Question

The mass of two pieces of brass is $60$ kg. The first piece contains $10$ kg of pure copper and the second piece contains $8$ kg of pure copper. What is the percentage of copper in the first piece of brass if the second piece contains $15$ per cent more copper than the first?

Answer 1

Let percentage of pure copper in first piece be $x\%$

Percentage of pure copper in second piece$=(x+15)\%$

Weight of first piece

$=\frac{10\times 100}{x}$

$=\frac{1000}{x}$

Weight of second piece

$=\frac{8\times 100}{x+15}$

$=\frac{800}{x+15}$

$\frac{1000}{x}+\frac{800}{x+15}=60$

$\frac{100}{x}+\frac{80}{x+15}=6$

$\frac{50}{x}+\frac{40}{x+15}=3$

$\frac{50(x+15)+40x}{x(x+15)}=3$

$=>50x+750+40x=3x(x+15)$

$=>3x^2-45x-750=0$

$x^2-15x-250=0$

$=>(x-25)(x+10)=0$

$=>x=25,-10$

But mass can not be negative

Percentage of pure copper in first piece$=25\%$