AMC 10 · 2024 · #5

Grade 7 arithmeticalgebra
mean-median-mode-rangeweighted-averagelinear-equations-one-varfraction-arithmetic convert-to-algebrawork-backwards ↑ Prerequisites: mean-median-mode-rangelinear-equations-one-var
📏 Short solution 💡 2 insights
Problem
A list of 20 numbers averages 45. Some of the entries are the number 6. Delete every 6 from the list; the numbers that survive average 66. Find how many 6s the original list contained.

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
8
How to solve
Strategy Convert to Algebra

The twenty numbers are never revealed, so no amount of listing or guessing at entries can help. What a mean does give is a total: Tool #11 (Work Backwards) runs the mean formula in reverse to recover the sum 20 · 45 = 900 before anything is deleted. Once totals are the currency, Tool #13 (Convert to Algebra) does the real work — name the number of 6s k, write the surviving count as 20 - k and the surviving sum as 900 - 6k, and the second mean becomes a single linear equation in k. Tool #15 (Organize Information in More Ways) closes the loop: regrouping the same twenty numbers into "the 6s" and "the rest" and balancing their distances from 45 reproduces k from a completely different direction, which is the check that the setup was right.

1STEP 1

Turn the mean into a total

Twenty times forty-five gives a total of 900.

45 = S/20 ⟹ S = 20 · 45 = 900
2STEP 2

Name the count of 6s

With k sixes the surviving sum is 900 minus 6k.

removed sum = 6k, remaining sum = 900 - 6k, remaining count = 20 - k
3STEP 3

Write the second mean as an equation

The surviving sum over the surviving count is 66.

(900 - 6k)/(20 - k) = 66 ⟹ 900 - 6k = 66(20 - k)
4STEP 4

Solve for k

It reduces to sixty k equals 420, so k is 7.

900 - 6k = 1320 - 66k ⟹ 60k = 420 ⟹ k = 420/60 = 7
5STEP 5

Confirm with the balance view

Sixes sit 39 below, survivors 21 above — again 7.

39k = 21(20 - k) ⟹ 39k + 21k = 420 ⟹ 60k = 420 ⟹ k = 7 → (D)
Answer
7
Rebuild the data set from k = 7 and test both stated means. Seven 6s contribute 7 · 6 = 42. The remaining 20 - 7 = 13 numbers must then sum to 900 - 42 = 858, and 858/13 = 66, matching the second mean exactly. Adding the two blocks back gives 858 + 42 = 900, and 900/20 = 45, matching the first mean. Both conditions hold, so 7 is correct. Direction check: the 6s are far below average, so deleting them should raise the mean, and 45 does rise to 66. Size check is tight too — the neighbouring choices fail. With k = 6 the survivors would average 864/14 ≈ 61.7, and with k = 8 they would average 852/12 = 71; only k = 7 lands on 66.
💡Key takeaway

This AMC 12 problem only needs Grade 6 "mean equals total divided by count" and a Grade 7 linear equation — stop thinking about the twenty hidden numbers and think about their total, and the number of 6s falls out in one line.

  • Turn the mean into a total
  • Name the count of 6s
  • Write the second mean as an equation
  • Solve for k
  • Confirm with the balance view