Add and Subtract Mixed Numbers 1

Topics

Pre - Algebra

Fractions

Add and Subtract Mixed Numbers

Add and Subtract Mixed Numbers 1

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Question 1 of 4

1. Question

Add the following:

3 \frac{1}{2} + 5 \frac{1}{5}

$3 1/2+5 1/5$

1. $2 \frac{7}{10}$ $2 7/10$
2. $7 \frac{4}{5}$ $7 4/5$
3. $8 \frac{3}{10}$ $8 3/10$
4. $8 \frac{7}{10}$ $8 7/10$

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A mixed number consists of a whole number and a fraction.

First, add the whole numbers

3 + 5

$3+5$

=

$=$

8

$8$

Notice that the fractions have different denominators.

Find the

L C D

$LCD$ of

2

$2$ and

5

$5$ so we can add the fractions.

Multiples of

2

$2$ :

24681012

$2\;\;4\;\;6\;\;8\;\;\color{#004ec4}{10}\;\;12$

Multiples of

5

$5$ :

5101520

$5\;\;\color{#004ec4}{10}\;\;15\;\;20$

The

L C D

$LCD$ of

2

$2$ and

5

$5$ is

10

$10$

Proceed with adding the fractions

		$\frac{1}{2} + \frac{1}{5}$ $1/2+1/5$

	$=$ $=$	$\frac{(\color{#004ec4}{10}\div2)+(\color{#004ec4}{10}\div5)}{\color{#004ec4}{10}}$

	$=$ $=$	$\frac{5+2}{10}$

	$=$ $=$	$\frac{7}{10}$

Finally, combine the sum of the whole numbers and the sum of the fractions

3 \frac{1}{2} + 5 \frac{1}{5}

$3 1/2+5 1/5$

=

$=$

8 \frac{7}{10}

$8 7/10$

8 \frac{7}{10}

$8 7/10$

Question 2 of 4

2. Question

Add the following:

4 \frac{3}{5} + 5 \frac{1}{8}

$4 3/5+5 1/8$

1. $9 \frac{23}{40}$ $9 23/40$
2. $10 \frac{1}{4}$ $10 1/4$
3. $9 \frac{29}{40}$ $9 29/40$
4. $9 \frac{3}{4}$ $9 3/4$

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A mixed number consists of a whole number and a fraction.

First, add the whole numbers

4 + 5

$4+5$

=

$=$

9

$9$

Add the fractions using cross method.

Start by multiplying the two denominators. Use the product as a denominator for a new fraction.

\frac{3}{5} + \frac{1}{8}

$3/5+1/8$

=

$=$

$☐/(5times8)$

$=$

$☐/40$

To get the numerator, cross multiply the given addition problem and add the products.

$\frac{\color{#00880A}{3}}{\color{#9a00c7}{5}}+\frac{\color{#9a00c7}{1}}{\color{#00880A}{8}}$	$=$	$\frac{(\color{#00880A}{3\times8})+(\color{#9a00c7}{5\times1})}{40}$

	$=$	$(24+5)/40$

	$=$	$29/40$

Finally, combine the sum of the whole numbers and the sum of the fractions

$4 3/5+5 1/8$	$=$	$9+29/40$

	$=$	$9 29/40$

$9 29/40$

Question 3 of 4

3. Question

Subtract the following:

$3 1/10-1 1/4$

1. $1 11/20$
2. $2 17/20$
3. $2 9/20$
4. $1 17/20$

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Transforming an Improper to Mixed Fraction

$\frac{\color{#007DDC}{b}}{\color{#9a00c7}{c}}=\color{#00880A}{Q}\frac{\color{#e65021}{R}}{\color{#9a00c7}{c}}$

$($ $b$

$-:$ $c$

$)=$ $Q$ and $R$ is the remainder

Transforming a Fraction from Mixed to Improper

$=$

$\frac{(\color{#9a00c7}{c}\times\color{#00880A}{A})+\color{#007DDC}{b}}{\color{#9a00c7}{c}}$

First, transform the mixed fractions to improper fractions

		$\color{#00880A}{3}\frac{\color{#007DDC}{1}}{\color{#9a00c7}{10}}-\color{#00880A}{1} \frac{\color{#007DDC}{1}}{\color{#9a00c7}{4}}$

	$=$	$\frac{(\color{#9a00c7}{10}\times\color{#00880A}{3})+\color{#007DDC}{1}}{\color{#9a00c7}{10}}-\frac{(\color{#9a00c7}{4}\times\color{#00880A}{1})+\color{#007DDC}{1}}{\color{#9a00c7}{4}}$

	$=$	$\frac{30+1}{10}-\frac{4+1}{4}$

	$=$	$\frac{31}{10}-\frac{5}{4}$

Notice that the fractions have different denominators.

