Difference between revisions of "2005 AMC 8 Problems/Problem 12"
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==Solution 3 (Lightning Quick)== | ==Solution 3 (Lightning Quick)== | ||
− | Simply realize that the middle term of the arithmetic sequence is the arithmetic mean of all terms, which is simply <math>100 | + | Simply realize that the middle term of the arithmetic sequence is the arithmetic mean of all terms, which is simply <math>\frac{100}{5}=20</math>. This means that the number of bananas the ape ate on May 5th is just <math>20+6*2=32</math>. Select <math>\boxed{D}</math>. |
You can do this quicker than you read it if you truly master arithmetic sequences. | You can do this quicker than you read it if you truly master arithmetic sequences. |
Latest revision as of 10:48, 4 May 2022
Problem
Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
Solution
There are days from May 1 to May 5. The number of bananas he eats each day is an arithmetic sequence. He eats bananas on May 5, and bananas on May 1. The sum of this arithmetic sequence is equal to .
Solution 2
There are days from May 1 to May 5. If we set the first day as , the second day can be expressed as , the third as , and so on, for five days. The sum is equal to , as stated in the problem. We can write a very simple equation, that is: . Now all we do is just solve. , so Big Al eats bananas on the first day. The fifth day, , is then , which is your answer.
Solution 3 (Lightning Quick)
Simply realize that the middle term of the arithmetic sequence is the arithmetic mean of all terms, which is simply . This means that the number of bananas the ape ate on May 5th is just . Select .
You can do this quicker than you read it if you truly master arithmetic sequences.
~hastapasta
See Also
2005 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.