Family Night Tables: Choosing and Using a Rule
The cafeteria is setting up square tables for family night. Each table seats 6, and one extra chair is placed at each end of the whole row, so a row of 1 table seats 8, a row of 2 tables seats 14, and a row of 3 tables seats 20. Sixty-two people have returned RSVPs. Three students propose rules for any number of tables n: Ella says “multiply n by 6,” Jon says “multiply n by 6 and add 2,” and Kim says “start at 8 and add 6 each time.” The custodian reports that the fire code allows no more than 12 tables in the row. The teacher must pick the rule the class will use, decide how many tables to set for 62 guests, and be ready for the office calling back to say 80 people are now expected.
1. Which student's rule should the class adopt? Test each proposal against the data and explain the difference between a rule and a pattern description.
Read the worked answer
Start with n = 1, because that is where the rules first disagree. Ella gives 6 × 1 = 6, but a one-table row seats 8, so her rule fails at the first test - no further checking is needed. Jon gives 6 × 1 + 2 = 8, 6 × 2 + 2 = 14, and 6 × 3 + 2 = 20, matching all three rows. Kim's instruction reproduces 8, 14, 20, but only by stepping from the previous row; it cannot produce the seats for row 9 without building rows 1 through 8. In fact 8 + 6(n − 1) simplifies to 6n + 2, so Kim and Jon agree. Adopt Jon's expression, 6n + 2, because it goes straight from n tables to the number of seats.
2. Use the chosen rule to find how many tables are needed for 62 guests, and check whether the fire code allows that row.
Read the worked answer
Solve 6n + 2 = 62 by undoing operations in reverse order: subtract 2 from both sides to get 6n = 60, then divide both sides by 6 to get n = 10. Check by substitution: 6 × 10 + 2 = 60 + 2 = 62. Since 10 tables is under the 12-table limit, the plan is legal, and the class can tell the custodian that 10 tables hold 60 plus the 2 end chairs. If a group reports n = 11, substitution catches it: 6 × 11 + 2 = 68 seats, which is too many, and 6 × 10 + 2 is exact - so 10 is the only whole-number solution.
3. The office calls back: 80 guests are now coming. Apply the rule, decide what the fire code means for the plan, and propose a workable arrangement.
Read the worked answer
6n + 2 = 80, so 6n = 78 and n = 13 tables, which exceeds the 12-table limit. The largest legal row seats 6 × 12 + 2 = 74, leaving 6 guests without a chair. The solution is a second row, but that changes the rule: each row needs its own 2 end chairs, so seats become 6n + 4. Splitting 13 tables as 7 and 6 gives 6 × 13 + 4 = 82 seats, enough for 80. The judgment being tested is that the constant in the rule counts end chairs, so adding a row changes the constant, not the coefficient.