Question

An alloy of copper and aluminum has 40% copper. Another alloy of copper and zinc has copper and zinc in the ratio 2: 7. These two alloys are mixed in ratio 5:3. Quantity of aluminum is what percent more/less than the quantity of copper in final alloy.

Answer 1

Quantity of aluminum is 4.2 percent more than the quantity of copper in final alloy.

Explanation:

Alloy-1 has 40% copper and 60% of aluminium.

Let the mass of alloy-1 mixed be 5x

Mass of aluminium in alloy-1 = $\frac{60}{100}\times 5x=3x$

Mass of copper in alloy-1 = $\frac{40}{100}\times 5x=2x$

Alloy-2 has copper and zinc in the ratio 2: 7.

percentage of copper = $\frac{2}{2+7}\times 100=22.2\%$

Let the mass of alloy-2 be 3x

Mass of copper in alloy-2 = $3x\times \frac{22.2}{100}=0.666 x$

After mixing both alloys, mass of th new alloy = 5x + 3x = 8x

Mass of copper = 2x+0.666x=2.666 x

Mass of aluminium = 3x

Percentage of copper in new alloy = $\frac{2.666x}{8x}\times 100=33.3\%$

Percentage of aluminium in new alloy = $\frac{3x}{8x}\times 100=37.5\%$

33.3% < 37.5%

Difference between quantity percentage of aluminium and copper = 37.5% - 33.3% = 4.2%

Quantity of aluminum is 4.2 percent more than the quantity of copper in final alloy.

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