2. Multiply Fractions
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Chapter 3
2.
Multiply Fractions
This lesson delves into the methods for multiplying fractions and mixed numbers, as well as simplifying fractions. It covers various techniques, including the use of area models and number lines, to make the concepts more understandable. The lesson also emphasizes the importance of estimation strategies to assess the reasonableness of answers. Practical examples are provided to demonstrate how these mathematical concepts can be applied in everyday situations like calculating time spent on activities or determining the area of objects.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Multiply Fractions
Slide of 10
This lesson will cover methods for multiplying fractions by whole numbers, fractions, and mixed numbers. Along with these methods, estimation strategies will be used to assess whether the products are reasonable.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Running Time
Paulina runs for two-sixths of her free time each day.
External credits: pikisuperstar
If Paulina has $5$ hours of free time per day, how many hours does she run in $4$ days? Give the exact answer.
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Discussion
Multiplying Fractions
The product of two fractions is equal to the product of the numerators divided by the product of the denominators. The resulting fraction is then simplified to its lowest terms, if possible.
$\dfrac{a}{b} \t \dfrac{c}{d} = \dfrac{a \t c}{b \t d}$
Here, $b$ and $d$ are not $0.$ When multiplying fractions, it makes no difference whether they are like or unlike fractions. Consider multiplying $\frac{5}{6}$ by $\frac{3}{4}.$ \begin{gathered} \dfrac{5}{6} \t \dfrac{3}{4} \end{gathered} The result of this multiplication can be found in three steps.
1
Multiply the Numerators
expand_more 2
Multiply the Denominators
expand_more The product of the denominators is $6 \t 4 = 24.$
3
Simplify if Possible
expand_more Note that $3$ is the greatest common factor of $15$ and $24.$ Divide both the numerator and the denominator by $3$ to simplify the fraction.
\(\dfrac{15}{24}\)
ReduceFrac
\ReduceFrac{3}
\(\dfrac{15/3 }{24/ 3 }\)
SimpQuot
\SimpQuot
\(\dfrac{5}{8}\)
Therefore, the product of $\frac{5}{6}$ and $\frac{3}{4}$ simplified to its lowest terms is $\frac{5}{8}.$
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Example
Finding What Fraction of the Water in a Bottle Is Drunk
Paulina drinks one-third of the water in her bottle before PE class. During class, she drinks ten-twelfths of the remaining water.
a Write a numeric expression to represent the amount of water in the bottle that Paulina drinks during class.
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b Which of the following is the best estimate for the value of the expression written in Part A?
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c Find the value of the expression found in Part A. Simplify the answer if possible.
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Hint
a What fraction of the bottle is full before the class? What mathematical operation must be performed to find ten twelfths of a number?
b If a fraction is greater than or equal to $\frac{1}{2},$ round the fraction to $1.$ If a fraction is less than $\frac{1}{2},$ round it to $0.$
c To multiply fractions, start with the multiplication of the numerators, followed by the multiplication of the denominators. Then, simplify the resulting fractions.
Solution
a Start by finding what fraction of the bottle is full before the physical education class. To do so, subtract one-third from $1.$
\(1-\dfrac{1}{3}\)
NumberToFrac
\NumberToFrac{3}
\(\dfrac{3}{3}-\dfrac{1}{3}\)
SubFrac
\SubFrac
\(\dfrac{3-1}{3}\)
SubTerm
\SubTerm
\(\dfrac{2}{3}\)
This represents the fraction of the bottle that is full before class. This means that Paulina drinks ten-twelfths of two-thirds of the bottle of water during class. To find this amount, these fractions will be multiplied. \begin{gathered} \textbf{Expression} \\[0.5em] \dfrac{10}{12} \times \dfrac{2}{3} \end{gathered} A number line can be used to help with this concept. First, divide the number line between $0$ and $1$ into thirds. Fill in $\frac{2}{3},$ representing the water that remains in the bottle before class. Then, divide this section into $12$ smaller equal parts and color in $10$ of them.
Notice that each third is divided into six smaller equal parts. The other third can also be divided into six parts. This will make it easier to find what fraction of the whole the red part represents.
