# Question 91298

Jun 25, 2017

The mass of the 30% alloy is $= 10 k g$ and the mass of the 70% alloy is $= 40 k g$

#### Explanation:

This is a mass balance with respect to copper.

Let the mass of the 30% be $= x k g$

The mass of the 70% is $= \left(50 - x\right) k g$

So,

$x \cdot \frac{30}{100} + \left(50 - x\right) \cdot \frac{70}{100} = 50 \cdot \frac{62}{100}$

$0.3 x + 35 - 0.7 x = 31$

$0.4 x = 35 - 31 = 4$

$x = \frac{4}{0.4} = 10 k g$

Jun 25, 2017

$10$ 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

$\left(50 - x\right)$ kg of 70% copper containing metal alloy were combined.

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

$x \cdot 0.3 + \left(50 - x\right) \cdot 0.7 = 50 \cdot 0.62 \mathmr{and} 0.3 x - 0.7 x = 31 - 35$ or

$- 0.4 x = - 4.0 \mathmr{and} 0.4 x = 4.0 \mathmr{and} x = 10 \therefore 50 - x = 50 - 10 = 40$

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

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

achieve $50 k g$ of 62% copper alloy. [Ans]

Jun 25, 2017

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. $\textcolor{b r o w n}{\text{This really does work!}}$

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:

$\left(\text{mass")/("copper content}\right) \to \frac{50}{70 - 30} \equiv \frac{x}{62 - 30}$

$x = \frac{50 \times 32}{40} = 40$

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%#