Difference between revisions of "2012 AMC 10A Problems/Problem 5"
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Half of the 100 adult cats are female, so there are <math>\frac{100}{2}</math> = <math>50</math> female cats. Half of those female adult cats have a litter of kittens, so there would be <math>\frac{50}{2}</math> = <math>25</math> litters. Since the average number of kittens per litter is 4, this implies that there are <math>25 \times 4</math> = <math>100</math> kittens. So the total number of cats and kittens would be <math>100 + 100</math> = <math>\boxed{\textbf{(B)}\ 200}</math> | Half of the 100 adult cats are female, so there are <math>\frac{100}{2}</math> = <math>50</math> female cats. Half of those female adult cats have a litter of kittens, so there would be <math>\frac{50}{2}</math> = <math>25</math> litters. Since the average number of kittens per litter is 4, this implies that there are <math>25 \times 4</math> = <math>100</math> kittens. So the total number of cats and kittens would be <math>100 + 100</math> = <math>\boxed{\textbf{(B)}\ 200}</math> | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/fAvTSrqCSus | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Latest revision as of 13:56, 1 July 2023
Problem
Last year 100 adult cats, half of whom were female, were brought into the Smallville Animal Shelter. Half of the adult female cats were accompanied by a litter of kittens. The average number of kittens per litter was 4. What was the total number of cats and kittens received by the shelter last year?
Solution
Half of the 100 adult cats are female, so there are 
= 
female cats. Half of those female adult cats have a litter of kittens, so there would be 
= 
litters. Since the average number of kittens per litter is 4, this implies that there are 
= 
kittens. So the total number of cats and kittens would be 
= 
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
See Also
| 2012 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
