Difference between revisions of "2011 AMC 10A Problems/Problem 8"
m (→Solution 2) |
m (→Solution 1) |
||
(One intermediate revision by one other user not shown) | |||
Line 8: | Line 8: | ||
=== Solution 1 === | === Solution 1 === | ||
− | 75% of the total birds were not swans. Out of that 75%, there was <math>30\% / 75\% = \boxed{40\%(C)}</math> of the birds that were not swans that were geese. | + | 75% of the total birds were not swans. Out of that 75%, there was <math>30\% / 75\% = \boxed{40\%\text{\textbf{ (C)}}}</math> of the birds that were not swans that were geese. |
=== Solution 2 === | === Solution 2 === | ||
WLOG, suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. <math>100-25 = 75</math> of the birds are not swans and 30 of these are geese, so the answer is <math>\frac{30}{75} \times 100 = \boxed{40 \%(C)}</math>. | WLOG, suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. <math>100-25 = 75</math> of the birds are not swans and 30 of these are geese, so the answer is <math>\frac{30}{75} \times 100 = \boxed{40 \%(C)}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/fFF5q_PuIZQ | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=7|num-a=9}} | {{AMC10 box|year=2011|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:12, 19 October 2023
Problem 8
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?
Solutions
Solution 1
75% of the total birds were not swans. Out of that 75%, there was of the birds that were not swans that were geese.
Solution 2
WLOG, suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. of the birds are not swans and 30 of these are geese, so the answer is .
Video Solution
~savannahsolver
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.