Correct option is (c) 1.75 km
Let the distance between the home & the school of the boy = x km.
In the first case, the speed is \(2\frac 12 \) km/hr = \(\frac 52\) km/hr.
∴ Time taken = x ÷ \(\frac 52\) hr.
\(= \frac{2x}5\) hr.
In the second case his speed = 1 + \(\frac 32\) km/hr = \(\frac 72\) km/hr
∴ Time taken = x ÷ \(\frac 72\) hr. = \( \frac{2x}7\) hr.
∴ The difference in time = \(\left(\frac{2x}5 - \frac{2x}7\right)\)hr.
\(= \frac{4x}{35} \)hr
\(= \frac{4x}{35} \times 60\) min
\(= \frac{48x}7 \) min
Now , in the first case he is 6 min late = −6 min. and
In the second case, he is 6 min early = +6 min.
∴ The difference in time= [6 − ( − 6)]min. = 12 min.
∴ \(\frac{48x}7 \) min = 12 min
Or 4x = 7
Or x = \(\frac 74\) = 1.75.
∴ The distance between the home & the school of the boy = 1.75 km.
