0
$\begingroup$

If mixture A has 20% ethyl alcohol and rest water, and mixture B has 20 % ethyl alcohol, rest water too. When both are mixed in, the resulting mixture formed is called Delta. Upon mixing these three mixtures, another solution is prepared called Gamma. And it is given that Gamma has 20% ethyl alcohol too.

How to prove that the ratio in which all three mixtures are mixed together is irrelevant as mixture Delta will be having 20% ethyl alcohol too.

$\endgroup$
2
  • 1
    $\begingroup$ In all cases, the amount of alcohol added to the total is 20% of the rest of the volume of the constituent. If $C_i$ is the amount of constituent $i$ in the mixture then the percentage of alcohol in the total is $\Sigma_{i=1}^{n}\frac{0.2C_1 +...+0.2C_n}{C_1+...C_n}=0.2$. $\endgroup$
    – John Douma
    Commented Nov 16, 2021 at 16:47
  • $\begingroup$ @JohnDouma Thanks mate. This was very helpful. $\endgroup$
    – Lakshay Khanna
    Commented Nov 16, 2021 at 16:55

1 Answer 1

Reset to default
0
$\begingroup$

The proof is straight forward.

Suppose you take $a$ litters of A, $b$ litters of B and mix them up. Your will have $\frac{a+b}{5}$ litter of ethyl alcohol, which is $20\%$ of total volume of Delta.

Now suppose, we take $a_1$ litters of A, $b_1$ litters of B and $c_1$ litters of Delta. The volume of ethyl alcohol will be $\frac{a_1+b_1+c_1}{5}$, which is $20\%$ of total volume of Gamma.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .