// Numbas version: finer_feedback_settings {"name": "proporcionalidad Inversa", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"maxRuns": 100, "condition": ""}, "variable_groups": [{"name": "Unnamed group", "variables": []}], "preamble": {"js": "", "css": ""}, "statement": "
$\\simplify{{a}}\\,$ llaves que entregan la misma cantidad de agua por minuto llenan un estanque en $\\simplify{{k/a}}\\,$ minutos, ¿Cuánto demorarán $\\,\\simplify{{b}}\\,$ de estas llaves?
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"}], "extensions": [], "tags": [], "advice": "Se trata de un problema de proporcionalidad inversa, ya que si una variable disminuye la otra aumenta de manera proporcional.
\n\\begin{align}
\\begin{array}{ccc}
\\text{Cantidad de Llaves} & & \\text{Minutos} \\\\
\\simplify{{a}} & \\longrightarrow & \\simplify{{k/a}} \\\\
\\simplify{{b}} & \\longrightarrow & \\simplify{x}
\\end{array}
\\end{align}
Como se trata de un problema de proporcionalidad inversa, tenemos que invertir la segunda columna:
\n\\begin{align}
\\begin{array}{ccc}
\\simplify{{a}} & \\longrightarrow & \\simplify{x} \\\\
\\simplify{{b}} & \\longrightarrow & \\simplify{{k/a}}
\\end{array}
\\end{align}
Calculamos el valor de $\\,x$
\n\\begin{align}
\\simplify[std]{x}=\\dfrac{\\simplify{{k/a}}\\cdot\\simplify{{a}}}{\\simplify{{b}}}=\\simplify{({(k/a)}{a})/{b}}
\\end{align}
Se demoran $\\,\\simplify{{k/b}}\\,$ minutos.
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