Question 1170323: Formulate a quadratic equation then answer what is asked in the problem.
If train travels 63km and then 72km at an average speed of 6kph more than its original speed. What is its original average speed if it takes 3 hours to complete the total journey of the train?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rate * time = distance.
you have 3 equations.
x * t1 = 63
(x + 6) * t2 = 72
t1 + t2 = 3
solve for t1 and t2 to get:
t1 = 63/x
t2 = 72/(x+6)
since t1 + t2 = 3, you get:
63/x + 72/(x+6) = 3
put everything on the left side of the equation under a common denominator to get:
(63 * (x + 6) + 72 * x) / (x * (x + 6) = 3
multiply both sides of the equation by x * (x + 6) to get:
63 * (x + 6) + 72 * x = 3 * x * (x + 6)
simplify to get:
63 * x + 378 + 72 * x = 3 * x^2 + 18 * x
combine like terms to get:
135 * x + 378 = 3 * x^2 + 18 * x
subtract 135 * x and 378 from both sides of the equation to get:
0 = 3 * x^2 + 18 * x - 135 * x - 378
combine like terms to tget:
0 = 3 * x^2 - 117 * x - 378
factor this quadratic equation to get:
x = 42 or x = -3
x has to be positive, so x = 42
this makes x + 6 = 48
go back to the 3 original equations of ...
x * t1 = 63
(x + 6) * t2 = 72
t1 + t2 = 3
and replace x with 42 and x + 6 with 48 to get:
42 * t1 = 63
solve for t1 to get:
t1 = 63/42
48 * t2 = 72
solve for t2 to get:
t2 = 72/48
since t1 + t2 = 3, then 63/42 + 72/48 should be equal to 3.
use your calculator to find that 63/42 + 72/48 does equal 3, confirming the values for x and x + 6 are good.
the average speed for the first leg of the trip was 42 kilometers per hour and the average speed for the second leg of the trip was 48 kilometers per hour.
the average speed for the whole trip was 135/3 = 45 kilometers per hour.
|
|
|
