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

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

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

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