Mixture and Alligation Questions in English

(C) Total volume of the mixture $= 75 \, litres$.
The ratio of milk to water is $2:1$.
Amount of milk $= \frac{2}{2+1} \times 75 = \frac{2}{3} \times 75 = 50 \, litres$.
Amount of water $= \frac{1}{2+1} \times 75 = \frac{1}{3} \times 75 = 25 \, litres$.
Let the amount of water to be added be $x \, litres$.
After adding water,the new amount of water becomes $(25 + x) \, litres$.
The new ratio of milk to water is given as $1:2$.
Therefore,$\frac{50}{25+x} = \frac{1}{2}$.
Cross-multiplying,we get $50 \times 2 = 1 \times (25 + x)$.
$100 = 25 + x$.
$x = 100 - 25 = 75 \, litres$.

(A) Let the cost price $(CP)$ of the mixture be $x$ $P$ per $kg$.
Given that the selling price $(SP)$ is $40$ $P$ per $kg$ and the gain is $25 \%$.
$CP = \frac{100}{100 + \text{Gain } \%} \times SP = \frac{100}{125} \times 40 = \frac{4}{5} \times 40 = 32$ $P$ per $kg$.
Using the method of alligation:
Cost of first salt = $42$ $P/kg$
Cost of second salt = $24$ $P/kg$
Mean price = $32$ $P/kg$
Difference for first salt = $|32 - 24| = 8$
Difference for second salt = $|42 - 32| = 10$
Ratio of quantities = $8 : 10 = 4 : 5$.
Let the quantity of the first salt be $x$ $kg$.
Then,$\frac{x}{25} = \frac{4}{5}$.
$x = \frac{4}{5} \times 25 = 20$ $kg$.

(C) Initial amount of solution $= 300 \ gm$.
Amount of sugar $= 40 \%$ of $300 = \frac{40}{100} \times 300 = 120 \ gm$.
Amount of water $= 300 - 120 = 180 \ gm$.
Let the amount of sugar to be added be $x \ gm$.
After adding sugar,the new amount of sugar $= 120 + x$.
The total amount of solution becomes $= 300 + x$.
We want the new concentration of sugar to be $50 \%$,so:
$\frac{120 + x}{300 + x} = \frac{50}{100} = \frac{1}{2}$.
Cross-multiplying gives: $2(120 + x) = 300 + x$.
$240 + 2x = 300 + x$.
$2x - x = 300 - 240$.
$x = 60 \ gm$.

(D) Let the cost price of $1$ litre of milk be $1$ unit.
Let the quantity of milk be $100$ litres.
Since the milkman sells at cost price but gains $25 \%$,it implies that the selling price of the mixture is equal to the cost price of $125$ litres of milk.
Since the milkman is selling the mixture at the cost price of pure milk,the extra $25$ litres must be water added to the $100$ litres of milk.
Total mixture quantity $= 100 \text{ (milk)} + 25 \text{ (water)} = 125 \text{ litres}$.
Percentage of water $= \frac{\text{Quantity of water}}{\text{Total mixture}} \times 100$.
Percentage of water $= \frac{25}{125} \times 100 = \frac{1}{5} \times 100 = 20 \%$.

(C) Initial volume of mixture $= 200 \text{ litres}$.
Water content $= 15 \% \text{ of } 200 = \frac{15}{100} \times 200 = 30 \text{ litres}$.
Milk content $= 200 - 30 = 170 \text{ litres}$.
Let $x$ litres of milk be added.
New total volume $= 200 + x$.
New milk content $= 170 + x$.
According to the problem,the new mixture contains $87.5 \%$ milk.
So,$\frac{170 + x}{200 + x} = \frac{87.5}{100} = \frac{875}{1000} = \frac{7}{8}$.
$8(170 + x) = 7(200 + x)$.
$1360 + 8x = 1400 + 7x$.
$8x - 7x = 1400 - 1360$.
$x = 40 \text{ litres}$.

(B) Let the number of rabbits be $x$ and the number of pigeons be $y$.
Since each animal has one head,we have:
$x + y = 200$ $...(i)$
Rabbits have $4$ legs and pigeons have $2$ legs. Therefore,the total number of legs is:
$4x + 2y = 580$ $...(ii)$
To solve the system of equations,multiply equation $(i)$ by $2$:
$2x + 2y = 400$ $...(iii)$
Subtract equation $(iii)$ from equation $(ii)$:
$(4x + 2y) - (2x + 2y) = 580 - 400$
$2x = 180$
$x = 90$
Substitute $x = 90$ into equation $(i)$:
$90 + y = 200$
$y = 110$
Thus,the number of pigeons is $110$.

(B) Total quantity of mixture = $66$ $litres$.
Ratio of milk to water = $5:1$.
Sum of ratio parts = $5 + 1 = 6$.
Quantity of milk = $(5/6) \times 66 = 55$ $litres$.
Quantity of water = $(1/6) \times 66 = 11$ $litres$.
Let the quantity of water added be $x$ $litres$.
New ratio of milk to water = $5:3$.
Therefore,$\frac{55}{11 + x} = \frac{5}{3}$.
Cross-multiplying,we get: $55 \times 3 = 5(11 + x)$.
$165 = 55 + 5x$.
$5x = 165 - 55 = 110$.
$x = 110 / 5 = 22$ $litres$.

