Question

Question #91298

Answer 1

The mass of the `30%` alloy is `=10kg` and the mass of the `70%` alloy is `=40kg`

Explanation:

This is a mass balance with respect to copper.

Let the mass of the `30%` be `=xkg`

The mass of the `70%` is `=(50-x)kg`

So,

`x*30/100+(50-x)*70/100=50*62/100`

`0.3x+35-0.7x=31`

`0.4x=35-31=4`

`x=4/0.4=10kg`

Answer 2

`10` kg of `30%` copper containing metal alloy and

`40` kg of `70%` copper containing metal alloy were combined.

Explanation:

Let `x` kg of `30%` copper containing metal alloy and

`(50-x)` kg of `70%` copper containing metal alloy were combined.

Then , balancing the input and output of copper content , we get ,

`x*0.3 +(50-x)*0.7 = 50*0.62 or 0.3x-0.7x =31-35` or

`-0.4x = -4.0 or 0.4x = 4.0 or x=10 :. 50-x=50-10=40`

Hence `10` kg of `30%` copper containing metal alloy and

`40` kg of `70%` copper containing metal alloy were combined to

achieve `50 kg` of `62%` copper alloy. [Ans]

Answer 3

Just for the hell of it this is a different approach:

40 kg at 70% content
10 Kg at 30% content

Explanation:

Set material 1 `M_1 -> 30%`copper
Set material 2 `M_2->70%`copper
Set target as `T->62%` copper

Set the amount of 70% alloy be `x`

The mass is fixed at 50Kg thus if you increase the amount of `M_1` then the amount of `M_2` must reduce to maintain this weight. Thus if we consider just one of these the other is directly implied. Consequently we can and may use a straight line graph situation to represent the blend. `color(brown)("This really does work!")`

It models the numbers only approach. It is just that it looks different.

The gradient of part is the same as the gradient of the whole:

`("mass")/("copper content")->50/(70-30)-=x/(62-30)`

`x=(50xx32)/(40)=40`

So we have:

40 kg at 70% content
10 Kg at 30% content

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Check:

`10/(40+10)xx30% = color(white)(5)6%`
`40/(40+10)xx70% = ul(56% larr" Add")`
`" "62%`