The red part shows what fraction of the water bottle Paulina drinks during class.
b Estimation is a great tool to get a rough idea of the result of an operation — in this case, the result of a multiplication. Consider the value of each fraction in the expression separately.
\begin{gathered} \dfrac{10}{12} \times \dfrac{2}{3} \end{gathered} Now compare each fraction with $\frac{1}{2}.$ These fractions do not have the same denominator. The denominators of the fractions are $12,$ $3,$ and $2.$ The least common denominator of these numbers is $12.$ Each fraction can be rewritten as an equivalent fraction with the common denominator to make it easier to compare them.
| Rewrite | Compare with $\frac{1}{2},$ or $\frac{6}{12}$ | |
|---|---|---|
| $\dfrac{10}{12}$ | $\dfrac{10}{12}$ | $\dfrac{10}{12} > \dfrac{6}{12}$ |
| $\dfrac{2}{3}$ | $\dfrac{2\t4}{3\t4} = \dfrac{8}{12}$ | $\dfrac{8}{12} > \dfrac{6}{12}$ |
Both fractions are greater than $\frac{1}{2}.$ An estimate for the product can then be $1$ because the fractions can be rounded to $1.$ \begin{array}{rccc} \textbf{Product:}\! & \colIII{\dfrac{10}{12}} & \! \times \! & \dfrac{2}{3} \\[0.5em] & \downarrow & & \downarrow \\ \textbf{Estimate:} \! & \colIII{1} & \! \times \! & 1 \end{array} Since $1$ is the identity element of multiplication, this product is also equal to $1.$ However, that does not say much about the original product, and $1$ is not one of the possible answer options. Now think of rounding only one of the fractions to $1$ one at a time.
| Estimate for $ \frac{10}{12}\times \frac{2}{3}$ | |
|---|---|
| Round $\frac{10}{12}$ to $1$ | $1 \times \dfrac{2}{3} = \dfrac{2}{3}$ |
| Round $\frac{2}{3}$ to $1$ | $ \dfrac{10}{12} \times 1 = \dfrac{10}{12}$ |
The fraction $\frac{10}{12}$ can be simplified to $\frac{5}{6}.$ The answer is either $\frac{2}{3}$ or $\frac{5}{6}.$ The fractions in the options are in simplest form and $\frac{2}{3}$ is among the options. Therefore, the answer is $\frac{2}{3}.$
c To multiply the fractions, start with the multiplication of the numerators, followed by the multiplication of the denominators. Then, simplify the resulting fraction.
\(\dfrac{10}{12} \times \dfrac{2}{3}\)
MultFrac
\MultFrac
\(\dfrac{10 \times 2}{12 \times 3}\)
Multiply
\Multiply
\(\dfrac{20}{12 \times 3}\)
Multiply
\Multiply
\(\dfrac{20}{36}\)
Notice that the numerator and denominator both have $4$ as a factor. This suggests that the resulting fraction can be reduced to its lowest form by dividing the numerator and denominator by this factor.
\(\dfrac{20}{36}\)
ReduceFrac
\ReduceFrac{4}
\(\dfrac{20 /4}{36 / 4}\)
CalcQuot
\CalcQuot
\(\dfrac{5}{9}\)
The product of the fractions is $\frac{5}{9}.$ This is the fraction of the bottle of water that Paulina drinks throughout the lesson. Notice also that this value is close the estimate found in Part B because $\frac{2}{3}$ is equivalent to $\frac{6}{9}.$ \begin{array}{ccc} \textbf{Result} & & \textbf{Estimate} \\ \dfrac{5}{9} & \approx & \dfrac{6}{9} \end{array} Since the actual result and the estimate are close to each other, the answer $\frac{5}{9}$ is reasonable.
Extra
Using an Area Model to Represent the Product of FractionsIn addition to a number line, the product can also be represented by an area model. In this model, the denominators are used to divide a rectangular diagram into smaller parts. Consider modeling the following product.
\begin{gathered}
\dfrac{10}{12} \times \dfrac{2}{3}
\end{gathered}
The denominators are $12$ and $3.$ The rectangle is then divided into $12$ columns and $3$ rows.
The numerators determine which parts will be shaded. Since $10$ is the numerator of $\frac{10}{12},$ $\textcolor{hotpink}{10}$ of the $12$ columns will be shaded. Similarly, $\textcolor{dodgerblue}{2}$ of the $3$ rows will be shaded.
