Difference between revisions of "2017 AMC 10B Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | By simply finding the volume of the larger block, we see that its area is <math>18</math>. The volume of the smaller block is <math>4</math>. Dividing the two, we see that only a maximum of <math>4</math> <math>2</math>in x<math>2</math>in x<math>1</math>in blocks can fit inside a <math>3</math>-in by <math>2</math> in by <math>3</math>in box. <math>\qquad\textbf{(B)}\ | + | By simply finding the volume of the larger block, we see that its area is <math>18</math>. The volume of the smaller block is <math>4</math>. Dividing the two, we see that only a maximum of <math>4</math> <math>2</math>in x<math>2</math>in x<math>1</math>in blocks can fit inside a <math>3</math>-in by <math>2</math> in by <math>3</math>in box. <math>\qquad\textbf{(B)}\ 4</math> |
− | {{AMC10 box|year=2017|ab= | + | {{AMC10 box|year=2017|ab=B|num-b=5|num-a=7}} |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:27, 16 February 2017
Problem
What is the largest number of solid by
by
blocks that can fit in a
by
by
box?
Solution
By simply finding the volume of the larger block, we see that its area is . The volume of the smaller block is
. Dividing the two, we see that only a maximum of
in x
in x
in blocks can fit inside a
-in by
in by
in box.
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 AMC 10 Problems and Solutions |
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