(B) Total amount $= 41 \text{ rupees} = 4100 \text{ paise}$.
Total number of children $= 50$.
Average amount per child $= \frac{4100}{50} = 82 \text{ paise}$.
Using the method of alligation:
Boys ($90$ paise) : Girls ($65$ paise)
Mean value = $82$ paise
Difference for boys $= |82 - 65| = 17$
Difference for girls $= |90 - 82| = 8$
Ratio of boys to girls $= 17 : 8$.
Total parts $= 17 + 8 = 25$.
Number of boys $= \frac{17}{25} \times 50 = 34$.

(A) In vessel $A$,the fraction of milk is $\frac{5}{7}$.
In vessel $B$,the fraction of milk is $\frac{8}{13}$.
The required fraction of milk in the final mixture is $\frac{9}{13}$.
Using the rule of alligation:
Difference for vessel $A = |\frac{8}{13} - \frac{9}{13}| = \frac{1}{13}$.
Difference for vessel $B = |\frac{5}{7} - \frac{9}{13}| = |\frac{65 - 63}{91}| = \frac{2}{91}$.
Required ratio of quantities from vessel $A$ and vessel $B = \frac{1}{13} : \frac{2}{91}$.
To simplify,multiply both sides by $91$: $(\frac{1}{13} \times 91) : (\frac{2}{91} \times 91) = 7 : 2$.

(B) Let the number of coins of $1$ rupee be $3x$,$50$ paise be $8x$,and $25$ paise be $20x$.
The value of $1$ rupee coins is $3x \times 1 = 3x$ rupees.
The value of $50$ paise coins is $8x \times 0.50 = 4x$ rupees.
The value of $25$ paise coins is $20x \times 0.25 = 5x$ rupees.
Given the total value is $Rs. 372$,we have:
$3x + 4x + 5x = 372$
$12x = 372$
$x = 31$
The total number of coins is $3x + 8x + 20x = 31x$.
Substituting $x = 31$,we get:
Total coins $= 31 \times 31 = 961$.

(B) Let the original quantity of wine be $x$ litres.
After $1$ operation,the quantity of wine left is $x(1 - 8/x)$.
After $4$ operations (the initial one plus three more),the quantity of wine left is $x(1 - 8/x)^4$.
The total volume of the mixture remains $x$ litres.
The ratio of wine to water is $16:65$,so the ratio of wine to total mixture is $16:(16+65) = 16:81$.
Thus,$\frac{x(1 - 8/x)^4}{x} = \frac{16}{81}$.
$(1 - 8/x)^4 = (2/3)^4$.
Taking the fourth root on both sides,$1 - 8/x = 2/3$.
$8/x = 1 - 2/3 = 1/3$.
$x = 8 \times 3 = 24$ litres.

(C) Let the initial quantity of liquid $A$ be $7x$ and liquid $B$ be $5x$ litres.
Total initial quantity $= 7x + 5x = 12x$ litres.
When $9$ litres of mixture are drawn off,the amount of $A$ removed is $\frac{7}{12} \times 9 = \frac{21}{4}$ litres,and the amount of $B$ removed is $\frac{5}{12} \times 9 = \frac{15}{4}$ litres.
Remaining quantity of $A = 7x - \frac{21}{4}$ litres.
Remaining quantity of $B = 5x - \frac{15}{4}$ litres.
After adding $9$ litres of liquid $B$,the new quantity of $B = 5x - \frac{15}{4} + 9 = 5x + \frac{21}{4}$ litres.
The new ratio is $7:9$,so:
$\frac{7x - 21/4}{5x + 21/4} = \frac{7}{9}$
$9(7x - 5.25) = 7(5x + 5.25)$
$63x - 47.25 = 35x + 36.75$
$28x = 84$
$x = 3$
Initial quantity of $A = 7x = 7 \times 3 = 21$ litres.

(D) The formula for the amount of substance remaining after $n$ operations is $A = x(1 - y/x)^n$,where $x$ is the initial quantity,$y$ is the quantity replaced,and $n$ is the total number of operations.
Here,$x = 80$,$y = 8$,and the process was performed once initially and repeated $3$ more times,so $n = 1 + 3 = 4$.
Remaining milk $= 80(1 - 8/80)^4$
$= 80(1 - 0.1)^4$
$= 80(0.9)^4$
$= 80 \times 0.6561$
$= 52.488 \text{ litre}$.
Rounding to the nearest provided option,the answer is $52.48$.

