Aptitude - Time and Distance - Discussion

@Shivasai.

I think here the difference between the two times I. E12noon and 2pm is 2, and also time =d÷s, so we should take distance as "X" and speeds which they are given so x÷10-x÷15=2.

How did 60/10 came in the last? Can anyone explain after that?

Just take average speed.

Required speed.
= 2xy/(x+y),
=(2 x 10 x 15)/(10+15),
= 300/25.
= 12Km/hr.

That's it.

What is the logical and easiest method for solving this problem?

Well I would make it simple !!
first, let us find the time (in hrs)
let t be the time

distance/speed = time, distance we don't know so take it as x.

here there are two references given with respect to time.
he once reached at 2pm which can be taken as t .
he again reached at 12pm so that can be taken as t - 2.

So now there are two times , (t) and (t-2).
substitute this in formula (above distance\speed = time).
x/10 - x/15 = (t) - (t-2) (difference of two times, two distances etc),
now, (15x - 10x) / 150 = 2,
5x = 300,
x=60km.

Now we have found the distance travelled

Again substitute this in formula
speed = d/t.

1st case
10= 60 / t = 6hrs.

2nd case
15 = 60 / t = 4hrs
so if in the frst case he has travelled for 6 hours and reached at 2pm. so he would have started at 8am.

Now the main problem, at what speed he must travel so that he can reach at 1pm?

he starts at 8 am and he should reach at 1 pm so his travel must be 5 hrs.

Now substitute 5hrs in the formula speed =d/t.

speed = 60/5,
speed = 12 kmph.

Distance = speed * time.

For time ,let say time taken to travel to point A = t.
d = 15*t ------- eqn.1
d = 10(t+2) ------eqn. 2
d = s* (t+1)--------- eqn. 3.

From eqn 1 and eqn 2
15* t = 10(t+2).
after solving, t = 4 means time to travel is 4 hours.
now,
we put value of t = 4 in eqn 3 and eqn. 2.
60 = s(4+1).
s = 12 kmph.

Consider starting time as x.

We know that time = (distance/speed),
When he reach at 2 P.M i.e.,14.00 (railway time)
(14-x) = (distance/ 10)

When he reach at 12 P.M
(12-x) = (distance/15)

As we know that., distance is the same.
So, from above equations we get x=8. so he started cycling at 8 A.M.

So from this, we can calculate distance = 60 km.
To reach at 1 P.M, the time taken is 5 hrs.
Speed = (60/5) = 12 kmph.

Thanks for the answer @Raja.

Then, x/10 - x/15 = 2 why we have taken this?

What is the logic behind it

How 60/5 comes, does any one please tell me?