# A man normally takes 5 hrs to travel at a certain speed from city A to B. One day, he increases his speed by 4 km/h and finds that the journey ...from A to B takes half an hour less than the normal...

A man normally takes 5 hrs to travel at a certain speed from city A to B. One day, he increases his speed by 4 km/h and finds that the journey ...

from A to B takes half an hour less than the normal time. Find the normal speed.

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Let the distance from A to B be D.

==> The time is T1 = 5 hrs.

==> Then, let the speed be S1 = D/T1 = D/ 5

==> S1 = D/ 5

==> D = 5*S1 .............(1)

After increasing the speed ( the distance remains the same.)

==> S2 = S1 + 4

==> T2 = T1 - (0.5)

==> D = S2* T2 = (S1 + 4) * (5 - 1/2)

==> D = (S1 + 4)(9/2)

==> D = (9/2)S1 + 18

But D = 5S1

==> 5S1 = (9/2)S1 + 18

==> (5-9/2) S1 = 18

==> (1/2) S1 = 18

==> S1 = 36 km/ h

**Then, the normal speed is 36 km / h**

Let the speed with which the man normally travels be V. It takes 5 hours for him to go from A to B.

So the distance from A to B is 5V.

When the speed is increased by 4 km/h it takes .5 hours less.

So we have 5V / (V + 4) = 5 - 0.5

=> 5V / ( V + 4) = 4.5

=> 5V = 4.5 V + 18

=> .5 V = 18

=> V = 18/.5

=> V = 36 km/h.

**The normal speed is 36 km/h**

Let the normal speed for the journey from A to B be v

Let the distance between A ab B be s.

Therefore s = 5v.

On a particular day speed of the man = (4+v )km.

So the tine taken for the journey = 5v/(v+4) which is les than 1/ hour from normal time for the journey = 5- 1/2 = 9/2 hrs.

5v/(v+4) = 9/2

10v = 9(v+4).

10v = 9v+36.

10v-9v = 36.

v = 36km/h.

**Therefore the normal speed of the journey is 36 kms/h.**