Aptitude - Time and Distance - Discussion

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.

(30/Sa-30/Ss) = 2.
(30/2Sa-30/Ss) = -1.

30/Sa - 2 = 30/2Sa +1.
Sa = 5.

Thanks @Shanti.

Thanks @Devi.

Superb and useful explanation, thanks @Devi.

S1= s = d/t = d/t+2.
S2 = 2s = d/t-1.
S/2s = 1/t+2/1/t-1.
1/2 = t-1/t+2.
t+2 = 2t-2.
t = 4.
S = D/T = 30/6 = 5.

One more method.

Let Sameer's speed is y kmph and time taken by Sameer is t.
So Abhay takes t+2 time more than summer's time at y kmph.

Then,
Abhay doubles his speed so the speed is 2y so the time taken by Sameer is t-1hr less than Sameer.

Speed = dis/time formula when abhay has normal speed.
y = 30km / t+2 ............... (1).
Speed = dis / time formula when abhay doubles his speed.
2y = 30km / t-1 .................(2).

Put the value of y from 1st equation into 2nd equaton.
2(30/t+2) = 30/t-1.
2/t+2 = 1/t-1.
2t-2 = t+2.
2t-t = 2+2.
t = 4 ................... (3).

Put the value of t in eq. 2.
2y = 30/ 4-1.
2y = 30/3.
2y = 10.
y = 5 kmph.

You can put the value of t in eq.1 also you will get the same answer 5kmph.