AMC 10 · 2004 · #11
Grade 8 geometry-3dA company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by 25% without altering the volume, by what percent must the height be decreased?
Pick an answer.
AMC 10 2004 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 peanut butter jar is a cylinder. Its diameter is made $25\%$ larger, but the volume stays exactly the same. Find the percent by which the height must be decreased.
Givens: The jar is a cylinder, so its volume equals the base area times the height.; The diameter of the base is increased by $25\%$.; The volume is unchanged after the diameter is increased.; Answer choices (as percents): (A) $10$, (B) $25$, (C) $36$, (D) $50$, (E) $60$.
Unknowns: The percent by which the height must be decreased so the volume stays the same.
Understand
Restated: A peanut butter jar is a cylinder. Its diameter is made $25\%$ larger, but the volume stays exactly the same. Find the percent by which the height must be decreased.
Givens: The jar is a cylinder, so its volume equals the base area times the height.; The diameter of the base is increased by $25\%$.; The volume is unchanged after the diameter is increased.; Answer choices (as percents): (A) $10$, (B) $25$, (C) $36$, (D) $50$, (E) $60$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #9 Solve an Easier Related Problem
Nothing in the problem gives an actual radius or height, only how they change, so name the original radius $r$ and original height $h$ (Tool #4) and write the volume as $\pi r^2 h$. Then the two changes can be tracked as multipliers on those letters: the diameter growing $25\%$ multiplies the radius, squaring turns that into the base-area multiplier, and forcing the volume to stay equal pins down the height multiplier. Because the volume formula is the same shape before and after, the $\pi$ and the letters $r$ and $h$ cancel, leaving a clean number for the new height as a fraction of the old. Once the algebra is set up, plugging in easy concrete numbers (Tool #9) is a fast independent check that the percent is right.
Execute — Answer: C
8.G.C.9 Step 1 Name the sizes and write the volume
- Let the original jar have base radius $r$ and height $h$.
- A cylinder's volume is the area of its circular base times its height, so the volume is $\pi r^2 h$.
- Writing it this way lets us follow exactly how each change to $r$ and $h$ moves the volume, instead of juggling percents in the air.
💡 Naming the unknown sizes turns a vague "by what percent" question into an equation you can actually balance.
7.RP.A.3 Step 2 Turn the wider diameter into a radius multiplier
- The diameter grows by $25\%$.
- Since the radius is exactly half the diameter, the radius grows by the same $25\%$.
- A $25\%$ increase means multiplying by $1 + 0.25 = 1.25 = \tfrac{5}{4}$, so the new radius is $\tfrac{5}{4}r$.
💡 A percent increase is just a multiplier, and halving a diameter to get a radius keeps that same multiplier.
7.G.B.4 Step 3 Square the radius to get the new base area
- The base is a circle, and a circle's area is $\pi r^2$, so it depends on the square of the radius.
- Multiplying the radius by $\tfrac{5}{4}$ multiplies the area by $\left(\tfrac{5}{4}\right)^2 = \tfrac{25}{16}$.
- So the wider base is $\tfrac{25}{16} = 1.5625$ times as large, a $56.25\%$ increase, from only a $25\%$ wider diameter.
💡 Area lives in two dimensions, so stretching a length by a factor stretches the area by that factor squared.
7.RP.A.3 Step 4 Force equal volume and read off the height cut
- The new volume is the new base area times the new height $h'$: $\tfrac{25}{16}\pi r^2 \cdot h'$.
- Set it equal to the old volume $\pi r^2 h$.
- The $\pi r^2$ cancels from both sides, giving $\tfrac{25}{16}h' = h$, so $h' = \tfrac{16}{25}h = 0.64\,h$.
- The new height is $64\%$ of the old, which is a decrease of $1 - 0.64 = 0.36 = 36\%$.
- That is choice (C).
💡 If a product must stay fixed and one factor grows, the other factor shrinks by exactly the reciprocal amount.
8.G.C.9 Let the original jar have base radius $r$ and height $h$. A cylinder's volume is 7.RP.A.3 The diameter grows by $25\%$. Since the radius is exactly half the diameter, the 7.G.B.4 The base is a circle, and a circle's area is $\pi r^2$, so it depends on the squ 7.RP.A.3 The new volume is the new base area times the new height $h'$: $\tfrac{25}{16}\p Review
Reasonableness: The base got wider, so the height has to shrink — and because area depends on the square of the radius, the base grew a lot ($56.25\%$), so the height must fall by more than the $25\%$ the diameter rose. That rules out (A) $10$ and (B) $25$ as too small, while $0.64$ of the old height is nowhere near a half-cut, ruling out (D) $50$ and (E) $60$ as too big. A $36\%$ decrease sits exactly where it should. The trap answer is (B) $25$, from wrongly assuming the height drops by the same percent the diameter rose, forgetting the squaring.
Alternative: Use easy concrete numbers instead of letters. Take a base radius of $4$ and a height of $25$, so the volume is $\pi \cdot 4^2 \cdot 25 = 400\pi$. A $25\%$ wider diameter makes the radius $5$, so the new base area is $\pi \cdot 5^2 = 25\pi$. To keep the volume at $400\pi$, the new height is $400\pi \div 25\pi = 16$. The height fell from $25$ to $16$, a drop of $9$ out of $25$, which is $\tfrac{9}{25} = 36\%$ — the same answer, (C).
CCSS standards used (min grade 8)
8.G.C.9Know the formulas for volumes of cones, cylinders, and spheres (Writing the jar's volume as $\pi r^2 h$ (base area times height) so the changes can be balanced in an equation.)7.G.B.4Know the formulas for area and circumference of a circle (Using area $=\pi r^2$ to see the base area scales by the square of the radius multiplier, $\left(\tfrac{5}{4}\right)^2 = \tfrac{25}{16}$.)7.RP.A.3Use proportional relationships to solve multi-step ratio and percent problems (Converting the $25\%$ diameter increase into the multiplier $\tfrac{5}{4}$ and reading the height ratio $0.64$ back as a $36\%$ decrease.)
⭐ When a circle's width grows, its area grows by the square of that factor, so to keep the same volume the height must shrink by more than the width grew.
⭐ When a circle's width grows, its area grows by the square of that factor, so to keep the same volume the height must shrink by more than the width grew.
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