From a vessel containing $1$ litre of pure acid, $100$ ml pure acid was drawn out in each of the beakers A and B. The acid in both the beakers was diluted by adding water in different proportions. After that, the contents of A and B were added back to the vessel. The concentration of acid in the vessel now is $80$%. Had the contents of beakers A and B be mixed with each other instead of adding into the vessel, what would be the concentration of acid in that mixture?
A. $28.57$%
B. $55.55$%
C. Insufficient data
My approach:-
Initially vessel=1000 ml
After transferring the contents,
Vessel=800 ml Beaker A=100 ml Beaker B=100 ml
Now beaker A and B are diluted, let their new concentrations be x% and y% respectively, and their combined concentration of 200 ml be a%
Now this 200 ml solution is being transferred back to the vessel of a% concentration
so if I apply alligation over here , I would get the following ratio
(100-80) / (80-a) = 1/4
giving a=0 %, and this the thing being asked in question too, but I cant find such option here , and also a=0 looks weird too as how can we get 0% acid concentration on mixing 2 acid solutions
Please guide me where I am going wrong
I think answer-B is a typo and should be $44.44\%$.
The acid put into A and B is returned so contents of the original container $(\space 1\text{ liter })\space $ plus the water-added are what make $\space 80\%.\quad$
$$\dfrac{1000\space\text{ml}} {80\%}=1,250\space\text{ml}$$ The combined volume of $A + B +\text{water}$ is $\space 200+250=450 \text{ ml}.$
$$\dfrac{200}{450}=44.44\%$$