Difference between revisions of "2017 AMC 10A Problems/Problem 4"
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<math>\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5</math> | <math>\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | 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 | + | 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 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> |
==Solution 2== | ==Solution 2== | ||
Revision as of 11:08, 31 January 2021
Problem
Mia is "helping" her mom pick up 
toys that are strewn on the floor. Mia’s mom manages to put 
toys into the toy box every seconds, but each time immediately after those seconds have elapsed, Mia takes 
toys out of the box. How much time, in minutes, will it take Mia and her mom to put all toys into the box for the first time?
Solution 1
Every seconds, 
toys are put in the box, so after 
seconds, there will be 
toys in the box. Mia's mom will then put toys into the box, and we have our total amount of time to be 
seconds, which equals 
minutes. 
Solution 2
Though Mia's mom places toys every seconds, Mia takes out toys right after. Therefore, after seconds, the two have collectively placed 
toy into the box. Therefore by 
minutes, the two would have placed toys into the box. Therefore, at minutes, the two would have placed toys into the box. Though Mia may take toys out right after, the number of toys in the box first reaches by minutes.
Video Solution
~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. 
