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Algebra rate problem

by daigo
Tags: algebra, rate
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daigo
#1
Feb13-12, 10:54 PM
P: 27
1. The problem statement, all variables and given/known data
A long dragon that is 100 feet long is walking.
A car starts driving from the rear of the dragon to the head of the dragon, then drives back to the tail of the dragon.
If the car drives three times as fast as the dragon can walk, how many feet has the dragon walked by the time the car has returned to the tail of the dragon?
2. Relevant equations

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?
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cepheid
#2
Feb13-12, 11:20 PM
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Quote Quote by daigo View Post
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?
This solution is not hard to understand. You're breaking the problem up into two parts. The first part is when the car moves from tail to head, which takes a time "t." If "r" is the walking speed of the dragon, then 3r is the driving speed of the car (this is given in the problem). In time t, the car moves a distance of speed*time = 3rt. HOWEVER, the distance travelled by the car must be equal to the length of the dragon PLUS the distance travelled by the dragon in time t, which is rt. Hence:

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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