Ametalworker has a metal alloy that is 20% copper and another alloy that is 60% copper. how many kilograms of each alloy should the metalworker combine to create 100 kg of a 52% copper alloy? the metalworker should use of the metal alloy that is 20% copper and kilograms of the metal alloy that is 60% copper (type whole numbers.)

I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.

That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.

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Using an equation

If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...

... 0.60x + 0.20(100 -x) = 0.52·100

... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20

... x = 32/0.40 = 80 . . . . . kg of 60% alloy

... (100 -80) = 20 . . . . . . . .kg of 20% alloy

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %