2016 AMC 10B Problems/Problem 13

Revision as of 12:28, 21 February 2016 by Shrummy (talk | contribs) (Solution)

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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png