Question 1163467
Sam drives from City A to City B, and he takes 8 hours for the entire trip.
Richard drives from City B to City A, and he takes 10 hours for the trip
assuming they are driving on the same road. If Sam leaves City A and Richard
leaves City B at the same time, how many hours will they meet at a place 40
km away from the halfway? Express your answer as a mixed number.
<pre>Oh well, someone beat me to this one!!

<b><u>Finding TIME, IMMEDIATELY</b></u>
Since Sam takes 8 hours to complete the trip, while Richard takes 10 hours, “needless to say,” Sam’s speed is greater 
than Richard’s, which means that Sam will get to the 40-km mark, on Richard’s side, after passing the ½-way mark

Let time taken by both to get to 40 kms beyond the ½-way point, from A, be T, and let distance from A to B, or from B to A, be 2D
Then the ½-way point is 2D/2 = D, and Sam will travel a distance of D + 40, while Richard will travel a distance of D - 40

Since Sam takes 8 hours to complete the trip, then Sam’s speed = {{{matrix(1,3, 2D/8, "=", D/4 )}}}
Since Richard takes 10 hours to complete the trip, his speed, from B to A, is {{{matrix(1,3, 2D/10, "=", D/5)}}}

We then get the following SPEED equations for:
      <b><u>Sam</b></u>                       <b><u>and</b></u>                  <b><u>Richard</b></u> 
{{{matrix(5,3, (D + 40)/T, "=", D/4,
4(D + 40), "=", TD,
4D + 160, "=", TD,
160, "=", TD  -  4D,
160, "=", D(T - 4))}}}                             {{{matrix(5,3, (D - 40)/T, "=", D/5,
5(D - 40), "=", TD,
5D - 200, "=", TD,
5D - TD, "=", 200,
D(5 - T), "=", 200))}}}
   {{{matrix(1,3, 160/(T  -  4), "=", D)}}} ------ eq (i)                           {{{matrix(1,3, D, "=", 200/(5  -  T))}}} ------ eq (ii)

With {{{matrix(1,3, D, "=", 160/(T  -  4))}}}, and {{{matrix(1,3, D, "=", 200/(5  -  T))}}}, we can say that: {{{matrix(1,3, 160/(T  -  4), "=", 200/(5  -  T))}}}
200(T - 4) = 160(5 - T) ------ Cross-multiplying
40(5)(T - 4) = 40(4)(5 - T)
5(T - 4) = 4(5 - T)
5T - 20 = 20 - 4T
5T + 4T = 20 + 20
9T = 40
T, or time taken by both to get to 40 kms beyond the ½-way point, from A is: {{{highlight_green(matrix(1,4, 40/9, or, 4&4/9, hours))}}} 

OR

<b><u>Finding Sam’s SPEED, in order to get the answer: TIME</b></u>
Since Sam takes 8 hours to complete the trip, while Richard takes 10 hours, “needless to say,” Sam’s speed is greater
than Richard’s, which means that Sam will get to the 40-km mark, on Richard’s side, after passing the ½-way mark

Let Sam’s speed be S
Since Sam takes 8 hours to complete the trip, then distance, from A to B, or from B to A, is 8S
Since Richard takes 10 hours to complete the trip, his speed, from B to A, is {{{matrix(1,3, 8S/10, "=", 4S/5)}}}

As the distance is 8S, the ½-way point is: {{{matrix(1,3, 8S/2, "=", 4S)}}}, and with Sam’s speed being greater than Richard’s, Sam will travel to the 40-km mark, on “Richard’s side,” after passing the ½-way point. 

As such, distance Sam will have traveled to get to 40 kms past the half-way point: 4S + 40, AND
         distance Richard will have traveled to get to “Sam’s point”: 8S - (4S + 40) = 4S - 40

Time Sam takes to travel to 40 kms past the half-way point (on Richard’s side): {{{(4S + 40)/S}}}
Time Richard takes to travel to “Sam’s point”: {{{(4S  -  40)/(4S/5)}}}

Since they left at the same time, then their times will be the same, to get to "Sam's point." Therefore, we get: 
{{{matrix(5,3,(4S + 40)/S, "=", (4S - 40)/(4S/5), (4S + 40)/S, "=", (4S - 40) * (5/(4S)), (4S + 40)/S, "=", 4(S - 10) * (5/(4S)), (4S + 40)/S, "=", cross(4)(S - 10) * (5/(cross(4)S)), (4S + 40)/S, "=", 5(S  -  10)/S)}}}
4S + 40 = 5(S - 10) ------ Denominators are equal and so are the numerators
4S + 40 = 5S - 50
4S - 5S = - 50  -  40
- S = - 90
Sam’s speed, or {{{matrix(1,6, S, "=", (  -  90)/(  -  1), "=", 90, "km/h")}}}

Travelling 40 kms past the halfway mark (on “Richard’s side) - a distance of 4S + 40, or 4(90) + 40 = 400 kms, at S (90) km/h, or 400 km @ 90 km/h - time Sam takes is: {{{highlight_green(matrix(1,6, 400/90, "=", 40/9, "=", 4&4/9, hours))}}}
<b><u>OR</b></u>
Travelling a distance of 4S - 40, or 4(90) - 40 = 320 kms, at {{{matrix(1,5, 4S/5, "=", 4(90)/5, "=", 72)}}}, or 320 kms @ 72 km/h - time Richard takes is: {{{highlight_green(matrix(1,6, 320/72, "=", 40/9, "=", 4&4/9, hours))}}}</pre>