(C) Using the method of alligation,we consider the fraction of milk in each mixture.
In vessel $A$,the fraction of milk is $\frac{4}{4+5} = \frac{4}{9}$.
In vessel $B$,the fraction of milk is $\frac{5}{5+1} = \frac{5}{6}$.
The required fraction of milk in the final mixture is $\frac{5}{5+4} = \frac{5}{9}$.
Applying the alligation rule:
Difference between milk in $B$ and final mixture: $\frac{5}{6} - \frac{5}{9} = \frac{15-10}{18} = \frac{5}{18}$.
Difference between milk in $A$ and final mixture: $\frac{5}{9} - \frac{4}{9} = \frac{1}{9}$.
The ratio of quantities taken from $A$ and $B$ is $\frac{5}{18} : \frac{1}{9}$.
Multiplying both sides by $18$,we get $5 : 2$.

(B) Let the initial quantity of liquid $A$ be $4x$ and liquid $B$ be $7x$. The total quantity is $11x$.
When $6 \text{ litres}$ of the mixture are removed,the amount of $A$ removed is $\frac{4}{11} \times 6 = \frac{24}{11}$ and the amount of $B$ removed is $\frac{7}{11} \times 6 = \frac{42}{11}$.
After adding $6 \text{ litres}$ of $B$,the new quantity of $A$ is $4x - \frac{24}{11}$ and the new quantity of $B$ is $7x - \frac{42}{11} + 6$.
The new ratio is given as $3:7$,so:
$\frac{4x - \frac{24}{11}}{7x - \frac{42}{11} + 6} = \frac{3}{7}$
$\frac{44x - 24}{77x - 42 + 66} = \frac{3}{7}$
$7(44x - 24) = 3(77x + 24)$
$308x - 168 = 231x + 72$
$77x = 240 \Rightarrow x = \frac{240}{77}$
The initial amount of liquid $A$ was $4x = 4 \times \frac{240}{77} = \frac{960}{77} \approx 12.467 \text{ litres}$.

(C) Initial volume of mixture $= 100 \text{ litres}$.
Water $= 20 \%$ of $100 = 20 \text{ litres}$.
Milk $= 100 - 20 = 80 \text{ litres}$.
Let $x$ litres of milk be added.
New volume of milk $= 80 + x$.
New total volume of mixture $= 100 + x$.
According to the problem,the new mixture contains $87.5 \%$ milk.
Therefore,the water content is $100 \% - 87.5 \% = 12.5 \%$.
Since the amount of water remains constant at $20 \text{ litres}$,we have:
$12.5 \% \text{ of } (100 + x) = 20$
$0.125 \times (100 + x) = 20$
$100 + x = \frac{20}{0.125} = 160$
$x = 160 - 100 = 60 \text{ litres}$.

(A) Let the three kinds of wheat be $W_1, W_2$ and $W_3$ with costs $1.20, 1.44$ and $1.74$ respectively.
By the rule of Alligation:
For $W_1$ and $W_2$ with mean price $1.41$:
$W_1 : W_2 = (1.74 - 1.41) : (1.41 - 1.20) = 0.33 : 0.21 = 11 : 7$.
For $W_2$ and $W_3$ with mean price $1.41$:
$W_2 : W_3 = (1.74 - 1.41) : (1.41 - 1.44) = 0.33 : (-0.03) = 11 : 1$ (taking absolute differences).
To combine these,we equate the ratio of $W_2$ in both parts:
$W_1 : W_2 = 11 : 7$
$W_2 : W_3 = 77 : 7$ (multiplying the second ratio by $7$ to match $W_2 = 77$)
$W_1 : W_2 = 121 : 77$
Thus,$W_1 : W_2 : W_3 = 121 : 77 : 7$.
Wait,re-evaluating the alligation steps from the provided image:
$W_1(1.20)$ and $W_2(1.44)$ with mean $1.41$ gives $W_1:W_2 = (1.44-1.41) : (1.41-1.20) = 0.03 : 0.21 = 1 : 7$.
$W_2(1.44)$ and $W_3(1.74)$ with mean $1.41$ gives $W_2:W_3 = (1.74-1.41) : (1.41-1.44) = 0.33 : 0.03 = 11 : 1$.
Equating $W_2$: $W_1:W_2 = 1:7 = 11:77$ and $W_2:W_3 = 11:1 = 77:7$.
Therefore,$W_1:W_2:W_3 = 11:77:7$.

(D) First,find the cost price $(CP)$ of the initial mixture of $T_{1}, T_{2}$,and $T_{3}$ in the ratio $1:2:4$:
$CP_{mix} = \frac{(1 \times 74) + (2 \times 68) + (4 \times 63)}{1+2+4} = \frac{74 + 136 + 252}{7} = \frac{462}{7} = Rs. 66$ per kg.
Now,the seller adds $z$ kg of $T_{1}$ (priced at $Rs. 74$ per kg) to $4$ kg of the initial mixture (priced at $Rs. 66$ per kg).
The selling price of the new mixture is $Rs. 84$ per kg with a profit of $20\%$.
Therefore,the cost price of the new mixture is $CP_{new} = 84 \times \frac{100}{120} = Rs. 70$ per kg.
Using the alligation method for the new mixture:
Quantity of $T_{1}$ ($z$ kg) at $Rs. 74$ per kg and quantity of initial mixture ($4$ kg) at $Rs. 66$ per kg result in a mixture at $Rs. 70$ per kg.
$\frac{74 - 70}{70 - 66} = \frac{4}{z}$
$\frac{4}{4} = \frac{4}{z}$
$1 = \frac{4}{z} \Rightarrow z = 4$ kg.

