Aptitude - Time and Distance - Discussion

Good explanation Devi.

Another Method.

Let Abhay's speed :- x kmph
Let Sameer's speed:- y kmph

According to the question

30/x - 30/y = 2 -- 1st equation

&

30/y - 30/2x = 1 -- 2nd equation

Solving above two we get:-

(y-x)/xy = 1/15 --3rd equation

&

(2x -y)/xy = 1/15 -- 4th equation

since the rhs of both the equations is equal equating the lhs of the both equations is also correct: so

y -x = 2x - y

i.e

2y = 3x or y = 3x/2

now equate the value of y in 3rd or the 4th equation as follows:-

(2x - (3/2)x)/(x*(3/2)x) = 1/15

we get

x/3x^2 = 1/15

so

1/x = 1/15

so

x = 5 km/hr

Hope this helps.

Thanks Devi

Thanks Abdus :).

Lets abhays speed x km/hr and sameer takes T hour to complete 30 km journey
X=30/T+2

2X= 30/T-1

THEN 2(30/T+2)=30/T-1
2T-2=T+2
T=4 hour
X= 30/T+2
30/4+2= 5 HOURS

Superb explanation devi, anoo, amit sakshi.

Please explain how 3 comes shortly?

thanks

Good Explanation Devi.. Thanx.. :)

Good explanation by rahul and amith.

Reason for subtracting: Then only you're getting the difference in time of both.