Difference between revisions of "1957 AHSME Problems/Problem 8"

(Problem)
m (see also box)
 
(One intermediate revision by one other user not shown)
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 50p box|year=1957|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:07, 25 July 2024

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 AHSC (ProblemsAnswer KeyResources)
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 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