(D) Let the initial capacity of the container be $x$ litres of water.
After the first operation,the amount of water left is $(x - 9)$ litres.
After the second operation,the amount of water left is $x(1 - 9/x)^2$ litres.
The ratio of water to the total volume is $16 : (16 + 9) = 16 : 25$.
Therefore,the fraction of water remaining is $\frac{x(1 - 9/x)^2}{x} = \frac{16}{25}$.
$(1 - 9/x)^2 = (4/5)^2$.
$1 - 9/x = 4/5$.
$9/x = 1 - 4/5 = 1/5$.
$x = 45$ litres.
The container holds $45$ litres of liquid.

(A) Let the amount of mixture released each time be $y \text{ litres}$.
The initial volume of oxygen is $16 \%$ of $8 \text{ litres}$,which is $0.16 \times 8$.
After the first replacement,the remaining oxygen is $8 \times 0.16 \times (1 - y/8)$.
After the second replacement,the remaining oxygen is $8 \times 0.16 \times (1 - y/8)^2$.
According to the problem,the final oxygen content is $9 \%$ of the total volume $(8 \text{ litres})$,so:
$8 \times 0.16 \times (1 - y/8)^2 = 0.09 \times 8$
$(1 - y/8)^2 = 0.09 / 0.16$
$(1 - y/8)^2 = (0.3 / 0.4)^2 = (3/4)^2$
$1 - y/8 = 3/4$
$y/8 = 1 - 3/4 = 1/4$
$y = 8 / 4 = 2 \text{ litres}$.

(A) Let the cost price of the second liquid be $x$ per litre.
Then,the cost price of the first liquid is $(x + 2)$ per litre.
The selling price of the mixture is $Rs. 11$ per litre at a profit of $10 \%$.
Therefore,the cost price $(CP)$ of the mixture is:
$CP = 11 \times \frac{100}{110} = Rs. 10$ per litre.
Using the method of alligation:
Ratio of first liquid to second liquid = $(10 - x) : (x + 2 - 10) = 3 : 2$
$\frac{10 - x}{x - 8} = \frac{3}{2}$
$2(10 - x) = 3(x - 8)$
$20 - 2x = 3x - 24$
$5x = 44$
$x = 8.8$
Thus,the cost per litre of the second liquid is $Rs. 8.80$.

(C) Total cost of $5 \ kg$ of potatoes $= 5 \times 5 = Rs. 25$.
Total cost of $3 \ kg$ of potatoes $= 3 \times 4 = Rs. 12$.
Total weight of the mixture $= 5 + 3 = 8 \ kg$.
Total cost of the mixture $= 25 + 12 = Rs. 37$.
Average cost per $kg = \frac{\text{Total cost}}{\text{Total weight}} = \frac{37}{8} = Rs. 4.625$.
Rounding to two decimal places,the average cost is approximately $Rs. 4.63$ per $kg$.

(B) The initial $20\, \text{litre}$ mixture contains milk and water in the ratio $3:2$. Thus,the amount of milk is $12\, \text{litres}$ and water is $8\, \text{litres}$.
Step $1$: When $10\, \text{litres}$ of the mixture is removed,the amount of milk removed is $(3/5) \times 10 = 6\, \text{litres}$ and water removed is $(2/5) \times 10 = 4\, \text{litres}$. Remaining: $6\, \text{litres}$ milk and $4\, \text{litres}$ water. After adding $10\, \text{litres}$ of pure milk,the container has $6+10 = 16\, \text{litres}$ of milk and $4\, \text{litres}$ of water.
Step $2$: When $10\, \text{litres}$ of this new mixture is removed,the ratio of milk to water is $16:4$,which is $4:1$. The amount of milk removed is $(4/5) \times 10 = 8\, \text{litres}$ and water removed is $(1/5) \times 10 = 2\, \text{litres}$. Remaining: $16-8 = 8\, \text{litres}$ milk and $4-2 = 2\, \text{litres}$ water. After adding $10\, \text{litres}$ of pure milk,the container has $8+10 = 18\, \text{litres}$ of milk and $2\, \text{litres}$ of water.
The final ratio of milk to water is $18:2$,which simplifies to $9:1$.

