Question

Insert < , > , =
$\begin{array}{cc}& i)\frac{1}{4}+\frac{1}{7}◯\frac{1}{12}+\frac{5}{6}\\ & ii)\frac{5}{6}-\frac{1}{6}◯\frac{8}{9}-\frac{2}{3}\\ & iii)6\frac{2}{3}-2\frac{1}{3}◯4\\ & iv)7\frac{1}{12}+3\frac{1}{3}◯11\end{array}$

Answer 1

$1)LHS=\frac{1}{4}+\frac{1}{7}=\frac{11}{28}$

$RHS=\frac{1}{12}+\frac{5}{6}=\frac{11}{12}$

$LHS<RHS⟹\frac{1}{4}+\frac{1}{7}<\frac{1}{12}+\frac{5}{6}$

$2)LHS=\frac{5}{6}-\frac{1}{6}=\frac{2}{3}=\frac{6}{9},RHS=\frac{8}{9}-\frac{2}{3}=\frac{2}{9}$

$LHS>RHS⟹\frac{5}{6}-\frac{1}{6}>\frac{8}{9}-\frac{2}{3}$

$3)6\frac{2}{3}-2\frac{1}{3}=6-2+\frac{1}{3}=4+\frac{1}{3}>4$

$4)7\frac{1}{12}+3\frac{1}{3}=\frac{85}{12}+\frac{10}{3}=\frac{125}{12}=10+\frac{5}{12}<11$