# Question bf080

May 19, 2017

Nickel content solution: you may not have seen this approach before.

The added pure nickel is $9 \frac{1}{3}$ pounds

#### Explanation:

Let the amount of added nickel as a percentage be $x$
Let the final blend mass be $b$

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

$\implies \left(\text{change in nickel content")/("change in added pure nickel}\right) \to \frac{100 - 15}{100} \equiv \frac{25 - 15}{x}$

Turn it all upside down:

$\frac{100}{100 - 15} \equiv \frac{x}{25 - 15}$

x=(100xx10)/85=11.76470....%

Thus 100%-x% represents the ${70}^{\text{lb}}$ of original the 15% content blend.

=>100%-x%" of "b=70^("lb")

$\left(\frac{100}{100} - \frac{11.76470 . .}{100}\right) b = 70$

$\implies \frac{88.23529 . .}{100} b = 70$

$b = 79.33333 \ldots \to 79 \frac{1}{3} {\textcolor{w h i t e}{}}^{\text{lb}}$

Thus the added pure nickel is $9 \frac{1}{3}$ pounds
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:

(25%xx79 1/3)/(15%xx70+9 1/3) =1

The nickel in each is the same so the answer is correct

May 19, 2017

Grain question

$\text{ corn } \to 50$ bushels

$\text{ barley } \to 100$ bushels

#### Explanation:

Gradient for part is the same as the gradient for the whole.

$\frac{3.80 - 2.6}{100} = \frac{3 - 2.6}{x}$

Turn up the other way

$\frac{100}{1.2} = \frac{x}{0.4}$

$\implies x = 33 \frac{1}{3}$

Thus the blend has 33 1/3% corn
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Total batch weight is to be 150 bushels

$\implies \text{ corn } \to \frac{33 \frac{1}{3}}{100} \times 150 = 50$ bushels

$\implies \text{ barley } \to 100$ bushels

May 20, 2017

Alternative method for the corn/silage question

The amount of corn is 50 bushels
Thus the amount of silage is 100 bushels

#### Explanation:

Let the amount of corn be x%

Target cost $3.00 Silage cost$2.60
Corn cost $3.80 As we have x% corn then there is the related silage amount of 100%-x% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ [x%xx$3.80]+[(100%-x%)xx$2.60]=$3.00#

$\left[\frac{x \times 3.80}{100}\right] + \left[\frac{100 \times 2.60}{100} - \frac{x \times 2.60}{100}\right] = 3$

$\frac{1}{100} \left(3.80 x + 260 - 2.6 x\right) = 3$

$3.80 x + 260 - 2.6 x = 300$

$1.2 x + 260 = 300$

$x = \frac{300 - 260}{1.2} = 33 \frac{1}{3}$

So the amount of corn is $\frac{33 \frac{1}{3}}{100} \times 150 = 50$ bushels
Thus the amount of silage is $150 - 50 = 100$ bushels
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~