Difference between revisions of "2015 AMC 8 Problems/Problem 17"

Latest revision as of 16:05, 17 January 2024

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?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Solutions

Solution 1

For starters, we identify d as distance and v as velocity (speed)

Writing the equation gives us: $\frac{d}{v}=\frac{1}{3}$ and $\frac{d}{v+18}=\frac{1}{5}$ .

This gives $d=\frac{1}{5}v+3.6=\frac{1}{3}v$ , which gives $v=27$ , which then gives $d=\boxed{\textbf{(D)}~9}$ .

Solution 2

$d = rt$ , $d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)$

$\frac{r}{3} = \frac{r}{5} + \frac{18}{5}$

$10r = 270$ so $r = 27$ , plug into the first one and it's $\boxed{\textbf{(D)}~9}$ miles to school.

Solution 3

We set up an equation in terms of $d$ the distance and $x$ the speed In miles per hour. We have $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ , giving $(5)(x)=(3)(x+18)$ $5x=3x+54$ $2x=54$ $x=27$ Hence, $d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}$ .

Solution 4

Since it takes $\frac{3}{5}$ of the original time for him to get to school when there is no traffic, the speed must be $\frac{5}{3}$ of the speed in no traffic or $\frac{2}{3}$ more. Letting $x$ be the rate and we know that $\frac{5}{3}x = x + 18$ , so we have $\frac{2x}{3} = 18$ miles per hour. Solving for $x$ gives us $27$ miles per hour. Because $20$ minutes is a third of an hour, the distance would then be $9$ miles $\boxed{\textbf{(D)}~9}$ .

Solution 5

When driving in rush hour traffic, he drives $20$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $3$ distances ( $3d$ ) to the school. When driving in no traffic hours, he drives $12$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $5$ distances ( $5d$ ) to the school. Subtracting these two situations, it gives us $5d-3d = 18 = 2d$ , then $d=\frac{18}{2}=9$ . So the distance to the school would be $\boxed{\textbf{(D)}~9}$ miles. ----LarryFlora

Solution 6

(Ratios) $\textbf{requires edits}$

In rush hour traffic, we can create this ratio for each distance we are going to try supposing the distance is $d$ :

$d$ : $20$ minutes

In no traffic, we can do the same:

$d$ : $12$ minutes

We want the ratio to be the distance to $60$ minutes:

$d$ : $20$ minutes = $3$ d : $60$ minutes

$d$ : $12$ minutes = $5$ d : $60$ 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

$3d+18 = 5d \cdot d=9$ , or $\boxed{\textbf{(D)}~9}$ .

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/J15jiF1mwdU

~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

https://youtu.be/4wZ4ToIyrnw

~savannahsolver

@@ Line 47: / Line 47: @@
 ===Solution 6===
 (Ratios)
+<math>\textbf{requires edits}</math>
 In rush hour traffic, we can create this ratio for each distance we are going to try supposing the distance is <math>d</math>:
@@ Line 64: / Line 65: @@
 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 = </math>5d \cdot d=9<math> </math>\boxed{\textbf{(D)}~9}$.
+<math>3d+18 = 5d \cdot d=9</math>, or <math>\boxed{\textbf{(D)}~9}</math>.
 ==Video Solution (HOW TO THINK CRITICALLY!!!)==

2015 AMC 8 (Problems • Answer Key • Resources)
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