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Milkman A adds some quantity of water to a milk-water solution, and thereby the ratio of milk to water in the resulting solution gets inverted. Milkman B realizes this and adds some amount of pure milk to the resulting solution such that the ratio of milk to water in the final solution is the same as that in the original solution. If the value of antecedent and consequent in the ratio of milk and water in the original solution were natural numbers, what could be a possible value of the ratio of milk in the original and final solutions?

A) 1 : 5

B) 1 : 4

C) 1 : 3

D) 1 : 2

My approach:-

Let in the original mixture milk be 1 Liter and water be x Liter, ratio = 1:x Now water is added in this (say k liters) and by adding this the ratio reverses that is it becomes x:1, Now while adding water the amount of milk remains constant so basically the ratio changes from $x:x^2$ to $x:1$, therefore $k=1-x^2$

Now after this pure milk is added (lets say y liters), and ratio becomes back to 1:x, now by previous logic this, time water quantity remains constant, therefore our ratio is being changed from $x^2:x$ to $1:x$ giving an addition of $1-x^2$ liter of milk

So now we have , milk in original solution = 1 Liter and milk in final solution = $1+1-x^2$=$2-x^2$

the ratio becomes $$(1/(2-x^2))$$

Now checking each option:-

(A) $1/(2-x^2$) = 1/5 x coming out to be imaginary

(D) $1/(2-x^2$) = 1/2 here x=0 which is also not possible

Where am I going wrong ?

Fin27
  • 958
  • To milk:water = $1:x$ you add $k$ liters of water to obtain $1:(x+k) = x:1$. So $1 = x(x+k)$. This is not equivalent to $k = 1-x^2$. Here, $k = \frac{1-x^2}{x}$. – Eric Towers Nov 13 '21 at 23:43
  • I think the final ratio is not meant to be (quantity of milk in initial solution):(quantity of milk in final solution). It's asking about the ratio of milk:water which is the same in the initial solution and final solution. Also, adding $(1-x^2)$ water to $1$ milk and $x$ water gives $1$ milk and $(1+x-x^2)$ water, which does not have the inverted ratio. – aschepler Nov 13 '21 at 23:43
  • I do not know where you are wrong. For every case you have a solution - just add a quantity to the nominator and the same multiplied by the denominator ratio to the denominator. All have proper solutions. – Moti Nov 14 '21 at 17:01

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