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Two boats are heading towards each other at constant speeds [#permalink]
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01 May 2012, 21:29
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Two boats are heading towards each other at constant speeds of 5 miles/hr and 20 miles/hr respectively. They begin at a distance 20 miles from each other. How far are they (in miles) one minute before they collide ? A. 1/12 B. 5/12 C. 1/6 D. 1/3 E. 1/5
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Re: Two boats are heading towards each other at constant speeds [#permalink]
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01 May 2012, 22:52
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Re: Two boats are heading towards each other at constant speeds [#permalink]
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02 May 2012, 22:34
Thanks Bunnel for the great link  awesome questions!!
For this one (two boats heading towards each other)  i calculated the time they would collide with each other (48th min) and then "tried" calculating from there on, taking eons to solve  how can the problem be solved with this apporach ? is 25*1/60 the only way to solve such problems ?



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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03 May 2012, 12:23



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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03 May 2012, 19:47
I first calculated the time at which they collide, assuming they started at the same time. but i did not even need it.



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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04 May 2012, 20:44
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gmihir wrote: Thanks Bunnel for the great link  awesome questions!!
For this one (two boats heading towards each other)  i calculated the time they would collide with each other (48th min) and then "tried" calculating from there on, taking eons to solve  how can the problem be solved with this apporach ? is 25*1/60 the only way to solve such problems ? Bunuel has already given you a great and most direct approach for this question. But if you did go the round about way, you could have solved it in this way: Since their combined speed is 20 + 5 = 25 miles/hr, they cover 20 miles in \(\frac{20}{25}\) hrs i.e. \(\frac{20}{25} * 60\) mins = 48 mins Now, in 48 mins, they cover 20 miles. So in 1 min, they must have covered 20/48 = 5/12 miles
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Re: Two boats are heading towards each other at constant speeds [#permalink]
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25 Aug 2012, 11:53
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Hi Bunuel, Your explanation is always the best. I have a slightly different Logic to solve this question. Hope you will like the logic The question asks: How much distance will both travel in 1 minute? Since the combined rate of the boats is 5+20=25 mph = 25/60 mile/min = 5/12 mile/min Hence Answer: B.
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Re: Two boats are heading towards each other at constant speeds [#permalink]
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06 Nov 2013, 19:21
Two boats are heading towards each other at constant speeds of 5 miles/hr and 20 miles/hr respectively. They begin at a distance 20 miles from each other. How far are they (in miles) one minute before they collide?
I probably did this a bit backwards but here is how I figured it out...
The boats travel at a combined rate of 25 miles/hour. If they need to cover 20 miles, it will take them 20/ (5/12) or roughly 48 minutes to reach one another. Either way, one minute before they collide, they will be 5/12ths of a mile away from one another, the distance they travel combined every minute. In this way, it is almost a trick question.
ANSWER B. 5/12



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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19 Nov 2013, 13:07
Just form this simple equation : 20  (20/25 1/60)X25
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Re: Two boats are heading towards each other at constant speeds [#permalink]
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19 Nov 2013, 18:38
Bunuel wrote: gmihir wrote: Two boats are heading towards each other at constant speeds of 5 miles/hr and 20 miles/hr respectively. They begin at a distance 20 miles from each other. How far are they (in miles) one minute before they collide ?
A. 1/12 B. 5/12 C. 1/6 D. 1/3 E. 1/5 The question asks: how far apart will they be 1 minute=1/60 hours before they collide? Since the combined rate of the boats is 5+20=25 mph then 1/60 hours before they collide they'll be rate*time=distance > 25*1/60=5/12 miles apart. Answer: B. Similar question to practice: 12easypiecesornot126366.html#p1033924Hope it helps. Can you tell me why my method didn't work: I figured that 5*T+20*T=20 (since they'll have traveled 20 miles combined when they collide). so 25T=20, thus T=4/5. So if they collide at the 48 minute mark, then the distance they travel in 44 minutes combined, subtracted from 20 should give you the answer. so 5*(47/60)+20*(47/60)=D 235/60+940/60=D 1175/60=D 19 \(\frac{35}{60}\)=D So if they had traveled 19 35/60 miles in 44 minutes combined, then one minute before they collide they should be 25/60 miles apart. EDIT: Just edited my work. I had a glaring error. My method did work. I keep doing this. I don't know what's happening to me. I'm starting to become concerned. I make errors in calculations on roughtly 80% of the problems I do. I read the questions, and by the second line of the question my mind is already wandering and I don't see or remember anything after the first few words of the question. UGH. Just today I've done ~100 problems, and I made errors in simple addition, or just didn't remember a constraint (answer must be positive or something) on 89 of them. I feel like my brain is dying or something.



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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18 Sep 2014, 06:12
Sorry I don't get it The question states that we must find out the distance 1 min from the time of collision. So if they collided at the 48th minute, question asks us to find the distance of the 2 boats from each other in the 47th minute right?



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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18 Sep 2014, 21:55
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alphonsa wrote: Sorry I don't get it The question states that we must find out the distance 1 min from the time of collision. So if they collided at the 48th minute, question asks us to find the distance of the 2 boats from each other in the 47th minute right? Yes, there are two ways we can find the distance between the boats in the 47th minute: One way is we find the distance traveled by the boats in 47 mins and then subtract that out of the total distance. Another way is to find the distance they will cover in the last 1 min. That must be the distance between them in the 47th minute. Because at the end of the 48th minute, the boats meet  i.e. there is no distance between them. Knowing their speed, we know the distance traveled by them in 1 min and hence know the distance between them 1 min before they meet. The solutions above use this approach to get the answer.
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Re: Two boats are heading towards each other at constant speeds [#permalink]
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25 Oct 2014, 01:48
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what is the problem with my solution
after calculating the time of collision i.e 4/5=48 mins.
since we need to calculate the distance b/w them in 47 mins,can`t we do like this ..
boat 1 48 mins=4miles 47 mins=(4/48) *47
similarly for boat 2 47 mins=(16/48)*47
then subtracting the distance of boat 2 from boat 1.. (16/48)*47(4/48) *47=answer.
kindly correct me on this.
thank you



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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25 Oct 2014, 01:49
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what is the problem with my solution
after calculating the time of collision i.e 4/5=48 mins.
since we need to calculate the distance b/w them in 47 mins,can`t we do like this ..
boat 1 48 mins=4miles 47 mins=(4/48) *47
similarly for boat 2 47 mins=(16/48)*47
then subtracting the distance of boat 2 from boat 1.. (16/48)*47(4/48) *47=answer.
kindly correct me on this.
thank you



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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25 Oct 2014, 05:53



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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30 Oct 2014, 11:49
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thank you for the valuable reply but why are we subtracting the total from 20 and not subtracting them by each other(as per my solution), as that will give us how far are they from each other.. confused.



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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31 Oct 2014, 06:34



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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06 Nov 2014, 01:19
okay now it`s clear. a very stupid mistake from my side. Thanx Bunuel.



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Re: Two boats are heading towards each other at constant speeds [#permalink]
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30 Jan 2015, 06:51
One question,
I thought that when 2 objects are moving towards each other we subtract the rates, as they cover this distance. When they more away from each other we add the rates because they create the distance. In this case, why did we add the rates?




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