Find the

$LCD$ of

$10$ and

$4$ so we can subtract the fractions.

Multiples of

$10$ :

$10\;\;\color{#004ec4}{20}\;\;30\;\;40$

Multiples of

$4$ :

$4\;\;8\;\;12\;\;16\;\;\color{#004ec4}{20}\;\;24$

The

$LCD$ of

$10$ and

$4$ is

$20$

Next, we get the same denominator and subtract the fractions

	$31/10-5/4$

$=$	$\frac{31\times\color{#CC0000}{2}}{10\times\color{#CC0000}{2}}-\frac{5\times\color{#CC0000}{5}}{4\times\color{#CC0000}{5}}$	Multiply each fraction so that the denominator becomes $20$

$=$	$\frac{62}{\color{#004ec4}{20}}-\frac{25}{\color{#004ec4}{20}}$

$=$	$\frac{37}{20}$

Transform the fraction back to a mixed fraction

Start by dividing the numerator by the denominator

Arrange the numbers for long division

$20$ goes into

$37$ once. So write

$1$ above the line.

Multiply

$1$ to

$20$ and write the answer below

$37$

Subtract

$20$ from

$37$ and write the answer one line below

Since

$20$ cannot go into

$17$ anymore, $17$ is left as the Remainder and $1$ is the Quotient

Substitute values into the given formula

$\frac{\color{#007DDC}{b}}{\color{#9a00c7}{c}}$	$=$	$\color{#00880A}{Q}\frac{\color{#e65021}{R}}{\color{#9a00c7}{c}}$

$\frac{\color{#007DDC}{37}}{\color{#9a00c7}{20}}$	$=$	$\color{#00880A}{1}\frac{\color{#e65021}{17}}{\color{#9a00c7}{20}}$

$1 17/20$

Question 4 of 4

4. Question

Subtract the following:

$12-8 3/10$

1. $7 3/10$
2. $3 7/10$
3. $10 3/10$
4. $3 3/10$

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Transforming an Improper to Mixed Fraction

$\frac{\color{#007DDC}{b}}{\color{#9a00c7}{c}}=\color{#00880A}{Q}\frac{\color{#e65021}{R}}{\color{#9a00c7}{c}}$

$($ $b$

$-:$ $c$

$)=$ $Q$ and $R$ is the remainder

Transforming a Fraction from Mixed to Improper

$=$

$\frac{(\color{#9a00c7}{c}\times\color{#00880A}{A})+\color{#007DDC}{b}}{\color{#9a00c7}{c}}$

First, transform the mixed fractions to improper fractions

		$12-\color{#00880A}{8} \frac{\color{#007DDC}{3}}{\color{#9a00c7}{10}}$

	$=$	$\frac{12}{1}-\frac{(\color{#9a00c7}{10}\times\color{#00880A}{8})+\color{#007DDC}{3}}{\color{#9a00c7}{10}}$

	$=$	$\frac{12}{1}-\frac{80+3}{10}$

	$=$	$\frac{12}{1}-\frac{83}{10}$

Use cross method to subtract the fractions.

Start by multiplying the two denominators. Use the product as a denominator for a new fraction.

$12/1-83/10$

$=$

$☐/(1times10)$

$=$

$☐/10$

To get the numerator, cross multiply the given addition problem and subtract the products.

$\frac{\color{#00880A}{12}}{\color{#9a00c7}{1}}-\frac{\color{#9a00c7}{83}}{\color{#00880A}{10}}$	$=$	$\frac{(\color{#00880A}{12\times10})-(\color{#9a00c7}{1\times83})}{10}$

	$=$	$(120-83)/10$

	$=$	$37/10$

Transform the fraction back to a mixed fraction

Start by dividing the numerator by the denominator

Arrange the numbers for long division

$10$ goes into

$37$ three times. So write

$3$ above the line.

Multiply

$3$ to

$10$ and write the answer below

$37$

Subtract

$30$ from

$37$ and write the answer one line below

Since

$10$ cannot go into

$7$ anymore, $7$ is left as the Remainder and $3$ is the Quotient

Substitute values into the given formula

$\frac{\color{#007DDC}{b}}{\color{#9a00c7}{c}}$	$=$	$\color{#00880A}{Q}\frac{\color{#e65021}{R}}{\color{#9a00c7}{c}}$

$\frac{\color{#007DDC}{37}}{\color{#9a00c7}{10}}$	$=$	$\color{#00880A}{3}\frac{\color{#e65021}{7}}{\color{#9a00c7}{10}}$

$3 7/10$

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