SOLUTION: Two trains were 332 miles apart and heading towards each other at 11 am. If the trains meet at 1 pm and one train was traveling 36 miles per hour faster than the other, then how

Algebra ->  Proportions -> SOLUTION: Two trains were 332 miles apart and heading towards each other at 11 am. If the trains meet at 1 pm and one train was traveling 36 miles per hour faster than the other, then how       Log On


   

Question 749529:
Two trains were 332 miles apart and heading towards each other at 11 am. If the trains meet at 1 pm and one train was traveling 36 miles per hour faster than the other, then how fast was each train moving?
Found 2 solutions by josmiceli, checkley79:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I like to think of one train as standing still
and the other one approaching at the sum of their speeds
from 11 AM to 1 PM is 2 hrs
Let +s+ = the speed of the slower train
+s+%2B+36+ = the speed of the faster train
-----------------
+332+=+%28+s+%2B+s+%2B+36+%29%2A2+
+332+=+%28+2s+%2B+36+%29%2A2+
+332+=+4s+%2B+72+
+4s+=+260+
+s+=+65+
+s+%2B+36+=+101+
The trains were traveling at 65 mi/hr and 101 mi/hr
check:
+332+=+%28+2s+%2B+36+%29%2A2+
+332+=+%28+2%2A65+%2B+36+%29%2A2+
+332+=+166%2A2+
+332+=+332+
OK

Answer by checkley79(3341) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
332=R*3
R=332/3
R=110.67 TOTAL SPEED OF THE 2 CARS.
X+X+36=110.67
2X=110.67-36
2X=74.67
X=74.67/2
X=37.335 MPH FOR THE SLOWER CAR.
37.335+36=73.335 MPH FOR THE FASTER CAR.
PROOF:
332=(37.335+73.335)*3
332=110.67*3
332=332