(C) The selling price of the mixture is $Rs. 30/kg$ and the profit is $20\%$. The cost price of the mixture is $\frac{30}{1.2} = Rs. 25/kg$.
Let the quantities of the three varieties be $Q_A, Q_B,$ and $Q_C$ with costs $20, 24,$ and $30$ respectively.
Using the alligation rule for the mixture:
For $Q_A$ and $Q_C$: $\frac{Q_A}{Q_C} = \frac{30 - 25}{25 - 20} = \frac{5}{5} = \frac{1}{1}$.
For $Q_B$ and $Q_C$: $\frac{Q_B}{Q_C} = \frac{30 - 25}{25 - 24} = \frac{5}{1}$.
To combine these,we equate the $Q_C$ parts. In both ratios,$Q_C$ is $1$ unit. Thus,the ratio $Q_A : Q_B : Q_C = 1 : 5 : 1$.
Wait,the total $Q_C$ is the sum of parts from both alligations: $Q_C = 1 + 1 = 2$ units.
So,the ratio is $Q_A : Q_B : Q_C = 1 : 5 : 2$.
Given $Q_C = 2$ $kg$,the quantity of the second variety $Q_B = 5$ $kg$.

(C) The selling price of the mixture is $Rs. 96/kg$ with a profit of $20\%$.
$\text{Cost Price (CP)} = \frac{\text{Selling Price}}{1 + \text{Profit}\%} = \frac{96}{1.20} = Rs. 80/kg$.
Let the three varieties be $A$ $(Rs. 60/kg)$,$B$ $(Rs. 75/kg)$,and $C$ $(Rs. 100/kg)$.
We use the alligation method to find the ratio of quantities $Q_A, Q_B, Q_C$ to achieve a mean price of $Rs. 80/kg$.
Step $1$: Compare $A$ and $C$ with the mean price $80$.
$\frac{Q_A}{Q_C} = \frac{100 - 80}{80 - 60} = \frac{20}{20} = \frac{1}{1}$.
Step $2$: Compare $B$ and $C$ with the mean price $80$.
$\frac{Q_B}{Q_C} = \frac{100 - 80}{80 - 75} = \frac{20}{5} = \frac{4}{1}$.
Step $3$: Equate the parts of $Q_C$ in both ratios. Since $Q_C$ is $1$ part in both,we combine them:
$Q_A : Q_B : Q_C = 1 : 4 : (1 + 1) = 1 : 4 : 2$.

(A) Total cost of $4 \, kg$ of potato $= 4 \times 5 = Rs. \, 20.$
Total cost of $8 \, kg$ of potato $= 8 \times 6 = Rs. \, 48.$
Total cost of the mixture $= 20 + 48 = Rs. \, 68.$
Total weight of the mixture $= 4 + 8 = 12 \, kg.$
Average price of the mixture $= \frac{\text{Total Cost}}{\text{Total Weight}} = \frac{68}{12} = Rs. \, 5.666... \approx Rs. \, 5.67.$
Since $5.67$ is not explicitly listed,the closest option is $5.66$.

(B) Step $1$: Calculate the total cost price of the onions.
Total cost price $= (20 \times 6.50) + (30 \times 7) = 130 + 210 = Rs. 340$.
Step $2$: Calculate the total selling price.
Total profit $= Rs. 60$.
Total selling price $= \text{Total cost price} + \text{Total profit} = 340 + 60 = Rs. 400$.
Step $3$: Calculate the selling price per $kg$.
Total quantity of onions $= 20 + 30 = 50 \ kg$.
Selling price per $kg = \frac{400}{50} = Rs. 8$ per $kg$.

(B) In the first cask ($48$ $litres$): Wine $= \frac{13}{20} \times 48 = 31.2$ $litres$,Water $= \frac{7}{20} \times 48 = 16.8$ $litres$.
In the second cask ($42$ $litres$): Wine $= \frac{18}{35} \times 42 = 21.6$ $litres$,Water $= \frac{17}{35} \times 42 = 20.4$ $litres$.
Total wine $= 31.2 + 21.6 = 52.8$ $litres$.
Total water $= 16.8 + 20.4 = 37.2$ $litres$.
After adding $20$ $litres$ of water,total water $= 37.2 + 20 = 57.2$ $litres$.
The ratio of wine to water $= 52.8 : 57.2 = 528 : 572$.
Dividing both by $44$,we get $12 : 13$.

(A) Total volume of the mixture $= 2 + 5 + 9 = 16 \, L$.
Quantity of milk in the first glass $= 90 \% \text{ of } 2 \, L = 0.9 \times 2 = 1.8 \, L$.
Quantity of milk in the second glass $= 80 \% \text{ of } 5 \, L = 0.8 \times 5 = 4.0 \, L$.
Quantity of milk in the third glass $= 70 \% \text{ of } 9 \, L = 0.7 \times 9 = 6.3 \, L$.
Total quantity of milk $= 1.8 + 4.0 + 6.3 = 12.1 \, L$.
Total quantity of water $= \text{Total volume} - \text{Total milk} = 16.0 - 12.1 = 3.9 \, L$.
Therefore,the ratio of milk to water $= 12.1 : 3.9 = 121 : 39$.

