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Mixture allegation. Please tell how to solve if any resource avilable please send.

Let ,in the mixture $x$ litre water and $y$ litre milk was present at first.

Now we added 1 litre of water in the mixture.

Now ,the in the mixture $(x+1)$ litre water and $y$ litre milk present .

Now percentage of milk is $\large =\frac{y}{x+1+y}$

Now it is given as $25$% .

$\large \frac{y}{x+1+y}=\frac{1}{4}$

or, $\large x-3y=-1$ ……………………….(1)

Now we added again 1 litre of milk in this mixture.

Now ,the in the mixture $(x+1)$ litre water and $(y+1)$ litre milk present .

Now percentage of milk is $\large =\frac{y+1}{x+1+y+1}$

Now it is given as $40$% .

$\large \frac{y+1}{x+1+y+1}=\frac{2}{5}$

or,         $\large 2x-3y=1$ ……………………….(2)

Now $(2)-(1)$

$\large 2x-3y-x+3y=1-(-1)$

so , $x=2$

Putting value of $x$ in $(2)$ we get,

$y=1$

So, initially in the mixture 2 litre water and 1 litre milk was present .

So percentage of milk initialy is=$\large \frac{1}{2+1}*100%$=$\large 33\frac{1}{3}%$%

### 1 comment

@Amit Mehta bro for this question we don’t need to do any calculation as the options are quite obvious because after adding 1 litre water the percentage of milk was 25% which means if we don’t add that 1 litre water the percentage of milk would be greater than 25% which was the answer . In option C only the percentage is greater than 25% so directly you can answer for this.

In youtube you can get many freely available resources on this .