Two boats are heading towards each other at constant speeds

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 ?

It's not the only way but for sure the shortest way to solve this problem.

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.

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

Just form this simple equation :-

20 - (20/25 -1/60)X25

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.

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.

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

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

It should be \(20 - ((\frac{16}{48})*47 + (\frac{4}{48}) *47)\).

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.

In 47 minutes boat 1 covers (16/48)*47 = 47/3 miles and boat 2 covers (4/48) *47 = 47/12 miles. Together in 47 minutes they cover 47/3 + 47/12 = 235/12 miles. Therefore after 47 minutes, so 1 minute before they meet, they will be 20 - 235/12 = 5/12 miles apart.

Hope it's clear.

P.S. 30-second approach to deal with this problem is given here: two-boats-are-heading-towards-each-other-at-constant-speeds-131737.html#p1080945

okay now it`s clear.
a very stupid mistake from my side.
Thanx Bunuel.

In these types of Combined Rate questions, you have to stay focused on what the question asks you to figure out and what the two vehicles are doing.

In this question, they two vehicles are heading TOWARDS one another, so the movement of BOTH vehicles decreases the distance between them. As such, we have to ADD their rates so that we can figure out how much the distance decreases with each passing time period (second, minute, hour, etc.).

In a different type of question, you might have one vehicle chasing another. In these situations, the vehicle "in front" is moving AWAY from the vehicle "in back." So even though the "back vehicle" might be moving faster, the "front vehicle" is still moving AWAY from it. As such, we have to SUBTRACT their rates to figure out how long the "back vehicle" takes to catch up to the "front vehicle".

Another quick way to think about this: go backwards. Assume they are at the same point and are going in opposite directions. One at t/12 miles per hour and the other at t/2 miles per hour. After one minute, they are

\(\frac{1}{12} + \frac{1}{3} = \frac{15}{36} = \frac{5}{12}\) miles apart.

There are some helpful and quick solutions to it but I thought of an alternate approach to solve it:

Relative speed for the boats = 5+20=25 mph
Total Distance = 20 miles
Time Taken=distance/speed=20/25=4/5 hrs=4*60/5=48 mins

So,
One minute before the collision, the distance between those boats would be the distance traveled by these boats in 1 minute which is
=20/48 miles (As it takes 48 mins to travel 20 miles so in 1 minute it would travel 20/48 miles)
=5/12 miles