Aptitude - Time and Distance - Discussion

A person covers a particular distance with a speed x in time t1 and in his second trip if he covers the same distance with speed y at time t2. Shortcut formula:

Distance covered= time gap X product of speed/difference in speed.

In this case: (t1+t2)X(xy)/(y-x).

Distance = 30.
Abhays normal speed = x.
Abhay doubles speed = 2x.
Sameers time = t.
t + 2 = 2 hrs + sameers time.
t - 1 = 1 hr less than sameers time.

[30/x] - [30/2x] = [t+2] - [t-1].
[30/x] - [30/2x] = 3.
[1/x] - [1/2x] = 1/10.
(2-1)/2x = 1/10.
1/2x = 1/10.
2x = 10.
x = 5 km/hr.

Let speed of Abhay's be X kmph.

And time taken by Sameer be Y.

So time taken by Abhay's is Y+2.
So 30/X = Y+2.

Again speed doubles means 2X.
So 30/2X = Y-1.

By subtraction of two equation we get.

30/X - 30/2X = (Y+2) - (Y-1) = 3.

=> X=5.

> LET THE TIME TAKEN BY SAMEER = T.
> LET THE SPEED OF ABHAY = X.

TIME = DIST/SPEED.

>Abhay takes 2 hours more than,
Sameer TO COVER 30 KM = T+2 = 30/X.

>Abhay doubles his speed, then he Would take 1 hour less,
Than Sameer TO COVER 30 KM = T-1 = 30/2X.

NOW CROSS MULTIPLY THE BELO EQUATIONS:
T+2 = 30/X.
T-1 = 30/2X.
(T+2)30/2X = (T-1)30/X.
(2TX+4X+30)/2X = (TX-X+30)/X.
(2TX+4X+30)/2 = (TX-X+30).

2TX+4X+30 = 2TX-2X+60.
2TX+4X+30-2TX+2X-60=0.
6X-30=0 => 6X=30 .

Ans) X = 5KMPH.

How 3 comes is the confusion to many of us.

Say, in 1st case- Sameer reaches at 10:00 Hrs, so Abhay must reach at 12:00 Hrs.

In 2nd case-Sameer reaches at 10:00 Hrs but Abhay reaches at 09:00 Hrs.

So, Abhay's time difference is 3 hrs due to change of speed.

Lets take distance = 30.

Speed=S;.
Time=T;.

D = S*T;.
30 = S*T;.

HERE,

30 = S*(T+2).
30 = ST+2S------------------1.

Next equation,

30 = 2S*(T-1).
30 = 2ST-2S-------------------2.

TO SOLVE 1 AND 2 EQUATION.

60 = 2ST+4S-------------------1.
30 = 2ST-2S-------------------2 (-).
----------------------------------.
30 = 0+6S.
-----------------------------------.

THEREFORE,

6S = 30.
S = 5.

Ans = 5.

Best one is T+2 T-1.

Hello I'm not getting from were do that "3" comes in second line. Please explain.

Let the speed be x, we will move with Abhay only at the outset.

A/Q D=S*T or, 30 = (t+2)*x or, 30/x = t+2.

If he doubles his speed then,

30 = (t-1)*2x or, 30/2x = t-1.

REMEMBER t is the time taken by Sameer.

D/S = T => 30/x-30/2x = 3 => x = 5.

Let Sammeer original time = x.

Abhay speed: y = 30/x+2 (let Abhay speed be y, x+2 is written because Abhay takes 2 hrs more than Sammer). This equation can be derived when you analyses the first statement alone.

So, let y = 30/x+2....> (1).

Case 2:

Abhay doubles his speed.

2y = 30/x-1(x-1 is because Abhay takes 1 hr less than Sammer).

So, 2y = 30/x-1....> (2).

From (2) , get the value of x and substitute in (1) , you will get the value of why which is Abhay speed.

2y = 30/x-1.
2y(x-1) = 30.
x-1 = 30/2y.

So, x = (30/2y)+1.

Substitute the x value in (1).

y = 30/x+2.
y =30/((30/2y)+1)+2.
y = 5.