Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 10A Problems/Problem 4"

m (→‎Problem)
(→‎Video Solutions)
 
(14 intermediate revisions by 8 users not shown)
Line 6: Line 6:
 
==Solution==
 
==Solution==
  
−
Every <math>30</math> seconds <math>3-2=1</math> toys are put in the box, so after <math>27\cdot30</math> seconds there will be <math>27</math> toys in the box. Mia's mom will then put <math>3</math> toys into to the box and we have our total amount of time to be <math>27\cdot30+30=840</math> seconds, which equals <math>14</math> minutes. <math>\boxed{(\textbf{B})\ 14}</math>
+
Every <math>30</math> seconds, <math>3</math> toys are put in the box and <math>2</math> toys are taken out, so the number of toys in the box increases by <math>3-2=1</math> every <math>30</math> seconds. Then after <math>27 \times 30 = 810</math> seconds (or <math>13 \frac{1}{2}</math> minutes), there are <math>27</math> toys in the box. Mia's mom will then put the remaining <math>3</math> toys into the box after <math>30</math> more seconds, so the total time taken is <math>27\times 30+30=840</math> seconds, or <math>\boxed{(\textbf{B})\ 14}</math> minutes.
  
−
==Solution 2==
+
Note: During the last time Mia's mom will complete picking up the 30 toys(before Mia can take 2 out) which is the reason that you calculate up to 27 and then the rest.
−
Though Mia's mom places <math>3</math> toys every <math>30</math> seconds, Mia takes out <math>2</math> toys right after. Therefore, after <math>30</math> seconds, the two have collectively placed <math>1</math> toy into the box. Therefore by <math>13.5</math> minutes, the two would have placed <math>27</math> toys into the box. Therefore, at <math>14</math> minutes, the two would have placed <math>30</math> toys into the box. Though Mia may take <math>2</math> toys out right after, the number of toys in the box first reaches <math>30</math> by <math>14</math> minutes. <math>\boxed{(\textbf{B})\ 14}</math>
+
 
 +
==Video Solutions==
 +
https://youtu.be/str7kmcRMY8
 +
-Video Solution by TheBeautyOfMath
 +
 
 +
https://youtu.be/1F0IB0y6578
 +
 
 +
~savannahsolver
  
 
==See also==
 
==See also==

Latest revision as of 14:15, 26 August 2022

Problem

Mia is "helping" her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?

$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$

Solution

Every $30$ seconds, $3$ toys are put in the box and $2$ toys are taken out, so the number of toys in the box increases by $3-2=1$ every $30$ seconds. Then after $27 \times 30 = 810$ seconds (or $13 \frac{1}{2}$ minutes), there are $27$ toys in the box. Mia's mom will then put the remaining $3$ toys into the box after $30$ more seconds, so the total time taken is $27\times 30+30=840$ seconds, or $\boxed{(\textbf{B})\ 14}$ minutes.

Note: During the last time Mia's mom will complete picking up the 30 toys(before Mia can take 2 out) which is the reason that you calculate up to 27 and then the rest.

Video Solutions

https://youtu.be/str7kmcRMY8 -Video Solution by TheBeautyOfMath

https://youtu.be/1F0IB0y6578

~savannahsolver

See also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&oldid=177536"