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Difference between revisions of "2017 AMC 12A Problems/Problem 6"

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==Solution==
 
==Solution==
  
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The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than <math>15 - (3 + 7) = 5</math> and shorter than <math>15 + 3 + 7</math> = 25. This means Joy can use the 19 possible integer rod lengths that fall into <math>[6, 24]</math>. However, she has already used the rods of length <math>7</math> cm and <math>15</math> cm so the answer is <math>19 - 2 = 17</math> <math>\boxed{\textbf{(B)}}</math>
+
The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than <math>15 - (3 + 7) = 5</math> and shorter than <math>15 + 3 + 7 = 25</math>. This means Joy can use the <math>19</math> possible integer rod lengths that fall into <math>[6, 24]</math>. However, she has already used the rods of length <math>7</math> cm and <math>15</math> cm so the answer is <math>19 - 2 = 17</math> <math>\boxed{\textbf{(B)}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 12:58, 19 December 2020

Problem

Joy has $30$ thin rods, one each of every integer length from $1 \text{ cm}$ through $30 \text{ cm}$. She places the rods with lengths $3 \text{ cm}$, $7 \text{ cm}$, and $15 \text{cm}$ on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?

$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 19 \qquad\textbf{(E)}\ 20$

Solution

The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7 = 25$. This means Joy can use the $19$ possible integer rod lengths that fall into $[6, 24]$. However, she has already used the rods of length $7$ cm and $15$ cm so the answer is $19 - 2 = 17$ $\boxed{\textbf{(B)}}$

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 12 Problems and Solutions

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