10% of a 50% alcohol solution is replaced with water

The quick answer is: (50%)(0.9)(0.9)(0.9). Here's why:

To set this up, imagine that you have 100 ml of solution, and imagine that the alcohol and the water are separated. So, you have 50 ml of water and 50 ml of alcohol.

When you remove 10% of the solution, you're removing 10% of the water and 10% of the alcohol. So, you're removing 5 ml of water and 5 ml of alcohol, to get 45 ml of water and 45 ml of alcohol remaining. From here, you'll add 10ml of water.

IMPORTANT #1: Each time we remove 10 ml of solution and replace it with 10 ml of water, the resulting solution is 100 ml.

IMPORTANT #2: All we need to do is keep track of the alcohol each time. Also notice that removing 10% of the alcohol is the same as leaving 90%.

So, we begin with 50 ml of alcohol.

Step 1: Remove 10% (i.e., keep 90%)
This leaves us with (50)(0.9) ml of alcohol

Step 2: Remove 10%
This leaves us with (50)(0.9)(0.9) ml of alcohol

Step 3: Remove 10%
This leaves us with (50)(0.9)(0.9)(0.9) ml of alcohol
(50)(0.9)(0.9)(0.9) equals approximately 36 ml.

So, our final mixture has a volume of 100 ml of which approximately 36 ml are alcohol. So, the concentration of alcohol is approximately 36%

Answer: D

Cheers,
Brent

Final concentration of alcohol = (50%) *(9/10) *(9/10) *(9/10) = 36.45%

This answer doesn't make sense.

Assume 100g of solution. 50 alcohol / 50 water.

Replace 10% of solution = 10 grams

10 grams will contain - 5 grams of other solution and 5 grams of alcohol

New solution - 45 grams of alcohol 55 grams of water

Replace another 10 grams , 4.5 grams of alcohol and 5.5 grams of water

Final concentration 40.5 grams of alcohol over a total of 100 grams

Not exactly 36% realized I missed a step

Yes, the replacement step is repeated 2 times.

Hi Brent,
I followed your way in the in the beginning of the example but it gave me different answer. Please comment on my way.

Start with 100 ml. Remover 10%........> result in 45 W & 45 A.......> then add 10% water=10 ml Water.........> result in 55 W & 45 A= 100 ml

Remove 10%.......................................> result in 50 W & 40 A........> then add 10 ml water...........................> result in 60 W & 40 A= 100 ml

Remove 10%.......................................> result in 55 W & 35 A.........> then add 10 ml water...........................> result in 65 W & 35 A= 100 ml

The final concentration is 35/100= 35%. I know it close to Answer 36 but why is it difference than the answer of the way you used above? is there is any mistake?

Thanks

Solution:

The question states that we have a solution with 50% alcohol and 50% water. Lets assume the solution is 100ml. This would mean that the solution has 50ml alcohol and 50ml water.

Step1: 10% of the solution is replaced with water.....If we remove 10% of the 100ml solution, we would be left with 90ml soultion which would result in 45ml of alchohol and 45ml of water. We now add 10ml of water to the soution. The resulting mix would contain 45ml alcohol and 55ml water

Step2: We replace 10 % of the new solution with water. If we now remove 10% of the solution we will again be left with 90ml of the solution but in a different mix. We would be left with 0.9(45ml) = 40.5 ml of alchol and 0.9(55ml ) = 49.5ml of water. We now add 10 ml of water to the solution. The new mix would be 40.5ml of alcohol and 59.5ml of water.

Step 3: We now need to repeat the above step one last time. Removing 10% of the new solution we will be left with 0.9(40.5ml)= 36.45 ml of alcohol and 0.9( 59.5) = 53.55ml of water. If we now add 10 ml of water we would be left with 36.45ml of alcohol an 63.55ml of water in a 100ml solution.

Now it is easy to calculate the concentration of alcohol in the final solution which will be 36.45/100 * 100 = 36.45%. The answer is D.

