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 "1957 AHSME Problems/Problem 8"

(→‎Problem)
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as <math>x</math> would have <math>2</math> parts, <math>y</math> would have <math>3</math>, and <math>z</math> would have <math>5</math>). One part, after some algebra, equals <math>10</math>, so <math>x</math>, <math>y</math>, and <math>z</math> are <math>20</math>, <math>30</math>, and <math>50</math>, respectively.
 +
 +
We can plug <math>x</math> and <math>y</math> into the equation given to us: <math>30 = 20a-10</math>, and then solve to get <math>a = \boxed{\textbf{(A)}2}</math>.
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1957|num-b=7|num-a=9}}
 
{{AHSME box|year=1957|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:40, 12 October 2020

The numbers $x,\,y,\,z$ are proportional to $2,\,3,\,5$. The sum of $x, y$, and $z$ is $100$. The number y is given by the equation $y = ax - 10$. Then a is:

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ \frac{5}{2}\qquad \textbf{(E)}\ 4$

Solution

In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as $x$ would have $2$ parts, $y$ would have $3$, and $z$ would have $5$). One part, after some algebra, equals $10$, so $x$, $y$, and $z$ are $20$, $30$, and $50$, respectively.

We can plug $x$ and $y$ into the equation given to us: $30 = 20a-10$, and then solve to get $a = \boxed{\textbf{(A)}2}$.

See Also

1957 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
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=1957_AHSME_Problems/Problem_8&oldid=134980"