How much of​ a(n) 70% orange juice drink must be mixed with 11 gallons of a 10% orange juice drink to obtain a mixture that is 50% orange​ juice?

Question

in an integer or a decimal

Answer 1

.7x + 11(.1) = .5(11+x)

.7x- .5x = (.5-.1)11

.2x = .4(11) = 4.4

x = 4.4/.2 = 22 gallons of 70% mixed with 11 gallons of 10% yield 33 gallons of 50%

check the answer

.7(22)+.1(11)= 15.4 +1.1 = 16.5= .5(33)

7-5 = 2

5-1 =4

7 is closer to 5

so 4/2 = 2 times as 7% as 2%

2 x11 =22

Answer 2

0.7j + 0.1(11) = 0.5(j + 11)

0.7j + 1.1 = 0.5j + 5.5

0.2j = 4.4

j = 22 gallons