1985 AHSME Problems/Problem 7

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 1

The expression would be grouped as $a\div(b-(c+d))$. This is equal to $\frac{a}{b-c-d}, \boxed{\text{E}}$.

Solution 2

First of all, let's start of we the right most part, $c+d$. So after the first step, we have $a\div b-(c+d)$.

Keep going: $a\div (b-(c+d))$.

More: $a\div (b-(c+d))$.

Simplify: $\frac{a}{b-c-d}$. Select $\boxed{E}$.

~hastapasta

See Also

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