how many grams of pure salt must be added to 40 grams of a 20% salt mix to make a solution that is 36% salt?

Answer 1

The original 40 g mixture is 20% salt. Amount of salt in original mixture equals 40 g x 0.20= 8g, therefore water = 40g-8g=32 g.

new mixture percent salt is 36% . where addl salt equals S,

0.36= (S+8)/(40 + S)

multiplying both sides by (40 + S)

.36(40+S)=S+8

14.4 + .36S= S+8

14.4 - 8 = S-.36S

6.4 = .64 S

S=10

Total salt in new mix = 8g +10g = 18g
Total new mix = 40g + 10g = 50g

18/50= .36 (36%)

Answer 2

x/40=20%
x=8
8+y / 40+y =35%
y= 120/13

Answer 3

Let x = grams of salt to be added
40*.2 + x = (40+x).36
8 + x = 14.4 =.36x
x-.36x = 14.4-8
.64x = 6.4
x = 10 grams must be added

Answer 4

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Answer 5

40/0.2 = x/0.36

-> x = 72

Which means 32 grams would needed to be added.

Answer 6

10grams of pure salt