Aptitude - Time and Distance - Discussion

Anyone find the original speed of the above question?

@Sivaram.

Is it the correct way and suitable to all this type of questions?

Let distance = D, speed = S, time taken = T (formula, we know S*T=D).

Let's assume the man covered the distance twice(2D), one D with S+3 kmph and another D with S-2 kmph.

So, D + D = (S + 3) (T - 40) + (S - 2) (T + 40).
2D = ST + 3T - 40S - 120 + ST - 2T + 40S - 80.
=> 2D = 2ST + T - 200.
But ST = D,
=> T = 200 min.

Now with S + 3 speed, time taken = 200 - 40 = 160.
and with S - 2 speed, time was taken = 200 + 40.
= 240.
But we have a problem in our earlier format, solve it. You will get D, distance = 40 km.

Please explain the way way of getting the value of y =12.

@Sivaram,

Here 3 is kmhr, but 40 is minutes.

Our requirement is Distance in KM.

So Please explain clearly?

Can anyone tell me the shortcut method to solve it?

Let distance = x km and usual rate = y kmph.

Then,

x-x = 40

2y (y+3) = 9x.....(i).

y y + 3 60.

And, x - x = 40 y (y - 2) = 3x.....(ii).

y-2 y 60.

On dividing (i) by (ii) , we get: x = 40.

X/y-2+40/60=x/y+3-40/60 is a wrong equation.

@Teja it's much better explanation.

Thanks.

2y(y+3) = 9x --- equation 1.

y(y-2) = 3x --- equation 2.

Equation 1 we can also write that in this form.

2y(y+3) = 3*3x --- equation 3.

Then substitute equation 2 in equation 3.

Then 2y(y+3) = 3*(y(y-2)).

Cancel y on both sides then 2y+6 = 3y-6.

y = 12.