Difference between revisions of "2017 AMC 10A Problems/Problem 4"
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==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>3-2=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> |
==See also== | ==See also== | ||
Revision as of 14:12, 9 February 2017
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
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 to the box and we have our total amount of time to be 
seconds, which equals 
minutes. 
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. 
