Difference between revisions of "1985 AHSME Problems/Problem 7"

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==Problem==
 
==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, <math> a\times b-c </math> in such languages means the same as <math> a(b-c) </math> in ordinary algebraic notation. If <math> a\div b-c+d </math> is evaluated in such a language, the result in ordinary algebraic notation would be
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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, <math>a \times b - c</math> in such languages means the same as <math>a(b-c)</math> in ordinary algebraic notation. If <math>a\div b-c+d</math> is evaluated in such a language, the result in ordinary algebraic notation would be
  
 
<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>

Latest revision as of 18:06, 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 $a\div (b-(c+d)) = \boxed{\text{(E)} \ \frac{a}{b-c-d}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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