HIGHLINE SCHOOL DISTRICT |
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MATH OLYMPIAD |
April 7, 2001 |
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The Great EscapeProblem Scoring Guide | ||||

Correct Answer

Points | Look for the following: |
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Correct answer, 84 coins, is given. | |

Answer is not correct |

Strategy

Points | Look for the following: |
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- Chose a valid strategy.(See discussion of valid strategies at the end of this rubric)
- Applied the strategy correctly and completely
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- Chose a valid strategy
- Applied the strategy correctly but with small omissions
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- Chose a valid strategy
- Strategy was partially or incompletely developed
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No attempt at response or inappropriate strategy |

Problem Understanding

Points | Look for the following: |
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- Shows understanding of the concept of
*fractional parts* - Understood that each fractional reduction was on
*remaining*coins
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- Understood concept of
*fractional parts* - Either did not understand that each fractional reduction was on
*remaining*coins or computed remaining fractional parts incorrectly
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No attempt at response. |

Communication

Points | Look for the following: |
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- Described strategy and solution clearly and completely, step-by-step
- Correct presentation of solution, with correct labels, terminology, and symbols
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- Described strategy correctly
- One step may not be described, requiring you to guess that it was done
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- Describe a strategy in some fashion, but discussion is incomplete.
- Steps are missing (they jumped to the answer)
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No attempt at response. |

Reasonable Result

Points | Look for the following: |
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- Reworked problem using a different strategy
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- Used their answer to run through each of the calculations forward to arrive at the final number of coins, 7.
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- Made some statement about checking their answer
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No attempt to demonstrate reasonableness. |

Valid strategies:

**Visual representation**: Drew a rectangle similar to the one below representing the transactions as they occur:1 2 3 1 2 4 1 2 4 1 2 **84**

Note: a pie chart or other similar representation works also.**Guess and Check**: Here is a typical guess-and-check scenario:

Guess that Spavadio had 75 coins.

Gave 1/3 to the first guard, leaving 50 coins.

Gave 1/2 to the second guard leaving 25 coins.

Gave 1/4 to the third guard...whoops! Can't take 1/4 of 25

New guess: 84 coins.

Gave 1/3 to first guard, leaving 56 coins.

Gave 1/2 to second guard, leaving 28 coins.

Gave 1/4 to third guard, leaving 21 coins.

Gave 2/3 to fourth guard, leaving 7 coins.It worked!**Work backwards**:

- Gave 2/3 to last guard and had 7 remaining, so he must have had 21 coins before he got to the last guard. (2/3 of 21 is 14,21-14 = 7)
- Gave 1/4 to third guard, so he must have had 28 coins before he got to the third guard (1/4 of 28 is 7, 28-7 = 21)
- Gave 1/2 to second guard, so he must have had 56 coins before he got to the second guard(1/2 of 56 = 28,56-28=28)
- Gave 1/3 to first guard, so he must have had 84 coins before he got to the first guard (1/3 of 84 = 28, 84-28=56)

**Use equations**: The above working backwards strategy can be used with equations to produce the results at each backward step:- Let X be number of coins after 3rd guard. So:

(1/3)X = 7, so X = 21;(After giving 2/3, 1/3 X is left) - Let Y be the number of coins after 2nd guard. So:

(3/4)Y= 21, so Y = 28; (after giving 1/4, 3/4 Y is left) - Let Z be the number of coins after the 1st guard. So:

(1/2)Z = 28, so Z = 56; (after giving 1/2, 1/2 Z is left) - Let C be the number of coins
**before**the 1st guard. So:

(2/3)C = 56, so C = 84, the correct answer.

- Let X be number of coins after 3rd guard. So: