An alloy $P$ has copper and zinc in the proportion of $5:2$ (by weight), while another alloy $Q$ has the same metals in the proportion of $3:4$ (by weight). If these two alloys are mixed in the proportion of $a:b$ (by weight), a new alloy $R$ is formed, which has equal contents of copper and zinc. Then, the proportion of copper and zinc in the alloy $S$, formed by mixing the two alloys $P$ and $Q$ in the proportion of $b:a$ (by weight) is

Understanding the given information:

Alloy $P$: Copper : Zinc = $5:2$ (by weight)

Alloy $Q$: Copper : Zinc = $3:4$ (by weight)

Alloy $R$: Made by mixing $P$ and $Q$ in ratio $a:b$, has equal copper and zinc

Alloy $S$: Made by mixing $P$ and $Q$ in ratio $b:a$

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Finding the fraction of copper in each alloy:

In Alloy $P$: Total parts = $5 + 2 = 7$

Copper fraction = $\dfrac{5}{7}$

In Alloy $Q$: Total parts = $3 + 4 = 7$

Copper fraction = $\dfrac{3}{7}$

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Setting up equation for Alloy $R$:

When mixing $P$ and $Q$ in ratio $a:b$, copper fraction in mixture:

$\dfrac{a \times \dfrac{5}{7} + b \times \dfrac{3}{7}}{a + b}$

Since $R$ has equal copper and zinc, copper fraction = $\dfrac{1}{2}$

$\dfrac{a \times \dfrac{5}{7} + b \times \dfrac{3}{7}}{a + b} = \dfrac{1}{2}$

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Solving for ratio $a:b$:

Cross multiplying:

$a \times \dfrac{5}{7} + b \times \dfrac{3}{7} = \dfrac{1}{2}(a + b)$

Multiplying by $14$:

$10a + 6b = 7a + 7b$

$3a = b$

Therefore: $a:b = 1:3$

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Finding composition of Alloy $S$:

Alloy $S$ mixes $P$ and $Q$ in ratio $b:a = 3:1$

Copper fraction in $S$:

$\dfrac{3 \times \dfrac{5}{7} + 1 \times \dfrac{3}{7}}{3 + 1} = \dfrac{\dfrac{15}{7} + \dfrac{3}{7}}{4} = \dfrac{\dfrac{18}{7}}{4} = \dfrac{18}{28} = \dfrac{9}{14}$

Zinc fraction in $S$:

$1 - \dfrac{9}{14} = \dfrac{5}{14}$

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Copper : Zinc = $\dfrac{9}{14} : \dfrac{5}{14} = 9:5$

The proportion of copper and zinc in alloy $S$ is $9:5$