Some may have seen this before, but I thought it was a good one.

 

 

Person A left Town X at 10:18 am. He walked at a constant speed and arrived at town Z at 1:30 pm. On the same day, Person B left town Z at 9 am. Person B walked the same route in the opposite direction at a constant speed. Person B arrived at town X at 11:40 am. The road crosses a wide river. By coincidence, both arrived at the bridge on opposite sides of the river at the same instant. Person A left the bridge 1 minute later than Person B. At what time did they arrive at the bridge?

 

does this look right?

 

 

I make it 11:00

 

slight typo on second last eq

 

 

 

imatfaal

does this look right?

 

Well your answer is correct, I didn't wade through your maths yet.

 

Well done.

 

 

 

speeds = 0.83333:1

1 minute = 1 - 0.83333 = .16666

bridge trip A = 6 minutes

bridge trip B = 5 minutes

 

relative bridge distance = bridge time B / entire time B = 5/160 = 1/32

 

X/Z = distance from x/z to bridge

x+z+0.03125 = 1

1-x = z+0.03125

-x = z−0.96875

x = -z + 0.96875

 

In B-relative time, A began 78/160 after B.

 

z/1 = time for B to reach

x/.83333 = time for A to reach

z - 78/160 = x/.83333

x = (z-.4875)*.83333

 

I need the intersection of these two lines, but my calculator won't find it for me. I'll see whether I can do it with wxMaxima.

 

The value of z at the intersection multiplied by 160 should give the amount of minutes it took B to reach the bridge.

 

 

 

My calculator won't finish it, and wxMaxima can only plot 1 line at a time.

Edited October 6, 201411 yr by MonDie

Interesting iterative approach, Mondie.

 

Looks promising.

Your thoughtfulness is noted.

 

I think "Solve" is the wxMaxima tool I wanted.

 

 

 

(10999987÷14666640)×160 = 120 = 2 hours

 

B (and A) reached the bridge at 11:00. Before looking at imatfaal's post, I acidentally put 10:20 as if there were 100 minutes in an hour!!

 

 

Edited October 6, 201411 yr by MonDie