What Time Does Bill Catch Up with Mike?

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Discussion Overview

The discussion revolves around a problem involving two individuals, Mike and Bill, traveling towards each other at different speeds and starting times. Participants explore how to determine the time at which Bill catches up with Mike, considering various methods such as ratios and equations of motion. The scope includes mathematical reasoning and problem-solving techniques.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant asks for a formula to determine when Bill meets Mike, suggesting the use of a ratio method or another simple approach.
  • Another participant emphasizes the need to find the distance first and questions the core of the problem regarding the timing of Bill's pursuit.
  • Several participants derive equations based on the distances traveled by Mike and Bill, indicating that they need to relate their travel times.
  • There is confusion among participants regarding the wording of the problem, with some suggesting that it may be improperly stated, leading to different interpretations of the meeting time.
  • One participant asserts that the answer of 9:15 am is incorrect based on their calculations, proposing that they should meet at 9:00 am instead.
  • Another participant agrees with the assertion that the answer of 9:15 am is wrong and asks for clarification on how to apply the LCM method to the problem.
  • Participants engage in correcting each other's mathematical steps and reasoning, indicating a collaborative effort to arrive at a solution.

Areas of Agreement / Disagreement

There is no consensus on the correct answer to the problem, with some participants arguing for 9:00 am as the meeting time while others maintain that 9:15 am is correct based on the initial information provided. The discussion remains unresolved regarding the final answer.

Contextual Notes

Participants express uncertainty about the wording of the problem and the implications of the speeds and start times. There are unresolved mathematical steps and differing interpretations of the problem's conditions.

zidan3311
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hi all..

i need some formula to figure out this below :

"Mike riding a bike from city A to city B. He departed at 07.00 am with speed of 40 kmph . From town A, Bill followed with a speed of 80 kmph start at 08.00 am, what time Bill met Mike ?

The answer is 09.15 am

how to easy identify that's answer above.

Possible using ratio method? Or another simple way?

thanks a lot in advance..

zidan3311
 
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the distance should be find at first, but the problem core is what's time Bill can follow Mike?
 
Sorry I read the question wrong. I thought they were starting on opposite ends and meeting in the middle. I'll post how to do it momentarily. :o
 
So they meet when the total distance traveled is the same.

We have the equations: $d=rt$. There will be two of them.

1) Mike: $d_m= 40t_m$.

2) Bill: $d_b = 80t_b$.

From here we need to figure out how do $t_m$ and $t_b$ relate to each other. Any ideas? Put another way, can we write the time Bill has been going in terms of how long Mike has (or the other way around)?
 
Jameson said:
So they meet when the total distance traveled is the same.

We have the equations: $d=rt$. There will be two of them.

1) Mike: $d_m= 40t_m$.

2) Bill: $d_b = 80t_b$.

From here we need to figure out how do $t_m$ and $t_b$ relate to each other. Any ideas? Put another way, can we write the time Bill has been going in terms of how long Mike has (or the other way around)?

hi Jameson,

i'm really blank ideas..

would you help me out?
 
Ok, who started first, Bill or Mike? How much time passed between when the first person leaves and the second person leaves?
 
Jameson said:
Ok, who started first, Bill or Mike? How much time passed between when the first person leaves and the second person leaves?

Mike 07.00
Bill 08.00

the deference is 1 hrs, then...
 
zidan3311 said:
Mike 07.00
Bill 08.00

the deference is 1 hrs, then...

Good! Here's how we use it. We can say that $t_m=t_b+1$ or $t_m-1=t_b$. They are both equivalent statements that use the fact we know Bill left 1 hour after Mike.

Going back to our two equations we are trying to solve:

$40t_m=80t_b$.

Now we can make a substitution using either one of the two equations above. Any ideas from here?
 
The question is not worded properly. Sometimes city is mentioned sometimes town.
If Mike starts from City A at 7 a.m. at the speed of 40 km/h, then in one hour, he covers 40 km and in two hours, 80 km.
Ben starts from City A at 8 a.m. (one hour later) with the speed of 80 km/h, so, in one hour, he travels 80 km.
Find the L.C.M.
They should meet at 9 a.m.? Either something is wrong in my answer or in the question.
 
