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Question 69535: One hour out of the station, a train engine develops trouble that slows the train to 3/5 of its average speed to the time of the failure. Continuing at this reduces speed,it reaches its destination two hours late. Had the trouble occurred 50 miles beyond, the delay would have been reduced by 40 minutes. Find the distance from station to the destination.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! ne hour out of the station, a train engine develops trouble that slows the train to 3/5 of its average speed to the time of the failure. Continuing at this reduces speed,it reaches its destination two hours late. Had the trouble occurred 50 miles beyond, the delay would have been reduced by 40 minutes. Find the distance from station to the destination.
Let T = normal time it takes to make the trip
Let D = distance of the trip
Let S = normal speed
Then normal Time = D/S
Distance = S*T
Let (3/5) = .6
Let 40 min = .67 hrs
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Two equations to go the same distance
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Eq1:
S*T = D; normal trip
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Substitute S*T for D in eq 2
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Eq2:
1S + .6S(T+2-1) = S*T
Divide equation by S
1 + .6(T+1) = T
1 + .6T + .6 = T
.6T - T = -.6 - 1
-.4T = -1.6
T = -1.6/-.4
T = + 4 hrs is the normal time to take the trip
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We then also know that D = 4S
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Equation for the addition 50 mi at normal speed:
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1S + 50 + .6S(4 + 2 - 1 - .67) = D
1S + .6S(4.33) = D - 50
Substitute 4S for D, multiply what's inside the brackets:
1S + 2.6S = 4S - 50
3.6S - 4S = - 50
-.4S = - 50
S = -50/-.4
S = 125 mph is normal speed
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.6 * 125 = 75 mph is reduced speed
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Distance = 4(125) = 500 miles is the trip
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Check:
Normal speed for 1 hr + .6 normal speed for 5 hrs = 6 hrs for the trip
1(125) + 5(75) = 500 miles
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Check using additional 50 mi at normal speed
1(125) + 50 + (5 - .67)* 75 =
125 + 50 + 325 = 500 mi
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This was a toughie, hope it makes sense to you.
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