Print preview
Worksheet Fractions
2.
3.
Evaluate each expression if \(a=-{\frac{1}{3}},\ b={\frac{3}{5}},\ c= -4 \frac{1}{3},\
d=3\frac{1}{5}\text{.}\)
Your answer should be a reduced fraction or a mixed number.
a) \(4a=\)
b) \(a+cd=\)
c) \(9d+\frac{7}{8}=\)
d) \(d(b+6)=\)
Solution.
Solution
a) \(4a=4\times -{\frac{1}{3}}=-1{\textstyle\frac{1}{3}}\)
b) \(a+cd= -{\frac{1}{3}} - 4{\textstyle\frac{1}{3}}\times 3{\textstyle\frac{1}{5}}= -14{\textstyle\frac{1}{5}}\)
c) \(9d+\frac{7}{8}=\(9\times 3{\textstyle\frac{1}{5}}+\frac{7}{8} = 29{\textstyle\frac{27}{40}}\)
d) \(d(b+6)=3{\textstyle\frac{1}{5}}({\frac{3}{5}}+6) = 21{\textstyle\frac{3}{25}}\)
4.
Put in simplest form.
For example if \(\frac{3}{7}\) were the answer you would put 3/7 in the answer box.
1) \(\frac{12}{14}\)=
2) \(\frac{32}{40}\)=
2) \(\frac{25}{5}\)=
Solution.
Solution
1) To put \(\frac{12}{14}\) into simplest form we must divide both numerator and denominator by 2
(which is their greatest common factor) to get answer \(\frac{6}{7}\) which we put in the answer box as \(6/7\text{.}\)
2) To put \(\frac{32}{40}\) into simplest form we must divide both numerator and denominator
5.
Write the following rational number in simplest form:
\(\displaystyle\frac{4420000}{37908000} =\) /
Hint:
6.
7.
Add the following and express your answer as a single fraction. No decimals or mixed fractions.
\(\displaystyle\frac{22}{7} - \frac{15}{7} =\)
Remember, to enter a fraction of the form \(\frac{a}{b}\text{,}\) type a / b.
8.
Add the following and express your answer as a single fraction. No decimals or mixed fractions.
\(\displaystyle\frac{23}{7} - \frac{11}{6} =\)
Remember, to enter a fraction of the form \(\frac{a}{b}\text{,}\) type a / b.
9.
i) \(\frac{1}{4}-\frac{1}{5}\) =.
ii) \(\frac{1}{3}-\frac{1}{4}\) =.
iii) \(\frac{1}{6}-\frac{1}{7}\) =.
The three problems above all satisfy a rule. It is
-
\(\displaystyle \frac{1}{a}-\frac{1}{a+1}= \frac{1}{a(a+1)}\)
-
\(\displaystyle \frac{1}{a}-\frac{1}{a+1}=\frac{1}{2a+1}\)
-
\(\displaystyle \frac{1}{a}-\frac{1}{a+1}=\frac{2a+1}{a (a+1)}\)
-
\(\displaystyle \frac{1}{a}-\frac{1}{a+1}=\frac{a}{2a+1}\)
10.
Write each sum in simplest form (as a mixed number).
\(17 \frac {7}{8} + 14 \frac{5}{6}\)=.
\(\frac {2}{5} + 3 \frac{2}{7}\)=.
\(13 \frac {7}{8} + 4 \frac{1}{4}\)=.
\(28 \frac {3}{5} + 16 \frac{2}{5}\)=.
\(5 \frac {2}{3} + 25 \frac{8}{9}\)=.
\(16 \frac {8}{9} + 18 \frac{4}{9}\)=.
\(19 \frac {5}{9} + 32 \frac{8}{15}\)=.
\(74 \frac {1}{3} + 56 \frac{1}{4}\)=.
11.
\(4\frac{4}{24}+2\frac{7}{24}+ 2\frac{15}{24}=\)
Solution.
Solution
Since the denominators of the fractions are all the same ( 24) we can start by changing each mixed number
to an improper fraction with denominator 24 and then add them
\(4\frac{4}{24}+2\frac{7}{24}+ 2\frac{15}{24}= \frac{100}{24}+ \frac{55}{24}+ \frac{63}{24}=\frac{218}{24}=9 \frac{2}{24}\)
12.
