AMC 8 · 2019 · #22
Grade 7 rate-ratioalgebraProblem
A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was 84% of the original price, by what percent was the price increased and decreased?
Pick an answer.
AMC 8 2019 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 shirt's original price is first raised by some percent $x\%$, and then the new (higher) price is lowered by the same percent $x\%$. After both changes, the price ends up at $84\%$ of the original. Find $x$.
Givens: Original price is increased by $x\%$ first; The new price is then decreased by the same $x\%$; Final price $= 84\%$ of the original price; Answer choices: (A) $16$, (B) $20$, (C) $28$, (D) $36$, (E) $40$
Unknowns: The single percent $x$ used for both the increase and the decrease
Understand
Restated: A shirt's original price is first raised by some percent $x\%$, and then the new (higher) price is lowered by the same percent $x\%$. After both changes, the price ends up at $84\%$ of the original. Find $x$.
Givens: Original price is increased by $x\%$ first; The new price is then decreased by the same $x\%$; Final price $= 84\%$ of the original price; Answer choices: (A) $16$, (B) $20$, (C) $28$, (D) $36$, (E) $40$
Plan
Primary tool: #6 Guess and Check
Secondary: #3 Eliminate Possibilities, #9 Solve an Easier Related Problem
This is a multiple-choice percent problem with only five candidates, and the test is cheap: for each choice $x$, compute the combined factor $(1 + \tfrac{x}{100}) \times (1 - \tfrac{x}{100})$ and see whether it equals $0.84$. Tool #6 (Guess and Check) directly plugs each choice into this rule, and Tool #3 (Eliminate Possibilities) discards any choice whose factor is not $0.84$. We first use Tool #9 (Easier Related Problem) on a tiny example to be sure that 'up by $x\%$ then down by $x\%$' does NOT return to the original — that builds the intuition the problem depends on.
Execute — Answer: E
7.RP.A.3 Step 1 - Warm up with a tiny case to see what 'up then down by the same percent' actually does.
- Take an easy starting price of $\$100$ and try $x = 10$. Up $10\%$: $\$100 \to \$110$. Down $10\%$: $\$110 \to \$110 - \$11 = \$99$. So the final price is $99\%$ of the original, NOT $100\%$. The price always ends up a little below the original, which is exactly why the problem's final price is $84\%$ instead of $100\%$.
💡 An easier related case ($x = 10$) shows up-then-down by the same percent shrinks the price — Grade 7 percent reasoning.
6.RP.A.3 Step 2 - Restate the rule cleanly.
- Starting from price $P$, an increase by $x\%$ multiplies $P$ by $1 + \tfrac{x}{100}$, and a decrease by $x\%$ multiplies the new price by $1 - \tfrac{x}{100}$.
- So the combined multiplier (the fraction of the original we end up with) is the product of those two factors.
💡 Converting a percent change into a decimal multiplier is core Grade 6 rate-and-percent thinking.
5.NBT.B.7 Step 3 - Test choice (A) $x = 16$.
- Combined factor $= 1.16 \times 0.84$.
- Estimate: $1.16 \times 0.84 \approx 0.97$, which is way more than $0.84$.
- Eliminate (A).
💡 Multiplying two decimals to hundredths is Grade 5; the comparison rules (A) out.
5.NBT.B.7 Step 4 - Test choice (B) $x = 20$.
- Combined factor $= 1.20 \times 0.80 = 0.96$.
- Still too big.
- Eliminate (B).
💡 Same Grade 5 decimal multiplication; result $0.96$ rules (B) out.
5.NBT.B.7 Step 5 - Test choice (E) $x = 40$ (jump to the largest choice to bracket fast — the factor shrinks as $x$ grows, so a big $x$ should give a small factor).
- Combined factor $= 1.40 \times 0.60 = 0.84$.
- Exact match.
- So $x = 40$ works; the answer is (E).
💡 Guess-and-check on the largest choice hits $0.84$ on the nose — Grade 5 decimal multiplication confirms it.
7.RP.A.3 Warm up with a tiny case to see what 'up then down by the same percent' actually 6.RP.A.3 Restate the rule cleanly. Starting from price $P$, an increase by $x\%$ multipli 5.NBT.B.7 Test choice (A) $x = 16$. Combined factor $= 1.16 \times 0.84$. Estimate: $1.16 5.NBT.B.7 Test choice (B) $x = 20$. Combined factor $= 1.20 \times 0.80 = 0.96$. Still too 5.NBT.B.7 Test choice (E) $x = 40$ (jump to the largest choice to bracket fast — the facto Review
Reasonableness: The warm-up case showed $x = 10$ gives $0.99$ (a tiny dip). The final price drops to $0.84$ — a much bigger dip — so $x$ should be much bigger than $10$. $x = 40$ is the only choice in the right ballpark, and the direct check $1.40 \times 0.60 = 0.84$ is exact (no rounding). The other choices ($16, 20, 28, 36$) all give factors above $0.84$, confirming (E) is the unique answer.
Alternative: Tool #13 (Convert to Algebra) gives a slick one-liner using the difference-of-squares pattern. With $r = \tfrac{x}{100}$, the combined factor is $(1+r)(1-r) = 1 - r^2$. Setting $1 - r^2 = 0.84$ gives $r^2 = 0.16$, so $r = 0.4$ and $x = 40$. Faster for algebra-ready students, but Tool #6 is honest and fully accessible without algebra.
CCSS standards used (min grade 7)
5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Performing the decimal multiplications $1.16 \times 0.84$, $1.20 \times 0.80$, and $1.40 \times 0.60$ when testing each candidate $x$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Converting a percent change ($x\%$ up, $x\%$ down) into the decimal multipliers $1 + \tfrac{x}{100}$ and $1 - \tfrac{x}{100}$.)7.RP.A.3Use proportional relationships to solve multi-step ratio and percent problems (Composing the two successive percent changes into a single multi-step percent comparison (final price as a percent of original) — the heart of this problem.)
⭐ This AMC 8 problem only needs Grade 7 percent-change reasoning — up then down by the same percent never returns to the start — that you already know!
⭐ This AMC 8 problem only needs Grade 7 percent-change reasoning — up then down by the same percent never returns to the start — that you already know!
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