**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.**