AMC 10 · 2011 · #7

Grade 7 rate-ratio
mean-median-mode-rangeratio-proportiondigit-constraints optimization ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 2 insights
Problem
Two two-digit numbers have a fixed average. Find the largest possible ratio between them.

Pick an answer.

(A)
3
(B)
$\frac{33}{7}$
(C)
$\frac{39}{7}$
(D)
9
(E)
$\frac{99}{10}$
How to solve
Strategy Extreme Principle

Tool #13 (Convert to Algebra): "mean 60" is a sentence, and turning it into x+y=120 is what exposes the link between the two numbers. Tool #15 (Organize Information in More Ways): rewriting x/y so that only y appears turns a two-unknown maximum into a one-unknown minimum, which is the whole difficulty of the problem. Tool #14 (Extreme Principle): once the ratio depends on a single variable and moves in one direction only, the answer sits at the boundary of the legal range for y — so the real work is finding which condition sets that boundary. Tool #6 (Guess and Check): a boundary value is only useful if a legal pair sits on it, so I test the candidate pair against every condition before reporting it.

1STEP 1

Mean 60 means the sum is fixed

A fixed average fixes the sum.

(x+y)/2=60 ⟹ x+y=120
2STEP 2

Rewrite the ratio using y alone

The ratio rewrites with one letter.

x/y=(120-y)/y=120/y-1
3STEP 3

Smaller y always gives a bigger ratio

It always grows as that letter shrinks.

0 < y₁ < y₂ ⟹ 120/y₁-1 > 120/y₂-1
4STEP 4

The floor on y is 21, not 10

The real floor is 21, not ten.

y ≥ 10 and 120-y ≤ 99 → y ≥ 21; max(10,21)=21
5STEP 5

Check the boundary pair and compute

That boundary pair gives 33/7, choice (B).

y=21, x=99 → x/y=99/21=33/7 → (B)
Answer
33/7
Probe the neighbours of the boundary. Going one step lower, y=20 forces x=100, which has three digits and is illegal, confirming y=21 really is the floor. Going one step higher, y=22 gives x=98 and a ratio of 98/22=49/11≈ 4.45, which is below 33/7≈ 4.71, matching the claim that larger y is worse. The two rejected large choices are exactly the traps this structure sets: (E) 99/10 pairs 99 with 10, but 99+10=109 gives mean 54.5, not 60; (D) 9 needs y=12 and x=108, and 108 is not two-digit. Both come from treating "biggest x" and "smallest y" as independent choices, which the fixed sum 120 forbids.
💡Key takeaway

When the sum is locked, one number decides the other, so a maximum with two unknowns is really a minimum with one — and the limit that stops you often comes from the number you were not watching.

  • Mean 60 means the sum is fixed
  • Rewrite the ratio using y alone
  • Smaller y always gives a bigger ratio
  • The floor on y is 21, not 10
  • Check the boundary pair and compute