2015 AMC 8 Problems/Problem 17
Contents
Problem
Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
Solutions
Solution 1
For starters, we identify d as distance and v as velocity (speed)
Writing the equation gives us: and .
This gives , which gives , which then gives .
Solution 2
,
so , plug into the first one and it's miles to school.
Solution 3
We set up an equation in terms of the distance and the speed In miles per hour. We have , giving Hence, .
Solution 4
Since it takes of the original time for him to get to school when there is no traffic, the speed must be of the speed in no traffic or more. Letting be the rate and we know that , so we have miles per hour. Solving for gives us miles per hour. Because minutes is a third of an hour, the distance would then be miles .
Solution 5
When driving in rush hour traffic, he drives minutes for one distance () to the school. It means he drives minutes for distances () to the school. When driving in no traffic hours, he drives minutes for one distance () to the school. It means he drives minutes for distances () to the school. Subtracting these two situations, it gives us , then . So the distance to the school would be miles. ----LarryFlora
Solution 6
(Ratios)
In rush hour traffic, we can create this ratio for each distance we are going to try supposing the distance is :
: minutes
In no traffic, we can do the same:
: minutes
We want the ratio to be the distance to minutes:
: minutes = d : minutes
: minutes = d : minutes
This shows the speed in miles per hour. This means that the speed in the 2nd ratio has to be 18 faster than the first ratio. We get
, or .
Video Solution (HOW TO THINK CRITICALLY!!!)
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Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=3033
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Video Solution
https://www.youtube.com/watch?v=TFm1jNgB4QM ~David
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See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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