AMC 10 · 2012 · #6

Grade 7 logicalgebra
estimationinterval-arithmetic convert-to-algebra ↑ Prerequisites: interval-arithmetic
📏 Short solution 💡 2 insights
Problem
One number is rounded up and the other down by the same amount before subtracting. Decide what the estimate must do.

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)
$\text{Her estimate is } 0$
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

One letter names the rounding amount.

k > 0, rounded x=x+k, rounded y=y-k
2STEP 2

Write and simplify her estimate

Simplifying shows both changes add.

(x+k)-(y-k)=x+k-y+k=(x-y)+2k
3STEP 3

Compare to the true value and choose

So the estimate is always larger, choice (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