Question

4. Sanyam travels to his office by a car at a speed of 40 km/hr and reaches
office 9 minutes late. If he drives his car at a speed of 50 km/hr, he reaches
6 minutes early. What is the distance of his office from his home?​

Answer 1

$</p><p>let \: the \: time \: taken \: be \: t. \\ Distance{1} \: =speed1 × Time1 \\ = 40 km/h × (t + 9÷60) </p><p> \\ = 40 km/h × t + \frac{3}{20} \\ = 40 km/h × \frac{20t + 3}{20} \\ = 20(20t + 3) \\ \\ \ Distance{2} \: =speed2× Time2 \\ = 50 km/h × (t-6÷60) </p><p> \\ = 50 km/h × t - \frac{1}{10} \\ = 50 km/h × \frac{10t - 1}{10} \\ = 5(10t - 1) \\ \\ </p><p>distance1 = distance2 \\ 20(20t + 3) = 5(10t - 1) \\ 4(20t + 3) = (10t - 1) \\ 80t + 12 = 10t - 1 \\ 80t - 10t = - 12 - 1 \\ 70t = - 13 \\ t = \frac{ - 13}{70}$