Starting from his house one day, a student walks at a speed of $2\frac{1}{2}$ km/hour and reaches his school $6$ minutes late. Next day he increased his speed by $1$ km/hr and reaches the school $6$ minutes early. How far is the school from his house?

The correct option is A $1\frac{3}{4}$ km
Student's speed is $2.5$ km/hr.

Let time taken be $t$ hrs.

So, time taken on first day is $(t+0.1)$

Distance covered is $2.5(t+0.1)=2.5t+0.25$

Similarly, speed on the next day is $3.5$ km/hr.

Let time taken be $t$ hrs.

So, time taken on next day is $(t-0.1)$

Distance covered is $3.5(t-0.1)=3.5t-0.35$

Equating both distances,

$2.5t+0.25=3.5t-0.35$

Therefore, $t=0.6$

So, distance covered is $2.5\left(0.6\right)+025=1.75$ km.