Answer
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Hint: In this question we have to solve the difference between two mixed fractions. For this, we will first convert the mixed fraction into improper fraction using the form $a\dfrac{b}{c}=\dfrac{\left( c\times a \right)+b}{c}$. After that, we will convert both the fraction into like fraction to calculate the difference between them. For this, we will convert their denominator into the LCM of both denominators i.e. we will find their equivalence fraction whose denominator are the same. Then we will find the required difference.
Complete step-by-step solution:
Here we are given the two mixed fractions as $4\dfrac{1}{8}-8\dfrac{1}{3}$.
We need to find a difference between them i.e. $4\dfrac{1}{8}-8\dfrac{1}{3}$.
Let us first convert this mixed fraction into improper fraction. We know a mixed fraction of the form $a\dfrac{b}{c}$ can be converted by into improper fraction of the form $\dfrac{\left( c\times a \right)+b}{c}$ so using this we get,
\[\begin{align}
& 4\dfrac{1}{8}=\dfrac{\left( 8\times 4 \right)+1}{8}=\dfrac{32+1}{8}=\dfrac{33}{8} \\
& 8\dfrac{1}{3}=\dfrac{\left( 3\times 8 \right)+1}{3}=\dfrac{24+1}{3}=\dfrac{25}{3} \\
\end{align}\]
Therefore we need to find $\dfrac{33}{8}-\dfrac{25}{3}$.
For subtraction, we need fraction to be like fraction i.e. we need denominator of both fraction same.
For this let us take LCM of 3 and 8 which is 24.
To convert $\dfrac{33}{8}$ having denominator as 24 we need to multiply the numerator and the denominator by 3, we get $\dfrac{33\times 3}{8\times 3}=\dfrac{99}{24}$.
Similarly to convert $\dfrac{25}{3}$ having denominator as 24, we need to multiply the numerator and the denominator by 8 we get $\dfrac{25\times 8}{3\times 8}=\dfrac{200}{24}$.
Therefore we need to find $\dfrac{99}{24}-\dfrac{200}{24}$.
It can be written as $\dfrac{99-200}{24}$.
Solving the numerator we get $\dfrac{-101}{24}$.
Hence our required answer is $\dfrac{-101}{24}$.
Note: Students should take care of signs while solving the subtraction. Make sure that the denominator is the same for performing subtraction or addition. Student can write mixed fraction in the following form also,
$4+\dfrac{1}{8}-\left( 8+\dfrac{1}{3} \right)\Rightarrow 4+\dfrac{1}{8}-8-\dfrac{1}{3}\Rightarrow -4+\dfrac{1}{8}-\dfrac{1}{3}$.
After that they could take LCM as 24 and solve it as,
$\dfrac{-4\times 24+1\times 3-1\times 8}{24}=\dfrac{-96+3-8}{24}=\dfrac{-101}{24}$.
Which gives us the same answer.
Complete step-by-step solution:
Here we are given the two mixed fractions as $4\dfrac{1}{8}-8\dfrac{1}{3}$.
We need to find a difference between them i.e. $4\dfrac{1}{8}-8\dfrac{1}{3}$.
Let us first convert this mixed fraction into improper fraction. We know a mixed fraction of the form $a\dfrac{b}{c}$ can be converted by into improper fraction of the form $\dfrac{\left( c\times a \right)+b}{c}$ so using this we get,
\[\begin{align}
& 4\dfrac{1}{8}=\dfrac{\left( 8\times 4 \right)+1}{8}=\dfrac{32+1}{8}=\dfrac{33}{8} \\
& 8\dfrac{1}{3}=\dfrac{\left( 3\times 8 \right)+1}{3}=\dfrac{24+1}{3}=\dfrac{25}{3} \\
\end{align}\]
Therefore we need to find $\dfrac{33}{8}-\dfrac{25}{3}$.
For subtraction, we need fraction to be like fraction i.e. we need denominator of both fraction same.
For this let us take LCM of 3 and 8 which is 24.
To convert $\dfrac{33}{8}$ having denominator as 24 we need to multiply the numerator and the denominator by 3, we get $\dfrac{33\times 3}{8\times 3}=\dfrac{99}{24}$.
Similarly to convert $\dfrac{25}{3}$ having denominator as 24, we need to multiply the numerator and the denominator by 8 we get $\dfrac{25\times 8}{3\times 8}=\dfrac{200}{24}$.
Therefore we need to find $\dfrac{99}{24}-\dfrac{200}{24}$.
It can be written as $\dfrac{99-200}{24}$.
Solving the numerator we get $\dfrac{-101}{24}$.
Hence our required answer is $\dfrac{-101}{24}$.
Note: Students should take care of signs while solving the subtraction. Make sure that the denominator is the same for performing subtraction or addition. Student can write mixed fraction in the following form also,
$4+\dfrac{1}{8}-\left( 8+\dfrac{1}{3} \right)\Rightarrow 4+\dfrac{1}{8}-8-\dfrac{1}{3}\Rightarrow -4+\dfrac{1}{8}-\dfrac{1}{3}$.
After that they could take LCM as 24 and solve it as,
$\dfrac{-4\times 24+1\times 3-1\times 8}{24}=\dfrac{-96+3-8}{24}=\dfrac{-101}{24}$.
Which gives us the same answer.
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