Aptitude - Time and Distance - Discussion

@All.

Simply, the solution is;

Abhay's doubled speed is = to the speed required to cover 30km in 3 hrs.
2x=30/3,
x= 10/2,
x= 5 km/hr,

Another way to solve is;
(30/x)-(30/2x) = 3.
30((1/x)-(1/2x)) = 3.
1/2x = 1/10.
2x = 10.
x = 5 km/hr.

Thanks @Devi.

Thanks all.

Well explanation, Thanks @Devi.

Superb explanation, Thanks @Devi.

Thanks @Devi. Superb explanation.

Let the speed of Abhay be X kmph.
Let the Time taken by Sameer to cover a distance of 30 km be T hr.
According to problem;

Abhay takes 2hr more than Sameer i.e, (T+2)
Speed =Distance/Time .
Abhay's speed = total distance/ time taken by Abhay.

X=30/(T+2) --------> eqn1.

Now,
If Abhay doubles his speed i.e, 2X he would take 1 hr less than Sameer i.e (T-1)
2X=30/(T-1)--------> eqn1.

Now divide eqn1/ eqn2.
You will get T = 4hr.
Now put T=4 in eqn1. You will get X = 5kmph.

Thanks, @Prakash. Understood the common factor.

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.

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.