Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1959 AHSME Problems/Problem 20"
m (formatting fix)
 
(One intermediate revision by the same user not shown)
Line 2: Line 2:
 
It is given that <math>x</math> varies directly as <math>y</math> and inversely as the square of <math>z</math>, and that <math>x=10</math> when <math>y=4</math> and <math>z=14</math>. Then, when <math>y=16</math> and <math>z=7</math>, <math>x</math> equals:
 
It is given that <math>x</math> varies directly as <math>y</math> and inversely as the square of <math>z</math>, and that <math>x=10</math> when <math>y=4</math> and <math>z=14</math>. Then, when <math>y=16</math> and <math>z=7</math>, <math>x</math> equals:
  
−
<math>\textbf{(A)}\ 180\qquad \textbf{(B)}\ 160\qquad \textbf{(C)}\ 154\qquad \textbf{(D)}\ 140\qquad \textbf{(E)}\ 120</math>
+
<math>\textbf{(A)}\ 180\qquad \textbf{(B)}\ 160\qquad \textbf{(C)}\ 154\qquad \textbf{(D)}\ 140\qquad \textbf{(E)}\ 120</math>
  
 
== Solution ==
 
== Solution ==
Line 22: Line 22:
 
Simplifying this we get,
 
Simplifying this we get,
  
−
<math>\fbox{B) 160}</math>
+
<math>\fbox{\textbf{(B) } 160}</math>
  
 
~lli, awanglnc
 
~lli, awanglnc

Latest revision as of 11:32, 21 July 2024

Problem 20

It is given that $x$ varies directly as $y$ and inversely as the square of $z$, and that $x=10$ when $y=4$ and $z=14$. Then, when $y=16$ and $z=7$, $x$ equals:

$\textbf{(A)}\ 180\qquad \textbf{(B)}\ 160\qquad \textbf{(C)}\ 154\qquad \textbf{(D)}\ 140\qquad \textbf{(E)}\ 120$

Solution

$x$ varies directly to $\frac{y}{z^2}$ (The inverse variation of y and the square of z)

We can write the expression

$x = \frac{ky}{z^2}$

Now we plug in the values of $x=10$ when $y=4$ and $z=14$.

This gives us $k=490$

We can use this to find the value of $x$ when $y=4$ and $z=14$

$x=\frac{490\cdot4}{14^2}$

Simplifying this we get,

$\fbox{\textbf{(B) } 160}$

~lli, awanglnc

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_AHSME_Problems/Problem_20&oldid=223300"