  • #10
phymat said:
The question is not worded properly. Sometimes city is mentioned sometimes town.
If Mike starts from City A at 7 a.m. at the speed of 40 km/h, then in one hour, he covers 40 km and in two hours, 80 km.
Ben starts from A at 8 a.m. (one hour later) with the speed of 80 km/h, so, in one hour, he travels 80 km.
Find the L.C.M.
They should meet at 9 a.m.? Either something is wrong in my answer or in the question.

I also get that answer but was trying to get there together with the OP. Even if the answer we get isn't the same as the book's, the method is still good to practice.

Either the speeds are incorrectly given or the answer is wrong, but with the data we have the time they meet is definitely not 9:15am.
 
  • #11
Jameson said:
Good! Here's how we use it. We can say that $t_m=t_b+1$ or $t_m-1=t_b$. They are both equivalent statements that use the fact we know Bill left 1 hour after Mike.

Going back to our two equations we are trying to solve:

$40t_m=80t_b$.

Now we can make a substitution using either one of the two equations above. Any ideas from here?

it's correct? than next...
 

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  • #12
That's not quite correct.

This part is right: $40(t_b+1)=80t_b$. You distributed to get $40t_b+40=80t_b$, but from there you have a mistake. How do we solve for $t_b$?
 
  • #13
Jameson said:
That's not quite correct.

This part is right: $40(t_b+1)=80t_b$. You distributed to get $40t_b+40=80t_b$, but from there you have a mistake. How do we solve for $t_b$?

hi Jameson, i confuse, I'm quiet!

how do the final answer?
 
  • #14
Ok, we have this:

$40t_b+40=80t_b$

If you had to solve $40x+40=80x$, how would you do it? You would first subtract $40x$ from both sides. We want to get the variable on one side and everything else on the other side. Same idea here. We're going to subtract $40t_b$ from both sides and we'll get.

$40t_b=40$. How to we solve for $t_b$ now?
 
  • #15
phymat said:
The question is not worded properly. Sometimes city is mentioned sometimes town.
If Mike starts from City A at 7 a.m. at the speed of 40 km/h, then in one hour, he covers 40 km and in two hours, 80 km.
Ben starts from City A at 8 a.m. (one hour later) with the speed of 80 km/h, so, in one hour, he travels 80 km.
Find the L.C.M.
They should meet at 9 a.m.? Either something is wrong in my answer or in the question.

hi phymat,

LCM (40,80) or LCM (7;8) etc..

could you show me?

- - - Updated - - -

Jameson said:
Ok, we have this:

$40t_b+40=80t_b$

If you had to solve $40x+40=80x$, how would you do it? You would first subtract $40x$ from both sides. We want to get the variable on one side and everything else on the other side. Same idea here. We're going to subtract $40t_b$ from both sides and we'll get.

$40t_b=40$. How to we solve for $t_b$ now?

hi..

tb = 1 hrs but my question where is .15

your answer just approx? or exactly number?
 
  • #16
No it's exact. What phymat was saying (and I agree with him) is that the answer of 9:15am is not correct given the info you provided. What is the answer with the numbers we have?
 
  • #17
Jameson said:
No it's exact. What phymat was saying (and I agree with him) is that the answer of 9:15am is not correct given the info you provided. What is the answer with the numbers we have?

yep, you are right..

the final answer is 09.00 am..

could you show me how to use phymat method with LCM??
 
  • #18
Hello, zidan3311!

Mike riding a bike from city A to city B.
He departed at 7:00 am with speed of 40 km/h.
From city A, Bill followed with a speed of 80 kmph start at 8:00 am.
At what time will Bill meet Mike?

The answer is 9:15 am . Wrong!
Mike had a one-hour headstart.
At 8 am, he was 40 km ahead.

Then Bill followed at 80 km/h.
The difference of their speeds is: \:80-40 \,=\,40 km/h.

It is as if Mike stopped, and Bill approached him at 40m/h.

To cover the 40-km distance, Bill will take one hour.

He overtakes Mike at 9:00 am.

 

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