In February I told you the Trottier math team placed 6th in the MathCounts chapter competition to qualify for the state competition. States were held on March 7th at Wentworth Institute of Technology, and the math whizzes from Trottier placed a respectable 16th out of 27 teams.
Trottier eighth-grader Tommy Yu ranked 8th among all mathletes in the competition.
Coach and Trottier math teacher Tom Griffin put Yu’s performance in perspective. “It started with 839 kids in Massachusetts, and 140 made the state competition – Tommy placed 8 – a really amazing accomplishment.”
And since you all had fun testing your math skills last time, here’s a question from last year’s state competition to puzzle over. Make your guesses in the comments, and I’ll post the answer there tomorrow.
A man has a 100-acre ranch that he wishes to stock with cows and sheep. Cows require ten acres of grazing land per animal and sheep require three acres of grazing land per animal. If he wishes to fully utilize the land, how many ways are there to stock the ranch so that it includes at least one cow and at least one sheep?
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Are you smarter than an eigth grader?
Well, apparently no one wanted to risk public humiliation by posting their guess, but as promised, here’s the solution from the MathCounts answer key.
The answer is three.
We have a 100-acre ranch. Cows require ten acres each of grazing land. Sheep require three acres each of grazing land. We are asked to find how many ways there are to stock the ranch so that it includes at least one cow and one sheep.
We certainly can’t have 10 cows because then we’d have no sheep.If we had 9 cows they would use 90 acres but we couldn’t fully utilize the rest of the land with sheep because we could put 3 sheep in using 9 acres. But that would leave 1 acre unutilized.
The same would be true with 8 cows because they would use 80 acres and 100 – 80 = 20 is not divisible by 3. 30 acres is so we could certainly have 7 cows and 10 sheep.
Similarly, we can’t have 6 or 5 cows but we could have 4 cows using 40 acres and that would leave 60 acres for 20 sheep. That’s the second way.
Similarly, we also can’t have 3 or 2 cows but we could have 1 cow and 30 sheep for the third way.