In this model, the overlapping region represents the product. For this example, the product is $\frac{20}{36}=\frac{5}{9}.$
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Pop Quiz
Finding the Product of Fractions
The product of two fractions is equal to the product of the numerators divided by the product of the denominators. Practice finding the product of fractions. Simplify the answer to its lowest terms.
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Discussion
Multiplying Fractions by Whole Numbers and By Mixed Numbers
When multiplying fractions by whole numbers or mixed numbers, both factors should be in the form of a proper fraction or an improper fraction.
Multiplying Fractions By Whole Numbers
To multiply fractions with whole numbers, the whole number is written as a fraction with a denominator is $1.$ The following steps are identical to those for multiplying fractions. Consider multiplying $9$ and $\frac{2}{27}.$
\(9 \times \dfrac{2}{27}\)
Rewrite
\Rewrite{9}{\dfrac{9}{1}}
\(\dfrac{9}{1} \times \dfrac{2}{27}\)
\(\dfrac{18}{27}\)
ReduceFrac
\ReduceFrac{9}
\(\dfrac{18 /9}{27 /9}\)
SimpQuot
\SimpQuot
\(\dfrac{2}{3}\)
Multiplying Fractions By Mixed Numbers
Recall that mixed numbers are fractions that consist of a whole number and a proper fraction. \begin{gathered} \col{3} \colIII{\tfrac{4}{5}} \ \WriteSysEqnIIb{\text{whole number: } \ \col{3}}{\text{proper fraction:} \ \colIII{\frac{4}{5}}} \end{gathered} The mixed fraction must be converted into an improper fraction before it can be multiplied by a fraction. The multiplication process can be better understood with the help of an example.
\(3\tfrac{4}{5} \times \dfrac{1}{3}\)
▼
\MMMixedToFrac
MixedToFrac
\MixedToFrac
\(\dfrac{3 \t 5 + 4}{5} \times \dfrac{1}{3}\)
Multiply
\Multiply
\(\dfrac{15 + 4}{5} \times \dfrac{1}{3}\)
AddTerms
\AddTerms
\(\dfrac{19}{5} \times \dfrac{1}{3}\)
\(\dfrac{19}{15}\)
\(\dfrac{19}{15}\)
▼
\MMMixedToFrac
Rewrite
\Rewrite{19}{15+4}
\(\dfrac{15+4}{15}\)
WriteSumFrac
\WriteSumFrac
\(\dfrac{15}{15}+\dfrac{4}{15}\)
CalcQuot
\CalcQuot
\(1 + \dfrac{4}{15}\)
Rewrite
\Rewrite{1+\dfrac{4}{15}}{1\tfrac{4}{15}}
\(1\tfrac{4}{15}\)
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Example
Finding Times Allocated For Activities
Paulina's PE class lasts for $1\frac{5}{12}$ hours. The table shows what fraction of the class time is allocated for various activities.
| Activity | Fraction |
|---|---|
| Warm-up | $\dfrac{1}{5}$ |
| Instruction | $\dfrac{1}{2}$ |
| Playing a game | $\dfrac{1}{10}$ |
| Cool-down | $\dfrac{1}{5}$ |
a Find how many minutes are devoted to warming up and cooling down.
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b Find how many minutes are devoted to the activities other than instruction.
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Hint
a Start by determining what fraction of the class time is allocated for warm-up and cool-down. To do so, add the fractions corresponding to the activities. Multiply the portion of the lesson spent on those activities by the total class time. Finally, use the fact that $1$ hour is $60$ minutes.
b Determine what fraction of the class time is allocated for the activities other than instruction.
Solution
a The fractions for both the warm-up and the cool-down to find the amount of time spent on them. Take a look at the given table.
| Part of $\bm{1\frac{5}{12}}$-hour Class | ||
|---|---|---|
| Activity | Fraction | |
| $\colIII{\text{Warm-up}}$ | $\colIII{\dfrac{1}{5}}$ | |
| Instruction | $\dfrac{1}{2}$ | |
| Playing game | $\dfrac{1}{10}$ | |
| $\col{\text{Cool-down}}$ | $\col{\dfrac{1}{5}}$ | |
The sum of the fractions is $\frac{2}{5}.$ \begin{gathered} \colIII{\dfrac{1}{5}} + \col{\dfrac{1}{5}} = \dfrac{2}{5} \end{gathered} Next, the time spent on warming up and cooling down will be found by multiplying this fraction number by the lesson time $1\frac{5}{12}.$ This will give the amount of time in hours. \begin{gathered} \dfrac{2}{5}\times 1\tfrac{5}{12} \end{gathered} This is a multiplication of a fraction by a mixed number. Before performing the multiplication, the mixed number will be rewritten as an improper fraction.
