Competition · AMC preparation · step 4 of 4
AMC 8 · 2023 · #20
Grade 6 arithmeticPick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The new range of 50 can come from three different sources: the original 3 stays as the minimum and a new large y becomes the maximum, OR the original 28 stays as the maximum and a new small x becomes the minimum, OR BOTH endpoints are new numbers. Tool #2 (Systematic List) handles this cleanly: list the three cases, solve each one in isolation, then compare. Inside each case the median and mode constraints squeeze x and y into a narrow integer band, and Tool #6 (Guess & Check) picks the extreme integer in that band so the sum is as big as it can be. Tool #3 (Eliminate) finally checks the winning sum against the five answer choices so a typo cannot slip through.
Read off the three statistics
Sorted 3,3,8,11,28 has range 25, mode 3, median 8 — the trio we must preserve.
Summarizing a small data set with its range, mode, and median is exactly the Grade 6 statistics standard.
6.SP.B.5Make A Systematic ListSqueeze x and y with the conditions
Median forces one insert below 8 and one above; mode bans 8,11,28 and ties — so x ＜ 8 ＜ y with y≠11,28.
Writing each "stays the same" condition as an inequality x ＜ 8, y ＞ 8 is the Grade 6 inequality standard.
6.EE.B.8Make A Systematic ListSplit into range cases
Apply new range = 50 three ways: keep min 3, or keep max 28, or make both ends new — three cases.
Listing every way the new max and min can be assembled is the Grade 6 "summarize the data" mindset, just applied to a hypothetical list.
6.SP.B.5Make A Systematic ListWork case 1
Case 1: min stays 3, so y=53; x can be as big as 7 (below 8), giving 7+53=60.
Solving y-3=50 for the unknown y is exactly Grade 6 "find the value that makes the equation true."
In the case where 3 stays the smallest number, the two inserted integers are forced to be 7 and 53.
▸ Why?
The larger inserted number must be 53: the smallest value is still 3 and the new range must be 50, so the largest value is 3 + 50 = 53.
▸ Why?
The smaller inserted number must be 7: it has to stay below the median 8 (otherwise the middle value or the mode would change) and be at least 3 (otherwise 3 stops being the smallest), and 7 is the largest whole number in that band.
▸ Why?
With seven numbers placed in order the median is the single center value, the 4th one, so keeping that center at 8 forces the newly added small number to land before it — under 8.
Work case 2
Case 2: max stays 28, so x=-22; y at most 27, giving -22+27=5 — far below Case 1.
Handling a negative integer like -22 as a real value on the number line is the Grade 6 "positives and negatives describe quantities" standard.
6.NS.C.5Guess And CheckWork case 3
Case 3: both ends new, x ＜ 3 and y=x+50; biggest x is 2, so y=52 and 2+52=54.
Picking the largest integer that satisfies x ＜ 3 is a Grade 6 "solve the inequality" move.
6.EE.B.5Guess And CheckTake the largest sum
Compare 60, 5, 54: Case 1 wins with 60, which is choice (D).
Picking the largest summary value from a short list of candidates is the Grade 6 data-summary skill in action.
6.SP.B.5Eliminate PossibilitiesThis AMC 8 problem only needs Grade 6 statistics (range, mode, median) and a little inequality reasoning you already know!
- Read off the three statistics
- Squeeze x and y with the conditions
- Split into range cases
- Work case 1
- Work case 2
- Work case 3
- Take the largest sum
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