A shopkeeper has two brands of coffee?

Question

A at the price 19.50 kr/kg and B at 10.50 kr/kg. He wants to make a 50 kg mixture of these two, that can be sold for 11.30 kr/kg. How much of each brand shall he take.

Answer 1

suppose that the quantities he will mix of each brand is x&y
so
x + y = 50
19.5x + 10.5y = 50 * 11.30

19.5x + 10.5y = 565
x + y = 50
_________________solve the 2 equations together
x = 4 4/9 kg
y = 45 5/9 kg

Answer 2

If he starts with "B", for every 10-percent addition of "A", he will have charge 0.90 kr/kg more to break even (assuming he is honest). In this case, he want to have a blend that is 0.80 kr/kg more. Thus, he has to substitute
0.80/0.90 x 10 percent or 8.89 percent.
If he started with 50 kg "B", he would add 4.45 kg "A".
for 54.45 kg. Scale back to 50 kg to get answer.

Alternately, weight x cost/weight= cost.
A+B= 50 (weight constraint)
Ax19.50 + Bx10.50 = 56.50 (kr to sell 50 kg)
Substitute A or B from the weight constraint into the cost equation to solve.

Answer 3

Let x be the amount of brand A coffee. So 50-x is the amount of brand B coffee.

19.50x+10.50(50-x) = 11.30(50)

Solve for x,
x = 4.44 kg
y = 45.56 kg

Answer 4

Let x = amount of Type A
Then 50 -x = amount of type B
19.5x +10.5(50-x) = 50*11.3
19.5x+ 525 - 10.5x = 565
19.5x-10.5x = 565-525
9x = 40
x = 4 4/9 kg type A
50-x = 45 5/9 kg type B