# 9.3 Simplification of denominate numbers

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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to simplify denominate numbers. By the end of the module students should be able to convert an unsimplified unit of measure to a simplified unit of measure, be able to add and subtract denominate numbers and be able to multiply and divide a denominate number by a whole number.

## Section overview

• Converting to Multiple Units
• Adding and Subtracting Denominate Numbers
• Multiplying a Denominate Number by a Whole Number
• Dividing a Denominate Number by a Whole Number

## 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.

## Simplified 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.

## Sample set a

Simplify 19 in.

Since , and $\text{19}=\text{12}+7$ ,

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

Simplify 4 gal 5 qt.

Since , and $5=4+1$ ,

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

Simplify 2 hr 75 min.

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

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

Simplify 43 fl oz.

Since $\text{8 fl oz}=\text{1 c}$ (1 cup), and $\text{43}÷8=5\text{R3}$ ,

$\begin{array}{ccc}\hfill \text{43 fl oz}& =& \text{40 fl oz}+\text{3 fl oz}\hfill \\ & =& 5\cdot \text{8 fl oz}+\text{3 fl oz}\hfill \\ & =& 5\cdot \text{1 c}+\text{3 fl oz}\hfill \\ & =& \text{5 c}+\text{3 fl oz}\hfill \end{array}$

But, $\text{2 c}=\text{1 pt}$ and $5÷2=2\text{R1}$ . So,

$\begin{array}{ccc}\hfill \text{5 c}+\text{3 fl oz}& =& \text{2}\cdot \text{2 c}+\text{1 c}+\text{3 fl oz}\hfill \\ & =& 2\cdot \text{1 pt}+\text{1 c}+\text{3 fl oz}\hfill \\ & =& \text{2 pt}+\text{1 c}+\text{3 fl oz}\hfill \end{array}$

But, $\text{2 pt}=\text{1 qt}$ , so

$\text{2 pt}+\text{1 c}+\text{3 fl oz}=\text{1 qt 1 c 3 fl oz}$

## Practice set a

Simplify each denominate number. Refer to the conversion tables given in [link] , if necessary.

18 in.

1 ft 6 in.

8 gal 9 qt

10 gal 1 qt

5 hr 80 min

6 hr 20 min

8 wk 11 da

9 wk 4 da

86 da

12 wk 2 da

## 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.

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

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

Subtract 5 da 3 hr from 8 da 11 hr.

Subtract 3 lb 14 oz from 5 lb 3 oz.

We cannot directly subtract 14 oz from 3 oz, so we must borrow 16 oz from the pounds.

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

Subtract 4 da 9 hr 21 min from 7 da 10 min.

$\begin{array}{c}\hfill \text{7 da 0 hr 10 min}\\ \hfill \underline{-\text{4 da 9 hr 21 min}}\end{array}$ Borrow 1 da from the 7 da.

Borrow 1 hr from the 24 hr.

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