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2010 Math Kangaroo

Problem 17

Problem 17 · 2010 Math Kangaroo Hard
Logic & Word Problems casework

In a box are 50 counters: white ones, blue ones, and red ones. There are eleven times as many white ones as blue ones. There are fewer red ones than white ones, but more red ones than blue ones. By how much is the number of red counters less than the number of white counters in the box?

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Answer: C — 19
Show hints
Hint 1 of 2
Let blue = b; then white = 11b, and the total is 50.
Still stuck? Show hint 2 →
Hint 2 of 2
Use that red sits strictly between blue and white to pin down b.
Show solution
Approach: set up totals, test integer cases
  1. With blue = b, white = 11b, red = r: 12b + r = 50 and b < r < 11b.
  2. b = 3 gives r = 14 (and 3 < 14 < 33, valid); b = 4 gives r = 2, breaking r > b.
  3. So white − red = 33 − 14 = 19.
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