AMC 10 · 2014 · #11
Grade 7 arithmeticFor the consumer, a single discount of n% is more advantageous than any of the following discounts:
(1) two successive 15% discounts
(2) three successive 10% discounts
(3) a 25% discount followed by a 5% discount
What is the smallest possible positive integer value of n?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Deal (1): two successive $15\%$ discounts; Deal (2): three successive $10\%$ discounts; Deal (3): a $25\%$ discount then a $5\%$ discount; The single $n\%$ discount must give a strictly lower final price than each of the three deals; Answer choices: (A) $27$, (B) $28$, (C) $29$, (D) $31$, (E) $33$
Unknowns: The smallest positive integer $n$ that makes the single discount cheaper than all three deals
Understand
Restated: 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.
Givens: Deal (1): two successive $15\%$ discounts; Deal (2): three successive $10\%$ discounts; Deal (3): a $25\%$ discount then a $5\%$ discount; The single $n\%$ discount must give a strictly lower final price than each of the three deals; Answer choices: (A) $27$, (B) $28$, (C) $29$, (D) $31$, (E) $33$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #13 Convert to Algebra
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.
Execute — Answer: C
6.RP.A.3 Step 1 Anchor the price at $100
- Let the original price be $\$100$. A discount of $p\%$ leaves you paying $\left(1-\tfrac{p}{100}\right)$ of the price, so a single $n\%$ discount leaves a final price of $100-n$ dollars.
- Because the starting price is the same for every deal, comparing final prices is the same as comparing deals, and the smaller the final price the better the deal.
💡 Starting from $\$100$ makes every percent read off directly as dollars, so deals become numbers you can line up.
6.NS.B.3 Step 2 Price out the three competing deals
- Each deal multiplies the $\$100$ by one factor per discount. Two $15\%$ discounts multiply by $0.85$ twice; three $10\%$ discounts multiply by $0.90$ three times; a $25\%$ then a $5\%$ discount multiply by $0.75$ then $0.95$.
- Work each product out to a final price in dollars.
💡 Successive discounts just stack as repeated multiplication, so each deal collapses to one final price.
7.RP.A.3 Step 3 Find the toughest deal to beat
- The single discount must be cheaper than all three, so it must be cheaper than the lowest of the three prices.
- Among $72.25$, $72.90$, and $71.25$, the smallest is $71.25$, coming from the $25\%$-then-$5\%$ deal.
- That deal is the champion — the deepest effective discount, $100-71.25=28.75\%$.
- If the single discount beats this one, it automatically beats the other two.
💡 Only the strongest rival sets the bar; clear the hardest one and the easier ones are already cleared.
7.EE.B.4 Step 4 Turn "cheaper than the champion" into an inequality
- The single discount's final price $100-n$ must be strictly less than the champion's $71.25$.
- Write that as an inequality and solve for $n$.
- Subtracting and flipping gives a lower bound on $n$.
💡 "Lower final price" is exactly "bigger discount," which is one clean inequality in $n$.
6.EE.B.5 Step 5 Take the smallest whole number past the bound
- We need the smallest positive integer $n$ with $n>28.75$.
- The integer $28$ fails (it only gives a $28\%$ discount, weaker than the champion's $28.75\%$), and every integer at or below $28$ fails too.
- The first integer above $28.75$ is $29$, and $29\%$ off leaves $\$71$, which is below all of $72.25$, $72.90$, and $71.25$. So the smallest value is $n=29$, choice (C).
💡 Just past a boundary of $28.75$, the first legal whole number is the next integer up, $29$.
6.RP.A.3 Let the original price be $\$100$. A discount of $p\%$ leaves you paying $\left( 6.NS.B.3 Each deal multiplies the $\$100$ by one factor per discount. Two $15\%$ discount 7.RP.A.3 The single discount must be cheaper than all three, so it must be cheaper than t 7.EE.B.4 The single discount's final price $100-n$ must be strictly less than the champio 6.EE.B.5 We need the smallest positive integer $n$ with $n>28.75$. The integer $28$ fails Review
Reasonableness: 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 $\left(1-\tfrac{n}{100}\right)P<0.7125P$, and $P$ cancels.
Alternative: Work purely in effective-discount percentages instead of dollars. Two $15\%$ discounts give $1-0.85^2=27.75\%$ off; three $10\%$ give $1-0.90^3=27.1\%$ off; the $25\%$-then-$5\%$ gives $1-0.75\cdot0.95=28.75\%$ off. The single discount must exceed the largest of these, $28.75\%$, so $n>28.75$ and the least integer is $n=29$, again (C).
CCSS standards used (min grade 7)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Reading a $p\%$ discount as multiplying the price by $\left(1-\tfrac{p}{100}\right)$ and fixing the price at $\$100$.)6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Multiplying the discount factors to get the three final prices $72.25$, $72.90$, and $71.25$.)7.RP.A.3Use proportional relationships to solve multi-step ratio and percent problems (Comparing the three multi-step percent deals and identifying the deepest effective discount ($28.75\%$) as the one to beat.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Writing and solving the inequality $100-n<71.25$ to get $n>28.75$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Selecting the smallest positive integer that satisfies $n>28.75$, namely $n=29$.)
⭐ 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 $29\%$.
⭐ 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 $29\%$.
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