Mixture and Alligation Questions in English

(C) Let the time spent traveling by bus be $t_1$ hours and the time spent traveling by train be $t_2$ hours.
Total time $t_1 + t_2 = 6 \text{ hours}$.
Total distance $= 40t_1 + 55t_2 = 285 \text{ km}$.
Using the alligation method:
Speed of bus $= 40 \text{ km/h}$,Speed of train $= 55 \text{ km/h}$.
Average speed $= \frac{285}{6} \text{ km/h} = 47.5 \text{ km/h}$.
Difference for bus $= |55 - 47.5| = 7.5$.
Difference for train $= |40 - 47.5| = 7.5$.
Ratio of time (bus : train) $= 7.5 : 7.5 = 1 : 1$.
Since the total time is $6 \text{ hours}$,time spent in each mode is $3 \text{ hours}$.
Distance traveled by train $= \text{Speed} \times \text{Time} = 55 \text{ km/h} \times 3 \text{ h} = 165 \text{ km}$.

(A) Using the method of alligation:
Profit percentage of first part = $8 \%$
Profit percentage of second part = $18 \%$
Mean profit percentage = $14 \%$
Difference for first part = $|18 - 14| = 4$
Difference for second part = $|8 - 14| = 6$
Ratio of quantities = $4 : 6 = 2 : 3$
Total quantity = $50 \text{ kg}$
Quantity sold at $18 \%$ profit = $\frac{3}{2 + 3} \times 50 = \frac{3}{5} \times 50 = 30 \text{ kg}$.

(D) Using the method of alligation:
Profit part = $10 \%$
Loss part = $-5 \%$
Mean profit = $7 \%$
Applying alligation:
$(10 - 7) = 3$ (for the loss part)
$(7 - (-5)) = 12$ (for the profit part)
Ratio of quantities sold at $10 \%$ profit to $5 \%$ loss = $12 : 3 = 4 : 1$.
Total quantity = $50 \text{ kg}$.
Quantity sold at $10 \%$ profit = $\frac{4}{4+1} \times 50 = \frac{4}{5} \times 50 = 40 \text{ kg}$.
Quantity sold at $5 \%$ loss = $\frac{1}{4+1} \times 50 = \frac{1}{5} \times 50 = 10 \text{ kg}$.

(A) Let the expenditure be $300$ and savings be $200$. Then,total income $= 300 + 200 = 500$.
After the increase,new income $= 500 \times 1.10 = 550$.
New expenditure $= 300 \times 1.12 = 336$.
New savings $= 550 - 336 = 214$.
Increase in savings $= 214 - 200 = 14$.
Percentage increase in savings $= (14 / 200) \times 100 = 7 \%$.
Alternatively,using the alligation method:
Let the percentage increase in savings be $x$.
By alligation rule:
$(12 - 10) / (10 - x) = 2 / 3$
$2 / (10 - x) = 2 / 3$
$10 - x = 3$
$x = 7$.
Thus,the savings increase by $7 \%$.

(C) Using the method of alligation:
Concentration of acid in solution $A = 10 \%$
Concentration of acid in solution $B = 30 \%$
Mean concentration of the mixture $= 25 \%$
By applying the alligation rule:
(Quantity of $A$) / (Quantity of $B$) = (Difference between $B$ and Mean) / (Difference between Mean and $A$)
(Quantity of $A$) / (Quantity of $B$) = $(30 - 25) / (25 - 10)$
(Quantity of $A$) / (Quantity of $B$) = $5 / 15$
(Quantity of $A$) / (Quantity of $B$) = $1 / 3$
Therefore,the required ratio is $1:3$.

(D) The mean cost price $(C.P.)$ per $kg$ of the alloy is calculated using the weighted average formula:
Mean $C.P. = \frac{(3 \times 2000) + (1 \times 400)}{3 + 1}$
Mean $C.P. = \frac{6000 + 400}{4} = \frac{6400}{4} = ₹ 1600/kg$
To find the cost of $8$ kilograms of this alloy,we multiply the cost per $kg$ by the total weight:
Total Cost $= 8 \times 1600 = ₹ 12800$.

