AMC 10 · 2016 · #5
Grade 6 geometry-3dPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The ratio 1:3:4 means the three sides move together — if the smallest is some length, the others are fixed multiples of it. Tool #4 (Introduce a Variable) names that smallest side k, so the sides become k, 3k, 4k and the volume collapses into a single formula in k. That formula reveals every volume the box can possibly have. Then tool #3 (Eliminate Possibilities) takes the five answer choices and tests them against the formula: only a number that fits the formula for some whole-number k can survive, and the rest drop out.
Name the smallest side
Let the smallest side be k; the ratio 1:3:4 then forces the other two sides to be 3k and 4k.
A ratio is just one shared multiplier in disguise — pin down k and all three sides are decided.
6.RP.A.1Use Matrix LogicWrite the volume
Multiply the three sides: the number parts give 1·3·4=12 and the three k's combine, so the volume is 12k³.
Volume of a box is the three sides multiplied together, so the lone scale factor shows up cubed.
5.MD.C.5Use Matrix LogicList the allowed volumes
Since k is a whole number, every volume is 12 times a perfect cube; k=1,2,3 give 12, 96, 324.
Whole-number k means the volume can only land on 12 times a cube — 12, 96, 324, … and nothing in between.
6.EE.A.1Use Matrix LogicTest the choices
Dividing each choice by 12 must leave a cube; only 96 passes (96/12=8=2³), with k=2 giving sides 2,6,8.
Strip off the 12 and the survivor must be a perfect cube — only 96 passes that test.
6.EE.A.1Eliminate PossibilitiesTurn the ratio into one letter: sides k, 3k, 4k make the volume 12k³, so the answer must be 12 times a perfect cube — and only 96 fits.
- Name the smallest side
- Write the volume
- List the allowed volumes
- Test the choices