Question

Abdul travels thrice the distance catherine travels, which is also twice the distance that binoy travels. catherine's speed is 1/3 of abdul's speed, which is also 1/2 of binoy's speed. if they start at the same time then who reaches first?

Answer 1

As t=distance/speed, Binoy has 1/2times lesser distance with 1/6 times greater speed regarding Catherine & Binoy has 1/6 times lesser distance with 1/2 times faster speed regarding Abdul hence Binoy will reach first on his destination.

Answer 2

Define x:

Let Binoy travels x km

⇒ Binoy = x km

Catherine travels twice the distance Binoy travels:

⇒ Catherine = 2x km

Abdul travels thrice the distance Catherine travels:

⇒ Abdul = 3(2x) = 6x km

Define y:

Let Catherine's speed be y

Catherine = y km/h

Catherine's speed is 1/3 of Abdul speed

Abdul = 3y km/h

Catherine's speed is 1/2 of Binoy's speed

Binoy = 2y km/h

Find the time needed:

Time = Distance ÷ Speed

Bionoy = x ÷ 2y = x/2y hour

Catherine = 2x ÷ y = 2x/y = 4x/2y hour

Abdul = 6x ÷ 2y = 6x/2y  hour

Compare:

6x/2y > 4x/2y > x/2y

⇒ Binoy is the fastest to reach because he needs the least number of hours.

Answer: Binoy will reach first