Aptitude - Time and Distance - Discussion

Here's Step by step explanation of the solution:

The Distance is equal for both --> D = 30 km,
Let the time taken by Sameer (T)= x.
So, time taken by Abhay = x+2 --> (2hrs more than sameer).

We know Distance and Time, so we can calculate speed i.e S=D/T.
Let the speed of Abhay (S) = 30/x+2-->[1]

Given that,
if Abhay doubles his speed --> new speed of Abhay = 2s.
He will take 1hr less than sameer --> now time taken by Abhay = x-1.
As S = D/T.

New speed of Abhay is calculated as -
2s = 30/x-1 -->[2]

Substitute the above 's' value of [1] in the '2s' value of [2]

2(30/x+2) = 30/x-1.
By solving, we get x=4 --> time taken by Sameer.

So, time taken by Abhay = x+2 = 4+2 = 6.
Now, speed of Abhay = 30/6 = 5kmph.

Thank you.

S*x = (x+2) * A -->(1).
S*x = (x-1) * 2A -->(2) {equate 1 and 2}
(x+2) = (2x-2).
x = 4.
A = 30/(x +2) => 30/6 = 5 kmph.

Speed => 1 : 2.
Time => 2 : 1 = 6 : 3.
So, Required Distance = 30/6 = 5 km/h (Ans.)

30/A-30/B = 2 ---> (1)
30/B-30/2A = 1 ---> (2)

By adding two equations
30/A - 30/2A = 3,
60 - 30/2A = 3,
30 = 6A.
A = 30/6.
A = 5km/h.

@Arun Kumar

Use 2 variables x and y, then it will be super easy.
Speed = distance/time.
Time taken by sameer = x,
Time taken by abhay = x + 2,
Speed of Abhay = y.
y = 30/x+2.
2y = 30/x -1.
as 30 i.e distance is constant,
xy + 2y = 2xy - 2y.
x = 4,
putting x = 4 in xy + 2y = 30,
y = 5.

I am not getting this. Anyone, please help me.

t is actual time (Let's take Sameer's speed as actual).

When Abhay travels normally;
30/x = t+2 , t = 30/x - 2 ----(> t)
when Abhay travels 2x speed.
30/2x = t-1.
30/2x +1 = t.
30/2x +1 = 30/x - 2.
30+2x/2x = 30 -2x/x.
30x+2x^2 = 60x - 4x^2.
6x^2 = 30x.
x = 30/6.
x = 5.

Abhay takes 2 hrs more than Sameer to cover 30 km.
Abhay's speed (s) = 30/t+2 ---> (1)

If Abhay doubles his speed he takes 1 hr less.

Abhay's double speed (2s) = 30/t-1 ---> (2)

Equate (1) and (2) we get,
2[30/t+2] = 30/t-1.
t = 4 hrs.
T = t+2 = 4+2 = 6 hrs.
S = 30/6 = 5 km/hr.

ta = x+2
ts = x
d = 30 km
Sa = 2Sa , ta = x-1.

Sa = 30/(x+2) --> xSa + 2Sa = 30.
2Sa = 30/(x-1) --> 2xSa - 2Sa = 30.

Since the distance is the same
xSa + 2Sa = 2xSa - 2Sa,
xSa = 4Sa.
x = 4,
ta = x+2,
ta = 6.
s = d/t = 30/6 = 5.

Let Abhay's speed=x, Sameer's time=y
Case 1: 30/x = 2+y.
Case 2: 30/2x = y-1.
30/x-30/2x = (2+y)-(y-1),
= 2+y-y+1,
= 3.
We get the equation:
30/x-30/2x = 3.

By solving we will get the answer.