7.81 lb of 21% aluminium and 33.19 lb of 42% aluminium must be use to make 41 lb of a third alloy containing 38% aluminium.
Given:
one alloy containing 21 % aluminum and another containing 42% aluminum.
41 pounds of a third alloy containing 38 % aluminum.
Let x represents the quantity containing of alloy 21% aluminium
Let y represents the quantity containing 42% of alloy aluminium.
From the question;
x + y = 41 ----------------------------(1)
21%x + 42%y = 38% (41)
⇒0.21x + 0.42y = 15.58 ------------------------------------(2)
We can now solve equation (1) and (2) using the substitution method
From equation (1);
x = 41 - y
Substitute into the equation equation and solve for y
0.21(41 - y) + 0.42y = 15.58
8.61 - 0.21y + 0.42y = 15.58
8.61 + 0.21y = 15.58
0.21y =6.97
Divide both-side by 0.21
y= 33.19
Substitute the value of y into x = 41 - y to deermine the x -value.
x = 41 - 33.19
x = 7.81
x=7.81 and y = 33.19
Hence, 7.81 lb of 21% aluminium and 33.19 lb of 42% aluminium must be use to make 41 lb of a third alloy containing 38% aluminium.