Difference between revisions of "2017 AMC 10A Problems/Problem 4"
Mathpro12345 (talk | contribs) (→Solution 1) |
Scrabbler94 (talk | contribs) (combine solutions 1 and 2 as they are essentially the same. also clarify that the number of toys increases by 1; not necessarily "1 toy is placed in the box" since different toys can be put in and taken out.) |
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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== |
− | Every <math>30</math> seconds, <math>3 | + | 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 increases by <math>3-2=1</math> every 30 seconds. Then after <math>27 \times 30 = 810</math> seconds (or <math>13 \frac{1}{2}</math> minutes), there are 27 toys in the box. Mia's mom will then put <math>3</math> toys into the box after 30 more seconds, so the total time taken is <math>27\cdot30+30=840</math> seconds, or <math>\boxed{(\textbf{B})\ 14}</math> minutes. |
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==Video Solution== | ==Video Solution== |
Revision as of 13:11, 3 February 2021
Contents
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 and toys are taken out, so the number of toys increases by every 30 seconds. Then after seconds (or minutes), there are 27 toys in the box. Mia's mom will then put toys into the box after 30 more seconds, so the total time taken is seconds, or 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.