Strength of solution A after mixing contents in 4 vessels

Question

Question:

A, B, C, D are $4$ vessels. B contains $40$L of $15\%$ salt solution, C has $30$L of $ 8\%$ salt solution and D has $50$L of $22\%$ salt solution.
$10$L of solution from A is transferred to B, then, $10$L of the solution from B is transferred to C, then, $10$L of the solution from C is transferred to D.
Transferring some quantity of solution from vessel D to A does not change A's strength. The initial strength of salt solution in vessel A was (in $\%$)?

Solution:

To make things "workable" the solution assumed the quantity of solution in vessel A as $10$L which wasn't too pleasing to me. Ultimately it arrived at $20\%$ as the answer.

I tried without assuming anything and was lost. Can somebody please guide me on how to do it without any assumptions?
And also what would be the shortest way to do it?

Answer 1

Alternative approach:

The volume of Vessel-A is irrelevant, because the concentration in Vessel-A is unchanged, from the beginning up to the point when some of the liquid in Vessel-D is about to be transferred back into Vessel-A.

The requirement in the original problem can be restated as follows:

The resulting concentration in Vessel-D, after the first $(3)$ transfers (i.e. Vessel-A to Vessel-B, Vessel-B to Vessel-C, Vessel-C to Vessel-D) must then match the original concentration in Vessel-A.

So, I would attack the problem as follows:

---

Throughout this answer, it will be assumed that the Volume being transferred, or amount of Salt is measured in Liters.

Let $P$ be the real number such that $0 \leq P \leq 1$ and $P$ represents the concentration of Vessel-A.

Once $10$ is transferred from Vessel-A to Vessel-B, the total volume :: amount of Salt in Vessel-B is

$$50 :: (.15 \times 40) + 10P = 6 + 10P.$$

So, when $10$ is transferred from Vessel-B to Vessel-C, $20\%$ of the salt in Vessel-B is also being transferred, because just before the transfer, Vessel-B has 50 (total).

So, after the transfer the Volume :: amount of Salt in Vessel-C is

$$40::(.08 \times 30) + [(.2) \times (6 + 10P)] = 3.6 + 2P.$$

Similarly, after the transfer from Vessel-C to Vessel-D, because this transfer represents $(25\%)$ of Vessel-C's volume, just before the transfer, you have that

in Vessel-D, after the transfer, the total::amount of salt is

$$60:: (0.22 \times 50) + [(0.25) \times (3.6 + 2P)] = 11.9 + .5P.$$

Since it is required that the concentration must equal $P$, you can solve for $P$ via the equation

$$\frac{11.9 + .5P}{60} = P \implies 11.9 + .5P = 60P \implies $$

$$11.9 = 59.5P \implies P = \frac{11.9}{59.5} = 0.2.$$

Answer 2

Let the volume of A be $V_A$ and its strength be $s_A$%. Then the total amount of the salt is $s_AV_A$ and the amount in $10L$ is $10s_A$ units.

New concentration in vessel B is:$\dfrac{10s_A+15\times40}{40+10}=\dfrac{10s_A+600}{50}=\dfrac{s_A}{5}+12$
and the amount in $10L$ from B now is $2s_A+120$ units (multiplied by 10).

New concentration in C:$\dfrac{(2s_A+120)+30\times8}{30+10}=\dfrac{s_A}{20}+9$
and the amount of salt in $10L$ of C is now $\dfrac{s_A}{2}+90$.

Finally, concentration in D is:$\dfrac{\tfrac{s_A}{2}+1190}{10+50}=\dfrac{s_A}{120}+\dfrac{119}{6}.$ Equate this expression with $s_A$ and solve.