Question 01
Starting from his house one day, a student walks at a speed of 2 ½ kmph and reaches his school 6 minutes late. Next day he increases his speed by 1 kmph and reaches the school 6 minutes early. How far is the school from his house?
Given: Speed of student on day 1 = 2 ½ kmph
And late by = 6 minutes
Increased speed on next day = 2 ½ + 1 = 3 ½ kmph
And reaches early by = 6 minutes
To find: distance between school and house
Let the distance be x km
Total difference in timings = 6 min + 6 min = 12 minutes
= 12/ 60 hour = 1/5 hour
Time during late – time during early = 1/5 hour
Distance / speed L – distance/ speed E = 1/5 hour
x/ (5/2) – x/(7/2) = 1/5 hour
2x/5 – 2x/7 = 1/5 hour
(14x -10x )/ 35 = 1/5
14x – 10x = 35/5
4x = 7
x = 7/4 km
Or according to option 1 ¾ km
Ans. The school 1 ¾ km from house.