Add these together: \(\displaystyle{ -5 + \frac{7}{10}}\)
When needed, use an improper fraction in your answer. Donβt use a mixed number.
Solution.
When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions:
\(\displaystyle{ -5 = \frac{-5}{1} }\)
Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator \(10\text{.}\) We will rewrite the first fraction:
\(\displaystyle{ \begin{aligned}\frac{-5}{1} \amp = \frac{-5 \cdot 10}{1 \cdot 10}\\ \amp = \frac{-50}{10} \end{aligned}}\)
Finally, we add the numerators and keep the denominator unchanged. The whole process is:
\(\displaystyle{\begin{aligned}
-5 + \frac{7}{10}
\amp = \frac{-50}{10} + \frac{7}{10} \\
\amp = \frac{-50 + 7}{10} \\
\amp = -\frac{43}{10}
\end{aligned}
}\)
The answer to this question is \({-{\frac{43}{10}}}\text{.}\)
13.
Reduce the fraction \(\displaystyle{ \frac{5}{25} }\text{.}\)
Solution.
There are at least two methods to reduce the fraction \(\displaystyle{ \frac{5}{25} }\text{.}\)
Method 1:We find a number that divides into both the numerator \(5\text{,}\) and the denominator \(25\text{.}\)
Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\)
In this case, \(5\) goes into both the numerator and denominator of \(\displaystyle{ \frac{5}{25} }\text{.}\) We divide \(5\) into both numbers, and we have:
\(\displaystyle{ \begin{aligned}
\frac{5}{25} \amp = \frac{5 \div 5}{25\div 5}\\
\amp = \frac{1}{5}\end{aligned} }\)
Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers.
In this case, \(\displaystyle{\frac{1}{5}}\) is the final answer.
Method 2:
A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
\(\displaystyle{\begin{aligned}[t]
\frac{5}{25} \amp = \frac{1 \cdot 5}{5 \cdot 5} \\
\amp = \frac{1}{5}
\end{aligned}
}\)
Notice that when the numeratorβs only prime factor, \(5\text{,}\) is canceled, we have to leave a \(1\) in the numerator.
14.
Evaluate the following.
-
\(\displaystyle{ \frac{-32}{-8}= }\)
-
\(\displaystyle{ \frac{25}{-5}= }\)
-
\(\displaystyle{ \frac{-20}{5}= }\)
Solution.
The rules for dividing positive numbers are the same as those for multiplication:
\(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
\(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
\(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
\(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
The solutions are:
-
\(\displaystyle \displaystyle{ \frac{-32}{-8}={4}, }\)
-
\(\displaystyle \displaystyle{ \frac{25}{-5}={-5}, }\)
-
\(\displaystyle \displaystyle{ \frac{-20}{5}={-4}. }\)
15.
16.
17.
Here is an expression with negative exponents.
\(\displaystyle\left(\frac{9}{2}\right)^{-2}=\)
Evaluate the expression; in other words, write the answer without using exponents.
Solution.
We evaluate the expression by remembering that \(x^{-n}\) is the same thing as \(\displaystyle{\frac{1}{x^n}}\) for any non-zero, real value of \(x\)
\(\begin{aligned}
\displaystyle\left(\frac{9}{2}\right)^{-2}\amp =\displaystyle\frac{9^{-2}}{2^{-2}}\\
\amp = \displaystyle\frac{\displaystyle\frac{1}{9^{2}}}{\displaystyle\frac{1}{2^{2}}} \\
\amp = \displaystyle\frac{1}{9^{2}}\cdot \frac{2^{2}}{1} \\
\amp = \frac{2^{2}}{9^{2}}\\
\amp = \displaystyle\frac{4}{81}
\end{aligned}\)
Remember that when dividing by a fraction we multiply by its reciprocal.
18.
Evaluate this expression:
\(\displaystyle{ {{\frac{8}{9}}}+2\cdot{{\frac{2}{9}}}= }\)
Solution.
