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

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==Solution==
 
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At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for <math>1000</math> 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 <math>1000</math> babies were in sets of quadruplets?
 
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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.
 
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.
 
We have: <math> 2a+3b+4c = 1000</math>
 
We have: <math> 2a+3b+4c = 1000</math>

Revision as of 13:28, 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

Let us have a system of equations where $a$ is the number of twins, $b$ is the number of triplets, and $c$ is the number of quadruplets. We have: $2a+3b+4c = 1000$ We also have $b=4c$ (Four times as many sets of triplets as quadruplets.) We also have $a=3b$ (Three times as many sets of twins as triplets.) Substituting we have: 

           $2a+a+a/3= 1000$
           $6a+3a+a = 3000$ 
           $10a= 3000$
           $a= 300$
           $a= 3b$
           $300=3b$
           $b=100$
           $b=4c$
           $100=4c$
           $c=25$

We have found our desire $(a,b,c)$=$(300,100,25)$ There are 25 sets of quadruplets: $A$

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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