(C) Using the rule of alligation:
Cost of first type = ₹ $2500$ per $kg$
Cost of second type = ₹ $1500$ per $kg$
Mean price of mixture = ₹ $2250$ per $kg$
Applying the alligation method:
(Quantity of first type) / (Quantity of second type) = (Mean price - Cost of second type) / (Cost of first type - Mean price)
Ratio = $(2250 - 1500) / (2500 - 2250)$
Ratio = $750 / 250$
Ratio = $3 / 1$
Therefore,the required ratio is $3:1$.

(C) Initial volume of ethanol $= 80 \text{ litres}$.
After the $1^{st}$ replacement,the amount of ethanol remaining is $80 \times (1 - \frac{20}{80}) = 80 \times \frac{60}{80} = 60 \text{ litres}$.
After the $2^{nd}$ replacement,the amount of ethanol remaining is $60 \times (1 - \frac{20}{80}) = 60 \times \frac{60}{80} = 45 \text{ litres}$.
The total volume of the liquid in the drum remains $80 \text{ litres}$.
Therefore,the amount of water present in the drum is $80 - 45 = 35 \text{ litres}$.

(B) Let the capacity of each bottle be $63$ litres (the $LCM$ of $7, 7, 9$).
The ratios of milk and water in the three bottles are $2:5$,$3:4$,and $4:5$.
For the $1^{\text{st}}$ bottle: Milk $= \frac{2}{7} \times 63 = 18$ litres,Water $= 63 - 18 = 45$ litres.
For the $2^{\text{nd}}$ bottle: Milk $= \frac{3}{7} \times 63 = 27$ litres,Water $= 63 - 27 = 36$ litres.
For the $3^{\text{rd}}$ bottle: Milk $= \frac{4}{9} \times 63 = 28$ litres,Water $= 63 - 28 = 35$ litres.
Total Milk $= 18 + 27 + 28 = 73$ litres.
Total Water $= 45 + 36 + 35 = 116$ litres.
Therefore,the required ratio of milk to water is $73:116$.

(C) Let the cost of the superior variety be $x$ per $kg$ and the cost of the inferior variety be $y$ per $kg$.
Let the cost of the mixture be $M$ per $kg$.
According to the problem:
$M = x - 0.50$ $(1)$
$M = y + 0.75$ $(2)$
From $(1)$,$x - M = 0.50$.
From $(2)$,$M - y = 0.75$.
Using the rule of alligation,the ratio of the superior variety to the inferior variety is given by:
$\text{Ratio} = \frac{M - y}{x - M} = \frac{0.75}{0.50}$.
Simplifying the ratio: $\frac{0.75}{0.50} = \frac{75}{50} = \frac{3}{2}$.
Thus,the ratio is $3:2$.

(A) Given:
Selling Price $(S.P.)$ of the mixture $= ₹ 42$ per $kg$.
Profit percentage $= 20 \%$.
First,calculate the Cost Price $(C.P.)$ of the mixture:
$C.P. = \frac{S.P. \times 100}{100 + \text{Profit } \%}$
$C.P. = \frac{42 \times 100}{100 + 20} = \frac{4200}{120} = ₹ 35$ per $kg$.
Now,apply the rule of Alligation:
Cost of first type of sugar $= ₹ 30$ per $kg$.
Cost of second type of sugar $= ₹ 45$ per $kg$.
Mean price $= ₹ 35$ per $kg$.
Using the alligation method:
Ratio of first sugar to second sugar $= (45 - 35) : (35 - 30)$
$= 10 : 5$
$= 2 : 1$.
Thus,the required ratio is $2:1$.

(C) Initial volume of mixture $= 80 \text{ litres}$.
Milk content $= 10 \% \text{ of } 80 = 8 \text{ litres}$.
Water content $= 80 - 8 = 72 \text{ litres}$.
Let $x$ be the amount of milk added.
In the new mixture,the amount of water remains $72 \text{ litres}$,which represents $80 \%$ of the new total volume.
Let the new total volume be $V$.
$0.80 \times V = 72 \implies V = \frac{72}{0.80} = 90 \text{ litres}$.
The amount of milk added $= V - 80 = 90 - 80 = 10 \text{ litres}$.

