Difference between revisions of "1954 AHSME Problems/Problem 3"

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== Solution ==
 
== Solution ==
 
<math>x=k\cdot y^3</math>, <math>y=j\cdot z^{\frac{1}{5}}\implies y^3=j^3\cdot z^{\frac{3}{5}}</math>, so <math>x=k\cdot j^3\cdot z^{\frac{3}{5}}</math>, <math>\fbox{C}</math>
 
<math>x=k\cdot y^3</math>, <math>y=j\cdot z^{\frac{1}{5}}\implies y^3=j^3\cdot z^{\frac{3}{5}}</math>, so <math>x=k\cdot j^3\cdot z^{\frac{3}{5}}</math>, <math>\fbox{C}</math>
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==See Also==
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{{AHSME 50p box|year=1954|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 20:34, 17 February 2020

Problem 3

If $x$ varies as the cube of $y$, and $y$ varies as the fifth root of $z$, then $x$ varies as the nth power of $z$, where n is:

$\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8$

Solution

$x=k\cdot y^3$, $y=j\cdot z^{\frac{1}{5}}\implies y^3=j^3\cdot z^{\frac{3}{5}}$, so $x=k\cdot j^3\cdot z^{\frac{3}{5}}$, $\fbox{C}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AHSME Problems and Solutions


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