Aptitude - Time and Distance - Discussion

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.

Take one option 5hrs if Abhay goes with 5km/hr then it will take 6 hours to reach means 4 hours for Sameer. If he doubles his speed means 30/10 = 3kmph and as usual Sameer speed is 4kmph so the right answer is A.

Thanks, @Prakash. Understood the common factor.

Thanks @Devi. Superb explanation.