Aptitude - Time and Distance - Discussion

Thank you @Devi.

Thanks @Devi.

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?

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.

Thanks @Devi.

Thanks for your explanation @Devi.

Thanks @Devi.

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

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

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.

Distance=30km.

Let the speeds of abhay be x kmph and sameer be y kmph.
then,
30/x - 30/y =2----i,
When abhay doubles his speed
30/y - 30/2x = 1-----ii,
By i+ii we get,
30/x - 30/2x=3,
6x=30,
x=5km/hr.