SOLUTION: Pure acid is to be added to a 10​% acid solution to obtain 45L of a 40​% acid solution. What amounts of each should be​ used?

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Question 1204463: Pure acid is to be added to a 10​% acid solution to obtain 45L of a 40​% acid solution.
What amounts of each should be​ used?
Found 4 solutions by josgarithmetic, math_tutor2020, greenestamps, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
If use v liters of the pure 100% acid, then use 45-v liters of the 10% acid.

To account for the amount of pure acid for the mixture,

-


----------think about this expression a bit, before continuing...




------and you know the quantity for the 10% acid is 30 L.
.
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answers:
15 liters of pure acid
30 liters of the 10% acid solution

============================================================================================

Explanation

There are at least two ways to solve this problem.
Method 1 is the standard algebraic approach.
Method 2 is what is taught to pharmacy students.



Method 1

Batch A = pure acid = 100% acid
Batch B = 10% acid solution

x = number of liters of batch A
45-x = number of liters of batch B
These two amounts combine to 45 liters total.

x = amount of pure acid from batch A
0.10(45-x) = 4.5 - 0.10x = amount of pure acid from batch B
x+(4.5 - 0.10x) = 0.90x+4.5 = total amount of pure acid

Divide this over the 45 total liters, and the goal is to arrive at a 40% concentration.

(amount of pure acid)/(total) = concentration
(0.90x+4.5)/(45) = 0.40
0.90x+4.5 = 45*0.40
0.90x+4.5 = 18
0.90x = 18-4.5
0.90x = 13.5
x = 13.5/0.90
x = 15
which leads to
45-x = 45-15 = 30

Therefore, we need 15 liters of pure acid to add to the 30 liters of the 10% acid solution.

In other words, we need 15 liters of batch A and 30 liters of batch B.

--------------------------------------------------------------------------------

Method 2

Batch A = 100% acid
Batch B = 10% acid
target solution = 40%

Gap from target to 100% is 60%
Gap from 10% to target is 30%

The ratio of 60% to 30% is 60:30 = 2:1

Meaning we'll need twice as much one batch compared to the other.

2:1 scales up to 2k:1k for some positive real number k.
2k+1k = 45
3k = 45
k = 45/3
k = 15
So 2k = 2*15 = 30

The amounts for batches A and B are 15 and 30. The order isn't clear just yet.

If we had 30 L of batch A, then we'll have 15*0.10 = 1.5 L of pure acid from batch B.
The total amount of pure acid would be 30+1.5 = 31.5
But notice that 31.5/45 = 0.70 = 70% which isn't the 40% target we're after.

If we had 15 L of batch A, then we'll have 30*0.10 = 3 L of pure acid from batch B.
15+3 = 18 L of pure acid total out of 45 L total
18/45 = 0.40 = 40%
We've reached the goal.

For more information, search out "Alligation Method Pharmacy".
The spelling of "Alligation" is correct. There isn't an "e" but an "i" instead. Think "alligator".
It's unfortunate the two words sound identical, and are spelled almost the same.

Here is a useful calculator
https://calculatormaths.com/alligation-calculator
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Here is a quick and easy way to solve any 2-part mixture problem like this without using the formal algebraic method shown by the other tutors.

You are starting with 10% acid and adding 100% acid, stopping when you get to 40% acid. The solution is then in a couple of easy steps.

40% is 1/3 of the way from 10% to 100%.
Therefore 1/3 of the mixture is the 100% acid that you are adding.
Since you want 45L of the mixture, the amount of 100% acid you are adding is 1/3 of 45L, or 15L, which means 30L of the 10% acid you started with.

ANSWERS: 30L of the 10% acid, 15L of the 100% acid
Answer by ikleyn(52782)   (Show Source): You can put this solution on YOUR website!
.
Pure acid is to be added to a 10​% acid solution to obtain 45L of a 40​% acid solution.
What amounts of each should be​ used?
~~~~~~~~~~~~~~~~~~~~


        There is an elegant method to solve this problem  (and many other similar problems)
        without using equations.  It uses your common sense and your logic,  ONLY. 


45 liters of the 40% acid contain  0.4*45 = 18 liters of the pure acid and  45 - 12 = 27 liters of water.


When we add pure acid to the 10% acid mixture, we do not add water.  

Hence, 27 liters of water was in 10% acid solution initially.


Again: the initial 10% acid mixture contained 27 liters of water, and water was 9 
of 10 parts of the volume of this 10% mixture.


Hence, the pure acid in this 10% acid mixture was 3 liters , while the water was 27 liters.


To get 18 liters of the pure acid in the final 40% mixture, 18-3 = 15 liters of the pure acid should be added
to the initial 10% mixture.


ANSWER.  15 liters of the pure acid and 3+27 = 30 liters of the 10% mixture shoud be used.

Solved.

--------------------

To see many other similar solved problems, look into the lessons
    - Special type mixture problems on DILUTION adding water
    - Increasing concentration of an acid solution by adding pure acid
    - Increasing concentration of alcohol solution by adding pure alcohol
    - How many kilograms of sand must be added to a mixture of sand and cement
    - How much water must be evaporated
in this site.



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