What Time Does Bill Catch Up with Mike?

[zidan3311]

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

 

[zidan3311]

the distance should be find at first, but the problem core is what's time Bill can follow Mike?

 

[Jameson]

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.

 

[Jameson]

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)?

 

[zidan3311]

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?

 

[Jameson]

Ok, who started first, Bill or Mike? How much time passed between when the first person leaves and the second person leaves?

 

[zidan3311]

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...

 

[Jameson]

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?

 

[phymat]

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.

 

[Jameson]

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.

 

[zidan3311]

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...

 

[Jameson]

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 $t_b$?

 

[zidan3311]

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?

 

[Jameson]

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?

 

[zidan3311]

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 - - -

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?

 

[Jameson]

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?

 

[zidan3311]

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??

 

[soroban]

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.