⚡ AZUCATION SHORTCUT:
Let $C$ and $K$ be costs of coffee and cocoa.
$0.16C + 0.84K = 240$ and $0.36C + 0.64K = 320$.
Solving these, $C = 600$ and $K = 171.4$.
Target mixture: $x(600) + (1-x)171.4 = 376 \implies x = 0.5$.
In $10$ kg, coffee content = $0.5 \times 10 = \mathbf{5}$ kg.

Step 1: Formulate equations for cost
Let the cost per kg of coffee be $x$ and cocoa be $y$.
From the first mixture: $0.16x + 0.84y = 240 \implies 16x + 84y = 24000 \quad \dots(1)$
From the second mixture: $0.36x + 0.64y = 320 \implies 36x + 64y = 32000 \quad \dots(2)$

Step 2: Solve for x and y
Divide (1) by 4: $4x + 21y = 6000 \implies 36x + 189y = 54000$.
Subtract (2) from this: $(189-64)y = 54000 - 32000 \implies 125y = 22000 \implies y = 176$.
Substitute $y=176$ in (1): $16x + 84(176) = 24000 \implies 16x = 24000 - 14784 = 9216$.
$x = 576$ (Cost of coffee per kg).

Step 3: Find Coffee percentage in new mixture
Let the new mixture have $c\%$ coffee.
$\frac{c}{100}(576) + \frac{100-c}{100}(176) = 376$
$576c + 17600 - 176c = 37600 \implies 400c = 20000 \implies c = 50\%$.

Step 4: Calculate final quantity
In $10$ kg of the new mixture, coffee quantity $= 50\%$ of $10 = \mathbf{5}$ kg.

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