(C) Let $y$ litres of the mixture be released each time.
Initially,the volume of oxygen is $40 \%$ of $12 \text{ litres} = 0.40 \times 12 = 4.8 \text{ litres}$.
The formula for the remaining amount of a substance after $n$ operations of replacing a volume $y$ from a total volume $V$ is: $A_{final} = A_{initial} \times (1 - \frac{y}{V})^n$.
Here,$A_{initial} = 0.40$,$A_{final} = 0.10$,$V = 12$,and $n = 2$.
Substituting the values: $0.10 = 0.40 \times (1 - \frac{y}{12})^2$.
Dividing both sides by $0.40$: $\frac{0.10}{0.40} = (1 - \frac{y}{12})^2$.
$\frac{1}{4} = (1 - \frac{y}{12})^2$.
Taking the square root of both sides: $\frac{1}{2} = 1 - \frac{y}{12}$.
$\frac{y}{12} = 1 - \frac{1}{2} = \frac{1}{2}$.
$y = 12 \times \frac{1}{2} = 6 \text{ litres}$.

(D) Let the capacity of each container be $L$ units. To make the calculations easier,we choose a common multiple of the sum of the parts in both ratios.
Sum of parts in the first container $= 3 + 1 = 4$.
Sum of parts in the second container $= 5 + 2 = 7$.
The least common multiple of $4$ and $7$ is $28$.
For the first container:
Milk $= (3/4) \times 28 = 21$ units,Water $= (1/4) \times 28 = 7$ units.
For the second container:
Milk $= (5/7) \times 28 = 20$ units,Water $= (2/7) \times 28 = 8$ units.
When mixed,total Milk $= 21 + 20 = 41$ units.
Total Water $= 7 + 8 = 15$ units.
Therefore,the ratio of milk to water in the mixture is $41:15$.

(B) The total volume of the mixture is $60\, L$.
The ratio of acid to water is $2:1$.
Quantity of acid $= \frac{2}{2+1} \times 60 = \frac{2}{3} \times 60 = 40\, L$.
Quantity of water $= 60 - 40 = 20\, L$.
Let the amount of water to be added be $x\, L$.
The new ratio of acid to water becomes $1:2$.
So,$\frac{40}{20+x} = \frac{1}{2}$.
Cross-multiplying,we get $40 \times 2 = 20 + x$.
$80 = 20 + x$.
$x = 80 - 20 = 60\, L$.
Therefore,$60\, L$ of water must be added.

(A) Total volume of the mixture $= 25\, L$.
The ratio of acid to water is $4: 1$.
Sum of ratio parts $= 4 + 1 = 5$.
Quantity of acid $= (4/5) \times 25 = 20\, L$.
Quantity of water $= (1/5) \times 25 = 5\, L$.
After adding $3\, L$ of water,the new quantity of water $= 5 + 3 = 8\, L$.
The quantity of acid remains $20\, L$.
New ratio of acid to water $= 20 : 8$.
Dividing both by $4$,we get $5 : 2$.

(A) The total quantity of the mixture is $729 \ L.$ The ratio of milk to water is $7:2.$
Quantity of milk $= \frac{7}{7+2} \times 729 = \frac{7}{9} \times 729 = 567 \ L.$
Quantity of water $= 729 - 567 = 162 \ L.$
Let the amount of water to be added be $x \ L.$
According to the problem,the new ratio of milk to water is $7:3.$
$\frac{567}{162 + x} = \frac{7}{3}$
Cross-multiplying,we get:
$567 \times 3 = 7 \times (162 + x)$
$1701 = 1134 + 7x$
$7x = 1701 - 1134$
$7x = 567$
$x = \frac{567}{7} = 81 \ L.$
Therefore,$81 \ L$ of water must be added.

(C) Total volume of the mixture = $75 \ L$.
The ratio of milk to water is $2:1$.
Amount of milk = $\frac{2}{2+1} \times 75 = \frac{2}{3} \times 75 = 50 \ L$.
Amount of water = $\frac{1}{2+1} \times 75 = \frac{1}{3} \times 75 = 25 \ L$.
Let $x \ L$ of water be added to the mixture.
The new amount of water becomes $(25+x) \ L$.
The new ratio of milk to water is given as $1:2$.
Therefore,$\frac{50}{25+x} = \frac{1}{2}$.
Cross-multiplying,we get $50 \times 2 = 25+x$.
$100 = 25+x$.
$x = 100 - 25 = 75 \ L$.

(B) Let the total volume of the mixture in the barrel be $4\,L$.
Since the ratio of wine to water is $3:1$,the quantity of wine is $3\,L$ and the quantity of water is $1\,L$.
Let $x$ be the fraction of the mixture drawn off.
When $x$ fraction of the mixture is removed,the amount of wine removed is $3x$ and the amount of water removed is $x$.
The remaining amount of wine is $3 - 3x$ and the remaining amount of water is $1 - x$.
After adding $x$ amount of water (since the total volume removed is $x$ fraction of $4\,L$,i.e.,$4x$),the new amount of water becomes $(1 - x) + 4x = 1 + 3x$.
According to the problem,the new ratio of wine to water is $1:1$,so:
$\frac{3 - 3x}{1 + 3x} = \frac{1}{1}$
$3 - 3x = 1 + 3x$
$6x = 2$
$x = \frac{2}{6} = \frac{1}{3}$
Thus,the required fraction of the mixture to be drawn off is $\frac{1}{3}$.