(C) Given:
Selling Price $(S.P)$ of the mixture = ₹ $324$ per $kg$
Profit percentage = $20 \%$
Step $1$: Calculate the Cost Price $(C.P)$ of the mixture.
$C.P = \frac{S.P \times 100}{100 + \text{Profit } \%}$
$C.P = \frac{324 \times 100}{120} = \frac{3240}{12} = ₹ 270$ per $kg$
Step $2$: Apply the Alligation method.
Cost of first type of tea = ₹ $240$ per $kg$
Cost of second type of tea = ₹ $280$ per $kg$
Mean price of mixture = ₹ $270$ per $kg$
Using the rule of alligation:
Ratio = $(280 - 270) : (270 - 240)$
Ratio = $10 : 30$
Ratio = $1 : 3$
Therefore,the tea should be mixed in the ratio $1:3$.

(C) The capacities of the three boxes are $24\, kg$,$36\, kg$,and $84\, kg$.
Total weight of the mixture $= 24 + 36 + 84 = 144\, kg$.
The ratio of wheat $A$,$B$,and $C$ in the mixture is $24 : 36 : 84$,which simplifies to $2 : 3 : 7$.
The fraction of type $A$ wheat in the total mixture is $\frac{2}{2+3+7} = \frac{2}{12} = \frac{1}{6}$.
Since the mixture is distributed back into the boxes,the third box (capacity $84\, kg$) will contain the same proportion of type $A$ wheat as the total mixture.
Quantity of type $A$ wheat in the third box $= \frac{1}{6} \times 84 = 14\, kg$.

(B) The initial volume of the sugar solution is $3$ $litre$,and it contains $60 \%$ sugar.
Amount of sugar $= 60 \% \text{ of } 3 \text{ litre} = 0.60 \times 3 = 1.8 \text{ litre}$.
When $1$ $litre$ of water is added,the total volume of the new solution becomes $3 + 1 = 4 \text{ litre}$.
The amount of sugar remains constant at $1.8 \text{ litre}$.
Percentage of sugar in the new solution $= \left( \frac{1.8}{4} \right) \times 100 = 0.45 \times 100 = 45 \%.$

(B) Initial volume of solution $= 32 \text{ litres}$.
Concentration of alcohol $= 20 \%$.
Amount of alcohol $= 32 \times 0.20 = 6.4 \text{ litres}$.
Water added $= 8 \text{ litres}$.
New total volume of solution $= 32 + 8 = 40 \text{ litres}$.
New concentration of alcohol $= \left( \frac{6.4}{40} \right) \times 100 = 16 \%$.

(A) Given:
Selling Price $(S.P.)$ of the mixture = ₹ $39$ per $kg$
Profit percentage = $30 \%$
Step $1$: Calculate the Cost Price $(C.P.)$ of the mixture.
$C.P. = S.P. \times \frac{100}{100 + \text{Profit } \%}$
$C.P. = 39 \times \frac{100}{130} = 39 \times \frac{10}{13} = 3 \times 10 = ₹ 30$ per $kg$
Step $2$: Use the rule of Alligation:
- Cost of first type of wheat = ₹ $32$ per $kg$
- Cost of second type of wheat = ₹ $24$ per $kg$
- Mean Price ($C.P.$ of mixture) = ₹ $30$ per $kg$
By Alligation:
(Price of 1st) $32$ --- (Price of 2nd) $24$
\ /
$30$
/ \
(Difference) $6$ --- (Difference) $2$
Ratio = $6 : 2 = 3 : 1$
Therefore,the required ratio is $3:1$.

