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Re: D01-24 [#permalink]
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The buses are travelling at the same speed. One bus is at the point 0.75d when the other bus starts. So, the distance between them is 0.75d and they are travelling at the same speed. Therefore, they will meet halfway through the journey. Hope that helps.
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Re: D01-24 [#permalink]
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its given that 1 bus starts 36 minutes late and the other 24 minutes early which means effectively 1 bus had head start of 1 hr. Since they are traveling at same speed and they meet 24 miles from where they met yesterday then the bus which had head start of 1 hr traveled 24 miles in 1 hr. Thus its speed is 24 miles. The speed of 2nd bus is also 24 (since both bus travels at constant speed)..
Thus both bus travels for 4 hrs hence distance traveled by 1 bus =24*4

Total distance= 2*24*4= 192.. Answer
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Re: D01-24 [#permalink]
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Really good question. Here's my approach:

Suppose both bus travels with V mph both days (as same constant speed).

Day1:
Bus A: Distance covered in 2 hours = 2v
Bus B: Distance covered in 2 hours = 2v
Total distance between city M and N = total distance covered by Bus A + Bus B = 2v + 2v = 4v.

Day2:
We need to account for delay to get proper time here. Bus A is delayed by 36 minutes, so it travels 36 minutes less. Bus B leaves 24 minutes early, so it travels 24 minutes extra. So we can say, bus B travels for 36 (delay of bus A) + 24 (early of bus B) = 60 minutes = 1 hour extra. So let t = Bus A travel time, t+1 = Bus B travel time.

Now, distance travelled by Bus A: vt
distance travelled by Bus B: v(t+1)

Total distance between City N and M = combined distance travelled of both bus A and bus B = vt + v(t+1).

Since total distance remains same on both days, we can say vt + v(t+1) = 4v. Solving this, we get t = 3/2 = 1.5 hours.

Now, we are given that both bus meet 24 miles from where they met yesterday (point P). Think logically, the bus who travels more (has more t+1 as V constant) will naturally travel 24 miles extra. And the bus who travels less, will travel 24 miles less. So Bus A travels 24 miles less than what it travelled on Day 1 (2v), and Bus B travels 24 miles extra than what it travelled on Day 2 (2v). Take any:

2v - 24 = v*3/2. Solving for V = 48.
2v + 24 = v*(3/2 + 1). Solving for V = 48.

So we got V = 48. Total distance = 4V, or V*3/2 + V(3/2 + 1). Put V = 48 in any. We get distance = 192 miles.

Your Kudos keeps us motivated to post more answers. Thanks!!
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Re: D01-24 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: D01-24 [#permalink]
I think this is a high-quality question and I agree with explanation.
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D01-24 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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D01-24 [#permalink]
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I had a different approach to the solution that is a bit more visual/logical but I don't think it is better or more reliable when you are under pressure. I guess it may work better for some.

1. We know that on day 2, the buses have started moving with a 1-hour difference.

2. This difference results into the meeting point moving by 24 miles. This point intrigued me and I immediately wanted to start approaching the question from this perspective to find out what speed they were moving at, and at what distance the cities are from each other.

3. If the distance is 24 miles away from the original point and one bus was driving an extra hour, does it mean they are moving 24 miles an hour? I felt that was a bit slow and unrealistic so likely it has to be more... and I think it is because we have 2 buses moving towards each other, so they are actually covering 2x the distance when moving towards each other, which means that if in 1 hour they covered 24 miles, they were actually driving 48 miles per hour (24x2). Honestly if the speed was more realistic, I may not have thought about the double speed effect.

4. We know they drove for 4 hours. 4x48 = 192

Curious if anyone else solved this way.
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Re D01-24 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: D01-24 [#permalink]
Hi Bunuel

I have a doubt in this question.....why is d/4 (the extra distance covered by the bus that left early) not equal to 24 Miles? Both the buses would ideally travel d/2 in 2 hrs, since this bus had a head start of an hour it covered some more miles = x than the other bus that started late. Why is x not equal to 24? Shouldn't that extra 1 hour account for the extra 24 miles travelled by one of the buses?

In this case the distance would be d/4= 24mile =>> d =96 miles?
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Re: D01-24 [#permalink]
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kop18 wrote:
Hi Bunuel

I have a doubt in this question.....why is d/4 (the extra distance covered by the bus that left early) not equal to 24 Miles? Both the buses would ideally travel d/2 in 2 hrs, since this bus had a head start of an hour it covered some more miles = x than the other bus that started late. Why is x not equal to 24? Shouldn't that extra 1 hour account for the extra 24 miles travelled by one of the buses?

In this case the distance would be d/4= 24mile =>> d =96 miles?


The early bus, in its one-hour head start, covers an extra distance which we can call "x". This "x" is a quarter of the total distance between the two cities.

When both buses eventually meet, they're 24 miles away from point P. This 24 miles doesn't represent the full extra distance "x" traveled by the early bus. Instead, it's only half of "x", because both buses are moving towards each other, effectively halving the advantage the early bus had.

So, while the early bus travels an extra distance equivalent to a quarter of the total distance in its head start, the point where they meet being 24 miles from point P shows that the extra advantage it maintains when they meet is only an eighth of the total distance, which is 24 miles.

Hope it's clear.
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Re: D01-24 [#permalink]
bb wrote:
I had a different approach to the solution that is a bit more visual/logical but I don't think it is better or more reliable when you are under pressure. I guess it may work better for some.

1. We know that on day 2, the buses have started moving with a 1-hour difference.

2. This difference results into the meeting point moving by 24 miles. This point intrigued me and I immediately wanted to start approaching the question from this perspective to find out what speed they were moving at, and at what distance the cities are from each other.

3. If the distance is 24 miles away from the original point and one bus was driving an extra hour, does it mean they are moving 24 miles an hour? I felt that was a bit slow and unrealistic so likely it has to be more... and I think it is because we have 2 buses moving towards each other, so they are actually covering 2x the distance when moving towards each other, which means that if in 1 hour they covered 24 miles, they were actually driving 48 miles per hour (24x2). Honestly if the speed was more realistic, I may not have thought about the double speed effect.

4. We know they drove for 4 hours. 4x48 = 192

Curious if anyone else solved this way.
-B



I was almost there with this method, but made mistake of assuming speed of bus at 24 miles & I marked 96 miles as answer. Thanks for providing additional insight into this.
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Re D01-24 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re D01-24 [#permalink]
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