Como o painel tem 12 azulejos, vou escrever as duas frações em doze avos: o 12 é múltiplo do 4 e do 3 ao mesmo tempo, então serve de denominador comum para as duas.

Reescrevo o $\mathbf{\frac{\textcolor{#ef4444}{1}}{\textcolor{#3b82f6}{4}}}$ em doze avos:

$\mathbf{\dfrac{\textcolor{#ef4444}{1}}{\textcolor{#3b82f6}{4}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{1} \times 3}{\textcolor{#3b82f6}{4} \times 3}}$

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

Reescrevo o $\mathbf{\frac{\textcolor{#ef4444}{1}}{\textcolor{#3b82f6}{3}}}$ em doze avos:

$\mathbf{\dfrac{\textcolor{#ef4444}{1}}{\textcolor{#3b82f6}{3}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{1} \times 4}{\textcolor{#3b82f6}{3} \times 4}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{4}}{\textcolor{#3b82f6}{12}}}$

Agora somo os numeradores e conservo o denominador:

$\mathbf{\dfrac{\textcolor{#ef4444}{3}}{\textcolor{#3b82f6}{12}} + \dfrac{\textcolor{#ef4444}{4}}{\textcolor{#3b82f6}{12}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{7}}{\textcolor{#3b82f6}{12}}}$

O total trocado é 7 dos 12 azulejos.

O número total de azulejos trocados é 7.

Questão Q12379