Question

please help me guys. .​

Answer 1

$Let \: \: the \: \: distance \: \: be \: \: d.$

Then,

${ \boxed{ \blue{ \boxed{(t_{1} )\: i.e \: \: time \: \: taken \: while \: going}}}} \\ { \boxed{ \blue{ \boxed{to \: school \: = > \frac{d}{20}}}}} \\ \\ \\ \\ { \boxed{ \purple{ \boxed{ (t_{2}) \: i.e \: \: time \: taken \: while \:coming}}}} \\{ \boxed{ \purple{ \boxed{ from \: school= \frac{d}{30} }}}}$

Therefore,

${ \large{Total \: \: time \: \: taken\: \: in \: \: trip }} \\ \\ = > \frac{d}{20} + \frac{d}{30} \\ \\ = > \frac{3d + 2d}{60} = \frac{5d}{60} = \frac{d}{12}$

Now,

$V_{av} \: for \: \: whole \: \: journey = > \frac{total \: distace}{total \: time \: taken} \\ \\ = > \frac{d + d}{ \frac{d}{12}}= \frac{2d}{ \frac{d}{12} } \\ \\ = > \frac{2d \times 12}{d} ={ \red{ 24 \: {kmh}^{ - 1} }}$