Aptitude - Time and Distance - Discussion

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.

In simple words:

The time taken to cover the distance of 30 km is always the same in both conditions.
let the speed of Abhay be x.

Now the time taken for 1st condition =time taken for 2nd condition.
Therefore,(30/x)+2 = (30/2x)-1.
(30/x)-(30/2x) = 3.
x = 5.