Mixture and Alligation question:

Question

There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How Much Milk is there is in the Vessel?

I have tried:

Initally Vessel containing 40 litres of Milk :

4 litres out means -> 36 litres

4 litres of water is poured in - > 4 litres

so Now total quantity is 40 litres

Mixture containing water and Milk in the ratio: 36:4 i.e 9:1

Again 5 litres of Mixture is replaced with the six litres of water

for that:

9x - 9/10*5 : x -1/10*5

Now the Ratio becomes:

90x - 45 : 10x -5 i.e 9x -9:2x -1

six litres of water is added

9x -9 :2x -5

again six litres of Mixture is replaced then

9x -9 - 9/10*6 : 2x -5 -9/10*6

that is

90x -144 :10x -84

after adding six litres of water again we got

90x -144 :10x -78

so Milk containing is

90x-144+10x -78 =41

100x -222 =41

100x = 263

x= 2.63

again substituting the value x=2.63 in 90x -144 then i am getting 92.7 milk

total itself 41 litres of Milk Please anyone guide me answer

what i am doing mistake please anyone guide me for the answer

Answer 1

Initial congifuration is $40$ litres of milk and no water, say $m=40$ and $w=0$.

Remove $4$ litres of milk and add $4$ litres of water: $m=36, w=4$.

The mixture is $36:4$ or $9:1$. $5$ litres of mixture contains $5\times \frac{9}{10}=\frac 92$ litres of milk and $5\times \frac{1}{10}=\frac 12$ litres of water.

Remove $5$ litres of mixture: $m=36-\frac 92=\frac{63}2, w=4-\frac 12 = \frac 72$.

Add $6$ litres water: $m=\frac{63}2, w=\frac 72+6=\frac{19}2$.

The mixture is now $63:19$. $6$ litres of mixture contains $6\times \frac{63}{82}=\frac{189}{41}$ litres of milk and $6\times \frac{19}{82}=\frac{57}{41}$ litres of water.

Remove $6$ litres of mixture: $m=\frac{63}2-\frac{189}{41}=\frac{2205}{82}, w=\frac{19}2-\frac{57}{41}=\frac{665}{82}$.

Add $6$ litres of water: $m=2205/82, w=\frac{665}{82}+6 = \frac{1157}{82}$.

Final configuration: $\frac{2205}{82}$ litres of milk (approx $26.9$) and $\frac{1157}{82}$ litres of water (approx $14.1$).

Answer 2

Think in the terms of %. In the first iteration, $10$% is removed, in the second iteration, $12.5$% is removed and in the third iteration, $15$% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back.

$40-10$%$=$$36$

$36-12.5$%$=$$31.5$

$31.5-15$%$=26.775$.

Therefore, after the third iteration, milk will be 26.775 liters.