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Difference between revisions of "2016 AMC 10B Problems/Problem 13"

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==Solution==
 
==Solution==
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Let us have a system of equations where <math>a</math> is the number of twins, <math>b</math> is the number of triplets, and <math>c</math> is the number of quadruplets.
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We can set up a system of equations where <math>a</math> is the sets of twins, <math>b</math> is the sets of triplets, and <math>c</math> is the sets of quadruplets.
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We have: <math> 2a+3b+4c = 1000</math>
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<cmath>\begin{split}
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We also have <math>b=4c</math> (Four times as many sets of triplets as quadruplets.)
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2a + 3b + 4c & = 1000 \\
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We also have <math>a=3b</math> (Three times as many sets of twins as triplets.)
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b & = 4c \\
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Substituting we have:
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a & = 3b
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            <math> 2a+a+a/3= 1000</math>
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\end{split}</cmath>
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            <math> 6a+3a+a = 3000</math>  
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            <math>          10a= 3000</math>
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Solving for <math>c</math> and <math>a</math> in the second and third equations and substituting into the first equation yields
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            <math>              a= 300  </math>
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<cmath>\begin{split}
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            <math>              a= 3b   </math>
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2 (3b) + 3b + 4 (0.25b) & = 1000 \\
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            <math>          300=3b    </math>
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6b + 3b + b & = 1000 \\
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            <math>              b=100   </math>
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b & = 100
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            <math>             b=4c    </math>
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\end{split}</cmath>
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            <math>         100=4c     </math>
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            <math>             c=25    </math>
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Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not <math>c</math>, but rather <math>4c</math>. Therefore, we strategically use the second initial equation to realize that <math>b</math> <math>=</math> <math>4c</math>, leaving us with the number of babies born as quadruplets equal to <math>\boxed{\textbf{(C)}\ 100}</math>.
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We have found our desire <math>(a,b,c)</math>=<math>(300,100,25)</math>
 
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There are 25 sets of quadruplets: <math>A</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2016|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:53, 21 February 2016

Problem

At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$

Solution

We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. \[\begin{split} 2a + 3b + 4c & = 1000 \\ b & = 4c \\ a & = 3b \end{split}\]

Solving for $c$ and $a$ in the second and third equations and substituting into the first equation yields \[\begin{split} 2 (3b) + 3b + 4 (0.25b) & = 1000 \\ 6b + 3b + b & = 1000 \\ b & = 100 \end{split}\]

Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not $c$, but rather $4c$. Therefore, we strategically use the second initial equation to realize that $b$ $=$ $4c$, leaving us with the number of babies born as quadruplets equal to $\boxed{\textbf{(C)}\ 100}$.

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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