Question

a boy increases his speed to 9/5th times his original speed. by this he reaches his school 30 minutes before the usual time. how much time does he takes actually? ​

Answer 1

$\huge \bf \dag \red {given \: that} :$

increase his speed = 9/5

$\huge \bf \dag \pink{to \: find }:$

time

$\huge \bf \dag \green {answer }:$

let the usual speed is s and time is t.

according to the question

time difference = 30 min

t = 5/9 t = 30 min

4t = 30 × 9

$t = \frac{30 \times 9}{4}$

$\bf \orange{t = g7.50min}$

$\bf \purple{hence \: the \: actual \: time \: is \: }$

$\bf \blue{g7.50min}$