Since the numbers in the answers are pretty far apart. I just simplified and rounded.

Alcohol = 50%
Total - 10% * 3 = 70%
.5 * .7 = .35 (so that means 70% of 50% of the total is 35%)

Answer: D) 36

I see the mistake you made.
When removing 10% for the second time you have to remove 10% of 55 W and 45 A .So you have to remove 5.5 from 55W and 4.5 from 45 A.
Which is different from 50 W and 40 A. It should be 40.5 A and 49.5 W= 90 L

Hope you understand your mistake.

Ans)
Suppose the vol of total solution is P liters. let x denote 10 % of P.
So, x= 10/100 * P = P/10

Let the amount of alcohol in the final solution be 'y'.

So,

50/100(P-3x) = y/100(P)
Replace x by P/10

50(7/10)=y
so, y= 35%

Is there a smart and quick way to calculate 50%*90%*90%*90%?

If the original solution is 100, we have 50% alcohol and 50% water. Once 10% is removed, we now have:

50 x 0 .9 = 45 water

50 x 0.9 = 45 alcohol

Once that 10% is replaced with water, we have now have 55 water and 45 alcohol. When another 10% is removed, we have:

55 x 0.9 = 49.5 water

45 x 0.9 = 40.5 alcohol

Once that 10% is replaced with water, we have 59.5 water and 40.5 alcohol. When another 10% is removed, we have:

59.5 x 0.9 = 53.55 water

40.5 x 0.9 = 36.45 alcohol

Once that 10% is replaced with water, we have 63.55 water and 36.45 alcohol.

Answer: D
_________________

The question should ask for the approximate amount, not the exact amount as implied.

36.45 is the concentration after the reductions.

Did i get it wrong cause I thought they replaced the water four times cause it says:
10% of a 50% alcohol solution is replaced with water. From the resulting solution, again 10% is replaced with water. This step is repeated once more. What is the concentration of alcohol in the final solution obtained?

"this step is repeated once more" means that the whole step is replace 10% of a 50% alcohol solution and again 10 % is replaced.

I calculated it like this:

50% x 10/100 = 45%
45% -(45% x10/100) = 40.5%
40.5% - (40.5% x 10/100) = 36.45%

That's it

Given the answer choices, the problem should read as follows:



Since \(\frac{1}{10}\) must be removed a total of 3 times, let the amount of alcohol = 10*10*10 = 1000 liters and the volume of the container = 2000 liters, yielding a container that is 50% alcohol.
With each removal, the amount of alcohol is reduced by 10%.

1st removal: 1000 - (10% of 1000) = 1000 - 100 = 900
2nd removal: 900 - (10% of 900) = 900 - 90 = 810
3rd removal: 810 - (10% of 810) = 810 - 81 = 729

Since the 2000-liter container now contains 729 liters of alcohol, we get:
\(\frac{alcohol}{total} = \frac{729}{2000} ≈ \frac{360}{1000} = \frac{36}{100} =\) 36%
_________________

How do we get 55 water and 45 alcohol after the 2nd step ? Can you please explain. I'm not getting it.

Though I got approx 36 as an answer but I wonder what this statement meant: this step is repeated once more:are we talking about doing the 10% thing once or twice !! Little bit confusing !

Posted from my mobile device

Response:

Alcohol and water form what is called a “homogenous solution”. This means that the concentration of any sample we take from this solution is the same as the concentration of the entire solution. In other words, if we remove 10 liters from a 50% solution, then exactly 50% of the 10 liters is alcohol and the rest is water.

Using the above reasoning, when we remove 10% from the solution, we actually remove 10% of the water and 10% of the alcohol. That’s why, if we begin with 100 liters, removing 10% of the solution will leave us with 45 liters of water and 45 liters of alcohol. After we add 10 liters of water, we will have 55 liters of water and 45 liters of alcohol.
_________________

I like the one-step formula (50%)(0.9)(0.9)(0.9)
but what if instead of water, we replaced with 25% alcohol-concentrated solution? is there a generic formula for such case?