A cyclist biked from Berkeley to a town 150km away. On her way back, she increased her average speed by 15 km/h and returned 30 minutes faster. Find the cyclist's average speed on the way to the other town.

 Jun 13, 2020
edited by BWStar  Jun 13, 2020

Alright, let's write a couple of equations.

First, let x equal the speed at which she went to the other town (what we're trying to find) and let y equal the time it takes her.

Then, because distance=speed*time, 150=xy

Our second equation is going to be on her way back. 150=(x+15)(y+\(\frac{1}{2}\)). We put \(\frac{1}{2}\)hr instead of 30 minutes because our speeds are given in km/hrs. 

Now we expand: 150=xy+\(\frac{1}{2}\)x+15y+\(\frac{15}{2}\)

Substitute 150=xy (our first equation). 150=150+\(\frac{1}{2}\)x+15y+\(\frac{15}{2}\)

This simplifies to \(\frac{1}{2}\)x+15y+\(\frac{15}{2}\)= 0

Multiply by two: x+30y+15=0

From our first equation, y=(150/x), and substituting this into our above equation:

x+30(150/x)+15= x+(4500/x)+15=0

Multiply by x: x2+15x+4500=0

Factorize to get (x+60)(x−75)=0

We get either -60 or 75. Speed must be positive, so the answer is 75. 

 Jun 13, 2020
edited by Guest  Jun 13, 2020
edited by thelizzybeth  Jun 13, 2020

hint: \(\text {distance} =\) \(\text {speed} \times \text {time}\)

 Jun 13, 2020

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