\(\dfrac{2}{5}\times 1\frac{5}{12}\)
MixedToFrac
\MixedToFrac
\(\dfrac{2}{5}\times \dfrac{1 \t 12 + 5}{12}\)
MultByOne
\MultByOne
\(\dfrac{2}{5}\times \dfrac{12 + 5}{12}\)
AddTerms
\AddTerms
\(\dfrac{2}{5}\times \dfrac{17}{12}\)
Now multiply the numerators and denominators.
\(\dfrac{2}{5}\times \dfrac{17}{12}\)
MultFrac
\MultFrac
\(\dfrac{2 \times 17}{5 \times 12 }\)
Multiply
\Multiply
\(\dfrac{34}{60}\)
Since a common factor exists between numerator and denominator, the fraction can be simplified. \begin{gathered} \dfrac{34 /2 }{60 / 2 } = \dfrac{17}{30} \end{gathered} The fraction simplifies to $\frac{17}{30}.$ This means that $\frac{17}{30}$ of an hour is spent on warming up and cooling down. Now the number of minutes spent on these activities will be found by multiplying the fraction by $60$ because $1$ hour is $60$ minutes. \begin{gathered} 60\times \dfrac{17}{30} \end{gathered} This is a multiplication of a whole number by an improper fraction. Start by rewriting $60$ as an improper fraction, then multiply the fractions.
\(60 \times \dfrac{17}{30}\)
Rewrite
\Rewrite{60}{\dfrac{60}{1}}
\(\dfrac{60}{1} \times \dfrac{17}{30}\)
MultFrac
\MultFrac
\(\dfrac{60 \times 17 }{1\times 30}\)
Multiply
\Multiply
\(\dfrac{1020}{30}\)
CalcQuot
\CalcQuot
\(34\)
The warm-up and cool-down activities last for $34$ minutes.
b Consider the given table again, this time focusing on the instruction time.
| Part of $\bm{1\frac{5}{12}}$-hour Class | ||
|---|---|---|
| Activity | Fraction | |
| $\text{Warm-up}$ | $\dfrac{1}{5}$ | |
| $\col{\text{Instruction}}$ | $\col{\dfrac{1}{2}}$ | |
| Playing game | $\dfrac{1}{10}$ | |
| $\text{Cool-down}$ | $\dfrac{1}{5}$ | |
As shown, half of the class time is spent on instruction. In other words, one-half of the lesson is not spent on instruction. With this in mind, the total lesson time will be multiplied by $\frac{1}{2}.$ \begin{gathered} \dfrac{1}{2} \times 1 \dfrac{5}{12} \end{gathered} Rewrite the mixed number as an improper fraction to find the product.
\(\dfrac{1}{2}\times 1\frac{5}{12}\)
▼
\MMMixedToFrac
MixedToFrac
\MixedToFrac
\(\dfrac{1}{2}\times \dfrac{1 \t 12 + 5}{12}\)
MultByOne
\MultByOne
\(\dfrac{1}{2}\times \dfrac{12 + 5}{12}\)
AddTerms
\AddTerms
\(\dfrac{1}{2}\times \dfrac{17}{12}\)
Next, multiply the numerators and denominators.
\(\dfrac{1}{2}\times \dfrac{17}{12}\)
MultFrac
\MultFrac
\(\dfrac{1 \times 17}{2 \times 12 }\)
Multiply
\Multiply
\(\dfrac{17}{24}\)
This fraction cannot be simplified further. It means that $\frac{17}{24}$ of an hour is spent on activities other than instruction. Finally, multiply $\frac{17}{24}$ by $60$ to write the amount of time in terms of minutes. \begin{gathered} 60\times \dfrac{17}{24} \end{gathered} This is a multiplication of a whole number by a fraction. An easy way to find this product is to move the whole number to the numerator of the fraction as a factor.
