Difference between revisions of "2016 AMC 10B Problems/Problem 13"
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<math> c=25 </math> | <math> c=25 </math> | ||
We have found our desire <math>(a,b,c)</math>=<math>(300,100,25)</math> | We have found our desire <math>(a,b,c)</math>=<math>(300,100,25)</math> | ||
− | There are 25 sets of quadruplets | + | 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 12:25, 21 February 2016
Problem
At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for 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 babies were in sets of quadruplets?
Solution
At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for 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 babies were in sets of quadruplets?
Let us have a system of equations where is the number of twins, is the number of triplets, and is the number of quadruplets. We have: We also have (Four times as many sets of triplets as quadruplets.) We also have (Three times as many sets of twins as triplets.) Substituting we have:
We have found our desire = There are 25 sets of quadruplets:
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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