Aptitude - Time and Distance - Discussion

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.

It's very easy by using relative speed concept there are two things up there.
So, now by using the formula 2xy/x+y.
where x=10 and y=15,
=> 2*10*15/10+15,
=>300/25,
=>12,
So the answer is 12kmph.

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?

Why don't you just find the mean value?

With 10kmph reach at 2 p.m.
Witn 15kmph reach at 12noon.
Find mean of both for 1p.m required time.
So, 10+15 = 25/2 = 12.5kmph should be the correct answer.