SOLUTION: what quantity x of an 80% solution must be added to a 15% solution to produce 600mL of a 40% solution

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what quantity x of an 80% solution must be added to a 15% solution to produce 600mL of a 40% solution
:
mix x amt with (600-x) amt
:
.8x + .15(600-x) = .4(600)
.8x + 90 - .15x = 240
.8x - .15x = 240 - 90
.65x = 150
x =
x = 230.77 ml of 80% solution
:
:
Check solution: (600 - 230.77 = 369.23 ml of 15% solution:
.8(230.77) + .15(369.23) = .4(600)
184.62 + 55.38 = 240

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what quantity x of an 80% solution must be added to a 15% solution to produce 600mL of a 40% solution
:
let x and y be amounts of 80% and 15% solutions respectively
:
x+y=600.....eq 1
.8x+.15y=.4(600)...eq 2
:
rewrite eq 1 to x=600-y and place that value into eq 2
:
.8(600-y)+.15y=.4(600)
:
480-.8y+.15y=240
:
-.65y=-240
:
ml of 15% solution
:
x=600-yml of 80% solution

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We know =ml of 80%
Let =ml of 15%
Then, ---------------------> EQN 1
Also, when mixed:
-----------> EQN 2
Goin back EQN 1 we get,
----------------------------> EQN 3, subst. in EQN 2:



---------->
needed for 15% sol'n.
Going back EQN 3:
, needed for the 80% sol'n.
In doubt? Go back EQN 2:



Thank you,
Jojo