Question

A man travel at 10 km/hr and reaches his house 10 minutes late. If he travels at the speed of 20 km/hr he reaches his house 15 minutes early. Find the original speed of man by which he should travel to his house on time

Answer 1

Answer:

12.5 km/hr.

Step-by-step explanation:

Let the distance to his home be d.

And the original time be t.

• In first case time is (t+10)mins when speed is 10km/hr.

d= 10×(x+10)

• In the second case time is (t-15)

d= 20×(x-15)

Equating, we get x= 40 mins.

d= 10×(40+10)/60

d= 50/6 km.

s= d/t. s= (50/6)/(40/60)

s=50/4= 12.5 km/hr.

Answer 2

Answer:

12.5 kmph

Step-by-step explanation:

Let distance be d kms

Original time be t hrs

case 1: 10kmph and 10mins late i.e.,

$t + \frac{10}{60}$

case 2: 20kmph and 15mins early

$t - \frac{15}{60}$

distance = speed * time

case 1 :

$d = 10 \times (t + \frac{10}{60} )$

case 2 :

$d = 20 \times (t - \frac{15}{60} )$

equating both, we get

t = 2/3 hrs = 40 mins

and substituting t in any of the above equations give d = 50/6 kms

Now calculate speed using t and d values,

speed = d/t

$= ( \frac{50}{6} ) \div ( \frac{2}{3}) \\ = ( \frac{50}{6} ) \times ( \frac{3}{2}) \\ = \frac{25}{2}$

= 12.5 kmph