Question

a student goes to school at the rate if 2 1/2 km/hr reaches 6 min late.
If he travels at the speed of 3 km/hr , he is 10 min early. what is the distance to school ?

Answer 1

Hi deepika !

Here is a shortcut to find the distance .
This formula can be applied when , the speed of a particular thing, person or vehicle varies with time , and distance is asked.

speed  :- 2 1/2 = 2.5 km/hr
6 min = 0.1 hour
10 min = 0.16 hour


distance = time difference*product of speeds / difference of speed

time taken = 6 min + 10 min = 16 min = 0.267 hour
product of speed = 2.5*3 = 7.5
difference of speed = 3 - 2.5 = 0.5



applying the formula ,


distance = 0.267*7.5/0.5
                = 2.0025 /0.5
                = 4.005 km
                = 4 km (approx.)

Answer 2

Hiya !

Q - A student goes to school at the rate if 2 1/2 km/hr reaches 6 min late.
If he travels at the speed of 3 km/hr , he is 10 min early. what is the distance to school ?

Answer -
when student travel with 2.5 km/hr speed he wil be late 

Speed of student  v1 = 2 and half km/h = 2.5 km/hr = 0.69 m/s 

t1 = 6 minutes = 360 seconds 

when student travel with 3 km/h speed he will be reach early 

v2 = 3 km/hr = 0.83 m/s 

t2 = 10 minutes = 600 seconds 

distance = speed × time 

let the distance be  " x " 

let the time be " t ''

in first case 

x =  0.69 ( t + 360) ----1

in second case 

x = 0.83 ( t - 600 ) ----2 

by 1 and 2 we got 

0.69 ( t + 360) = 0.83 ( t - 600 )

0.69 t + 248.4 = 0.83t - 498 

248.4 + 498    = 0.83t - 0.69t 

746.4              = 0.14t 

 t  = 746.4/0.14

t = 5331.4 seconds 

put its value in first equation 

we got 

x  = 0.69 ( 5331.4 + 360) 

x = 0.69 x 5691 .42

x = 3927.08 m or 3.927 km