AMC 10 · 2014 · #11

Grade 7 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps
percentageoptimizationdecimal-arithmetic bound-inequality-then-enumerate ↑ Prerequisites: percentage
📏 Medium solution 💡 2 insights
Problem
A store offers a single discount of n% that a shopper finds better (a lower final price) than every one of these three deals: two 15% discounts in a row, three 10% discounts in a row, and a 25% discount followed by a 5% discount. Find the smallest positive whole number n for which the single deal beats all three.

Pick an answer.

(A)
$\ 27$
(B)
28
(C)
29
(D)
31
(E)
33

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

Beating three deals at once sounds like three tasks, but Tool #14 (Extreme Principle) cuts it to one: whichever deal gives the lowest price is the hardest to beat, so if the single discount beats that champion it beats them all. Tool #4 (Introduce a Variable) pins the original price at a convenient $100 so each percent turns into a plain final price. Tool #7 (Identify Subproblems) computes the three final prices separately by multiplying discount factors. Then Tool #13 (Convert to Algebra) turns "cheaper than the champion" into one inequality, and the Extreme Principle again reads off the smallest whole number that clears the boundary.

1STEP 1

Anchor the price at $100

Set the original price to $100; then a single n% discount leaves a final price of 100-n dollars.

single n%: 100(1-n/100)=100-n
2STEP 2

Price out the three competing deals

Each discount multiplies 100 by its factor: 0.85 twice, 0.90 three times, and 0.75 then 0.95 give 72.25, 72.90, and 71.25.

(1) 100(0.85)²=72.25, (2) 100(0.90)³=72.90, (3) 100(0.75)(0.95)=71.25
3STEP 3

Find the toughest deal to beat

Only the cheapest rival matters: 71.25, from the 25%-then-5% deal — an effective 28.75% off.

min(72.25, 72.90, 71.25)=71.25 → effective discount =28.75%
4STEP 4

Turn "cheaper than the champion" into an inequality

So 100-n must be less than 71.25, and rearranging makes n greater than 28.75.

100-n ＜ 71.25 → n ＞ 100-71.25 → n ＞ 28.75
5STEP 5

Take the smallest whole number past the bound

The first integer past 28.75 is 29: 28 leaves 72, still above 71.25, while 29 leaves 71 — under all three.

n ＞ 28.75, ninZ^+ → n=29; 100-29=71 ＜ 71.25 → (C)
Answer
29
Check the boundary from both sides. At n=29 the price is $71, which is less than the champion $71.25 and less than $72.25 and $72.90, so 29% genuinely beats all three. At n=28 the price is $72, which is more than $71.25, so the 25%-then-5% deal is still cheaper and 28 fails. So 29 is the first integer that works, matching (C). Note the answer does not depend on the $100 choice: with price P the same comparison reads (1-n/100)P ＜ 0.7125P, and P cancels.
💡Key takeaway

To beat several stacked deals at once, just beat the single deepest one: here that is 28.75% off, so the smallest whole discount that wins is 29%.

  • Anchor the price at $100
  • Price out the three competing deals
  • Find the toughest deal to beat
  • Turn "cheaper than the champion" into an inequality
  • Take the smallest whole number past the bound

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