Saturday, March 29, 2014

Mathcounts Prep -- Number Sense

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Problems: (Solutions below)
#1: 2005 Chapter Team-- A standard deck of playing cards with 26 red cards and 26 black cards is split into two piles, each having at least one card. In pile A there are six times as many black cards as red cards. In pile B, the number of red cards is a multiple of the number of black cards. How many red cards are in pile B?

#2:  2000 State sprint #30. Joe bought a pumpkin that cost 10 cents more per pound than his sister's. Together, the two pumpkins weighed 20 pounds, but Joe's pumpkin was heavier. Joe paid \ 3.52 dollars and his sister paid 48 cents. How many pounds did Joe's pumpkin weigh?











 

Solutions :

#1: You know the total cards in pile A is a multiple of 7 because there are six times as many black cards as the red cards. (given)

6 Black, 1 Red on pile A gives you 20 Black and 25 Red cards on pile B. (doesn't work)
12 Black and 2 Red cards on pile A gives you 14 Black and 24 Red cards on pile B. (doesn't work)
18 Black and 3 Red cards on pile A gives you 8 Black and 23 Red cards on pile B. (doesn't work)
24 Black and 4 Red cards on pile A gives you 2 Black and 22 Red cards on pile B. Yes!!
The answer is 22 Red cards.

#2: 
Solution I : 
Let x dollars be the cost per pound for Joe's sister's pumpkin and x + .1 dollars are the cost per pound for Joe's pumpkin. Since the pounds of each pumpkin is the cost \div cost per pound, we have
\frac{3.52}{x+.1}+\frac{.48}{x}=20. Solving, x=.12. Cost per pumpkin for Joe is .22, so \frac{3.52}{.22}=\boxed{16 \text{ pounds}}

Solution II:  Make a list:
Joe's sister        Joe
1 lb.                 19 lb.       (doesn't work since 19 x 58 cents are too much)
2 lbs.                18 lbs      (doesn't work)
3 lbs.                17 lbs      (No)17 x (48/3 + 10) = 442 (still too much)
4 lbs                 16 lbs      16 x (48/4 + 10)= 352 (yes)
The answer is 16 pounds.

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