Aptitude - Time and Distance - Discussion

Then what is the speed of Sameer?

If suppose x speed abay cover in y+2 time.

Then 2x speed he covered in y-1. (here why is sameer to cover time).

Then different between time = (y+2) - (y-1) =3. So.

First covered time - second covered time = 3.

Distance = 30.

Abhay's normal speed = x.
Abhay doubles speed = 2x.
Sameer's time = t.

t+2 = 2 hrs + sameer's time.

t-1 = 1 hr less than sameer's time.

[30/x] - [30/2x] = [t+2] - [t-1].
[30/x] - [30/2x] = 3.

[1/x] - [1/2x] = 1/10.
(2-1) /2x = 1/10.

1/2x = 1/10.
2x = 10.

x = 5 km/hr.

Can any one use proper formula of maths to solve this question because it is all confusing?

Please explanation clearly.

Hey just stick with anyone solution then you will definitely sure about the solution, don't go for each and every solution. Because it will confuse you. And yes nice explanation by @Devi :).

How 3 comes?

Guys please solve this question by answer don't get confused.

Example: 1st option is 5 km/hr if we take then Abhaya speed 30/5 = 6 hr then Sameer has time 4 hr means Abay has 2 hr more then Sameer after then speed double means 10 km/hr, Abhay takes time 30/10 = 3 hr mean 1 hr less then Sameer. So 1st option is correct answer.

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?