AMC 10 · 2015 · #5

Grade 7 logicarithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
estimationfraction-arithmeticinterval-arithmetic logical-deductionextreme-principle ↑ Prerequisites: estimationfraction-arithmetic
📏 Medium solution 💡 2 insights
Problem
Each number in an expression is replaced by a larger or smaller one. Find the pattern that always raises the value.

Pick an answer.

(A)
She rounds all three numbers up.
(B)
She rounds } a \text{ and } b \text{ up, and she rounds } c \text{ down.
(C)
She rounds } a \text{ and } c \text{ up, and she rounds } b \text{ down.
(D)
She rounds } a \text{ up, and she rounds } b \text{ and } c \text{ down.
(E)
She rounds } c \text{ up, and she rounds } a \text{ and } b \text{ down.
How to solve
Strategy Introduce a Variable

The phrase 'rounds up' is vague until it is written as an inequality, so Tool #4 (Introduce a Variable) names the rounded numbers a', b', c' and records each rounding as a' ＞ a, b' ＜ b, or c' ＜ c. Tool #7 (Identify Subproblems) then splits the expression into two independent questions — which way does the quotient a/b move, and which way does the -c move — because the two parts respond to different variables. The quotient question is the one worth proving rather than asserting: Tool #14 (Extreme Principle) frames it as pushing each ingredient to the extreme that makes the value largest, and a common-denominator comparison settles the direction. Finally Tool #3 (Eliminate Possibilities) is not decoration here: the problem asks for a guarantee, so each rival choice needs an actual counterexample, and the counterexamples show why no wrong direction can be counted on to cancel out.

1STEP 1

Write rounding as inequalities

Each change is one plain inequality.

up: x' ＞ x, down: x' ＜ x, estimate = a'/b' - c'
2STEP 2

Split off the -c part

The subtracted term flips its direction.

c' ＜ c ⟹ -c' ＞ -c
3STEP 3

Prove the quotient rule, do not assume it

The quotient's direction is proved, not assumed.

a'/b' - a/b = (a'b - ab')/bb', a'b ＞ ab ＞ ab' ⟹ a'/b' ＞ a/b
4STEP 4

A wrong direction cannot be counted on to cancel

A wrong direction cannot be trusted to cancel.

1100/900 - 300 ≈ -298.78 ＜ 1050/950 - 250 ≈ -248.89
5STEP 5

Assemble the one surviving pattern

Only one pattern survives, choice (D).

a'/b' ＞ a/b and -c' ＞ -c ⟹ a'/b' - c' ＞ a/b - c → (D)
Answer
She rounds a up, and she rounds b and c down.
Test (D) on the same numbers used to break the other choices. With a=1050, b=950, c=250 rounded to the nearest hundred, (D) gives 1100/900 - 200 ≈ 1.22 - 200 = -198.78, comfortably above the exact -248.89. With a=60050, b=150, c=250, (D) gives 60100/100 - 200 = 601 - 200 = 401, above the exact 150.33. Both agree with the proof. One honest caveat about the word 'greater': the inequality is strict only because rounding actually moves each number. If a number were already sitting on the rounding mark and stayed put, that piece would contribute equality rather than strict increase, and the estimate would tie the exact value rather than beat it — the argument then still gives 'at least as large', which is why (D) is the only defensible choice either way.
💡Key takeaway

To push a/b - c upward, feed the fraction more on top and less on the bottom, and take away less at the end — and check the fraction rule with a common denominator instead of just believing it.

  • Write rounding as inequalities
  • Split off the -c part
  • Prove the quotient rule, do not assume it
  • A wrong direction cannot be counted on to cancel
  • Assemble the one surviving pattern

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