Question 282711: Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution. How many gallons of each solution must be used to obtain the desired mixture?
Found 6 solutions by richwmiller, oberobic, mananth, ikleyn, josgarithmetic, greenestamps:
Answer by richwmiller(17219)   (Show Source):
Answer by oberobic(2304)  (Show Source):
You can put this solution on YOUR website! When solving solution problems, you need to focus on how much 'pure' stuff you need.
In this case, you need to produce 4 gal. of a desired 45% solution.
That means you will need .45*4 = 1.8 gal. of the 'pure' stuff.
The simplest way would be to take 1.8 of 'pure' stuff and add 2.2 gal. of water, but you do not have that option.
Instead you have to use 90% and 30% solutions.
.
Defining the amounts of is critical.
x = amount of 90% solution
y = amount of 30% solution
4 = total amount of 45% solution
.
So y = 4-x, which will make the job of solving easier.
.
.9x + .3(4-x) = .45*4
.9x + 1.2 -.3x = 1.8
.6x + 1.2 = 1.8
.6x = .6
x = 1
.
That means you would propose to use 1 gal. of 90% solution + 3 gal. of 30% solution.
.
Check by determining if that is enough 'pure' stuff.
1*.9 = .9
3*.3 = .9
.9 + .9 = 1.8, which is what you needed.
.
Answer:
Mix 1 gal. of 90% solution with 3 gal. of 30% solution to obtain 4 gal. of 45% solution.
.
Done.
Answer by mananth(16949)  (Show Source):
You can put this solution on YOUR website! Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution. How many gallons of each solution must be used to obtain the desired mixture?
let the quantity of 90% solution added be x
the total solution obtained is 4 gallons
So the remaing solution will be 30% solution
=4-x gallons
The sum of the acid content of mixed solutions is equal to acid content in the mixture
0.9x+0.3*(90-x)= 0.45 *4
0.9x+0.27 - 0.3x= 1.80
0.6x= 1.80-0.27
0.6x= 1.53
x=1.53 / 0.6
x= 2.55 gallons which is 90% acid solution
Balance will be 30% acid solution.
4-2.55 = 1.45 gallons
Answer by ikleyn(53742)  (Show Source):
You can put this solution on YOUR website! .
Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution.
How many gallons of each solution must be used to obtain the desired mixture?
~~~~~~~~~~~~~~~~~~~~~
The solution by @mananth is fatally wrong due to wrong setup of the basic equation.
I came to bring a correct solution.
let the quantity of 90% solution added be x
the total solution obtained is 4 gallons
So the remaining volume will be 30% solution
=4-x gallons
The sum of the acid content of mixed solutions is equal to acid content in the mixture
0.9x + 0.3*(4-x) = 0.45*4
0.9x + 1.2 - 0.3x = 1.80
0.6x = 1.8 - 1.2
0.6x = 0.6
x = 0.6 / 0.6
x = 1 gallons which is 90% acid solution
Balance will be 30% acid solution.
4 - 1 = 3 gallons
ANSWER. 1 gallon of the 90% solution and 3 gallons of the 30% solution.
CHECK for concentration.  = 0.45, or 45%. ! Precisely correct !
Solved correctly.
Answer by josgarithmetic(39790) (Show Source):
Answer by greenestamps(13325)  (Show Source):
You can put this solution on YOUR website!
For a formal algebraic solution using the standard method for solving 2-part mixture problems like this, see the response from tutor @ikleyn.
If formal algebra is not required, the answer can be obtained quickly with some mental arithmetic, using the fact that the ratio in which the two ingredients need to be mixed is exactly determined by where the target percentage lies between the percentages of the two ingredients.
Having used a lot of words to explain how this method works, here is the short work needed to get the answer.
(1) use some mental arithmetic to find that 45% is "3 times as close" to 30% as it is to 90%;
(2) that means the mixture needs to contain 3 times as much of the 30% acid as it does the 90% acid
Since the total amount is 4 gallons, you need 3 gallons of the 30% acid and 1 gallon of the 90% acid
ANSWER: 3 gallons of the 30% acid, 1 gallon of the 90% acid
|
|
|
