Let 'd' be the distance between the two cities M & N.

Both the buses drive for 2 hrs and meet at P. Hence 'P' is in the centre of the meeting point.
Thus distance covered by each is d/2.
Time taken = 4 hrs.

Next day, one bus is delayed by 24 mins and other leaves early by 36 mins; thus net difference between the buses starting time is 24+36= 60mins.
In 60mins, distance covered is d/4 as total distance 'd' is covered in 4 hrs.

Now, they jointly cover a distance of (d-d/4 = 3d/4) and meet at a point which is 24 miles from point P. Point P is at d/2 distance from both the points.

Thus, we can frame the eqn as -

d/2 - 24 = (3d/4)/2
i.e. 0.5d - 24 = (0.75d/2)
i.e. d - 48 = 0.75d
i.e. 0.25d = 48
i.e. d = 192

Ans: Option (E)

shrey2287 wrote:A bus from city M is traveling to city N
at a constant speed while another bus is
making the same journey in the opposite
direction at the same constant speed.
They meet in point P after driving for 2
hours. The following day the buses do the
same trip again at the same constant
speed. One bus is delayed 24 minutes and
the other leaves 36 minutes earlier. If
they meet 24 miles from point P, what is
the distance between the two cities?
A. 48
B. 72
C. 96
D. 120
E. 192

M---------q----p------------N

let m be the bus starting from M. And n be the bus starting from N.

Let the distance between M and N be d. Therefore p=d/2 and speed=d/4

Suppose m is delayed by 24 minutes and n leaves early by 36 minutes. So, by the time n starts journey, N has traveled for 60 minutes. ie n is at d/4 from N. Distance between them is 3d/4

Again they are travelling with same speed, they'll meet at midpoint (let's call it q) which is (3d/4)*(1/2)= 3d/8 from M.

We are given that distance between p and q is 24 miles. Therefore (d/2) - (3d/8) = 24. Solving for d we get d=192. Ans E.

Lets consider D be the distance and S be the speed of each. Using relative speed
2S = D/2. In this case both bus will meet at the distance D/2.

When one bus is 36 minute earlier and another is 24 minutes late, it means by the time late bus start the other bus will already travel 36 + 24 = 60 minutes = 1 hr.

In one hour the bus will travel D/4 distance. the remaining distance = 3D/4
Using relative speed ,
3D/4 will be covered by both the bus.
hence, time taken = (3D/4)/ 2*S
= (3D/4)/D*2 = 3/2 = 1.5 hour.

Both bus will meet at 1.5 hour from the point the late bus was started.

Now check the option. let D = 120 , speed = 120/4 = 30.

D/2 where the bus meet earlier = 60

Distance in 1.5 hour = 1.5 * speed = 1.5 * 30 = 45. the difference between the earlier and later meeting point is 15 ( should be 24) wrong choice

check with 196, speed = 192/4 = 48
D/2 when the bus meet earlier = 96.

Distance in 1.5 hour = 1.5 * 48 = 72.

Difference between the earlier and later meeting point = 96 - 72 = 24 (which is correct). hence E is the correct choice.

Lets consider D be the distance and S be the speed of each. Using relative speed
2S = D/2. In this case both bus will meet at the distance D/2.

When one bus is 36 minute earlier and another is 24 minutes late, it means by the time late bus start the other bus will already travel 36 + 24 = 60 minutes = 1 hr.

In one hour the bus will travel D/4 distance. the remaining distance = 3D/4
Using relative speed ,
3D/4 will be covered by both the bus.
hence, time taken = (3D/4)/ 2*S
= (3D/4)/D*2 = 3/2 = 1.5 hour.

Both bus will meet at 1.5 hour from the point the late bus was started.

Now check the option. let D = 120 , speed = 120/4 = 30.

D/2 where the bus meet earlier = 60

Distance in 1.5 hour = 1.5 * speed = 1.5 * 30 = 45. the difference between the earlier and later meeting point is 15 ( should be 24) wrong choice

check with 196, speed = 192/4 = 48
D/2 when the bus meet earlier = 96.

Distance in 1.5 hour = 1.5 * 48 = 72.

Difference between the earlier and later meeting point = 96 - 72 = 24 (which is correct). hence E is the correct choice.