Aptitude - Time and Distance - Discussion

We know that:

D = ST.

Case 1 : Let's x be the time taken by Sameer.

So, 30 = s(x+2) ----- 1.

Case 2: If Abhay doubles the speed, 1 hr less than Sameer.

So 30 = 2s(x-1) ----- 2.

Equate 1 and 2.

We can get x+2 = 2x-1.

x = 4.

Put x = 4 in any of the equation.

So we will get s = 5 km/hr.

Please I'm not getting how that 3 came please explain anyone?

Please, anyone explain from where 3 came?

Let us assume first that the velocity of Abhay is V km/hr and say Sameer takes x hours to cover the distance 30km.

Now, we all know S = V*t
=> 30 = V*(x + 2) ; where (x + 2) is the time taken by Abhay to cover the distance 30 km with speed Vkm/hr.

For the second case, i.e, for V' = 2*V, the double speed of Abhay

The above relation changes to:
30 = V'*t = (2*V)*(x-1); where (x-1) is the time taken by Abhay to cover 30km.

Now equating both equations :
=> V*(x+2) = 2*V*(x - 1).
=> x + 2 = 2*x - 2
=> x = 4
Putting x=4 in any of the above two equations we get the velocity V=5 km/hr.

I still not understand how 3 came? Explain in a simple manner please.

Let the time is taken by Sameer to cover 30 km be t. Abhay will take t+2 time to cover the same. If he doubles speed he takes t-1 time.

We know time = distance/speed.

So t+2 = 30/x. Where x is the speed of Abhay? t-1 = 30/2x. Now subtract both sides you can get the solution for x.

@Devi.

Superb and clear explanation, Thanks.

Thank you, very clear solution @Devi.

Why can the two equations not be just divided, you'll be able to cancel all unwanted constants and find time directly, but it gives a different answer.

30/x = t + 2
30/2x = t - 1

Dividing both (30/x) / (30/2x) = (t + 2)/(t - 1).
t = 4.

Good explanation Thank you, @Devi.