Difference between revisions of "2001 AMC 10 Problems/Problem 10"
Pidigits125 (talk | contribs) (Created page with '== Problem == If <math>x</math>, <math>y</math>, and <math>z</math> are positive with <math>xy = 24</math>, <math>xz = 48</math>, and <math>yz = 72</math>, then <math>x + y + z<…') |
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<math>\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24</math> | <math>\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24</math> | ||
− | == Solution == | + | == Solution 1== |
− | + | The first two equations in the problem are <math>xy=24 </math> and <math>xz=48</math>. Since <math>xyz \ne 0</math>, we have <math>\frac{xy}{xz}=\frac{24}{48} \implies 2y=z </math>. We can substitute <math> z = 2y </math> into the third equation <math>yz = 72</math> to obtain <math> 2y^2=72 \implies y=6 </math> and <math> 2y=z=12 </math>. We replace <math>y</math> into the first equation to obtain <math> x=4 </math>. | |
− | <math> | + | Since we know every variable's value, we can substitute them in to find <math> x+y+z = 4+6+12 = \boxed{\textbf{(D) }22} </math>. |
− | + | == Solution 2 == | |
− | + | These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives <math>(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288</math>. We divide <math>xyz = 288</math> by each of the given equations, which yields <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = 22</math>, so the answer is <math>\boxed{\textbf{(D) } 22}</math>. | |
− | + | ==Solution 3(strategic guess and check)== | |
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− | + | Seeing the equations, we notice that they are all multiples of 12. Trying in factors of 12, we find that <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math> work. <math>4 + 6+ 12 = \boxed{\textbf{(D) } 22} </math> | |
− | <math> | ||
− | <math> | ||
− | + | ~idk12345678 | |
+ | |||
+ | ==Video Solution by Daily Dose of Math== | ||
+ | |||
+ | https://youtu.be/tiDp5E3rwfI?si=n2h6UvQUW-V-bLT2 | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2001|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:41, 15 July 2024
Contents
[hide]Problem
If , , and are positive with , , and , then is
Solution 1
The first two equations in the problem are and . Since , we have . We can substitute into the third equation to obtain and . We replace into the first equation to obtain .
Since we know every variable's value, we can substitute them in to find .
Solution 2
These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives . We divide by each of the given equations, which yields , , and . The desired sum is , so the answer is .
Solution 3(strategic guess and check)
Seeing the equations, we notice that they are all multiples of 12. Trying in factors of 12, we find that , , and work.
~idk12345678
Video Solution by Daily Dose of Math
https://youtu.be/tiDp5E3rwfI?si=n2h6UvQUW-V-bLT2
~Thesmartgreekmathdude
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.