(B) Total quantity of tea $= 49 \ kg$.
Ratio of Assam tea to Darjeeling tea $= 5:2$.
Sum of ratio parts $= 5 + 2 = 7$.
Quantity of Assam tea $= (5/7) \times 49 = 35 \ kg$.
Quantity of Darjeeling tea $= (2/7) \times 49 = 14 \ kg$.
Let $x \ kg$ of Darjeeling tea be added.
New quantity of Darjeeling tea $= 14 + x \ kg$.
According to the problem,the new ratio of Assam to Darjeeling tea is $2:1$.
So,$35 / (14 + x) = 2 / 1$.
$35 = 2(14 + x)$.
$35 = 28 + 2x$.
$2x = 35 - 28 = 7$.
$x = 7 / 2 = 3.5 \ kg$.

(B) Using the rule of alligation:
Cost price of first sugar = ₹ $12$ per $kg$
Cost price of second sugar = ₹ $7$ per $kg$
Mean price of the mixture = ₹ $8$ per $kg$
Applying the alligation formula:
(Quantity of first sugar) : (Quantity of second sugar) = (Mean price - Cost price of second) : (Cost price of first - Mean price)
Ratio = $(8 - 7) : (12 - 8)$
Ratio = $1 : 4$
Thus,the grocer must mix the two types of sugar in the ratio $1:4$.

(B) Let the volumes of the three containers be $2x, 3x,$ and $4x$ respectively.
$1^{st}$ container: Ratio of spirit to water is $4:1$. Total parts = $5$.
Spirit = $(4/5) \times 2x = 1.6x$,Water = $(1/5) \times 2x = 0.4x$.
$2^{nd}$ container: Ratio of spirit to water is $11:4$. Total parts = $15$.
Spirit = $(11/15) \times 3x = 2.2x$,Water = $(4/15) \times 3x = 0.8x$.
$3^{rd}$ container: Ratio of spirit to water is $7:3$. Total parts = $10$.
Spirit = $(7/10) \times 4x = 2.8x$,Water = $(3/10) \times 4x = 1.2x$.
Total Spirit = $1.6x + 2.2x + 2.8x = 6.6x$.
Total Water = $0.4x + 0.8x + 1.2x = 2.4x$.
Resultant ratio = $6.6x : 2.4x = 66 : 24 = 11 : 4$.

(D) Let the two types of brass be $A$ and $B$.
In brass $A$,the ratio of Copper to Zinc is $8:3$. The fraction of Copper is $\frac{8}{11}$ and the fraction of Zinc is $\frac{3}{11}$.
In brass $B$,the ratio of Copper to Zinc is $15:7$. The fraction of Copper is $\frac{15}{22}$ and the fraction of Zinc is $\frac{7}{22}$.
We mix these in the ratio $5:2$. Let us take $5$ units of brass $A$ and $2$ units of brass $B$.
Total Copper $= 5 \times \frac{8}{11} + 2 \times \frac{15}{22} = \frac{40}{11} + \frac{15}{11} = \frac{55}{11} = 5$.
Total Zinc $= 5 \times \frac{3}{11} + 2 \times \frac{7}{22} = \frac{15}{11} + \frac{7}{11} = \frac{22}{11} = 2$.
The ratio of Copper to Zinc in the new mixture is $5:2$.

(A) Let the cost price $(CP)$ of $1$ litre of petrol be $₹ 100$.
The cost price of $1$ litre of kerosene is $40 \%$ of $₹ 100 = ₹ 40$.
Rama mixes $20 \%$ kerosene with petrol. For $1000 \text{ ml}$ of petrol,he adds $200 \text{ ml}$ of kerosene.
The total cost price of the mixture $(1200 \text{ ml})$ is: $100 + (0.2 \times 40) = 100 + 8 = ₹ 108$.
The selling price $(SP)$ of $1200 \text{ ml}$ of the mixture (sold at the price of petrol) is: $100 + (0.2 \times 100) = ₹ 120$.
Profit $= SP - CP = 120 - 108 = ₹ 12$.
Profit percentage $= (\text{Profit} / CP) \times 100 = (12 / 108) \times 100 = 100 / 9 = 11.11 \%$.

