Difference between revisions of "2015 AMC 8 Problems/Problem 3"

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minutes for Jack to arrive at the pool.
 
minutes for Jack to arrive at the pool.
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\boxed{\textbf{(D)}~9}$
  
 
==Solution 2==
 
==Solution 2==

Revision as of 19:23, 15 September 2024

Problem

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

Solution

Using $d=rt$, we can set up an equation for when Jill arrives at the swimming pool:

$1=10t$

Solving for $t$, we get that Jill gets to the pool in $\frac{1}{10}$ of an hour, which is $6$ minutes. Doing the same for Jack, we get that

Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn is $15$ minutes. Thus, Jill has to wait $15-6=\boxed{\textbf{(D)}~9}$

minutes for Jack to arrive at the pool.


\boxed{\textbf{(D)}~9}$

Solution 2

T=D/s Jill: (1/10)x60 because in minutes, is equal to 6 min Jack:(1/4)x60 is 15 minutes.

15-6 is 9, so our answer is \boxed{\textbf{(D)}~9}$ - TheNerdWhoIsNerdy.

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/srGyMofBMsE

~Education, the Study of Everything


Video Solution

https://youtu.be/YvYq3iM4jP8

~savannahsolver

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

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