Difference between revisions of "2008 AMC 8 Problems/Problem 16"
(→Solution) |
MRENTHUSIASM (talk | contribs) (Tag: Undo) |
||
Line 18: | Line 18: | ||
==Solution== | ==Solution== | ||
− | The volume is of seven unit cubes which is <math>7</math>. The surface area is the same for each of the protruding cubes which is 5 | + | The volume is of seven unit cubes which is <math>7</math>. The surface area is the same for each of the protruding cubes which is <math>5\cdot 6=30</math>. The ratio of the volume to the surface area is <math>\boxed{\text{(D)}\ 7:30}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=15|num-a=17}} | {{AMC8 box|year=2008|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:24, 8 August 2021
Problem
A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?
Solution
The volume is of seven unit cubes which is . The surface area is the same for each of the protruding cubes which is . The ratio of the volume to the surface area is .
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.