Problem 9 · 2010 Math Kangaroo
Easy
Algebra & Patterns
ratiosum-constraint
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 ones in the box?
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Answer: C — 19
Show hints
Hint 1 of 2
Let the number of blue ones be small and write white as eleven times that.
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Hint 2 of 2
Use the clues that red is between blue and white to pin everything down.
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Approach: set up with the blue count
- If blue = b then white = 11b, and blue + white + red = 50 gives red = 50 - 12b.
- With b = 3: white = 33, red = 14, and indeed blue(3) < red(14) < white(33).
- White minus red = 33 - 14 = 19.
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