An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?

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Solution

Initial quantity of copper $= \frac{80}{100} \times 50 = 40$ g
And that of Bronze $= 50 - 40 = 10$ g
Let $^{'} p^{'}$ gm of copper is added to the mixture
$\Rightarrow (50 + p) \times \frac{90}{100} = 40 + p$
$\Rightarrow 45 + 0.9 p = 40 + p$
$\Rightarrow p = 50$ g
Hence, $50$ gms of copper is added to the mixture, so that the copper is increased to $90 %$ .

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An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?

Initial quantity of copper =80100×50=40 g And that of Bronze =50−40=10 gLet ′p′ gm of copper is added to the mixture⇒(50+p)×90100=40+p⇒45+0.9p=40+p⇒p=50 gHence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.