Difference between revisions of "1964 AHSME Problems/Problem 23"

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<math>\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96</math>
 
<math>\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96</math>
  
==Solution=
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==Solution==
  
 
Set the two numbers as <math>x</math> and <math>y</math>. Therefore,  
 
Set the two numbers as <math>x</math> and <math>y</math>. Therefore,  

Latest revision as of 03:00, 19 June 2019

Problem

Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24$. The product of the two numbers is:

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96$

Solution

Set the two numbers as $x$ and $y$. Therefore, $x+y=7(x-y),  xy=24(x-y)$, and $24(x+y)=7xy$. Simplifying the first equation gives $y=\frac{3}{4}x$. Substituting for $y$ in the second equation gives $\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$. Substituting $x=0$ back into the first equation yields $1=-7$ which is false, so $x=0$ is not valid and $x=8$. Substituting into $y=\frac{3}{4}x$ gives $y=6$ and $xy=\boxed{\textbf{(D) } 48}$.

See More

1964 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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

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