

• 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

Definition: Denominate Numbers

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.

Definition: Simplified Denominate Numbers

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.

Sample Set A

Simplify 19 in.

Solution

Since $\text{12 in. = 1 ft.}$, and $19 = 12 + 7$.

$\begin{array} {rcl} {\text{19 in.}} & = & {\text{12 in. + 7 in.}} \\ {} & = & {\text{1 ft + 7 in.}} \\ {} & = & {\text{1 ft 7 in.}} \end{array}$

Sample Set A

Simplify 4 gal 5 qt.

Solution

Since $\text{4 qt = 1 gal}$, and $5 = 4 + 1$.

$\begin{array} {rcl} {\text{4 gal 5 qt}} & = & {\text{4 gal + 4 qt + 1 qt}} \\ {} & = & {\text{4 gal + 1 gal + 1 qt}} \\ {} & = & {\text{5 gal + 1 qt}} \\ {} & = & {\text{5 gal 1 qt}}\end{array}$

Sample Set A

Simplify 2 hr 75 min.

Solution

Since $\text{60 min = 1 hr}$, and $75 = 60 + 15$.

$\begin{array} {rcl} {\text{2 hr 75 min}} & = & {\text{2 hr + 60 min + 15 min}} \\ {} & = & {\text{2 hr + 1 hr + 15 min}} \\ {} & = & {\text{3 hr + 15 min}} \\ {} & = & {\text{3 hr 15 min}}\end{array}$

Sample Set A

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:

1. writing the numbers vertically so that the like units appear in the same column.
2. adding or subtracting the number parts, carrying along the unit.
3. simplifying the sum or difference.

Sample Set B

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 $\text{12 in. = 1 ft.}$

$\begin{array} {rcl} {\text{8 ft + 12 in. + 5 in.}} & = & {\text{8 ft + 1 ft + 5 in.}} \\ {} & = & {\text{9 ft + 5 in.}} \\ {} & = & {\text{9 ft 5 in.}} \end{array}$

Sample Set B

Subtract 5 da 3 hr from 8 da 11 hr.

Solution

$\begin{array} {r} {\text{8 da 11 hr}} \\ {\underline{\text{- 5 da 3 hr}}} \\ {\text{3 da 8 hr}} \end{array}$

Sample Set B

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.

$\begin{array} {rcl} {\text{5 lb 3 oz}} & = & {\text{5 lb + 3 oz}} \\ {} & = & {\text{4 lb + 1 lb + 3 oz}} \\ {} & = & {\text{4 lb + 16 oz + 3 oz (Since 1 lb = 16 oz.)}} \\ {} & = & {\text{4 lb + 19 oz}} \\ {} & = & {\text{4 lb 19 oz}} \end{array}$

$\begin{array} {r} {\text{4 lb 19 oz}} \\ {\underline{\text{- 3 lb 14 oz}}} \\ {\text{1 lb 5 oz}} \end{array}$

Sample Set B

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.

Sample Set C

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 $\text{12 in. = 1 ft.}$

$\begin{array} {rcl} {\text{12 ft + 24 in.}} & = & {\text{4 yd + 2 ft}} \\ {} & = & {\text{4 yd 2 ft}} \end{array}$

Sample Set C

$\begin{array} {rcl} {8 \cdot \text{(5 hr 21 min 55 sec)}} & = & {8 \cdot 5 \text{ hr} + 8 \cdot 21 \text{ min} + 8 \cdot 55 \text{ sec}} \\ {} & = & {\text{40 hr + 168 min + 440 sec}} \\ {} & = & {\text{40 hr + 168 min + 7 min + 20 sec}} \\ {} & = & {\text{40 hr + 175 min + 20 sec}} \\ {} & = & {\text{40 hr + 2 hr + 55 min + 20 sec}} \\ {} & = & {\text{42 hr + 55 min + 20 sec}} \\ {} & = & {\text{24 hr + 18 hr + 55 min + 20 sec}} \\ {} & = & {\text{1 da + 18 hr + 55 min + 20 sec}} \\ {} & = & {\text{1 da 18 hr 55 min 20 sec}} \end{array}$

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.

Sample Set D

Perform the following divisions. Simplify if necessary.

$\text{(12 min 40 sec)} \div 4$

Solution

Thus $\text{(12 min 40 sec)} \div 4 = \text{3 min 10 sec}$

Sample Set D

$\text{(5 yd 2 ft 9 in.)} \div 3$

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