\(60 \times \dfrac{17}{24}\)
MoveLeftFacToNum
\MoveLeftFacToNum
\(\dfrac{60\times 17}{24}\)
Multiply
\Multiply
\(\dfrac{1020}{24}\)
CalcQuot
\CalcQuot
\(42.5\)
The activities other than instruction last for $42.5$ minutes.
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Example
Finding the Area of a Printed Photograph
Paulina loves a photo of her playing volleyball and prints it. The diagram shows the dimensions of the photograph.
External credits: pch.vector
a Estimate the area of the photograph.
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b What is the area of the photograph? Write the answer as a mixed number.
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c Paulina frames the photo using a frame with a $\frac{1}{5}\text{-}$centimeter border. What is the area of the photo including the frame? Write the answer as a mixed number.
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Hint
a Use the formula for the area of a rectangle. Round the mixed numbers to the nearest whole numbers
b To multiply fractions, multiply the numerators and denominators with each other.
c Add $2\times \frac{1}{5}$ to each side of the photo. Then repeat the process from Part A.
Solution
External credits: pch.vector
The length of the photo is $\col{8\tfrac{3}{4}}$ inches and the width of the photo is $\colIV{6\tfrac{2}{3}}$ inches. \begin{aligned} \text{Area } = \col{8\tfrac{3}{4}} \t \colIV{6\tfrac{2}{3}} \end{aligned} Since the fractional parts of the mixed numbers are greater than $\frac{1}{2},$ the mixed numbers can be rounded up. \begin{aligned} \text{Area } = & \ \col{8\tfrac{3}{4}} \t \colIV{6\tfrac{2}{3}} \\ & \ \ \downarrow \quad \ \downarrow \\ & \ \ \ \col{9} \ \t \ \colIV{7} \end{aligned} Therefore, the area of the photo is about $9\t 7,$ or $63$ square inches.
b Recall the expression for the area of the photo from Part A. It is the product of the two mixed numbers.
\begin{aligned} \text{Area } = \col{8\tfrac{3}{4}} \t \colIV{6\tfrac{2}{3}} \end{aligned} To multiply these mixed numbers, they first need to be rewritten as improper fractions.
\(8\frac{3}{4} \t 6\frac{2}{3}\)
MixedToFrac
\MixedToFrac
\(\dfrac{8\t 4 +3}{4} \t \dfrac{6 \t 3 +2}{3}\)
Multiply
\Multiply
\(\dfrac{32+3}{4} \t \dfrac{18 +2}{3}\)
AddTerms
\AddTerms
\(\dfrac{35}{4} \t \dfrac{20}{3}\)
Recall that the product of two fractions is equal to the product of the numerators divided by the product of the denominators.
\(\dfrac{35}{4} \t \dfrac{20}{3}\)
MultFrac
\MultFrac
\(\dfrac{35\t 20}{4\t 3}\)
Multiply
\Multiply
\(\dfrac{700}{12}\)
ReduceFrac
\ReduceFrac{4}
\(\dfrac{700/4}{12/4}\)
CalcQuot
\CalcQuot
\(\dfrac{175}{3}\)
▼
\MMFracToMixed
Rewrite
\Rewrite{175}{174+1}
\(\dfrac{174+1}{3}\)
WriteSumFrac
\WriteSumFrac
\(\dfrac{174}{3}+\dfrac{1}{3}\)
CalcQuot
\CalcQuot
\(58+\dfrac{1}{3}\)
Rewrite
\Rewrite{58+\dfrac{1}{3}}{58\tfrac{1}{3}}
\(58\tfrac{1}{3}\)
The area of the photo is $58\tfrac{1}{3}$ square inches. Note that this is about the same as the estimate found in Part A, so this answer is reasonable.
c In this part, start by adding a $\frac{1}{5}\text{-}$inch border on each side of the photo. This will extend each side by $2\t\frac{1}{5}$ inches.
