Question

pure tin was mixed with a4%in tin alloy to produce an ally that was 16%tin .How much pure in and how much 4% alloy were used to produce 32 kg of 16% alloy ?

Answer 1

Answer:

4 kg of pure tin

Step-by-step explanation:

Let x kg be the amount of pure tin and y kg be the amount of 4% tin alloy mixed.

1. The weight of the mixture is 32 kg, then

$x+y=32$

2. There are 4% of tin in y kg of 4% tin alloy, so there are $0.04y$ kg of tin in this alloy. There are 16% of tin in 32 kg of the mixture, then there are $0.16\cdot 32=5.12$ kg of tin in alloy. Thus,

$x+0.04y=5.12$

3. Solve the system of two equations:

$\left\{\begin{array}{l}x+y=32\\x+0.04y=5.12\end{array}\right.$

From the first equation

$x=32-y$

Substitute it into the second equation:

$32-y+0.04y=5.12\\ \\3,200-100y+4y=512\\ \\-100y+4y=512-3,200\\ \\-96y=-2,688\\ \\y=28\\ \\x=32-28=4$

Answer 2

Answer:

The amount of 4% alloy were used to produce 32 kg of 16% alloy is 4

Explanation:

let x kg be the amount of pure tin and y kg be the amount of 4% tin alloy mixed.

The weight of the mixture is 32 kg

then$x+y=32$  ----- -(1)

there are 4% of tin in y kg of 4% tin alloy, so there are 0.04y kg of tin in this alloy.

There are 16% of tin in 32 kg of the mixture, then there are 0.16x0.32 = 5.12 kg of tin in alloy;

$x+0.04y=5.12$ ------------------- - (2);

solving (1) and (2)

we get, $x+y=32$

$x+0.04y=5.12;$

$x=32-y$

substituting in eq-(2)

we get; $32-y+0.04y=5.12$

multiplying with 100 on both sides we get;

$3200-100y+4y=512$

$-96y=-2688$

y=28

$x = 32-y = 32-28 = 4$

X=4