Aptitude - Time and Distance - Discussion

Distance=30km/hr
time taken by Sameer is t
Abhay travelling with a Speed X

Case 1:
time taken by Abhay to cover 30km is 30/X = t+2

Case 2(doubles speed):
time taken by Abhay to cover 30km is 30/2X = t-1

(30/X)- (30/2X) = (t+2)-(t-1)
(30/X) - (30/2X) = t+2-t+1
(30*2X) - (30*X) = 3 (2X^2) //LCM
60X - 30X = 6 X^2
30X = 6 X^2
30 = 6 X
X = 5

How come 3 here? Please explain.

Let the speed of Abhay be X kmph.
Let the Time taken by Sameer to cover a distance of 30 km be T hr.
According to problem;

Abhay takes 2hr more than Sameer i.e, (T+2)
Speed =Distance/Time .
Abhay's speed = total distance/ time taken by Abhay.

X=30/(T+2) --------> eqn1.

Now,
If Abhay doubles his speed i.e, 2X he would take 1 hr less than Sameer i.e (T-1)
2X=30/(T-1)--------> eqn1.

Now divide eqn1/ eqn2.
You will get T = 4hr.
Now put T=4 in eqn1. You will get X = 5kmph.

By applying direct formulation speed - distance /time.
30/3 * 2,
= 5.

How that 6x came. Please can You explain. I Don't know That method.

I cant understand how it gets value 3 can any one make me clear to understand this please.

Anyone please explain in detail this problem.

Thanx DEVI you are great.

@thiru.g.

Thank you.

Thanks devi:).