External credits: pch.vector
The expression for the area of the photo with the frame is then the product of the side lengths shown in the diagram. \begin{gathered} \left(8\tfrac{3}{4}+2\t \dfrac{1}{5}\right) \t \left(6\tfrac{2}{3}+2\t \dfrac{1}{5}\right) \\ \Updownarrow \\ \left(8\tfrac{3}{4}+\dfrac{2}{5}\right) \t \left(6\tfrac{2}{3}+ \dfrac{2}{5}\right) \end{gathered} Once again, rewrite the fractions so that all of them are either proper or improper fractions.
\(\left(8\tfrac{3}{4}+ \dfrac{2}{5}\right) \t \left(6\tfrac{2}{3}+\dfrac{2}{5}\right)\)
▼
\MMMixedToFrac
MixedToFrac
\MixedToFrac
\(\left(\dfrac{8\t 4+3}{4}+ \dfrac{2}{5}\right) \t \left(\dfrac{ 6 \t 3+2}{3}+\dfrac{2}{5}\right)\)
Multiply
\Multiply
\(\left(\dfrac{32+3}{4}+ \dfrac{2}{5}\right) \t \left(\dfrac{18+2}{3}+\dfrac{2}{5}\right)\)
AddTerms
\AddTerms
\(\left(\dfrac{35}{4}+ \dfrac{2}{5}\right) \t \left(\dfrac{20}{3}+\dfrac{2}{5}\right)\)
▼
\AddFrac
ExpandFrac
\ExpandFrac{5}
\(\left(\dfrac{35 \t 5}{4\t5}+ \dfrac{2}{5}\right) \t \left(\dfrac{20}{3}+\dfrac{2}{5}\right)\)
ExpandFrac
\ExpandFrac{4}
\(\left(\dfrac{35 \t 5}{4\t5}+ \dfrac{2\t 4}{5\t 4}\right) \t \left(\dfrac{20}{3}+\dfrac{2}{5}\right)\)
ExpandFrac
\ExpandFrac{5}
\(\left(\dfrac{35 \t 5}{4\t5}+ \dfrac{2\t 4}{5\t 4}\right) \t \left(\dfrac{20\t5}{3\t5}+\dfrac{2}{5}\right)\)
ExpandFrac
\ExpandFrac{3}
\(\left(\dfrac{35 \t 5}{4\t5}+ \dfrac{2\t 4}{5\t 4}\right) \t \left(\dfrac{20\t5}{3\t5}+\dfrac{2 \t 3}{5\t 3}\right)\)
Multiply
\Multiply
\(\left(\dfrac{175}{20}+ \dfrac{8}{20}\right) \t \left(\dfrac{100}{15}+\dfrac{6}{15}\right)\)
AddFrac
\AddFrac
\(\dfrac{175+8}{20} \t \dfrac{100+6}{15}\)
AddTerms
\AddTerms
\(\dfrac{183}{20} \t \dfrac{106}{15}\)
When multiplying fractions, the product is equal to the product of the numerators divided by the product of the denominators.
\(\dfrac{183}{20} \t \dfrac{106}{15}\)
MultFrac
\MultFrac
\(\dfrac{183\t 106}{20\t 15}\)
Multiply
\Multiply
\(\dfrac{19\,398}{300}\)
ReduceFrac
\ReduceFrac{6}
\(\dfrac{19\,398/6}{300/6}\)
CalcQuot
\CalcQuot
\(\dfrac{3233}{50}\)
▼
\MMFracToMixed
Rewrite
\Rewrite{3233}{3200+33}
\(\dfrac{3200+33}{50}\)
WriteSumFrac
\WriteSumFrac
\(\dfrac{3200}{50}+\dfrac{33}{50}\)
CalcQuot
\CalcQuot
\(64+\dfrac{33}{50}\)
Rewrite
\Rewrite{64+\dfrac{33}{50}}{64\tfrac{33}{50}}
\(64\tfrac{33}{50}\)
The area of the photo with the frame is $64\tfrac{33}{50}$ square inches.
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Pop Quiz
Finding the Product of Mixed Numbers
To multiply a fraction by a whole number, the whole number is multiplied by the numerator of the fraction. To multiply mixed numbers, the mixed numbers can be converted into improper fractions before multiplying. Practice finding the product of fractions. Simplify the answer to its lowest terms.
