Starting from his house one day, a student walks at a speed of 5/2 km/hr and reaches his school 6 minutes late. next day he increases his speed by 1 km/hr and reaches the school 6 minutes early. how far is the school from his house?
Answers
Answered by
33
let the distance of the school = x km
time to cover the distance from his house = t mnts
for first day speed= d/t= x/(t+6) implies
5/2=x/((t+6)/60)..... (1)
next day he increases his speed by 1 km/hr
so speed in the nxt day = 5/2+1 km/hr
= 7/2 km/hr
so for second day .....7/2=x/((t-6)/60)..... (2)
from (1) x= 5/2*(t+6)/60
frm (2) x= 7/2*(t-6)/60
equating them ...5/2*(t+6)= 7/2*(t-6)
or,t= 36
so from (1)... x= 5/2*(36+6)/60= 7/4 km
time to cover the distance from his house = t mnts
for first day speed= d/t= x/(t+6) implies
5/2=x/((t+6)/60)..... (1)
next day he increases his speed by 1 km/hr
so speed in the nxt day = 5/2+1 km/hr
= 7/2 km/hr
so for second day .....7/2=x/((t-6)/60)..... (2)
from (1) x= 5/2*(t+6)/60
frm (2) x= 7/2*(t-6)/60
equating them ...5/2*(t+6)= 7/2*(t-6)
or,t= 36
so from (1)... x= 5/2*(36+6)/60= 7/4 km
Answered by
27
Answer:
Step-by-step explanation:
Difference of timing= 6+6=12min
Which is 12/60=1/5 hr
Initial speed is 5/2,so time must be 2/5
Same as after increase of 1 km/h it become 7/2
So in 2nd case time become 2/7
Now equation become 2x/5-2x/7=1/5
Solving this x=7/4
Similar questions