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

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==Solution==
 
==Solution==
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)</math>, 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}}</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)</math>, and thus the whole expression would be grouped as <math>a\div (b-(c+d)) = \boxed{\text{(E)} \ \frac{a}{b-c-d}}</math>.
  
 
==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 $a\div (b-(c+d)) = \boxed{\text{(E)} \ \frac{a}{b-c-d}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
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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