Question

Put $<\text{or}>$in box

$\frac{14}{5}÷9\left(\right)\frac{16}{7}÷8$

Answer 1

Step 1 : Solve LHS

$\frac{14}{5}÷9=\frac{14}{5}\times \frac{1}{9}=\frac{14}{45}$

Step 2 : Solve RHS

$\frac{16}{7}÷8=\frac{16}{7}\times \frac{1}{8}=\frac{16}{56}$

and $\frac{16}{56}=\frac{8\times 2}{8\times 7}=\frac{2}{7}$

Step 3 : On comparing LHS and RHS, we get

LCM of $(45,7)=315$

$\frac{14}{45}=\frac{14\times 7}{45\times 7}=\frac{98}{315}$

And

$\frac{2}{7}=\frac{2\times 45}{7\times 45}=\frac{90}{315}$

Here, $98>90$

LHS $>$RHS

Hence, $\frac{14}{5}÷9>\frac{16}{7}÷8$