Pulling my hair out=this math problem

[topdent1]

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Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 MPH and starts 3 hours after the first cyclist who is traveling at 6 MPH. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

a. 2hrs
b. 4.5 hrs
c. 5.75 hrs.
d. 6 hrs.
e. 7.5 hrs

 

[TheGreenWave]

Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 MPH and starts 3 hours after the first cyclist who is traveling at 6 MPH. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

a. 2hrs
b. 4.5 hrs
c. 5.75 hrs.
d. 6 hrs.
e. 7.5 hrs

If im reading this correctly the problem isnt that bad. So the biker that is going 6mph starts 3 hours before the one going 10mph. So in those 3 hours the first biker has gone 18 miles (6mph X 3hours). So the second biker has to make up 18miles. Now just think how fast he is going relative to the first biker. 10mph- 6pmh = 4mph. The second biker is "catching up" at a rate of 4mph. So if you divide the 18 miles by the 4mph you get the time : 4.5 hours

Answer: B

 

[Predentole]

From what I did, they meet up exactly at 4 hours. So the closest time post 4 hours is when he passes up the 1st cycler.

B

 

[topdent1]

If im reading this correctly the problem isnt that bad. So the biker that is going 6mph starts 3 hours before the one going 10mph. So in those 3 hours the first biker has gone 18 miles (6mph X 3hours). So the second biker has to make up 18miles. Now just think how fast he is going relative to the first biker. 10mph- 6pmh = 4mph. The second biker is "catching up" at a rate of 4mph. So if you divide the 18 miles by the 4mph you get the time : 4.5 hours

Answer: B

Awesome! Thanks so much.

Do you want to give a shot to this one:

Jim can fill a pool carrying buckets of water in 30 min. Sue can do the same job in 45 minutes. Tony can do the same job in 1.5 hours. How quickly can all three fill the pool together.

I have done this type of problem using 2 people filling something, but never with 3.

a. 12 min
b. 15 min
c. 21 min
d. 23 min
e. 28 min.

 

[baedero1]

the travel distance of 1st person = the distance of 2nd person

so, 6(3+x) = 10x

x=4.5hr

 

[topdent1]

nvm, i figured it out.

1/30 + 1/45 + 1/90 = 6/90. 90/5=15. I have no idea why i did this. This is the procedure I follow for all these types of problems.

 

[PooyaH]

From what I did, they meet up exactly at 4 hours. So the closest time post 4 hours is when he passes up the 1st cycler.

B

Nawh, it would be 4.5 I believe, as The GreenWave showed.
If you write down the miles each one travels you find out:

1st biker goes 6m/hr, so 42 miles after 7hrs
2nd biker goes 10m/hr, so 40 miles after 4 hours and 50 miles after 5hrs!
So it wouldn't be exactly 4! it would be somewhere between 4 and 5 hrs. in this case 4.5

 

[baedero1]

correct!!

 

[osimsDDS]

This question is a work problem, its the same for 2 ppl or 500 ppl, here use this formula....

If 1 person can do the job in 2 hours they can do 1/2 the job in 1 hour so you will have x/2. Another person can do the job in 3 hours so they will do 1/3 the job in 1 hour, so you will have x/3 for that person.

Now in this problem its the same concept, first change 1.5 hrs to 90 minutes so everything is constant.

x/30 + x/45 + x/90 = 1
multiply each side by 90 (LCD) to cancel out the fractions faster.
3x + 2x + x =90
6x=90
x=15 minutes

Answer: B 15 minutes all together

 

[PooyaH]

This question is a work problem, its the same for 2 ppl or 500 ppl, here use this formula....

If 1 person can do the job in 2 hours they can do 1/2 the job in 1 hour so you will have x/2. Another person can do the job in 3 hours so they will do 1/3 the job in 1 hour, so you will have x/3 for that person.

Now in this problem its the same concept, first change 1.5 hrs to 90 minutes so everything is constant.

x/30 + x/45 + x/90 = 1
multiply each side by 90 (LCD) to cancel out the fractions faster.
3x + 2x + x =90
6x=90
x=15 minutes

Answer: B 15 minutes all together

I agree 😀

 

[osimsDDS]

hopefully the DAT are full of these word problems but i have a feeeeeeling they are much much harder...

 

[Streetwolf]

nvm, i figured it out.

1/30 + 1/45 + 1/90 = 6/90. 90/5=15. I have no idea why i did this. This is the procedure I follow for all these types of problems.

That's the exact way to do it.

The first person does a job in 30 minutes. The second in 45 minutes. The third in 90 minutes. You first want to see how much they can do in the same amount of time (because the answer you get assumes they all work in the same amount of time). The most obvious thing here is to do it per minute. So in 1 minute, the first guy does 1/30 of a job, the second guy 1/45 of a job, and the third guy 1/90 of a job. So in 1 minute if they work together, they have done 6/90 of the job. They need to do 1 full job. So you need to know how many times they have to do 6/90 of a job in order to do the full job. Thus you take 1 divided by 6/90 which is 90/6 = 15. It will take 15 minutes for them to do the full job.

It does not matter how many people are involved. Make sure you bring it to the same time limit. It's easy to work with 1 minute like I did here because when you have the rate per minute you just flip the fraction to get your answer. I could have easily chosen per 30 minutes:

The first guy does 1 job in 30 minutes, the second guy does 2/3 a job in 30 minutes, and the third guy does 1/3 a job in 30 minutes. So every 30 minutes you have 2 full jobs done. That means you need 15 minutes to do 1 full job.

Or had I gotten lucky and chosen 15 minutes:

The first guy does 1/2 a job in 15 minutes, the second guy does 1/3 a job in 15 minutes, and the third guy does 1/6 a job in 15 minutes. So every 15 minutes you have 1 full job done.