To the geniuses.?

Question

Put this in two equations:
Pure salt was added to a 10% saline solution to produce 9cc of 20% saline solution. How much of each ingredient was used?

Answer 1

First, in a problem like this, you want to understand the problem:
Let x = amount of pure salt. Since the total solution is 9 cc, the amount of 10% saline solution added has to be (9 - x).

~100% Pure Salt ~
(A) x --- Number of cc's of solution
(B) 100% --- Percent pure salt
(C) 1.00(x) --- Number of cc's of pure salt

~10% Saline Solution~
(A) (9 - x) --- Number of cc's of solution
(B) 10% --- Percent of pure salt
(C) 0.10(9 - x) --- Number of cc's of pure salt

~Final 20% Solution~
(A) 9 --- Number of cc's of solution
(B) 20% --- Percent of pure salt
(C) 0.20(9) --- Number of cc's of pure salt

In the three "charts" above, (C) is obtained by multiplying the amount (A) by the percentage (B). So, using or (C) values:
1.00(X) + 0.10(9 - x) = 0.20(9) --- Multiply through...
x + 0.90 - 0.10(x) = 1.80 --- Combine like terms...
0.90x + 0.90 = 1.80 --- Subtract 0.90 from both sides...
0.90x = 0.90 --- Divide both sides by 0.90...
x = 1

If x = 1, then (9 - x) = 8. Thus, you need 1 cc of pure salt and 8 cc's of 10% saline solution to produce 9 cc's of 20% saline solution.

Answer 2

Well i'm not a genius but i'll trie :
x + 10/100 y = 9 + 20/100 y ( , the thing that i don't understand is
it 20% of 9 cc or out of 20% produced 9cc of saline solution )
In order of x :
( x - salt )
x + 10/100y = 9 + 20/100y
--------------------------------------------------------------------------
if 20 % is to 9 cc , 100 % is to X
X = 100 mulltipied by 9 and the result of that divided by 20
X = 900 / 20
X = 45
, and the same with the other ,
if 100% is to 45 cc , 10% is to Y
Y : 10 multiplied by 45 and the result of that divided by 100
Y : 450/100
Y : 4.5

Now , to find the salt is :
x = 9 + 45 + 4.5 or 450/100
100x = 900 + 4500 + 450
100x = 5850
x = 5850/100
x = 58.5 cc

Resume : There was 45 cc ( 10% is 4.5 cc and 20% is 9 cc ) of saline solution and 58.5 of salt .

I think that i am wrong , no let me rephrase that , i know that i am wrong but what the hell !!!

Answer 3

If you add salt to a volume of saline, then theoretically,
the volume of salt added needs to be taken into account,
as it most likely will increase the total volume, although
I think there may be substances that actually decrease
the volume. I'm not sure of that, but for practical purposes,
the volume of a salt added is not usually taken into account
because the amount added is generally small enough not to
make too much difference to the final concentration.
So for this exercise, I'll keep it simple.

10% saline is 10 grams of salt made up to a 100 cc with water.
So how much salt is required for 9 cc of water?
The equation for this is : N = 9 * 10 / 100 = 0.9 grams
So the 10% saline is 0.9 grams of salt made up to 9 cc with water.

Now if we add M grams of salt to this solution, the final
concentration of salt will be : (0.9 + M) / 9 = 20% = 20 / 100.

Cross-multiplying that equation gives :
(0.9 + M) * 100 = 9 * 20
90 + 100M = 180
100M = 90
M = 0.9 grams

In summary, to get 9 cc of 20% solution,
add 0.9 grams of salt to 9 cc of 10% saline.

Answer 4

I'm doing this using w to represent water and s to represent the amount of salt that was _added_, not the total amount of salt in the solution. (You can easily go back and figure the total salt once you solve for either of the other variables.)
w = .9 (9 - s)
.1 (9-s) + s = .2 (9)

Solving goes as follows:
.9 -.1s + s = 1.8
.9s = .9
s = 1

So, 1cc of salt was added.
Solving for w, we get:
w = .9 (9 - 1)
w = .9 (8)
w = 7.2

Thus, there was originally 7.2 cc of water and .8 cc of salt (10% salinity. After 1 cc of salt was added, you have 1.8 cc of salt with the same 7.2 cc water (20%).

Answer 5

It's kind of tricky because the salt will increase the density & volume.
9cc @ 10% contains about 1 g of salt. Off hand I don't know the density of a 10% salt solution. You can look it up in many places. 20% wil ne about 2g. If you have the density of 10% & 20% brines, I can give you an exact answer.