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# I need help really quick

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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

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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
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hint: $$\text {distance} =$$ $$\text {speed} \times \text {time}$$

Jun 13, 2020