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I'm new here, so kindly bear with me if this question is too trivial for the forum. I tried working it out but to no avail.

A solution contain wine and water in ratio 2:1. Out of it 36 litre solution is replaced by 15 litre water. Again 26 litre solution is replaced by 26 litres of water. Now wine to water ratio becomes 8:7. Find the difference between the quantity of water in the initial and the final solution.

Also, why does the ratio stay the same every time we take out liquid from a mixture of liquids. I mean not in this case because water is added here. I mean is there a proof of the intuitive fact?

  • The assumption is that the solution is uniform throughout the liquid so when you divide it into two parts the ratio is the same in both parts. This is not necessarily true (see the Wikipedia article on the halocline about sea salt concentration with depth) but is a more reasonable assumption for a well-mixed solution. – Henry Feb 23 '24 at 10:18
  • @Henry Thank you. I get that it should be the case but i can't work it out mathematically. Also, can you please work out the first part? –  Feb 23 '24 at 10:23
  • What have you tried on the first part? – Henry Feb 23 '24 at 10:27
  • The calculations are quite cumbersome. Hence, I'll make you understand the method. Here assume the total volume to be x, so vol of wine is $\frac{2x}{3}$ and vol of water is $\frac{x}{3}$. As they said 36 L of solution is removed, so remove 24 L of wine and 12L of water following the ratio 2:1 and later add 15L water. The problem comes here as once again 26L is removed from the solution and hence ratio becomes more complicated (becomes quadratic and longer to solve) – JEEAspirant Feb 23 '24 at 10:30
  • After the first operation, the quantity of wine left is 2X—24 litres. The quantity of water is X+3 litres(after adding 15 litres). Next we draw out (2X—24/3X—21)26 litres of wine and (X+3/3X—24)26 litres of water. Then I calculated how much wine and water remained in the container by subtracting these quantities from (2X—24) and (X+3). After this I equalled this ratio to 8/7 (after adding 26 litres to the second quantity got earlier. But it led me nowhere. Is there a niftier and clever way to solve this that skirts all this labor? –  Feb 23 '24 at 10:35
  • @MahendraVarma Thank you. Yes the calculation is tedious. –  Feb 23 '24 at 10:36
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    You are learning that some calculations are tedious. That is a valuable lesson. – Gerry Myerson Feb 23 '24 at 10:49
  • @GerryMyerson Is that so? How does one go about solving these kind of problems without having to toil with the quantities? –  Feb 23 '24 at 10:52
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    What makes you think all mathematical questions have easy solutions? Sometimes, you have to do pages & pages of computations to solve a question, and the sooner you learn that, the better. – Gerry Myerson Feb 23 '24 at 11:16
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    @GerryMyerson I agree with you that that must be the case for math is about patience and long calculations, but I was refering to this kind of questions. It's basic stuff, so if it can't be done quick one gets frustrated. –  Feb 23 '24 at 11:58

1 Answers1

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Let Wine = $2x$ and Water = $x $
So, total volume = $3x$ Litre

Step $1$ : $36$ litre solution is replaced by $15$ litre water.

Wine $= 2x - 24 = 2(x-12)$
Water $= x - 12 + 15 = x + 3 $

Total Volume $= 3x - 21 = 3(x - 7)$

Step $2$ : Now $26$ litre Solution is replaced by $26$ litre water

Wine $= 2(x - 12) - \frac{2(x - 12)}{3(x - 7)} \cdot 26 = \frac{2(x-12)(3x-47)}{3(x-7)}$

Water $ = x + 3 - \frac{(x + 3)}{3(x - 7)} \cdot 26 + 26 = \frac{(3x² + 40x - 687)}{3(x - 7)}$

Now we know the ratio should be $8:7$ So, we end up with $$\Rightarrow \frac{2(3x^2 -83x + 564)}{3x^2 + 40x - 687} =\frac{8}{7}$$

$$\Rightarrow \frac{3x^2 -83x + 564}{3x^2 + 40x - 687} = \frac{4}{7}$$

$$\Rightarrow 21x^2 - 581x + 3948 = 12x^2 + 160x - 2748$$

$$\Rightarrow 9x^2 - 741x + 6696 = 0$$

$$\Rightarrow 3x^2 - 247x + 2232 =0$$

$$\Rightarrow (3x - 31) (x - 72) = 0 $$

Therefore, $x = \frac{31}{3}$ or $x = 72$

$x = \frac{31}{3}$ is not possible as we are supposed to take out $36$ litres initially.

Hence, $x = 72$

Initial water volume = $72$

Final water volume $= \frac{(3x² + 40x - 687)}{3(x - 7)}$ , putting $x = 72$ gives $91$

Hence, the difference between water quantity in the initial and final solution is $19$ Litre

JEEAspirant
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  • Thank you very much. I tallied your answer with mine and I'm embarrassed to admit I've done the calculations(multiplication) wrong. @Mahendra Verma. –  Feb 23 '24 at 11:55