Aptitude - Time and Distance - Discussion

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.

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?

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.

30/x-30/2x=3.
2*30/2*x-30/2x=3.
60/2x-30/2x=3.
30/2x=3
30=3*2x
30=6x,
x=30/6,
x=5kmph.

Case 1: Abhay's speed=x.
Abhays time =t+2,
velocity(x)=30/t+2,

Case 2: abhays speed=2x.
Abhays time =t-1,
velocity doubled 2x=30/t-1.

now substitute the value of x in case 2.
t=4 substitute this value in any of the equations above.
x=5 is the ans.

Distance=30km/h0r.
time taken by Sameer is t,
Abhay travelling with a Speed X.

Case 1:
time taken by Abhay to cover 30km is 30/X = t+2.

Case 2(doubles speed):
time taken by Abhay to cover 30km is 30/2X = t-1.

(30/X)- (30/2X) = (t+2)-(t-1),
(30/X) - (30/2X) = t+2-t+1,
(30*2X) - (30*X) = 3 (2X^2) //LCM,
60X - 30X = 6 X^2,
30X = 6 X^2,
30 = 6 X,
X = 5.