(D) Let the common ratio multiplier be $x$.
Then,the quantity of spirit $= 3x$ and the quantity of water $= 2x$.
According to the problem,the mixture contains $3 \ L$ more spirit than water:
$3x - 2x = 3$
$x = 3$
Therefore,the quantity of spirit $= 3x = 3 \times 3 = 9 \ L$.

(B) In the first vessel,the fraction of milk is $\frac{3}{5}$.
In the second vessel,the fraction of milk is $\frac{7}{10}$.
In the final mixture,the fraction of milk is $\frac{2}{3}$.
Using the rule of alligation:
Difference $1$ (between second vessel and mixture) $= \frac{7}{10} - \frac{2}{3} = \frac{21-20}{30} = \frac{1}{30}$.
Difference $2$ (between mixture and first vessel) $= \frac{2}{3} - \frac{3}{5} = \frac{10-9}{15} = \frac{1}{15}$.
The required ratio is the ratio of these differences: $\frac{1}{30} : \frac{1}{15}$.
Multiplying both sides by $30$,we get $1 : 2$.

(C) Let the initial quantity of liquid $A$ be $7x$ and liquid $B$ be $5x$. Total quantity $= 12x$.
When $9\,L$ of mixture is removed,the amount of $A$ removed is $\frac{7}{12} \times 9 = \frac{21}{4}\,L$ and the amount of $B$ removed is $\frac{5}{12} \times 9 = \frac{15}{4}\,L$.
Remaining $A = 7x - \frac{21}{4}$.
Remaining $B = 5x - \frac{15}{4}$.
After adding $9\,L$ of liquid $B$,the new ratio is $\frac{7x - 21/4}{5x - 15/4 + 9} = \frac{7}{9}$.
$\frac{28x - 21}{20x - 15 + 36} = \frac{7}{9} \Rightarrow \frac{28x - 21}{20x + 21} = \frac{7}{9}$.
Dividing the numerator by $7$: $\frac{4x - 3}{20x + 21} = \frac{1}{9}$.
$36x - 27 = 20x + 21 \Rightarrow 16x = 48 \Rightarrow x = 3$.
Initial quantity of liquid $A = 7x = 7 \times 3 = 21\,L$.

(B) Let the fraction of acid in vessel $A$ be $\frac{4}{7}$ and in vessel $B$ be $\frac{5}{8}$.
The fraction of acid in the final mixture in vessel $C$ is $\frac{3}{5}$.
Using the rule of alligation:
Ratio of $A$ to $B = \left| \frac{5}{8} - \frac{3}{5} \right| : \left| \frac{3}{5} - \frac{4}{7} \right|$
$= \left| \frac{25 - 24}{40} \right| : \left| \frac{21 - 20}{35} \right|$
$= \frac{1}{40} : \frac{1}{35}$
$= 35 : 40 = 7 : 8$.
Thus,the mixtures must be mixed in the ratio $7:8$.

(A) In vessel $A$,the fraction of acid is $\frac{5}{7}$.
In vessel $B$,the fraction of acid is $\frac{8}{13}$.
The required mixture should have an acid fraction of $\frac{9}{13+4} = \frac{9}{13}$.
Using the rule of alligation:
Ratio of quantities = (Difference between acid in $B$ and mean) : (Difference between mean and acid in $A$)
Ratio = $|\frac{9}{13} - \frac{8}{13}| : |\frac{5}{7} - \frac{9}{13}|$
Ratio = $\frac{1}{13} : |\frac{65-63}{91}| = \frac{1}{13} : \frac{2}{91}$
To simplify,multiply both sides by $91$:
Ratio = $(\frac{1}{13} \times 91) : (\frac{2}{91} \times 91) = 7 : 2$.

(A) In vessel $A$,the fraction of acid is $\frac{4}{7}$.
In vessel $B$,the fraction of acid is $\frac{2}{5}$.
The target mixture in vessel $C$ should have half acid,so the fraction is $\frac{1}{2}$.
Using the rule of alligation:
Ratio of vessel $A$ to vessel $B = (\frac{1}{2} - \frac{2}{5}) : (\frac{4}{7} - \frac{1}{2})$
$= (\frac{5-4}{10}) : (\frac{8-7}{14})$
$= \frac{1}{10} : \frac{1}{14}$
$= 14 : 10$
$= 7 : 5$
Thus,the required ratio is $7:5$.

(A) Let the concentration of wine in the first mixture be $\frac{3}{5}$ and in the second mixture be $\frac{4}{9}$.
We want the final mixture to have equal quantities of wine and water,so the concentration of wine in the final mixture should be $\frac{1}{2}$.
Using the rule of alligation:
Ratio of first mixture to second mixture $= |\frac{4}{9} - \frac{1}{2}| : |\frac{3}{5} - \frac{1}{2}|$
$= |\frac{8-9}{18}| : |\frac{6-5}{10}| = \frac{1}{18} : \frac{1}{10} = 10 : 18 = 5 : 9$.
Given that $3 \, L$ of the first mixture is used,let $x$ be the quantity of the second mixture.
$\frac{5}{9} = \frac{3}{x}$
$5x = 27$
$x = \frac{27}{5} = 5 \frac{2}{5} \, L$.

