Question

White gold" alloy contains 90% gold and 10% silver. \Red gold" alloy contains 80% gold and 20% copper. How much \white gold" and how much \red gold" should we add to 60g of the alloy containing 45% gold, 50% silver and 5% copper, in order to obtain alloy with 70% gold, 15% silver and 15% copper concentration?

Answer 1

Let White gold = W

Let Red gold = R

Let gold = g

Let silver = s

Let copper = c

$\therefore$ W = 0.9g + 0.1s

$\therefore$ R = 0.8g + 0.2c

$xg(W) + yg(R) + 60g(0.45g+0.5s+0.05c) = (x+y+60)g\times(0.7g+0.15s+0.15c)$

where x = mass of White gold; and

where y = mass of Red gold

$x(0.9g+0.1s) + y(0.8g+0.2c) +60(0.45g+0.5s+0.05c) = (x+y+60)(0.7g+0.15s+0.15c)$

$0.9gx + 0.1sx + 0.8gy + 0.2cy + 27g + 30s + 3c = 0.7gx + 0.7gy + 42g + 0.15sx + 0.15sy + 9s + 0.15cx + 0.15cy + 9c$

collecting like terms;

$0.9gx + 0.8gy + 27g + 0.1sx + 30s + 0.2cy + 3c = 0.7gx + 0.7gy + 42g + 0.15sx + 0.15sy + 9s + 0.15cx + 0.15cy + 9c$

$g(0.9x + 0.8y + 27) + s(0.1x + 30) + c(0.2y + 3) = g(0.7x + 0.7y +42) +s(0.15x + 0.15y + 9) + c(0.15x +0.15y +9)$

Comparing both sides

Firstly;

$g(0.9x + 0.8y + 27) = g( 0.7x + 0.7y + 42)$

$0.9x + 0.8y + 27 = 0.7x + 0.7y + 42$

$0.2x + 0.1y = 15$ $(\times10)$

$2x + y = 150$ ---------(i)

Secondly;

$s(0.1x + 30) = s(0.15x + 0.15y +9)$

$0.1x + 30 = 0.15x + 0.15y +9$

$-0.05x - 0.15y = -21$ $(\times-20)$

$x + 3y = 420$ ---------(ii)

Thirdly;

$c(0.2y + 3) = c(0.15x + 0.15y +9)$

$0.2y + 3 = 0.15x + 0.15y +9$

$0.05y - 0.15x = 6$ $(\times-20)$

$y - 3x = 120$ ---------(iii)

Making equation (i) and equation (iii) into simultaneous equations

$2x + y = 150$ ---------(i)

$y - 3x = 120$ ---------(iii)

Subtracting equation (ii) from equation (i)

$2x-(-3x) +y -y = 150 -120$

$5x = 30$

$\therefore x = 6$

Substituting the value of x into equation (ii)

$y - 3x = 120$

$y = 120 + 3x$

$y = 120 + 3(6)$

$y = 138$

$\therefore$ 6g of White gold and 138g of Red gold should be added