According to the order of operations, the multiplication has higher priority than the addition.
You might need to go back to previous units to review how to add fractions.
\(\begin{aligned}[t]
{{\frac{8}{9}}}+2\cdot{{\frac{2}{9}}} \amp = {{\frac{8}{9}}} + {{\frac{4}{9}}} \\
\amp = {{\frac{4}{3}}}
\end{aligned}\)
19.
Evaluate the following expressions:
-
\(\displaystyle{ {{\frac{3}{8}}}-\left({{\frac{3}{4}}}\right)^{3}= }\)
-
\(\displaystyle{ {{\frac{3}{8}}}-\left(-{{\frac{3}{4}}}\right)^{3}= }\)
Solution.
You might need to go back to previous units to review how to do arithmetic with fractions.
-
\(\displaystyle \begin{aligned}[t] {{\frac{3}{8}}}-\left({{\frac{3}{4}}}\right)^{3} \amp = {{\frac{3}{8}}}-{{\frac{3}{4}}}\cdot{{\frac{3}{4}}}\cdot{{\frac{3}{4}}} \\ \amp = {{\frac{3}{8}}}-{{\frac{27}{64}}} \\ \amp = {-{\frac{3}{64}}} \end{aligned}\)
-
\(\displaystyle \begin{aligned}[t] {{\frac{3}{8}}}-\left(-{{\frac{3}{4}}}\right)^{3} \amp = {{\frac{3}{8}}}-\left(-{{\frac{3}{4}}}\right)\cdot\left(-{{\frac{3}{4}}}\right)\cdot\left(-{{\frac{3}{4}}}\right) \\ \amp = {{\frac{3}{8}}}-\left({-{\frac{27}{64}}}\right) \\ \amp = {{\frac{51}{64}}} \end{aligned}\)
20.
Evaluate the following expressions:
-
\(\displaystyle{ \left( -{{\frac{3}{10}}} \right) ^{2}= }\)
-
\(\displaystyle{ - \left( {{\frac{3}{10}}} \right) ^{2}= }\)
-
\(\displaystyle{ - \left( -{{\frac{3}{10}}} \right) ^{2}= }\)
Solution.
Notice the difference in the position of the negative symbol.
-
\(\displaystyle{\begin{aligned}[t] \left( -{{\frac{3}{10}}} \right) ^{2} \amp = \left( -{{\frac{3}{10}}} \right) \cdot \left( -{{\frac{3}{10}}} \right) \\ \amp = {{\frac{9}{100}}} \end{aligned}}\)In Part a, parentheses have higher priority than exponents.
-
\(\displaystyle{\begin{aligned}[t] -\left( {{\frac{3}{10}}} \right) ^{2} \amp = -\left( {{\frac{3}{10}}} \right) \cdot \left( {{\frac{3}{10}}} \right) \\ \amp = {-{\frac{9}{100}}} \end{aligned}}\)In Part b, exponents have higher priority than the negative symbol (which is, in essence, multiplication).
-
\(\displaystyle \displaystyle{\begin{aligned}[t] -\left( -{{\frac{3}{10}}} \right) ^{2} \amp = -\left( -{{\frac{3}{10}}} \right) \cdot \left( -{{\frac{3}{10}}} \right) \\ \amp = \left( {{\frac{3}{10}}} \right) \cdot \left( -{{\frac{3}{10}}} \right) \\ \amp = {-{\frac{9}{100}}} \end{aligned}}\)
The following pattern shows why the negative symbol means βnegative one timesβ:
\(\displaystyle{\begin{aligned}[t]
-2 \amp = -1 \cdot 2 \\
-3 \amp = -1 \cdot 3 \\
-4 \amp = -1 \cdot 4 \\
\amp \vdots
\end{aligned}
}\)
21.
Simplify. Your answer should be a reduced fraction, with no common factors in numerator and denominator.
| \(\displaystyle \frac{\frac{1}{5} + \frac{2}{3}}{\frac{3}{4} - \frac{1}{5}} =\) |
22.
Courtney walks 3 laps around a \(\frac{1}{4}\)-mile track.
[1 mile is 5280 feet].
