Question

The alloy of copper and tin with a mass of $8$ kg contains $p$ per cent of copper. What piece of the copper-tin alloy containing $40$ per cent of tin must be alloyed with the first piece in order to obtain a new alloy with the minimum percentage of copper if the mass of the second piece is $2$ kg?

Answer 1

Let amt of silver be $x$ and copper be $y$

ATP,

$\frac{x+3}{x+y+3}\times 100=90$

$=>(x+3)\times 100=90(x+y+3)$

$=>(x+3)\times 10=9(x+y+3)$

$x+3=9y------\left(1\right).$

Again,$90\%of2kg=1.8kg$

$\therefore \frac{x+1.8}{x+y+2}\times 100=84$

$=>\frac{x+1.8}{x+y+2}=\frac{84}{100}$

$=>\frac{x+1.8}{x+y+2}\times 100=\frac{21}{25}$

$25x+45=21x+21y+42$

$=>4x+3=21y------\left(2\right).$

from eqn$1$.,$x=9y-3$

Putting$x$ in eqn $2$.

$4(9y-3)+3=21y$

$=>36y-12+3=21y$

$15y=9$

$y=\frac{9}{15}$

$=0.6$

$\therefore$ $x=9\left(0.6\right)-3$

$=5.4-3$

$=2.4$

$\therefore$ weight of silver=$2.4kg$

$\%$of silver$=\frac{x}{x+y}\times$100$\%$

$=\frac{2.4}{2.4106}\ast 100\%$

$=\frac{2.4}{3}\times 100\%$

$=0.84\times 100=80\%$

$\therefore$$\%$of silver=$80\%$.