Question

Question #bf080

Answer 1

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

The added pure nickel is `9 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.

`=>("change in nickel content")/("change in added pure nickel") ->(100-15)/(100)-=(25-15)/x`

Turn it all upside down:

`100/(100-15)-=x/(25-15)`

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

Thus `100%-x%` represents the `70^("lb")` of original the 15% content blend.

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

`(100/100-(11.76470..)/100)b=70`

`=>(88.23529..)/100 b= 70`

`b=79.33333...->79 1/3 color(white)()^("lb")`

Thus the added pure nickel is `9 1/3` pounds
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Check:

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

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

Answer 2

Grain question

`" corn "-> 50` bushels

`" barley "-> 100` bushels

Explanation:

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

`(3.80-2.6)/100 = (3-2.6)/x`

Turn up the other way

`100/1.2=x/0.4`

`=>x=33 1/3`

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

`=>" corn "->(33 1/3)/100xx150=50` bushels

`=>" barley "-> 100` bushels

Answer 3

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%`
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`[x%xx$3.80]+[(100%-x%)xx$2.60]=$3.00`

`[(x xx3.80)/100]+[(100xx2.60)/100-(x xx2.60)/100]=3`

`1/100(3.80x+260-2.6x)=3`

`3.80x+260-2.6x=300`

`1.2x+260=300`

`x=(300-260)/(1.2)=33 1/3`

So the amount of corn is `(33 1/3)/100xx 150 = 50` bushels
Thus the amount of silage is `150-50=100` bushels
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