You had solve word problems in PE? We just played dodge ball, steal the pins, whiffle ball and stuff like that.
Easy Word Problem to Equation Question
- Thread starter StClone
- Start date
I won’t attempt to create the equation, but I’ll take a stab at the story portion … Here’s how I interpret the summary (I could be wrong):
Car A travels 20 miles. Car B, traveling 20 mph faster than Car A, traverses 30 miles in the same amount of time. How much time has elapsed when B catches up to A?
Car A travels 20 miles. Car B, traveling 20 mph faster than Car A, traverses 30 miles in the same amount of time. How much time has elapsed when B catches up to A?
Not necessarily when B catches A, we don’t have enough info for that. Just how long those journeys take.
That makes more sense, I think you're correct now that I re-examine the original.
Driver A is an elderly individual who forgets where they are going, takes a wrong turn and then ends up during around to get directions, so they never technically make it there.
That's how I solve it:
distance = rate × time
Let's call the rate of the first car ""r"" (in miles per hour) and the time it takes to travel 20 miles ""t"" (in hours). Then, we can write:
distance = r × t
Similarly, we can call the rate of the second car ""r + 20"" (since it's traveling 20 miles per hour faster) and the time it takes to travel 30 miles ""t"" (since we're told the two cars take the same amount of time). Then, we can write:
distance = (r + 20) × t
Now we have two equations:
distance = r × t
distance = (r + 20) × t
Since we're trying to find the time it takes, we can set these two equations equal to each other:
r × t = (r + 20) × t
We can simplify this equation by dividing both sides by ""t"":
r = r + 20
This is a contradiction - the equation is saying that ""r"" is equal to ""r + 20"", which is impossible. Therefore, there is no solution to this problem.
However, if we assume that the distance traveled by the first car is actually 30 miles instead of 20 miles, then we can solve for the time it takes. Using the same equations as before, we get:
distance = r × t
distance = (r + 20) × t
30 = r × t
30 = (r + 20) × t
We can solve for ""t"" by setting the two equations equal to each other:
r × t = (r + 20) × t
r = r + 20
Simplifying this equation, we get:
20 = r
Substituting this value of ""r"" into one of our earlier equations, we get:
30 = 20 × t
Solving for ""t"", we get:
t = 1.5 hours
Therefore, it takes 1.5 hours for the first car to travel 30 miles and the second car to travel 50 miles.
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One of the worst worded problems I’ve ever read. I wouldn’t expect any student to get that correct.
