Aptitude - Time and Distance - Discussion

Thanks everyone for explaining.

30/x - 30/2x = 3.

Here why should we subtract 30/x - 30/2x and how we will get 3?

Please can anyone explain?

Thanks for the given explanation @Sakshi.

How come 3 here?

Given:

d=30;
Sameer speed=s1;
Abay speed=s2;

Time taken for covering 30 km is =t;
Speed*time=distance
s1*t=30
s2*(t+2)=30 -----> 1
s2*(t-1)=30 -----> .2

Abay have a different speed.
when he travelled in speed 1 he needs two hours more to get the distance
when he travelled in speed 2 he needs 1 hour less then Sameer's speed to get the distance
So we need to find the difference of time for both speeds,
equation 1-2 provides the time different.
(30/s2)-(30/2s2) = (t+2)-(t-1),
(30/s2)-(30/2s2) = 3,
(15/s2) = 3,
s2=5.

I don't understand why we subtract 30/x-30/2x? Please explain me.

Thanks for the answer @Devi.

Thanks for the answer @Devi.

Case 1:
Abhay's speed = Y.
Sameer has taken time = T,
Then, Y = 30km/T + 2.

Case 2:
Abhay's speed = 2Y.
Abhay's taken time = T-1.
Then, 2Y=30km/T-1.

We put the value of Y in case 2.
Then , 2*30km/T+2 = 30km/T+1.
60T+60=30T+60
30T=120.
T=4.

Put the value of T in case 1.
Y=30km/4hrs + 2hrs.
= 30km/6hrs.
= 5 km/hrs.

Distance=30,
Time of sameer=s,
Time taken by abhay = 2hrs+s.

The Speed of Abhay= 30/(2hrs+s),
Time is taken by Abhay after doubling speed = s-1hr.
Speed of abhay= 60/(s-1hr),
We know that both are the speeds of Abhay, So both the speeds are equal.
30/(2hrs+s) = 60/(s-1hr),
30(s-1hr) = 60(2hr+s).
150hr = 30s,
s=5.

So, the final answer is 5.