A man riding a bicycle from his house at 10 km/h and reaches his office late by 6 min. He increases his speed by 2 km/h and reaches 6 min before. How far is the office from his house?

Time = \( \Large \frac{Distance}{Speed} \)

As per first condition, \( \Large \frac{d}{10} = t + \frac{6}{60} = d = 10t + 1 \) ......(i)

As per second condition, \( \Large \frac{d}{12}=t-\frac{6}{60}= 12t-\frac{6}{5} \) .... (ii)

From Eqs. (i) and (ii), we have

\( \Large 10t + 1 = 12t - \frac{6}{5} \)

\( \Large 2t = 1 + \frac{6}{5} = \frac{11}{5} = t = \frac{11}{10} \)

Therefore, From E (i), \( \Large d = 10 \times \frac{11}{10}+1 \)

d = 11 + 1 = 12

d = 12 km