Difference between revisions of "2015 AMC 8 Problems/Problem 17"
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===Solution 1=== | ===Solution 1=== | ||
+ | |||
+ | For starters, we identify d as distance and v as velocity (speed) | ||
Writing the equation gives us: <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>. | Writing the equation gives us: <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>. | ||
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===Solution 6=== | ===Solution 6=== | ||
− | (Ratios | + | (Ratios) |
+ | <math>\textbf{requires edits}</math> | ||
− | In rush hour traffic, we can create this ratio for each distance we are going to try supposing distance is d | + | In rush hour traffic, we can create this ratio for each distance we are going to try supposing the distance is <math>d</math>: |
− | d / | + | <math>d</math> : <math>20</math> minutes |
− | In no traffic, we can do the same | + | In no traffic, we can do the same: |
− | d : <math>12</math> minutes | + | <math>d</math> : <math>12</math> minutes |
− | We want the ratio to be the distance to <math>60</math> minutes | + | We want the ratio to be the distance to <math>60</math> minutes: |
− | d : <math>20</math> minutes = <math>3</math>d : <math>60</math> minutes | + | <math>d</math> : <math>20</math> minutes = <math>3</math>d : <math>60</math> minutes |
− | d : <math>12</math> minutes = <math>5</math>d : <math>60</math> minutes | + | <math>d</math> : <math>12</math> minutes = <math>5</math>d : <math>60</math> 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 | + | 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 |
− | + | <math>3d+18 = 5d \cdot d=9</math>, or <math>\boxed{\textbf{(D)}~9}</math>. | |
==Video Solution (HOW TO THINK CRITICALLY!!!)== | ==Video Solution (HOW TO THINK CRITICALLY!!!)== |
Latest revision as of 17:05, 17 January 2024
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!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=3033
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=TFm1jNgB4QM ~David
~savannahsolver
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.