SOLUTION: A----------60m----------- -C------40m-------B Alphonse starts at point A and runs at a constant rate towards point C. At the same time, Brigitte starts at point B and runs tow

Question 288054: A----------60m-------------C------40m-------B

Alphonse starts at point A and runs at a constant rate towards point C. At the same time, Brigitte starts at point B and runs towards point C also at
a constant rate. They arrive at C at exactly the same moment. If they continue running in the same directions, Alphonse arrives at B exactly 10 seconds before Brigitte arrives at A. How fast was Brigitte running?
(A) 3 m/s (B) 10/3 m/s (C) 13/3 m/s (D) 5 m/s (E) Not enough information
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A----------60m-------------C------40m-------B
Alphonse starts at point A and runs at a constant rate towards point C. At the same time, Brigitte starts at point B and runs towards point C also at
a constant rate. They arrive at C at exactly the same moment.
---
1st leg:
Alphonse DATA:
distance = 60m ; rate = a m/sec ; time = 60/a seconds
Brigitte DATA:
distance = 40m ; rate = b m/sec ; time = 40/b seconds
---
1st leg Equation:
60/a = 40/b
a = (3/2)b
------------
If they continue running in the same directions, Alphonse arrives at B exactly 10 seconds before Brigitte arrives at A. How fast was Brigitte running?
---------------
Total trip:
Alphonse DATA:
distance = 100m ; rate = a m/sec ; time = 100/a m/sec
Brigitte DATA:
distance = 100m ; rate = b m/sec ; time = 100/b m/sec
Total trip Equation:
Brit time - Alphon time = 10 sec.
100/b - 100/a = 10 sec
Multiply thru by ab to get:
100a - 100b = 10ab
---
Substitute for "a" and solve for "b":
100(3/2)b - 100b = 10(3/2)b^2
(1/2)100b = 15b^2
15b^2 - 50b = 0
b(15b-50) = 0
b = 0 or b = 50/15 = 10/3
---
Brigitte rate = 10/3 meters/sec
===================================
Cheers,
Stan H.

(A) 3 m/s (B) 10/3 m/s (C) 13/3 m/s (D) 5 m/s (E) Not enough information

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