Question

A man travel to a car park at an average speed of $40$ km/h, then he walks to his office at an average speed of $6$ km/h. The total time for journey is $25$ minutes. One day his car breaks down and he walked three times as far thereby arriving at his office $17$ minutes late. How far is the office?

• A. $10$ km
• B. $11$ km
• C. $12$ km
• D. $15$ km
• E. None of these

Answer 1

The correct option is B $11$ km

Total time take for Journey = 25 minutes. = $\frac{25}{60}$ hours.

Time taken by Car = $\frac{X}{40}$ kmph.

Time Taken by Walk = $\frac{Y}{6}$kmph.

$\therefore \frac{X}{40}+\frac{Y}{6}=\frac{25}{60}$

$3X+20Y=50-\left(1\right)$

One day when car breaks down, he walks $3$ times as he walks before,

$=3Y.$

So, Distance travelled by Car $=X-2Y.$
Now time taken $=25+17=42$ minutes = $\frac{42}{60}$ hours.

$\therefore \frac{(X-2Y)}{40}+\frac{3Y}{6}=\frac{42}{60}$.

$3X+54Y=84-\left(2\right)$

Equation (1) - Equation (2)

$3X+20Y-3X-54Y=50-84.$

$Y=1km.$

Putting the value of $Y$ in equation $1,$ we get,

$X=10Km.$
So, total distance $=X+Y=10+1=11km$.