Aptitude - Time and Distance - Discussion

Speed = distance/time.
Let x be Abhay's speed and y be Sameer's time.

* first condition
speed s = 30/(y+2) ; ------- a.
speed 2s=30/(y-1); ---------b.

Put a ((s=30/(y+2)) in b.
Now you get y=4.

put y=4 in either a or b.

You will get speed as 5kmph.

You can use this formula for this type of questions d=(s1*s2)/s1-s2*t.

Here d:distance(30) ,t: time difference( 3).

Thanks @Devi.

Thanks for your explanation @Devi.

Thanks @Devi.

In this problem we need to calculate two different 'Time values' of Abhay only(consider it like two different journeys of a single person)& Sameer is to get clue or value comparison.

Let's consider the time of Abhay and Sameer be x and y.
So,In Time(1) x = y+2hr
In Time(2) x= y-1hr (after speed doubles).

Now simply calculate the problem.
T1 - T2 = x
D/S - D/S = x

(y+2) - ( y-1) = 3

30/x - 30/2x = 3,
30 = 6x,
X=5.

Let total time saved by Abhay is ( 2+1=3).

Since, in first part of the question, Abhay takes 2 hours more
In the second part of the question Sameer takes 1 hour more, it means Abhay takes 1 hour less than Sameer.
When coming to remaining part of the question,
total distance= 30km
Let Abhay speed= x km/hr
In the second part, he doubles his speed= 2x km/hr.
by using formula t= D/S.
30/x - 30/2x = 3.
Here why we subtracting means the relative speed of the Abhay?

Thanks @Devi.

Thank you @Devi.

Let time taken by Sameer = T hrs.

Before doubling speed, time taken by Abhay = T+2 hrs --> (1).

After doubling the speed time taken = T-1 hrs --> (2).

(1) - (2) = (T+2) - (T-1) = 3 hrs.