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

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

Step-by-step 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

Answer 2

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`