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# A bus from city M is traveling to city N at a constant speed while ano

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Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
Vinit800HBS wrote:
Hoozan wrote:
KarishmaB Vinit800HBS

Reading the first half we get:

Let total distance = D
Let Rate of each bus = R

Since they meet at D/2 in 2 hours we can infer that the total time = 4hrs

Now, reading the latter we get

Bus 1 had an early start of 1hr. So for 1hr, the bus drove at a rate of R thereby covering D/4 the distance.

But we see that the extra 1 hour results in the extra 24 miles. So why can't we say that Bus 1 drove at a rate of 24 miles/hour? and hence:

D = R x T
D = 24 x 4 --> 96 miles.

Dear Hoozan,

I am glad that you figured out that the meeting will happen exactly in the middle (Point P). And hence, the deduction that total duration will be 4 hours is perfect.

But, I am afraid that you missed the second aspect.

Quote:
One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P

So, the meeting now is taking place at exactly in the middle of the remaining 3/4th Distance.

Check the attachment for the visualization

In your image, you mention that "case 2" both cover 1.5d distance each till P' But wouldn't one bus cover 2.5d and the other 1.5d?
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Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
Hoozan wrote:
Vinit800HBS wrote:
Hoozan wrote:
KarishmaB Vinit800HBS

Reading the first half we get:

Let total distance = D
Let Rate of each bus = R

Since they meet at D/2 in 2 hours we can infer that the total time = 4hrs

Now, reading the latter we get

Bus 1 had an early start of 1hr. So for 1hr, the bus drove at a rate of R thereby covering D/4 the distance.

But we see that the extra 1 hour results in the extra 24 miles. So why can't we say that Bus 1 drove at a rate of 24 miles/hour? and hence:

D = R x T
D = 24 x 4 --> 96 miles.

Dear Hoozan,

I am glad that you figured out that the meeting will happen exactly in the middle (Point P). And hence, the deduction that total duration will be 4 hours is perfect.

But, I am afraid that you missed the second aspect.

Quote:
One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P

So, the meeting now is taking place at exactly in the middle of the remaining 3/4th Distance.

Check the attachment for the visualization

In your image, you mention that "case 2" both cover 1.5d distance each till P' But wouldn't one bus cover 2.5d and the other 1.5d?

Hi Hoozan,

Each will cover 1.5d distance AFTER M has already traveled for initial 1 hour.
You are right about the part that M will cover 2.5d and N will cover 1.5d.

But as you can see, I have placed M not at the starting point but at a certain distance ‘D’ from its initial starting point.

Posted from my mobile device
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Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
Quote:
Hi Hoozan,

Each will cover 1.5d distance AFTER M has already traveled for initial 1 hour.
You are right about the part that M will cover 2.5d and N will cover 1.5d.

But as you can see, I have placed M not at the starting point but at a certain distance ‘D’ from its initial starting point.

Posted from my mobile device

Ahh! Got it. Thanks a lot
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Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
Solution
• Time taken by buses to complete dist. Between M & N: 4 hr.
• Let, the const. speed of both the buses be x miles/hr.
• The total length of the journey: is 4x miles.
• Meeting point P: 2x, reason: both buses have the same speed, so they will meet at the midpoint.
• During the return journey, one bus starts 36 min early -> 0.6 hr & the other bus starts 24 min late -> 0.4 hr. From these two statements: one bus covers dist : x*(0.6+0.4) -> x. Before the other bus starts heading back.
• So remaining dist that both will travel is: 4x-x = 3x.
• The meeting point will be half of this 3x miles dist. -> 1.5x
• We know that they meet 24 miles from point P,
that implies:
-> 2x-1.5x = 24;
-> 0.5x=24
-> x=48
• The required dist is 4x. Therefore, 4x= 192 miles (dist between two cities)
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Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
Think of this question in terms of a straight line divided in 4 parts as follows

M------------X------------P------------Y------------N

Both buses meet halfway at P because they are travelling at the same speed.
Next day, one bus leave one hour before the other bus (24+36)
Let's assume that bus that left 1 hour early left from M.
In 1 hour, the bus would've travelled 1/4th of the total distance and would be at point X.
Now, the bus at N starts its journey and the two buses meet halfway of the remaining 3/4th of the distance say at point R.
Point R would be half of 3/4th = 3/8
Now we are given that the distance between P and the return trip's meeting point (R) is 24

So, P - R = 24
P = x/2
R = 3x/8

x/2 - 3x/8 = 24
x/8 = 24
x = 192
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Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
Let the total distance be d.

As the speed of both buses is constant, they meet halfway through at d/2.

Speed of both the buses together = 1/2 => speed of each bus = (1/2)/2 = 1/4

On the second day, assume the bus (b2) from M starts 1 hour earlier, hence covers 1/4d in 1 hour. Remaining distance = 3d/4

Distance traveled by each bus (3d/4)/2 = 3d/8

Now the total distance d = 3d/8 + d/2 + 24 (you can infer this when you visualize a diagram) => 1d/8 = 24 => d = 192 [E]
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Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
They are traveling at the same speed so when they meet they each have covered D/2 (D being the distance)

Assume they are traveling with speed V. So after 2 hours when they meet we have:
2*v + 2*v = d -> 4v = d -> v = d/4 (1).

We are also told one train is delayed by 36 and the other one is ahead of the schedule by 24. So when they meet for the second time at time let's say t one of the trains has been on the road for an hour. And based on the problem, it has covered 24 more miles than last time, which was d/2. So we get:

(2) d/2 + 24 = v(t + 1)

And similarly the delayed train has covered: d - (d/2 + 24) = d - 24 miles when they meet for the second time.

(3) d/2 - 24 = v*t

from (2):
d/2 + 24 - v = vt

from (3):
d/2 - 24 = vt

-> d/2 + 24 - v = d/2 - 24
48 = v
and from (1) we know v = d/4.
-> D = 4 * 48 = 192.
Re: A bus from city M is traveling to city N at a constant speed while ano [#permalink]
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