Para comparar $\mathbf{\dfrac{3}{4}}$ e $\mathbf{\dfrac{5}{6}}$ , convertemos ambas em frações equivalentes com um denominador comum — um múltiplo dos 2 denominadores. Procurando o menor múltiplo comum de 4 e 6, percebemos que 12 está na tabuada de 4 (4 × 3 = 12) e na tabuada de 6 (6 × 2 = 12).

Convertendo $\mathbf{\dfrac{3}{4}}$ para denominador 12:

$\mathbf{\dfrac{\textcolor{#ef4444}{3}}{\textcolor{#3b82f6}{4}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{3 \times 3}}{\textcolor{#3b82f6}{4 \times 3}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{9}}{\textcolor{#3b82f6}{12}}}$

Convertendo $\mathbf{\dfrac{5}{6}}$ para denominador 12:

$\mathbf{\dfrac{\textcolor{#ef4444}{5}}{\textcolor{#3b82f6}{6}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{5 \times 2}}{\textcolor{#3b82f6}{6 \times 2}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{10}}{\textcolor{#3b82f6}{12}}}$

Comparando: $\mathbf{\dfrac{\textcolor{#ef4444}{9}}{\textcolor{#3b82f6}{12}}}$ < $\mathbf{\dfrac{\textcolor{#ef4444}{10}}{\textcolor{#3b82f6}{12}}}$ , então $\mathbf{\dfrac{\textcolor{#ef4444}{3}}{\textcolor{#3b82f6}{4}}}$ < $\mathbf{\dfrac{\textcolor{#ef4444}{5}}{\textcolor{#3b82f6}{6}}}$ .

Logo, Sara (com $\mathbf{\dfrac{5}{6}}$ ) percorreu mais do que Lara (com $\mathbf{\dfrac{3}{4}}$ ).

O resultado correto do problema é Sara.

Questão Q11417