A fidget day of a business man

Algebra Level 3

On a usual day, a man goes home from work by catching a train from Station A to Station B, then walks 10 minutes to go home.

Today, on his way back, the train stopped at Station K, located between A and B because of an accident ahead. He got down to the Station K and caught a taxi for home. The taxi took 30 minutes to arrived at his home. He looked at the clock and found that he was 5 minutes later than the usual day.

Question: What is the distance (in km) between Station K and his home?

Details and Assumptions:

  • The distance between K and his home is 3 times that of the distance between A and K.
  • The train ran faster 2 times as fast as the taxi.
  • The distance between A and B is 40 km.


The answer is 30.

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1 solution

Chew-Seong Cheong
Oct 26, 2018

Let the distance between K and the man's home be x x km. Then that between A and K is x 3 \dfrac x3 km. Let the speed of the train by v v . Then that of the taxi is v 2 \dfrac v2 .

The time taken for the man to reach home from work is t usual = 40 v train + 10 walk t_{\text{usual}} = \underbrace{\dfrac {40}v}_{\text{train}} + \underbrace{10}_{\text{walk}} minutes. Since today, he was 5 minutes late, t today = 40 v + 15 t_{\text{today}} = \dfrac {40}v + 15 . This time is the same as t today = x / 3 v A-K + x v / 2 taxi t_{\text{today}} = \underbrace{\dfrac {x/3}v}_{\text{A-K}} + \underbrace{\dfrac x{v/2}}_{\text{taxi}} . Therefore,

40 v + 15 = x 3 v + 2 x v Since taxi ride is 2 x v = 30 minutes 40 v + 15 = 5 + 30 v = 2 2 x v = 30 x = 30 \begin{aligned} \frac {40}v + 15 & = \frac x{3v} + \color{#3D99F6} \frac {2x}v & \small \color{#3D99F6} \text{Since taxi ride is }\frac {2x}v = 30 \text{ minutes} \\ \frac {40}v + 15 & = 5 + \color{#3D99F6} 30 \\ \implies v & = 2 \\ \frac {2x}v & = 30 \\ \implies x & = \boxed {30} \end{aligned}

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