(A) Initial volume of milk = $80 \text{ litre}$,Initial volume of water = $18 \text{ litre}$.
Total volume = $80 + 18 = 98 \text{ litre}$.
Ratio of milk to water = $80:18 = 40:9$.
When $49 \text{ litre}$ of mixture is removed,the amount of milk removed = $49 \times (40/49) = 40 \text{ litre}$.
The amount of water removed = $49 \times (9/49) = 9 \text{ litre}$.
Remaining milk = $80 - 40 = 40 \text{ litre}$.
Remaining water = $18 - 9 = 9 \text{ litre}$.
Let the quantity of milk added be $2x \text{ litre}$ and water added be $x \text{ litre}$.
According to the problem,the new ratio is $(40 + 2x) / (9 + x) = 4/1$.
$40 + 2x = 4(9 + x)$.
$40 + 2x = 36 + 4x$.
$4x - 2x = 40 - 36$.
$2x = 4 \Rightarrow x = 2$.
Quantity of milk added = $2x = 2(2) = 4 \text{ litre}$.

(C) Initial mixture $= 120$ $litre$,water $= 25 \%$.
Water $= 0.25 \times 120 = 30$ $litre$.
Milk $= 120 - 30 = 90$ $litre$.
After selling $20$ $litre$ of the mixture,the remaining mixture $= 120 - 20 = 100$ $litre$.
In the remaining $100$ $litre$ mixture,the ratio of milk to water remains the same ($75 \%$ milk and $25 \%$ water).
Water in $100$ $litre$ mixture $= 0.25 \times 100 = 25$ $litre$.
Milk in $100$ $litre$ mixture $= 0.75 \times 100 = 75$ $litre$.
After adding $16.2$ $litre$ of milk and $3.8$ $litre$ of water:
New milk quantity $= 75 + 16.2 = 91.2$ $litre$.
New water quantity $= 25 + 3.8 = 28.8$ $litre$.
Total new mixture volume $= 91.2 + 28.8 = 120$ $litre$.
Percentage of water $= (28.8 / 120) \times 100 = 24 \%$.

(D) Total volume of the mixture $= 729 \, ml$.
Ratio of milk to water $= 7:2$.
Sum of ratio parts $= 7 + 2 = 9$.
Value of one part $= 729 / 9 = 81 \, ml$.
Quantity of milk $= 7 \times 81 = 567 \, ml$.
Quantity of water $= 2 \times 81 = 162 \, ml$.
We need to add water such that the new ratio becomes $7:3$. Since the quantity of milk remains constant $(567 \, ml)$,the new ratio $7:3$ implies that $7$ parts correspond to $567 \, ml$.
Value of one part in the new ratio $= 567 / 7 = 81 \, ml$.
Required quantity of water $= 3 \times 81 = 243 \, ml$.
Water to be added $= 243 \, ml - 162 \, ml = 81 \, ml$.

(A) Step $1$: Calculate the total cost price of the mixture.
Total cost price $= (30 \times 70) + (20 \times 70.75) = 2100 + 1415 = ₹ 3515$.
Step $2$: Calculate the total selling price of the mixture.
Total quantity $= 30 + 20 = 50 \text{ kg}$.
Total selling price $= 50 \times 80.50 = ₹ 4025$.
Step $3$: Calculate the total gain.
Gain $= \text{Total Selling Price} - \text{Total Cost Price} = 4025 - 3515 = ₹ 510$.

(C) Initial amount of solution = $300$ $grams$.
Amount of sugar in the initial solution = $40 \%$ of $300 = 0.40 \times 300 = 120$ $grams$.
Amount of water in the initial solution = $300 - 120 = 180$ $grams$.
Let $x$ $grams$ of sugar be added to the solution.
New amount of sugar = $120 + x$ $grams$.
New total amount of solution = $300 + x$ $grams$.
According to the problem,the new concentration of sugar is $50 \%$,so:
$(120 + x) / (300 + x) = 50 / 100 = 1 / 2$.
Cross-multiplying gives: $2(120 + x) = 300 + x$.
$240 + 2x = 300 + x$.
$2x - x = 300 - 240$.
$x = 60$ $grams$.