(A) Let the fraction of zinc in the first alloy be $\frac{1}{1+2} = \frac{1}{3}.$
Let the fraction of zinc in the second alloy be $\frac{2}{2+3} = \frac{2}{5}.$
Let the fraction of zinc in the final alloy be $\frac{5}{5+8} = \frac{5}{13}.$
Using the rule of alligation:
Ratio of alloys = $\left| \frac{2}{5} - \frac{5}{13} \right| : \left| \frac{5}{13} - \frac{1}{3} \right|$
$= \left| \frac{26-25}{65} \right| : \left| \frac{15-13}{39} \right|$
$= \frac{1}{65} : \frac{2}{39}$
$= \frac{1}{5 \times 13} : \frac{2}{3 \times 13}$
$= \frac{1}{5} : \frac{2}{3} = 3 : 10.$

(C) In alloy $A,$ the ratio of gold to copper is $5:3.$ The total parts are $5+3 = 8.$
In alloy $B,$ the ratio of gold to copper is $5:11.$ The total parts are $5+11 = 16.$
To mix equal quantities,we make the total parts equal. The least common multiple of $8$ and $16$ is $16.$
For alloy $A,$ multiply the ratio by $2$: $(5 \times 2) : (3 \times 2) = 10:6.$
For alloy $B,$ the ratio remains $5:11.$
When equal quantities are melted to form alloy $C,$ the total gold is $10 + 5 = 15$ and the total copper is $6 + 11 = 17.$
Therefore,the ratio of gold to copper in alloy $C$ is $15:17.$

(A) Let us consider the fraction of gold in each alloy.
In the first alloy,the ratio of gold to silver is $7: 22$,so the fraction of gold is $\frac{7}{7+22} = \frac{7}{29}$.
In the second alloy,the ratio of gold to silver is $21: 37$,so the fraction of gold is $\frac{21}{21+37} = \frac{21}{58}$.
In the final mixture,the ratio of gold to silver is $25: 62$,so the fraction of gold is $\frac{25}{25+62} = \frac{25}{87}$.
Using the rule of alligation:
Ratio of first alloy to second alloy = $\left| \frac{21}{58} - \frac{25}{87} \right| : \left| \frac{25}{87} - \frac{7}{29} \right|$
$= \left| \frac{63 - 50}{174} \right| : \left| \frac{25 - 21}{87} \right|$
$= \frac{13}{174} : \frac{4}{87}$
$= \frac{13}{174} : \frac{8}{174}$
$= 13: 8$.

(A) Let the cost price of pure milk be $CP_m = 16 / L$ and the cost price of water be $CP_w = 0 / L$.
The selling price of the mixture is $SP = 15 / L$ with a profit of $25 \%$.
The cost price of the mixture $(CP_{mix})$ is calculated as:
$CP_{mix} = \frac{SP}{1 + \text{Profit}\%} = \frac{15}{1 + 0.25} = \frac{15}{1.25} = 12 / L$.
Using the rule of alligation:
Milk $(16)$ : Water $(0)$
Mean price = $12$
Difference (Milk side) = $|12 - 0| = 12$
Difference (Water side) = $|16 - 12| = 4$
Ratio of Milk to Water = $12 : 4 = 3 : 1$.

(A) Using the rule of alligation:
Concentration of first solution $= 15 \%$
Concentration of second solution $= 40 \%$
Mean concentration required $= 30 \%$
By alligation method:
Difference between second concentration and mean $= 40 - 30 = 10$
Difference between mean and first concentration $= 30 - 15 = 15$
Ratio of first solution to second solution $= 10 : 15 = 2 : 3$.

(B) Let the initial quantity of milk be $7x$ and water be $5x$.
According to the problem,when $15 \, L$ of water is added,the new ratio becomes $7:8$.
Therefore,the equation is: $\frac{7x}{5x + 15} = \frac{7}{8}$.
Cross-multiplying,we get: $7x \times 8 = 7(5x + 15)$.
Dividing both sides by $7$: $8x = 5x + 15$.
Subtracting $5x$ from both sides: $3x = 15$,which gives $x = 5$.
The initial quantity of water was $5x = 5 \times 5 = 25 \, L$.
The new quantity of water in the mixture is $25 \, L + 15 \, L = 40 \, L$.

(C) To find the container with the minimum quantity of milk relative to water,we calculate the fraction of milk in each mixture:
$1$. First container: $\frac{5}{5+3} = \frac{5}{8} = 0.625$
$2$. Second container: $\frac{2}{2+1} = \frac{2}{3} \approx 0.666$
$3$. Third container: $\frac{3}{3+2} = \frac{3}{5} = 0.600$
$4$. Fourth container: $\frac{7}{7+4} = \frac{7}{11} \approx 0.636$
Comparing the values: $0.600 < 0.625 < 0.636 < 0.666$.
Thus,the third container has the minimum quantity of milk relative to water.