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Difference between revisions of "1967 AHSME Problems/Problem 19"

(Created page with "== Problem == The area of a rectangle remains unchanged when it is made <math>2 \frac{1}{2}</math> inches longer and <math>\frac{2}{3}</math> inch narrower, or when it is made <m...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
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We are given <math>xy = (x+\frac{5}{2})(y-\frac{2}{3}) = (x - \frac{5}{2})(y + \frac{4}{3})</math>
 +
 
 +
FOILing each side gives:
 +
 
 +
<math>xy = xy - \frac{2}{3}x + \frac{5}{2}y - \frac{5}{3} = xy + \frac{4}{3}x - \frac{5}{2}y - \frac{10}{3}</math>
 +
 
 +
Taking the last two parts and moving everything to the left gives:
 +
 
 +
<math>-2x + 5y + \frac{5}{3} = 0</math>
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 +
Taking the first two parts and multiplying by <math>3</math> gives <math>-2x + \frac{15}{2}y - 5 = 0</math>
 +
 
 +
Solving both equations for <math>-2x</math> and setting them equal to each other gives <math>5y + \frac{5}{3} = \frac{15}{2}y - 5</math>, which leads to <math>y = \frac{8}{3}</math>
 +
 
 +
Plugging that in to <math>-2x + 5y + \frac{5}{3} = 0</math> gives <math>x = \frac{15}{2}</math>.
 +
 
 +
The area of the rectangle is <math>xy = 20</math>, or <math>\fbox{E}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=18|num-a=20}}   
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{{AHSME 40p box|year=1967|num-b=18|num-a=20}}   
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:40, 16 August 2023

Problem

The area of a rectangle remains unchanged when it is made $2 \frac{1}{2}$ inches longer and $\frac{2}{3}$ inch narrower, or when it is made $2 \frac{1}{2}$ inches shorter and $\frac{4}{3}$ inch wider. Its area, in square inches, is:

$\textbf{(A)}\ 30\qquad \textbf{(B)}\ \frac{80}{3}\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ \frac{45}{2}\qquad \textbf{(E)}\ 20$

Solution

We are given $xy = (x+\frac{5}{2})(y-\frac{2}{3}) = (x - \frac{5}{2})(y + \frac{4}{3})$

FOILing each side gives:

$xy = xy - \frac{2}{3}x + \frac{5}{2}y - \frac{5}{3} = xy + \frac{4}{3}x - \frac{5}{2}y - \frac{10}{3}$

Taking the last two parts and moving everything to the left gives:

$-2x + 5y + \frac{5}{3} = 0$

Taking the first two parts and multiplying by $3$ gives $-2x + \frac{15}{2}y - 5 = 0$

Solving both equations for $-2x$ and setting them equal to each other gives $5y + \frac{5}{3} = \frac{15}{2}y - 5$, which leads to $y = \frac{8}{3}$

Plugging that in to $-2x + 5y + \frac{5}{3} = 0$ gives $x = \frac{15}{2}$.

The area of the rectangle is $xy = 20$, or $\fbox{E}$.

See also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
All AHSME Problems and Solutions

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