The Question is:

From a milk jar with some amount of milk, $6$ litres of milk is drawn out and is replaced by water. This operation is repeated again. If the ratio of the milk and water after these two operations is $9:16$ then what was the initial quantity of milk present in the jar?

(A)$\ 15\hspace{150pt}$ (B)$\ 16$

(C)$\ 24\hspace{150pt}$ (D)$\ 112$

My approach:

Let, the initial quantity of milk was $x$L.

Now after 1st replacement the jar contains $(x-6)$L milk and $6$L water.

Again if $6$L milk is drawn out then individually $\dfrac{6(x-6)}{x}$L of milk and $\dfrac{6\times6}{x}=\dfrac{36}{x}$L of water will be drawn out from that jar.

So after 2nd replacement that the jar contains $\left((x-6)-\dfrac{6(x-6)}{x}\right)$L of milk and $\left(6-\dfrac{36}{x}+6\right)$L i.e. $\left(12-\dfrac{36}{x}\right)$L of water.

Now, \begin{align} \dfrac{(x-6)-\dfrac{6(x-6)}{x}}{12-\dfrac{36}{x}}&=\dfrac{9}{16}\\ \implies\dfrac{(x-6)^2}{3x-9}&=\dfrac94\\ \implies4x^2-75x+225&=0\\ \implies x&=15\quad \text{or, }3.75\text{ (Not possible)} \end{align} Therefore the initial quantity of milk in that jar was $15$L.

The ans. is alright but I'm trying to find out an alternate easy way so that it can be solved just arithmetically (i.e. without using $x$ and all that algebraic stuff). I've assumed the initial quantity as $x$ the problem is easier but there is a lot of calculation in it. Is there any alternate way to solve it?



1 Answer 1


While I could not find an arithmetic solution, there is a solution which is algebraically much simpler and, in my opinion, more elegant. The trick here is that the amount of fluid in the container stays constant, and that regardless of the composition of the fluid, a constant portion of milk is removed. More specifically, if we keep your definition of $x$, we will remove $\frac{x-6}{x}$ of the milk each time we make a change as described. From this, we can set up a simple equation: $$\left(\frac{x-6}{x}\right)^2=\frac9{9+16}$$where both sides of the equation are equal to the portion of the final fluid which is milk. Moving some things around, we get $$4x^2-75x+225=0$$ and then we can solve to once more arrive at $$x = 15, 3.75$$ While this solution is certainly not as beautiful as it maybe could have been, it is a step up from the ugly algebra which your method required.

Additionally, this solution can be used, and is far nicer when we have more than two switches, say for example three or four. With the method described above, going through the algebra would be a very long and tedious process, and not enjoyable for anyone. Here, we would simply need to increase the power on $\frac{x-6}x$.

Hope this helped.

Note: An entirely arithmetic solution to this is unlikely, as the solution at 3.75, while impractical, is mathematically correct and any arithmetic is unlikely to correctly produce 2+ solutions.


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