Difference between revisions of "1985 AHSME Problems/Problem 7"
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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> | ||
− | ==Solution== | + | == Solution 1 == |
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 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>. | ||
+ | |||
+ | == 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}} |
Revision as of 17:35, 2 May 2022
Contents
[hide]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, in such languages means the same as in ordinary algebraic notation. If is evaluated in such a language, the result in ordinary algebraic notation would be
Solution 1
The expression would be grouped as . This is equal to .
Solution 2
First of all, let's start of we the right most part, . So after the first step, we have .
Keep going: .
More: .
Simplify: . Select .
~hastapasta
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.