Question

Fill in the boxes.

$5\frac{3}{4}-\square =3\frac{5}{12}$

Answer 1

Step 1:

Given $5\frac{3}{4}-\square =3\frac{5}{12}$

Lets convert to improper fraction

Consider

$$\begin{gathered} 5\frac{3}{4} \\ =\frac{5\times 4+3}{4} \\ =\frac{20+3}{4} \\ =\frac{23}{4} \\ 3\frac{5}{12} \\ =\frac{3\times 12+5}{12} \\ =\frac{36+5}{12} \\ =\frac{41}{12} \end{gathered}$$

Hence, the improper fraction of $5\frac{3}{4}=\frac{23}{4}$ and $3\frac{5}{12}=\frac{41}{12}$

Step 2:

Let the fraction be $x$

Consider

$$\begin{gathered} \frac{23}{4}-x=\frac{41}{12} \\ \frac{41}{12}+\frac{23}{4}=x \\ \Rightarrow x=\frac{41\times 4}{12\times 4}+\frac{23\times 12}{4\times 12} \\ x=\frac{164}{48}+\frac{276}{48} \\ x=\frac{164+276}{48} \\ x=\frac{440}{48} \\ x=\frac{110}{12} \\ x=\frac{55}{6} \end{gathered}$$

Hence, the required fraction is $\frac{55}{6}$

To find the mixed fraction of $\frac{55}{6}$

Consider $\frac{55}{6}$

$695554-1$

Here remainder$=1$ and quotient$=9$

$$\begin{gathered} Mixedfraction=Quotient\frac{Remainder}{Divisor} \\ =9\frac{1}{6} \end{gathered}$$

Hence, Required fraction is $9\frac{1}{6}$