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 12:32, 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 (ProblemsAnswer KeyResources)
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