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Difference between revisions of "2017 AMC 10A Problems/Problem 4"
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==Solution 2==
 
==Solution 2==
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Though Mia's mom places <math>3</math> toys every <math>30</math> seconds, Mia takes out <math>2</math> toys right after. Therefore, after <math>30</math> seconds, the two have collectively placed <math>1</math> toy into the box. Thereforeby <math>13.5</math> minutes, the two would have placed <math>27</math> toys into the box. Therefore, at <math>14</math> minutes, the two would have placed <math>30</math> toys into the box. Though Mia may take <math>2</math> toys out right after, the number of toys in the box first reaches <math>30</math> by <math>14</math> minutes.  
+
Though Mia's mom places <math>3</math> toys every <math>30</math> seconds, Mia takes out <math>2</math> toys right after. Therefore, after <math>30</math> seconds, the two have collectively placed <math>1</math> toy into the box. Thereforeby <math>13.5</math> minutes, the two would have placed <math>27</math> toys into the box. Therefore, at <math>14</math> minutes, the two would have placed <math>30</math> toys into the box. Though Mia may take <math>2</math> toys out right after, the number of toys in the box first reaches <math>30</math> by <math>14</math> minutes. <math>\boxed{(\textbf{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 15:02, 14 February 2017

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 $3-2=1$ toys are put in the box, so after $27\cdot30$ seconds there will be $27$ toys in the box. Mia's mom will then put $3$ toys into to the box and we have our total amount of time to be $27\cdot30+30=840$ seconds, which equals $14$ minutes. $\boxed{(\textbf{B})\ 14}$

Solution 2

Though Mia's mom places $3$ toys every $30$ seconds, Mia takes out $2$ toys right after. Therefore, after $30$ seconds, the two have collectively placed $1$ toy into the box. Thereforeby $13.5$ minutes, the two would have placed $27$ toys into the box. Therefore, at $14$ minutes, the two would have placed $30$ toys into the box. Though Mia may take $2$ toys out right after, the number of toys in the box first reaches $30$ by $14$ minutes. $\boxed{(\textbf{B})\ 14}$

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

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