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Algebra rate problem 
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#1
Feb1312, 10:54 PM

P: 27

1. The problem statement, all variables and given/known data
distance = rate * time 3. The attempt at a solution I thought I could put it into a ratio like this: (r = rate) x / r = 100 ft / 3r Cross multiply: 3rx = 100r Divide by r on both sides: 3x = 100 x = 100/3 But then I figured that neither dragon nor car was travelling a certain distance, they were just travelling at a pace. I already know how to arrive at the answer, but I don't understand why I've arrived at the answer: distance = rate * time (r = rate, t = time) rt + 100 = 3rt 100 = 2rt 50 = rt (r = rate, y = time) 4ry = 100 ry = 25 distance = rate(t + y) distance = rate(t) + rate(y) (plug in from the solutions of above equations) distance = 50 + 25 distance = 75 I don't understand from the very first step of how the equations came to be. Is there a different way to solve this? So I can understand it intuitively instead of algebraically? 


#2
Feb1312, 11:20 PM

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PF Gold
P: 5,197

100+rt = 3rt This equation is saying "distance travelled by dragon + length of dragon = distance travelled by car" Now consider the second part, when the car moves from the head back to the tail, which takes an amount of time "y". In this case, the distance travelled by the car is the length of the dragon MINUS the distance travelled by the dragon. It's minus, because the car has to go less than 100 ft to reach the tail, because the tail is moving towards it at speed r. The distance travelled by the car is: 3ry The distance travelled by the dragon is ry Hence, from what we said above: 3ry = 100  ry 4ry = 100 Now you know where these equations came from. 


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