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 cents more per pound than his sister's. Together, the two pumpkins weighed pounds, but Joe's pumpkin was heavier. Joe paid dollars and his sister paid 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 cost per pound, we have

. Solving, . Cost per pumpkin for Joe is , so

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.