Problem 28 · 2015 Math Kangaroo
Stretch
Logic & Word Problems
caseworksum-constraint
On a board there are blue and red rectangles. Exactly 7 of the rectangles are squares. There are 3 more red rectangles than blue squares. There are also two more red squares than blue rectangles. How many blue rectangles are there on the board?
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Answer: B — 3
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Hint 1 of 2
Treat squares as a special kind of rectangle and name four counts: blue squares, blue non-square rectangles, red squares, red non-square rectangles.
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Hint 2 of 2
Translate each sentence into an equation, then require every count to be a non-negative whole number.
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Approach: set up four counts and solve with non-negativity
- Let TB be the total blue rectangles (squares included) and use: blue squares + red squares = 7; red rectangles = blue squares + 3; red squares = TB + 2.
- These give blue squares = 5−TB, blue non-square rectangles = 2·TB−5, red non-square rectangles = 6−2·TB.
- All must be ≥ 0, which forces TB = 3, so there are 3 blue rectangles (B).
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