BY K.H. “WICKS” WICKREMASINGHE

Seniors' Ages

Al, the youngest of six seniors (Al, Ben, Cam, David, Ezra and Fred), announced, “For each of us, I have calculated our average age before and then after including each of our ages. Fred's age is 50-percent higher than mine. My age dropped the average by 3.6 and Ezra's dropped it by 1.8. Cam's, Ben's and David's ages raised it by 0.6, 0.8 and 1.2, respectively.”
What are their six ages?

Grand Prize Winners
Congratulations to Edward J. Breitung of Hobbs Corp. (Springfield, Ill.) for solving CrossNumber 62, and to Peter Douglas of Custom Chip Connections Inc. (Huntsville, Ala.) for solving the May Day Party puzzle.

CrossNumber 64

Across

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1. 24 Down + 30 Across
4. (8 Across + 30 Across) ÷ 2
7. Square root of (23 Down x 28 Down)
8. 26 Down + 30 Across
10. 8 Down – 22 Down
11. 1 Across x 7 Across
13. (15 Down – 1 Across) ÷ 2
14. (2 Down + 28 Down) big ÷ 2
16. 1 Across + 4 Across
17. (1 Across x 22 Down) ÷ 2 – 24 Down
18. 22 Across – 8 Down
20. (10 Across + 18 Across) ÷ 2
22. 23 Down + 26 Down
25. (28 Across + 28 Down) ÷ 2
27. 4 x 22 Across
28. (7 Across + 22 Down + 25 Across) ÷ 3
29. (28 Down squared) ÷ 4
30. (8 Down x 28 Down) ÷ 4

Down

1. (2 x 29 Across) – 1
2. 2 x 26 Down squared ÷ 28 Down
3. (24 Down x 28 Across) ÷ 2
5. 2 x 26 Down
6. 19 Down – 24 Down
8. 24 Down – 23 Down
9. (2 Down + 5 Down) ÷ 2
11. 3 Down + 19 Down + 24 Down
12. 22 Down x (21 Down – 26 Down)
13. 15 Down – (19 Down ÷ 2)
15. 8 Across + 24 Down
19. (23 Down squared) ÷ 4
21. 22 Across + 24 Down
22. 7 Across – 25 Across
23. 2 x 28 Across
24. (2 x 23 Down) – 1
26. (24 Down – 1) ÷ 3
28. (7 Across + 20 Across + 22 Down) ÷ 3

Contest Rules: All entries must be received by July 30, 2001. The winners of the word puzzle and CrossNumber (can be the same person) will be drawn from all correct entries. Fax all entries to 847-634-4240. All entries must include name, complete address, company affiliation and daytime phone number to be considered.

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Solutions to May's Puzzles

CrossNumber 62

May Day Party
Answer: 4 and 5 miles
Solution: Let the starting point be denoted by A, the correct exit by E, the point at which she realized the mistake be B and the next exit be C, and the party is at D. Let AE = x, EB = y and BC = z. ED is given as 20 miles.
If time taken from A to D without a mistake is T, then T = x x 60 ÷ 60 + 20 x 60 ÷ 30.

i.e., T = x + 40 Equation (1)

Time with the mistake T + 6 = x x 60 ÷ 60 + y x 60 ÷ 60 + (2z + y)(60 ÷ 70) + 20 x (60 ÷ 40) therefor x + 130 y ÷ 70 + 120z ÷ 70 + 30 = T + 6 Equation (2)

Equation (2) – Equation (1) and simplifying 13y + 12z = 112. Because y and z are integers, this can be written as z = 9 – y + (4 – y) ÷ 12. I.e., z = 9 – y + k, where k is an integer. k = (4 – y) big ÷ 12; y = 4 – 12k. This will have a positive integer valued only when k = 0 and y = 4. Thus, z = 5.