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Closure
Finding Time Paulina Runs
The important point in multiplying fractions is to ensure that the fractions are either proper fractions or improper fractions. \begin{gathered} 1\tfrac{2}{6} \times 2 = \dfrac{8}{6} \times \dfrac{2}{1} \end{gathered} The final step usually involves simplifying the resulting fraction. However, to make calculations easier, first check if the two fractions are already in their lowest forms. If not, the fractions can be simplified first before multiplying them. \begin{aligned} \small\colIII{4} \ \, & \\ \dfrac{\cancel{8}}{\cancel{6}} &\times \dfrac{2}{1} = \dfrac{4}{3} \times \dfrac{2}{1} = \dfrac{8}{3}\\ \small\colIII{3}\ \, & \end{aligned} Consider the challenge presented at the beginning of the lesson. Paulina devotes two-sixths of her free time to exercise.
External credits: pikisuperstar
If Paulina has $5$ hours of free time per day, how many hours does she run in $4$ days?
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Hint
Start by finding the number of hours Paulina spends running in a day.
Solution
It is given that Paulina has $5$ hours of free time per day and that two-sixths of this time is devoted to physical activities. The time spent on running can be found by finding two-sixths of $5.$
\begin{gathered}
\dfrac{2}{6} \times 5
\end{gathered}
Notice that the fraction can be simplified to its lowest terms.
\(\dfrac{2}{6} \times 5\)
▼
\MMReduce{2}
\(\dfrac{1}{3}\times 5\)
MoveRightFacToNumOne
\MoveRightFacToNumOne
\(\dfrac{5}{3}\)
Paulina runs for $\frac{5}{3}$ hours every day. Since the number of hours she runs in $4$ days is required, the daily amount of activity should be multiplied by $4$ to find the total amount of time.
\(4 \times \dfrac{5}{3}\)
MoveLeftFacToNum
\MoveLeftFacToNum
\(\dfrac{4\times 5}{3}\)
Multiply
\Multiply
\(\dfrac{20}{3}\)
This final fraction can also be written as a mixed number.
\(\dfrac{20}{3}\)
▼
\MMFracToMixed
Rewrite
\Rewrite{20}{18+2}
\(\dfrac{18+2}{3}\)
WriteSumFrac
\WriteSumFrac
\(\dfrac{18}{3} + \dfrac{2}{3}\)
CalcQuot
\CalcQuot
\(6+\dfrac{2}{3}\)
Rewrite
\Rewrite{6+\dfrac{2}{3}}{6\tfrac{2}{3}}
\(6\tfrac{2}{3}\)
In $4$ days, Paulina runs $6\frac{2}{3}$ hours.
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Multiply Fractions
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We want to find the product of $\frac{3}{6}$ and $\frac{8}{11}.$ \begin{gathered} \dfrac{3}{6} \t \dfrac{8}{11} \end{gathered} When we multiply fractions, we need to remember that the product of two fractions is equal to the product of the numerators divided by the product of the denominators. Let's find the given product!
Now we will split the numerator and denominator into prime factors. This will enable us to find the greatest common factor (GCF) of the numbers.
| Number | Prime Factorization | GCF |
|---|---|---|
| $24$ | $2^3 \t 3$ | $2\t 3=6$ |
| $66$ | $2\t 3 \t 11$ |
We can simplify the fraction by dividing the numerator and denominator by their GCF.
We will start again by multiplying the numerators, followed by multiplying the denominators. Then we will reduce the fraction if necessary. Let's do it!
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We want to multiply a fraction and a whole number. To do so, we will first rewrite the whole number as a fraction with a denominator of $1.$ Then, we will multiply the numerators and denominators with each other. If necessary, we will also simplify the obtained product. Let's do it!
To find the given product, we will rewrite the whole number as a fraction. Then, we will multiply the numerators and denominators with each other like we did in Part A.
We can simplify this fraction because they have a common factor. \begin{aligned} 14 = 2 \t 7 \\ 21 = 3 \t 7 \end{aligned} The greatest common factor (GCF) of the numbers is $7.$ Let's dividing the numerator and denominator by the GCF.
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We want to find the product of a mixed number and a whole number. Let's start by rewriting them as improper fractions. Then, their product will be equal to the product of the numerators divided by the product of the denominators.
Now, we need to write it as a mixed number.
This time we are asked to find the product of two mixed numbers. To do so, we will first rewrite them as improper fractions. Let's do it!
