SOLUTION: A cyclist decides to ride from bugibba to mellieha ,a distance of six kilometres .The average speed is x kilometres per hour .On his return journey from mellieha to bugibba he rid

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Question 1201640: A cyclist decides to ride from bugibba to mellieha ,a distance of six kilometres .The average speed is x kilometres per hour .On his return journey from mellieha to bugibba he rides 12 kilometres per hour faster and completes the journey 15 minutes more quickly .Form and solve an equation in X(Ans =12km/hr)
So , : 15 minutes into hours is 0.25 hrs
6/x =6/x+12 -0.25
I worked the above out but I still got it wrong .
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
A cyclist decides to ride from bugibba to mellieha ,a distance of six kilometres .The average speed is x kilometres per hour .On his return journey from mellieha to bugibba he rides 12 kilometres per hour faster and completes the journey 15 minutes more quickly .Form and solve an equation in X(Ans =12km/hr)


bugibba to mellieha ,a distance of 6 km
speed x km/h
time = 6/x km/h
from mellieha to bugibba distANCE = 6KM
he rides 12 kilometres per hour faster = (x+12)
Time return = 6/(x+12)
Time onwards - time return = 0.25
%28+6%2Fx%29+-+6%2F%28x%2B12%29=+0.25
%28+6%28x%2B12%29-6x%29%2Fx%28x%2B12%29+=+0.25
%286x+%2B+72+-6x%29%2Fx%28x%2B12%29+=+0.25
72%2F%28x%28x%2B12%29%29+=0.25 (0.25=1/4)
72 = (1/4) *x(x+12)
4*72 = x^2 +12x
288 = x^2+12x
x^2+12x-288 =0
x^2+24x-12x-288 =0

x(x+24)-28(x+24) =0
(x+24)(x-12) =0
x = -24 or 12 speed cannot be negative

x=12 km/h



Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.

Hey, you were on the right track: your equation 

    6%2Fx = 6%2F%28x%2B12%29 - 0.25


was "almost" right. The error was only in the sign of 0.25 --- it should be +0.25.


When you write such equation, you should think, which term is greater:  6%2Fx  or  6%2F%28x%2B12%29.


Obviously, the greater denominator produces the smaller fraction.

So, in order to have an equality, you should  highlight%28highlight%28ADD%29%29  0.25 of an hour;  highlight%28highlight%28NOT%29%29  subtract.

Such problems require to concentrate your attention.