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Difference between revisions of "1985 AHSME Problems/Problem 7"
(Combined solutions and improved wording and formatting)
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<math> \mathrm{(A)\ } \frac{a}{b}-c+d \qquad \mathrm{(B) \ }\frac{a}{b}-c-d \qquad \mathrm{(C) \  } \frac{d+c-b}{a} \qquad \mathrm{(D) \  } \frac{a}{b-c+d} \qquad \mathrm{(E) \  }\frac{a}{b-c-d}  </math>
 
<math> \mathrm{(A)\ } \frac{a}{b}-c+d \qquad \mathrm{(B) \ }\frac{a}{b}-c-d \qquad \mathrm{(C) \  } \frac{d+c-b}{a} \qquad \mathrm{(D) \  } \frac{a}{b-c+d} \qquad \mathrm{(E) \  }\frac{a}{b-c-d}  </math>
  
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== Solution 1 ==
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==Solution==
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The expression would be grouped as <math> a\div(b-(c+d)) </math>. This is equal to <math> \frac{a}{b-c-d}, \boxed{\text{E}} </math>.
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The rightmost part of the expression is <math>c+d</math>, so <math>b-c+d</math> would be grouped as <math>b-(c+d), and thus the whole expression would be grouped as </math>a\div (b-(c+d)) = \frac{a}{b-c-d}<math>. Select </math>\boxed{\text{(E)} \ \frac{a}{b-c-d}}$.
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== Solution 2 ==
 
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First of all, let's start of we the right most part, <math>c+d</math>. So after the first step, we have <math>a\div b-(c+d)</math>.
 
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Keep going: <math>a\div (b-(c+d))</math>.
 
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More: <math>a\div (b-(c+d))</math>.
 
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Simplify: <math>\frac{a}{b-c-d}</math>. Select <math>\boxed{E}</math>.
 
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~hastapasta
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=6|num-a=8}}
 
{{AHSME box|year=1985|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:05, 19 March 2024

Problem

In some computer languages (such as APL), when there are no parentheses in an algebraic expression, the operations are grouped from right to left. Thus, $a\times b-c$ in such languages means the same as $a(b-c)$ in ordinary algebraic notation. If $a\div b-c+d$ is evaluated in such a language, the result in ordinary algebraic notation would be

$\mathrm{(A)\ } \frac{a}{b}-c+d \qquad \mathrm{(B) \ }\frac{a}{b}-c-d \qquad \mathrm{(C) \ } \frac{d+c-b}{a} \qquad \mathrm{(D) \ } \frac{a}{b-c+d} \qquad \mathrm{(E) \ }\frac{a}{b-c-d}$

Solution

The rightmost part of the expression is $c+d$, so $b-c+d$ would be grouped as $b-(c+d), and thus the whole expression would be grouped as$a\div (b-(c+d)) = \frac{a}{b-c-d}$. Select$\boxed{\text{(E)} \ \frac{a}{b-c-d}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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