Notice that the numerator of $\frac{21}{10}$ and the denominator of $\frac{15}{14}$ have a common factor of $\col{7}.$ Additionally, the numerator of $\frac{15}{14}$ and the denominator of $\frac{21}{10}$ have a common factor of $\colIII{5}.$ We can use this information to simplify the fractions before we multiply them. Let's do it by factoring the numbers! \begin{gathered} \dfrac{21}{10} \t \dfrac{15}{14} = \dfrac{3 \t \cancel{\col{7}}}{2 \t \cancel{\colIII{5}}} \t \dfrac{3\t \cancel{\colIII{5}}}{2 \t \cancel{\col{7}}} \end{gathered} The fractions in the product can be simplified to $\frac{3}{2}.$ Let's find this product!
We are asked to write the answer as a mixed number. Let's convert it!
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We want to estimate the product of a mixed number and a whole number. \begin{gathered} 5\tfrac{4}{5} \t 4 \end{gathered} To do so, we will round the mixed number. Recall that if the fractional part of a mixed number is greater than or equal to $\frac{1}{2},$ we round the mixed number up. If not, we round it down. Now let's check the fractional part of our mixed number! \begin{gathered} \col{\dfrac{4}{5}} >\dfrac{1}{2} \quad \text{because} \quad \dfrac{8}{10} > \dfrac{5}{10} \end{gathered} Since the fractional part $\frac{4}{5}$ is greater than $\frac{1}{2},$ we can round the mixed number up. This means that $5\frac{4}{5}$ is about $6.$ \begin{gathered} 5\col{\tfrac{4}{5}} \t 4 \\ \downarrow \quad \\ \ 6 \t 4 \end{gathered} Since $6\t4$ is equal to $24,$ the given product is about $24.$
We want to estimate the product of two mixed numbers.
\begin{gathered}
1\tfrac{6}{7} \t 6\tfrac{1}{6}
\end{gathered}
We can round the first mixed number up because $\frac{6}{7}$ is greater than $\frac{1}{2}.$ We can round the other mixed number down because its fractional part is less than $\frac{1}{2}.$
\begin{gathered}
\col{\dfrac{6}{7}} >\dfrac{1}{2} & \text{because} & \dfrac{12}{14} > \dfrac{7}{14} \\[0.8em]
\colIII{\dfrac{1}{6}} <\dfrac{1}{2} &\text{because} & \dfrac{1}{6} < \dfrac{3}{6}
\end{gathered}
Let's now multiply the rounded numbers.
\begin{gathered}
1\col{\tfrac{6}{7}} \t 6 \colIII{\tfrac{1}{6}}
\\ \downarrow \quad \, \downarrow \\
2 \, \t \, 6
\end{gathered}
The product of the mixed numbers is about the product of $2$ and $6,$ or $12.$
\begin{gathered}
1\tfrac{6}{7} \t 6 \tfrac{1}{6} \approx 12
\end{gathered}
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In area model, the shaded columns represents one fraction and the shaded rows represents another. If we break down the model into two parts, we can understand what this means. Let's do it!
The figure on the left is divided into $5$ parts and $3$ of them are shaded. This means that the shaded columns represent $\frac{3}{5}.$ In the same way, the figure on the right is divided into $4$ parts and $1$ of them is shaded, so the shaded rows represent $\frac{1}{4}.$ The diagram models the multiplication of these fractions. \begin{gathered} \dfrac{3}{5} \times \dfrac{1}{4} \end{gathered}
Let's multiply the fractions. Remember that we multiply the numerators with each other and the denominators with each other.
The product of the fractions is $\frac{3}{20},$ which cannot be simplified further.
We can verify that the product is $\frac{3}{20}$ by looking at the diagram. Let's focus on the region where the blue and yellow regions overlap.
This region consists of $3$ of the $20$ small squares. Notice also that $3$ and $20$ are the product of the numerators and denominators, respectively. \begin{gathered} \dfrac{3}{20} \end{gathered} The green region represents the product of the multiplication written in Part A and matches the product of the two fractions. \begin{gathered} \dfrac{3}{5}\times \dfrac{1}{4} = \dfrac{3}{20}\ \StatementTrue \end{gathered}
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