Question

Ravi starts for his school at 8:20 a.m. on his bicycle if he travels at a speed of 10 km per hour then he reaches School late by 8 minutes but on travelling by 16 km per hour he Reaches School 10 minutes earlier at what time the school starts. ​

Answer 1

SoluTion :-

Let the distance be x km.

Let the time taken be y.

Equation 1

$\sf {\frac{x}{10} =y+\frac{8}{60} }$

Equation 2

$\sf {\frac{x}{16} =y-\frac{10}{60} }$

Solving (1) and (2),

$\sf {\frac{x}{10} -\frac{x}{16} =\frac{8}{60} +\frac{10}{60}}\\\\\\\sf {\frac{3x}{80} =\frac{9}{30}}\\\\\\\sf {x=\frac{9 \times 80}{3 \times 30} }\\\\\\\sf {x=8 \ km}$

Now,

$\sf {\frac{8}{16} =y-\frac{1}{6}}\\\\\\\sf {y-\frac{1}{6} =\frac{8}{16} }\\\\\\\sf {y-\frac{1}{6} =\frac{1}{2}}\\\\\\\sf {y=\frac{1}{6} +\frac{1}{2} }\\\\\\\sf {y=\frac{2}{3} \ hours }\\\\\\\sf {Or}\\\\\\\sf {y=\frac{2}{3} \times 60 }\\\\\\\sf {y=40 \ minutes}$

School starts at 8:20 am + 40 minutes = 9:00 am

Answer 2

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