Difference between revisions of "2016 AMC 10B Problems/Problem 13"
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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{(D)}\ 100}</math>. | + | 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{(D)}\ 100}</math>. |
==Solution 2== | ==Solution 2== | ||
Say there are <math>12x</math> sets of twins, <math>4x</math> sets of triplets, and <math>x</math> sets of quadruplets. That's <math>12x\cdot2=24x</math> twins, <math>4x\cdot3=12x</math> triplets, and <math>x\cdot4=4x</math> quadruplets. A tenth of the babies are quadruplets and that's <math>\frac{1}{10}(1000)=\boxed{\textbf{(D) }100}</math> | Say there are <math>12x</math> sets of twins, <math>4x</math> sets of triplets, and <math>x</math> sets of quadruplets. That's <math>12x\cdot2=24x</math> twins, <math>4x\cdot3=12x</math> triplets, and <math>x\cdot4=4x</math> quadruplets. A tenth of the babies are quadruplets and that's <math>\frac{1}{10}(1000)=\boxed{\textbf{(D) }100}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/9XU2ITIJRpk | ||
| + | |||
| + | ~savannahsolver | ||
==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}} | ||
Latest revision as of 20:57, 18 April 2022
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 1
We can set up a system of equations where 
is the sets of twins, 
is the sets of triplets, and 
is the sets of quadruplets.
![\[\begin{split} 2a + 3b + 4c & = 1000 \\ b & = 4c \\ a & = 3b \end{split}\]](http://latex.artofproblemsolving.com/e/1/9/e1922ea77df0f34afd0ad8dd22675f0e1c25e3dd.png)
Solving for and in the second and third equations and substituting into the first equation yields
![\[\begin{split} 2 (3b) + 3b + 4 (0.25b) & = 1000 \\ 6b + 3b + b & = 1000 \\ b & = 100 \end{split}\]](http://latex.artofproblemsolving.com/9/3/9/93942a59ca8959ac88e8b5ae99818ebf315ca679.png)
Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not , but rather 
. Therefore, we strategically use the second initial equation to realize that 
, leaving us with the number of babies born as quadruplets equal to 
.
Solution 2
Say there are 
sets of twins, 
sets of triplets, and 
sets of quadruplets. That's 
twins, 
triplets, and 
quadruplets. A tenth of the babies are quadruplets and that's 
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
