2009 AMC 12A Problems/Problem 4: Difference between revisions
Yanchudeng (talk | contribs) |
|||
| (13 intermediate revisions by 2 users not shown) | |||
| Line 8: | Line 8: | ||
== Solution 1 == | == Solution 1 == | ||
Pre-Note: This solution is kinda just guessing, idk | Pre-Note: This solution is kinda just guessing, idk you decide. | ||
We can solve this problem by trying out numbers to get the answer choices and use the process of elimination. Thinking for a few seconds for 15 cents | We can solve this problem by trying out numbers to get the answer choices and use the process of elimination to get the answer. Thinking for a few seconds, you can see that for 15 cents there are no possible ways to get it. Now you may be inclined to chose <imath>\textbf{A}</imath>, but we may have missed some way to get 15 cents, so we try the other answer choices. 25 cents can be made with 3 nickels and 1 dimes, 35 cents can be made with 3 dimes and 1 nickel, 45 cents can be made with 1 quarter, 1 dime, and 2 nickels, and finally, 55 cents can be made with 1 quarter and 3 dimes. Therefore the answer is <imath>\boxed{\textbf{A}}</imath> | ||
[[User:Maxisw|MisW]] ([[User talk:Maxisw|talk]]) | |||
Minor Edits by [[yanchudeng]] | |||
Post-Note: This is my first time using LaTeX so it may look a little ugly. | |||
== See Also == | == See Also == | ||
Latest revision as of 18:51, 9 November 2025
- The following problem is from both the 2009 AMC 12A #4 and 2009 AMC 10A #2, so both problems redirect to this page.
Problem
Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?
Solution 1
Pre-Note: This solution is kinda just guessing, idk you decide.
We can solve this problem by trying out numbers to get the answer choices and use the process of elimination to get the answer. Thinking for a few seconds, you can see that for 15 cents there are no possible ways to get it. Now you may be inclined to chose 
, but we may have missed some way to get 15 cents, so we try the other answer choices. 25 cents can be made with 3 nickels and 1 dimes, 35 cents can be made with 3 dimes and 1 nickel, 45 cents can be made with 1 quarter, 1 dime, and 2 nickels, and finally, 55 cents can be made with 1 quarter and 3 dimes. Therefore the answer is 
MisW (talk) Minor Edits by yanchudeng
Post-Note: This is my first time using LaTeX so it may look a little ugly.
See Also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| 2009 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America. 
