Difference between revisions of "2018 AMC 10B Problems/Problem 4"
(→Solution 2) |
|||
Line 20: | Line 20: | ||
==Solution 2== | ==Solution 2== | ||
− | Simply | + | Simply guess and check to find that the dimensions are <math>4</math> by <math>6</math> by <math>12</math>. Therefore, the answer is <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math> |
==Solution 3== | ==Solution 3== |
Revision as of 10:46, 9 November 2022
Problem
A three-dimensional rectangular box with dimensions ,
, and
has faces whose surface areas are
,
,
,
,
, and
square units. What is
+
+
?
Solution 1
Let be the length of the shortest dimension and
be the length of the longest dimension. Thus,
,
, and
.
Divide the first two equations to get
. Then, multiply by the last equation to get
, giving
. Following,
and
.
The final answer is .
Solution 2
Simply guess and check to find that the dimensions are by
by
. Therefore, the answer is
.
Solution 3
If you find the GCD of ,
, and
you get your first number,
. After this, do
and
to get
and
, the other 2 numbers. When you add up your
numbers, you get
which is
.
Solution 4
Since the surface areas of the faces are the product of two of the dimensions. Therefore, ,
, and
. You can multiply
, which simplifies to
which means that the volume
equals
. The individual dimensions,
,
, and
can be found by doing
,
, and
, which yields
,
, and
. Adding this up, we have that
which is
- Solution by smartninja2000
Video Solution
~savannahsolver
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.