Question

A man cycles to his office from his house at a speed of 5 km/h and reaches 10 minutes late. If he cycles at a speed of 6 km/h, he reaches 10 minutes early. What is the distance between the office and his house?

• A. 40 km
• B. 20 km
• C. 35 km
• D. 10 km

Answer 1

The correct option is D 10 km

Let the distance between the office and his house is x km and 't hr' be the normal time taken.

Cycling from House to office:

Speed = 5 km/h and time taken = $(t+\frac{10}{60})hour$

So, distance = $5(t+\frac{10}{60})km$ ... (1)

Cycling from Office to house:

Speed = 6 km/h and time taken = $(t-\frac{10}{60})hour$

So, distance = $6(t-\frac{10}{60})km$ ...(2)

From (1) and (2), we get

$5(t+\frac{10}{60})=6(t-\frac{10}{60})⟹5(t+\frac{1}{6})=6(t-\frac{1}{6})⟹5(6t+1)=6(6t-1)⟹30t+5=36t-6⟹30t-36t=-6-5⟹-6t=-11⟹t=\frac{11}{6}hour$

Put value of $t=\frac{11}{6}hour$ in (1), we get

Distance = $5(\frac{11}{6}+\frac{10}{60})=5(\frac{11}{6}+\frac{1}{6})=5\left(\frac{12}{6}\right)=10km$

Hence, the distance between his house and office is 10 km.