SOLUTION: sir pls help me to solve this prblm.A train travels at a certain average for a didtance of 63km and then travels a distance of 72km at an average speed of 6km/hr more than its orig

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Question 392246: sir pls help me to solve this prblm.A train travels at a certain average for a didtance of 63km and then travels a distance of 72km at an average speed of 6km/hr more than its original speed.if it takes 3 hours to complete the total journey,what is its original average speed.kindly send me the solution to rexi_2004@yahoo.co.in
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A train travels at a certain average for a distance of 63km and then travels a distance of 72km at an average speed of 6km/hr more than its original speed.
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If it takes 3 hours to complete the total journey,what is its original average speed.
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Original-speed DATA:
distance = 63 km ; rate = x km/hr ; time = 63/x hrs.
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Faster-speed DATA:
distance = 72 km ; rate = (x+6)km/hr ; time = 72/(x+6) hrs.
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Equation:
time + time = 3 hrs
---
63/x + 72/(x+6) = 3 hrs
----
63(x+6) + 72x = 3x(x+6)
21(x+6) + 24x = x(x+6)
45x + 21*6 = x^2+6x
x^2-39x-21*6 = 0
---
x^2-39x-126 = 0
(x-42)(x+3) = 0
---
Positive solution:
x = 42 km/hr (original rate)
=========================
Cheers,
Stan H.
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi

Let x and (x + 6kmph) represent the speeds traveling 63km and 72km respectively

Question States***total time of trip 3hr.  t = D/r

 63/x  + 72/(x+6) = 3  |Multiplying each term by x(x+6) so as all denominators= 1

 63(x+6) + 72x = 3x(x+6)

 63x + 6*63 + 72x = 3x^2 + 18x

   3x^2 -117x - 378 = 0

  3(x^2 - 39x - 126) = 0

factoring

  (x+3)(x-42) = 0

  (x+3) = 0   |tossing out negative solution

  (x-42) = 0

    x = 42kmph, speed of train while traveling 63kmph, the original average speed



CHECKING our Answer ***

  63km/42kmph + 72km/48kmph = 1.5hr + 1.5hr =  3hr

  

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