Sideways Arithmetic From Wayside School (Wayside School 2.50)

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Problem 12: o is an even number.

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Problem 17: m = 5.

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Problem 18: m = 7.

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Problem 20: h = 1; e = 5.

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Problem 21: s = 1; o = 7.

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Problem 22: e = 6.

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Problem 23: s = 4.

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Problem 24: w = 8; s = 2.

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Problem 29: The best way to figure out this problem is to figure out who was the person who got all the answers correct. Remember that nobody missed all five. Also, notice that Paul and Bebe never gave the same answer. Therefore Paul could not have gotten them all right, because if he did, then Bebe would have gotten them all wrong. Similarly Bebe couldn’t have gotten them all right, because then Paul would have missed them all. Check out what the results would be if Todd, Sharie, or Jason got them all right.

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Problem 30: Nancy only missed one answer. The easiest way to figure out the problem is to figure out which answer Nancy missed.

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Problem 31: Deedee missed question 3.

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Problem 32: Compare Joe’s answers to Maurecia’s. They are all the same except for the last two. Since Joe got a better grade than Maurecia, what are the correct answers to problems 4 and 5? Now compare Allison’s answers to Stephen’s.

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Problem 33: Compare Dameon’s answers to D.J.’s. They are identical except for question 1. Since Dameon got a better grade than D.J., Dameon’s answer for question 1 must be correct. Now compare Terrence’s to Rondi’s. Rondi got a better grade than Terrence, but Terrence got the first question right, and Rondi missed it.

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Problem 34: Who got all five correct?

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Problem 35: No two people answered the same question correctly. Therefore the answer to question 1 can’t be John Kennedy. The answer to question 2 can’t be George Washington. Three can’t be Jimmy Carter. Four has to be Abe Lincoln. Five can’t be Abe Lincoln.