The volume of the water used in the original mixture:

$100 \% - 20 \% = 80 \% \Rightarrow 80 \% × 100 = 80 \ ml$

This remains the same 80 ml (since we only add milk, without adding any water), but will be the 100% - 25 % = 75% of the new mix.

Hence, the volume of the new mix:

$\frac {80 }{0.75} = 106.67 \ ml$

The amount of milk added:

$106.67 - 100 = 6.67 \ ml$

Therefore, our answer should be:

$\boxed { \text {More than 5 ml} }$

Let $x$ be the amount of milk we add.
We want:
$\frac{20+x}{100+x}100$ % $=25$ % , since there were 20ml of milk originally (20% of 100ml)
$\frac{20+x}{100+x}=\frac{1}{4}$
$80+4x=100+x$
$3x=20$
$x=\frac{20}{3}\approx6.66ml>5ml$

If you want a solution that tells you how much milk you need, go ahead and see Maximos's and Zee's solution. But here I am going to try to present a solution that does not begin with "Let $x$ be the amount of..."

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Attempt 1 (Iterative Repair)

There is 20 mL of of milk in an 100 mL solution. Our first guess would be to hope that adding 5 mL of milk makes it 25%. So, we go ahead and add 5 mL of milk.

How much milk do we have now? We whip out a calculator and plug the numbers in to see that

$\frac{20 mL + 5 mL} {100 mL + 5 mL} \approx 23.8 \%$

So, we definitely need to add more milk.

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Attempt 2 (Noticing Quantities)

We want $25\%$ of milk.

• If we had exactly 100 ml of solution, we'd have exactly 25 mL of milk in the solution.
• If we had more or less than 100 mL of the solution, we'd correspondingly need more or less than 25 mL.

But we already have 100 mL of the solution. So, we are definitely going to have more than 25 mL of milk. And that means, we are going to have to add more than 5 mL of milk, for sure.

Quick solution

Adding 5 mL alcohol will make it 25% of the original amount of liquid. However, since the total amount of liquid increases, the percentage will be less than 25%. We need to add $\boxed{\text{more}}$ to make up for this.

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Detailed conceptual solution

The first solution has a ratio of water:alcohol = 4:1. The second solution has a ratio of 3:1.

The amount of water stays the same, so I multiply both ratios with a constant to make them the same:

We go from a ratio of 12:3 to a ratio of 12:4. Thus the amount of alcohol that should be added is 1/12th of the amount of water there is. 1/12th of 80 is 6 2/3 mL.

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Mindless algebra solution

$\frac{20 + x}{100 + x} = 25\%\ \ \ \therefore\ \ \ 20 + x = 25 + \tfrac14 x \ \ \ \therefore \ \ \ \tfrac34 x = 5\ \ \ \therefore\ \ \ x > 5.$

If we add 5 ml alcohol to this solution, the fraction of alcohol in it will be (25/105). This is less than 25%. Thus more than 5 ml alcohol would need to be added.

It took me a couple of minutes to sort through the terrible grammar that results in the misleading nature of this question.

1) The mixture IS 20% isopropyl aclohol, not OF. When one says 20% blank, you don't say of. It gives the misleading idea that some solution that is 20% alcohol was poured into some mixture, meaning that the actual quantity of isopropyl aclohol in the entire mixture is less than 20%, which means it was a dilution.

2) Dangling modifier: The modifier, "which consists of 80 ml of water and 20 ml of isopropyl alcohol", is supposed to be right after the object it's describing. This is not the case here, making it seem like the isopropyl alcohol is 80 mL of water and 20 mL of isopropyl alcohol.

Both of these interpretations lead to different answers.

Please, whoever chooses these questions, please put the effort to choose questions that aren't ambiguous due to several interpretations of meaning, because the problem should be mathematical, and not a problem of grammar that the solver has to deal with. Right? Or at least put the effort to fix the ambiguity.

It can be simply solved by - $\frac{20+x}{100+x}$ = $\frac{25}{100}$ let x be the amount of alcohol further added to the mixture Then amount of alcohol in mixture will increase by 20+x and the total will be increased by 100+x comparing with $\frac{25}{100}$ we will get our answer !!

At 20% the ratio of water to alcohol is 4:1. At 25% the ratio is 3:1. The volume of water is not changing. The final volume of alcohol must be 80/3 = 26 2/3 ml, an additional 6 2/3 ml more. The answer is "More than 5 ml".

Who needs "x" here anyway?

Who needs isopropyl alcohol? It would be better with ethyl when one could mix a nice cool drink when it was all done. That detail rubs me the wrong way.

You would need 20/3 ml of milk.