Question

How to solve this math contest question?

Answer 1

I'm sorry, it's very rude to cheat in a contest. You should solve this on your own and be fair for other contestants.

Answer 2

this is a problem send for practise ;P its not cheating

Answer 3

If it's practice, it's given to you for practice, or a warm-up. There's really no reason to get answers from others on a "practice".

Answer 4

I mean that's why they call it practice. :P

Answer 5

i want to know how to solve it...

Answer 6

I guess someone can guide you through it, unfortunately I'm clueless. Sorry! :(

Answer 7

anyone ?

Answer 8

i think i can solve it

Answer 9

8km?

Answer 10

how do you even start

Answer 11

I've got the same, so it's probably right

Answer 12

ca you guys show me how to solve this please

Answer 13

The first clue you can use is that the biker takes 4 minutes to pass the walker after he has passed the jogger.

Answer 14

So you know that the distance between the jogger and the walker must have increased by enough so that the speed of the walker plus the speed of the biker are enough to close the gap between the jogger and walker in 4 minutes. Since the speed of the biker is 0.2333 km/min and the speed of the walker is 0.1km/min, the distance between the jogger and the walker needs to be 4*0.3333 km

Answer 15

My intuition would be to set up some equations. First, suppose the town of Pi is at location 0, and Epsilonville is at location x. Then the equation for the location of the jogger is 10t, and for the walker, it's 6t (where t is number of hours). Finally, the equation for the biker is x-14t.

Using these, you can set up a system of equations, and solve for x.

Answer 16

Then, since you now know the distance between the jogger and the walker at that moment, you can calculate how much time would have passed before they arrived at that point. (since the jogger travels faster, the distance increases)

Answer 17

once you know the time they would have needed to create a gap large enough, you can then measure the distance the biker and the jogger would have traveled in that period of time, and add them together. Since they meet at a certain point, that would be the distance.

Answer 18

That system of equations will be\[x-14t=10t\]\[x-14(t+\frac{1}{15})=6(t+\frac{1}{15})\]At time \(t\) the biker and the jogger pass, and at time \(t+1/15\) the biker and walker pass.

Just fyi, we use \(t+15\) since 4 minutes is \(1/15\) of an hour.

Answer 19

@KingGeorge I am getting 8t = 4/3

Answer 20

that does not seem right

Answer 21

x =28t from the first equation and we can sub it in the second equation

Answer 22

Well if I work it out, I get that \(x=10t+14t\), so substituting that into the second equation, and simplifying a little bit, I get\[10t+14t-14t-\frac{14}{15}=6t+\frac{6}{15}\]\[4t=\frac{20}{15}\]\[t=\frac{1}{3}\]
You should have \(x=24t\), not \(28t\).

Answer 23

@KingGeorge one question. why did you not keep 6t rather than making it 6(t+1/15)

Answer 24

We used \(t+1/15\) since the biker passed the walker 4 minutes after they passed the jogger. Since we're assuming they passed the jogger at time \(t\), I set the equations up so that \(t\) is in hours, and since 4 minutes is \(1/15\) hours, that means the biker passed the walker at time \(t+1/15\).

Answer 25

if x-14t = speed of the jogger then shouldnt
x-14(t+1/15) = speed of the walker because if we add t+1/15 for the walker then they are not equal.

Answer 26

which means distace of teh biker + 4 minutes = speed of walker

Answer 27

= distance of walker (not speed)

Answer 28

I think I understand now, we were adding 1/15t to both because we want equation 1 to equal equation 2

Answer 29

You could write it as

"distance of biker at time \(t\) + distance biker traveled in 4 minutes = distance of walker at time t+4 minutes."

Answer 30

That's correct. If we didn't add it to both, the biker and the walker wouldn't be at the same location.

Answer 31

One last question
I understand we want the equations to equal to each other so by adding 1/15 to the time of the biker, wouldnt he be 4 minutres ahead of the jogger? and 8 minutes aead of teh walker. so we should only add 1/15 to the equation in which the walker is being represented?

Answer 32

That problem is fixed by the minus sign in front of the 14. By putting the minus sign there, we assure that the biker is traveling from Epsilonville to Pi, while the walker and the jogger are traveling in the other direction.

Answer 33

we know that x-14t = 10t (the point where the biker and the jogger are the same) but the jogger is 4 minutes behind so i was thinking the equation should be

x-14t = 6(t+1/15) which is the same as 10t = 6(t+1/15) so now they are at the same point?

Answer 34

when I solve it I am getting x as 1.4km?

Answer 35

You need to put the \(t+1/15\) on both sides so the time will remain consistent for the walker and the biker.

The equation x-14t = 6(t+1/15) would only be correct if the biker immediately stopped moving as soon they encountered the jogger. Since they (presumably) keep moving, we must write x-14(t+1/15) = 6(t+1/15)

Answer 36

1.4 km doesn't seem right. Remember that we solved that \(t=1/3\) hours when the biker and the jogger passed.

Answer 37

so 28(1/3) = x
x = 9.333333333333333333

Answer 38

It should be \(24(1/3)\) and not \(28(1/3)\), since \(10+14=24\neq28\).

Answer 39

t = 1/3
x = 8

you must have made a mistake somewhere

Answer 40

Oh, thank you so much I really understand this know

Answer 41

you ROCK @KingGeorge

Answer 42

yep @zimmah I am getting 8 km now :D

Answer 43

this was not difficult at all, with your help once again thanks

Answer 44

You're welcome.