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1985 AHSME Problems/Problem 7: Difference between revisions

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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
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>


== Solution 1 ==
==Solution==
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>.
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>.
 
== Solution 2 ==
 
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>.
 
Keep going: <math>a\div (b-(c+d))</math>.
 
More: <math>a\div (b-(c+d))</math>.
 
Simplify: <math>\frac{a}{b-c-d}</math>. Select <math>\boxed{E}</math>.
 
~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}}

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 (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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