a boy covers a certain distance between his house and school on bicycle. having an average speed of 20 km/hr, he is late by 15 minuets. However with a speed of 25 km/hr l, he reaches his school 5 min earlier. Find the distance between his sachool ane home.
Answers
Answer:
33.33 km
Explanation:
Let the distance between the boy’s house and the school = x
Let the speed needed to travel = s
Let the time taken to travel = t
Case 1:
Speed s1 = 20km/hr
Time t1 = t + 15 minutes
i.e., t1 = t + (15/60) hr
[Since it has to be calculated at a speed measured in km/hr]
so,
t1 = t + (1/4)
Now,
Distance = Speed x Time
Distance (x) = s1 x t1
= 20 x [t + (1/4)]
Case 2:
Speed s2 = 25km/hr
Time t2 = t - 5 minutes
i.e., t2 = t + (5/60) hr
[Since it has to be calculated at a speed measured in km/hr]
so,
t2 = t - (1/12)
Now,
Distance = Speed x Time
Distance (x) = s2 x t2
= 25 x [t - (1/12)]
The speed and time to reach may vary but the distance (x) is constant.
The distance is the same in both the cases
20 x [t + (1/4)] = 25 x [t - (1/12)]
[ 20 x (4t + 1) ]/4 = [25 x (12t - 1)]/12
4 x (4t + 1) = [5 x (12t - 1)]/3
12 x (4t + 1) = 5 x (12t - 1)
48t + 12 = 60t - 5
60t - 48t = 12 + 5
12t = 17
t = 17/12
=> t1 = t + (1/4) = (17/12) + (1/4) = 5/3
Also, t2 = t - (1/12) = (17/12) - (1/12) = 4/3
Distance = Speed X Time
x = s1 x t1
= 20 x (5/3)
= 100/3
= 33.33 km
Therefore, the distance from the boy’s house to the school = 33.33 km