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olasank [31]
3 years ago
14

300 ml of pure alcohol is poured from a bottle containing 2 l of pure alcohol. Then, 300 ml of water is added into the bottle. A

gain 300 ml of the diluted alcohol is poured out and 300 ml of water is added into the bottle, Find the percentage of pure alcohol in the solution now.
Mathematics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

The present percentage of pure alcohol in the solution is 72.25% of pure alcohol

Step-by-step explanation:

The volume of pure alcohol poured from the 2 l bottle of pure alcohol = 300 ml of pure alcohol

The volume of water added into the bottle after pouring out the pure alcohol = 300 ml of water

The volume of diluted alcohol poured out of the bottle = 300 ml of diluted alcohol

The volume of water added into the bottle of diluted alcohol after pouring out the 300 ml of diluted alcohol = 300 ml of water

Step 1

After pouring the 300 ml of pure alcohol and adding 300 ml of water to the bottle, the percentage concentration, C%₁ is given as follows;

C%₁ = (Volume of pure alcohol)/(Total volume of the solution) × 100

The volume of pure alcohol in the bottle = 2 l - 300 ml = 1,700 ml

The total volume of the solution = The volume of pure alcohol in the bottle +  The volume of water added = 1,700 ml + 300 ml = 2,000 ml = 2 l

∴ C%₁ = (1,700 ml)/(2,000 ml) × 100 = 85% percent alcohol

Step 2

After pouring out 300 ml diluted alcohol from the 2,000 ml, 85% alcohol and adding 300 ml of water, we have;

Volume of 85% alcohol = 2,000 ml - 300 ml = 1,700 ml

The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol = 85/100 × 1,700 = 1,445 ml

The total volume of the diluted solution = The volume of the 85% alcohol in the solution + The volume of water added

∴ The total volume of the twice diluted solution = 1,700 ml + 300 ml = 2,000 ml

The present percentage of pure alcohol in the solution, C%₂ = (The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol)/(The total volume of the diluted solution) × 100

∴ C%₂ = (1,445 ml)/(2,000 ml) × 100 = 72.25 %

The present percentage of pure alcohol in the solution, C%₂ = 72.25%

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Answer:

The probability that the second ball drawn is a white ball if the second ball is drawn without replacing the first ball is \frac{1}{3} or 0.3333

Step-by-step explanation:

Probability is the greater or lesser possibility that a certain event will occur. In other words, the probability establishes a relationship between the number of favorable events and the total number of possible events. Then, the probability of any event A is defined as the ratio between the number of favorable cases (number of cases in which event A may or may not occur) and the total number of possible cases. This is called Laplace's Law:

P(A)\frac{number of favorable cases of A}{total number of possible cases}

Each of the results obtained when conducting an experiment is called an elementary event. The set of all elementary events obtained is called the sample space, so that every subset of the sample space is an event.

The total number of possible cases is 15 (10 black balls added to the 5 white balls).

As each extraction is without replacement, the events are dependent. For that, the dependent probabilities are defined first

Two events are dependent on each other when the fact that one of them is verified influences the probability of the other being verified.  In other words, the probability of A happening is affected because B has happened or not.

The probability of two events A and B of two successive simple experiments in a dependent compound experiment is:

A and B are dependent ⇔ P (A ∩ B) = P (A) · P (B / A)

                                              P (A ∩ B) = P (B) · P (B / A)

As the color of the first ball that is extracted is unknown, there are two cases: that the ball is black or that the ball is white.

<em> It will be assumed first that the first ball drawn is black</em>. Then the probability of this happening is \frac{10}{15} since the number of black balls in the urn is 10 and the total number of cases is 15. It is now known that the second ball extracted will be white. Then the number of favorable cases will be 5 (number of white balls inside the ballot urn), but now the number of total cases is 14, because a ball was previously removed that was not replaced.  So the probability of this happening is \frac{5}{14}

So the probability that the first ball is black and the second white is:

<em>\frac{10}{15} *\frac{5}{14} =\frac{5}{21}</em>

<em>It will now be assumed first that the first ball that is drawn is white.</em> Then the probability of this happening is \frac{5}{15} since the number of white balls in the urn is 5 and the total number of cases is 15. And it is known that the second ball drawn will be white. Then, the number of favorable cases will be 4 (number of white balls inside the urn, because when removing a white ball and not replacing it, its quantity will decrease), and the total number of cases is 14, same as in the previous case  So, the probability of this happening is  \frac{4}{14}

So the probability that the first ball is white and the second white is:

<em>\frac{5}{15} *\frac{4}{14} =\frac{2}{21}</em>

If A and B are two incompatible events, that is, they cannot occur at the same time, the probability of occurrence A or of occurrence B will be the sum of the probabilities of each event occurring separately.

These are events are incompatible, since I cannot, in a first extraction, extract a black and white ball at the same time. So:

<em>\frac{5}{21} +\frac{2}{21} =\frac{7}{21}=\frac{1}{3} =0.3333</em>

Finally, <u><em>the probability that the second ball drawn is a white ball if the second ball is drawn without replacing the first ball is \frac{1}{3} or 0.3333</em></u>

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