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Re: two buses, same speed... head spinning [#permalink]
27 Feb 2012, 22:45
Bunuel wrote:
T740qc wrote:
Bunuel wrote:
\(\frac{d}{2}-24=\frac{0.75d}{2}\)
why do you assume -24 instead of +24?
You can derive this either my reasoning or simply by noticing that d/2>0.75d/2, so it should be d/2-24=0.75d/2 (greater value minus 24 equals to smaller value).
Hope it's clear,
do you mind explaining the reasoning? think it'll help me set up similar equations more easily. thanks!
Re: A bus from city M is traveling to city N at a constant speed [#permalink]
12 Apr 2012, 05:51
Help me please to realize whats wrong with my solution? @what point exactly I'm making a mistake?
form the first part of data i can get (1 bus) X*2=2X (2 bus) X*2=2X -->the total distance is 4X
Taking into account the following assumptions, 1)the 1st bus began to drive later for 24 min 2)both buses have the same rate --> the 1st has to travel less distance than the 2d, since it started to travel later but the second earlier 3)the total distance is 4X i set up two equations: (1 bus) X*(t+24)=2X-24 (2 bus) X*(t-36)=2X+24
since 2X is a half of the distance i can rewrite the two equations: X(t+24)+24=X*(t-36)-24 --->60X=-48 what makes me crazy and i don't know where is the flaw in my logic (excluding the fact, that I'm a woman)
Re: A bus from city M is traveling to city N at a constant speed [#permalink]
09 Jul 2013, 03:40
Let s be the speed of the buses. Thus total distance betn the two cities is 4s. On the first day, thus they meet P, which is at a distance of 4s/2= 2s from their starting points. The next day , in effect one bus has travelled for 1 hour before the other starts. Distance covered in 1 hour =4s/4= s. Remaining distance is 3s. Point at which the two buses meet now, is at a distance of 3s/2 which is also a distance of 24 from P. Thus, 3s/2 = 2s- 24 or s =48 . therefore distance = 4 * 48 =192.
Re: A bus from city M is traveling to city N at a constant speed [#permalink]
05 Aug 2013, 10:04
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 return trip 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?
let d=distance (what the question is looking for)
On the first day, bus a and bus b travel at the same constant speed. Because they travel at the same constant speed, when they meet at point P they have each traveled for 2 hours. Therefore, the total time (at that constant speed) from A to B is 4 hours and P is the mid distance between those two points.
On the second day each bus travels at the same constant speed. The bus that leaves first spends a total of one hour on the road before the second bus leaves (if bus a and bus b normally leave at the same time and today one leaves 36 minutes earlier and 24 minutes later respectively)
Here is where I get thrown off (using Bunuel's explanation to structure mine)
The wording is a bit ambiguous (at least to me) regarding the constant speed of bus A and bus B. On the second day, do they travel the same constant speed they did the day before or did they each travel the same constant speed that day that is different from the day before?
If they each traveled at the same rate they did before (which is 1/4 of the distance every hour) and today one bus traveled for an hour before the other set off, then today one bus traveled .25d of the way
(d/2) - 24 = (.75d/2)
(I'm not sure as to why that is)
Maybe this is why:
(d/2) represents the speed each bus travels and the midpoint at which bus A and B meet on the first day. (d/2)-24 would represent the bus reaching a point 24 miles before the midpoint. This is equal to a distance that's 3/4ths normal???? Help!!!
Bunuel, I read the link you posted offering an explanation of how that formula represents the point at which each bus meets the second day but I am still confused. Why is the halfway .75d?
Here is another way to look at it. Let's say bus A starts first and B, second. When bus A travels one hour bus B starts. When bus A has traveled for two hours, bus B has traveled for one. When bus A and B meet, A has been on the road for 2.5 hours and B, for 1.5 at a place 24 miles away from the mid distance of the journey.
We can break the journey up into four identical blocks, each representing an hour long portion of the journey (see attached image) when bus A and B meet they meet in the exact middle of one of those 4 identical blocks meaning that 24 miles represents 1/2 of one block. This means there are two 24 mile portions in each of the four blocks. 2*24*4=192.
Re: two buses, same speed... head spinning [#permalink]
31 Dec 2013, 05:18
Bunuel wrote:
T740qc wrote:
Bunuel wrote:
\(\frac{d}{2}-24=\frac{0.75d}{2}\)
why do you assume -24 instead of +24?
You can derive this either my reasoning or simply by noticing that d/2>0.75d/2, so it should be d/2-24=0.75d/2 (greater value minus 24 equals to smaller value).
Hope it's clear,
Here's the way I did it. Let's call X the total distance
Let's focus on the second part of the problem
One bus had a head start of 1 hour. In that hour we traveled 1/4 of the distance. (Since he traveled 1/2 of the distance in 2 hours from question stem)
Now, there are 3/4x of the distance to be traveled
Since they both have the same rate, they will meet at half the distance of 3/4x, so they will meet at 1/4x + (3x/4/2) = 5/8x
Now remember from the first part of the question that point P is exactly the mid point (Same rates, same time)
So we are now at point 5/8x. The question also says that this distance is equal to 24 miles.
So we have 5/8x - 1/2x =1/8x = 24 miles
And so, we get 24*8 = 192 miles for the total distance 'x'
Re: A bus from city M is traveling to city N at a constant speed [#permalink]
17 Jan 2015, 22:46
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A bus from city M is traveling to city N at a constant speed [#permalink]
12 Oct 2015, 13:12
R=speed of each bus 2R=distance from each city to P 4R=distance between two cities bus1 distance=2R+24 4R-(2R+24)=2R-24 bus2 distance=2R-24 2R+24-(2R-24)=48 miles bus1 drives 48 miles more than bus2 in 1 extra hour 48 mph=R 4R=192 miles
A bus from city M is traveling to city N at a constant speed [#permalink]
19 Dec 2015, 04:58
gmattokyo 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 return trip 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
The source is GMATClub's diagnostic test... would look forward to see some innovative approach to this. Thanks!
I just don't know, why my brain is in the standby mode during the cat, now I've solved this one very quickly ~ 1m 20sec ))) P is the midpoint or Total \(\frac{Distance}{2}\), so one of them have traveled 24 miles more than the half of the distance, and guess which bus has made it (the one, that started his trip 36+24=60 min earlier, let's say it was Bus B) Bus A:\(r*t=p-24\) Bus B \(r*(t+1)=p+24\) => \(r*t+24=r*(t+1)-24\) Rate = 48, so if each of them traveled 2 hours on the first day, than the whole distance can be covered in 4 hours by each of them --> \(4*48=192\) Answer E _________________
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