Difference between revisions of "2017 AMC 10A Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | Every <math>30</math> seconds <math>3-2</math> toys are put in the box, so after <math>27\ | + | Every <math>30</math> seconds <math>3-2</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{(B) 14}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:15, 8 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.