AMC 10 · 2012 · #6

Grade 7 logicalgebra

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

estimationinterval-arithmetic convert-to-algebra ↑ Prerequisites: interval-arithmetic

📏 Short solution 💡 2 insights

Problem

To estimate x-y, where x ＞ y ＞ 0, Xiaoli makes x a little bigger by rounding it up, makes y a little smaller by rounding it down using the same small amount, and then subtracts her two rounded numbers. Decide which statement about her estimate must always be true.

Pick an answer.

(A)

$\text{Her estimate is larger than } x-y$

(B)

$\text{Her estimate is smaller than } x-y$

(C)

$\text{Her estimate equals } x-y$

(D)

$\text{Her estimate equals } y-x$

(E)

$\text{Her estimate is } 0$

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The problem gives no numbers, so the safe move is Tool #4 (Introduce a Variable): call the small rounding amount k and write the two rounded numbers as x+k and y-k. Tool #13 (Convert to Algebra) then turns “rounded x minus rounded y” into a single expression that can be simplified and compared to x-y. Tool #3 (Eliminate Possibilities) reads that comparison against the five statements and keeps the one that must hold for every allowed x and y.

1STEP 1

Name the rounding amount

Name the unstated rounding amount k, with k ＞ 0; the two rounded numbers are then x+k and y-k.

k ＞ 0, rounded x=x+k, rounded y=y-k

2STEP 2

Write and simplify her estimate

Her estimate is (x+k)-(y-k)=x+k-y+k, which collapses to (x-y)+2k — the true value plus 2k.

(x+k)-(y-k)=x+k-y+k=(x-y)+2k

3STEP 3

Compare to the true value and choose

Since k ＞ 0, the gap 2k is positive, so (x-y)+2k ＞ x-y for every x and y — the estimate overshoots, leaving (A).

2k ＞ 0 → (x-y)+2k ＞ x-y → (A)

Answer

Her estimate is larger than x-y

Try concrete numbers: let x=10, y=3, and round by k=1. The true difference is 10-3=7. Xiaoli computes (10+1)-(3-1)=11-2=9, which is larger than 7 by 2, exactly 2k. Any other choice of k ＞ 0 gives the same pattern: the estimate beats the true value by 2k. This matches the algebra and confirms the estimate is always too high, so (A) is consistent.

💡Key takeaway

Making the front number bigger and the number you subtract smaller both push a difference up, so this estimate is always too high.

Name the rounding amount
Write and simplify her estimate
Compare to the true value and choose

Name the rounding amount

Write and simplify her estimate

Compare to the true value and choose

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