Aptitude - Time and Distance - Discussion

(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.

Superb explanation, Thanks @Devi.