(D) Let the volume of each glass be $1$ unit.
In the first glass,the amount of acid is $\frac{2}{5}$ and water is $\frac{3}{5}$.
In the second glass,the amount of acid is $\frac{3}{7}$ and water is $\frac{4}{7}$.
In the third glass,the amount of acid is $\frac{4}{9}$ and water is $\frac{5}{9}$.
Total acid in the large vessel $= \frac{2}{5} + \frac{3}{7} + \frac{4}{9} = \frac{126 + 135 + 140}{315} = \frac{401}{315}$.
Total water in the large vessel $= \frac{3}{5} + \frac{4}{7} + \frac{5}{9} = \frac{189 + 180 + 175}{315} = \frac{544}{315}$.
Therefore,the required ratio of acid to water $= \frac{401}{315} : \frac{544}{315} = 401:544$.

(D) In $60\, kg$ of alloy $A$,the ratio of lead to tin is $3:2$.
Amount of tin in alloy $A = \frac{2}{3+2} \times 60 = \frac{2}{5} \times 60 = 24\, kg$.
In $100\, kg$ of alloy $B$,the ratio of tin to copper is $1:4$.
Amount of tin in alloy $B = \frac{1}{1+4} \times 100 = \frac{1}{5} \times 100 = 20\, kg$.
Total amount of tin in the new alloy = (Tin in alloy $A$) + (Tin in alloy $B$) = $24\, kg + 20\, kg = 44\, kg$.

(D) Let the weight of the first alloy be $3x$ and the second alloy be $4x$.
In the first alloy $(A)$,the ratio of tin to iron is $1:2$. Total parts = $1+2=3$.
Amount of tin in $A = \frac{1}{3} \times 3x = x$.
Amount of iron in $A = \frac{2}{3} \times 3x = 2x$.
In the second alloy $(B)$,the ratio of tin to iron is $2:3$. Total parts = $2+3=5$.
Amount of tin in $B = \frac{2}{5} \times 4x = \frac{8x}{5}$.
Amount of iron in $B = \frac{3}{5} \times 4x = \frac{12x}{5}$.
Total tin in the new mixture = $x + \frac{8x}{5} = \frac{13x}{5}$.
Total iron in the new mixture = $2x + \frac{12x}{5} = \frac{22x}{5}$.
The ratio of tin to iron in the new alloy = $\frac{13x}{5} : \frac{22x}{5} = 13:22$.

(B) In $20$ litres of the initial mixture:
Alcohol $= 20 \times \frac{20}{100} = 4$ litres.
Water $= 20 - 4 = 16$ litres.
When $4$ litres of water are added:
New quantity of water $= 16 + 4 = 20$ litres.
New total quantity of the mixture $= 20 + 4 = 24$ litres.
The percentage of alcohol in the new mixture is calculated as:
$\text{Percentage} = \left( \frac{\text{Quantity of Alcohol}}{\text{Total Quantity of Mixture}} \right) \times 100$
$\text{Percentage} = \left( \frac{4}{24} \right) \times 100 = \frac{1}{6} \times 100 = \frac{50}{3} = 16 \frac{2}{3} \%$.

(D) Let the capacity of each container be $x$ litres.
In the first container:
Milk $= \frac{3}{4}x$ litres
Water $= \frac{1}{4}x$ litres
In the second container:
Milk $= \frac{5}{7}x$ litres
Water $= \frac{2}{7}x$ litres
On mixing both,the total quantity of milk is:
Milk $= \frac{3}{4}x + \frac{5}{7}x = \frac{21x + 20x}{28} = \frac{41x}{28}$ litres
The total quantity of water is:
Water $= \frac{1}{4}x + \frac{2}{7}x = \frac{7x + 8x}{28} = \frac{15x}{28}$ litres
Therefore,the required ratio of milk to water is:
$\frac{41x}{28} : \frac{15x}{28} = 41:15$

(B) Initially:
Milk in Vessel $A = 40$ litres
Water in Vessel $B = 22$ litres
After the first operation (transferring $8$ litres of milk from $A$ to $B$):
Milk in Vessel $A = 40 - 8 = 32$ litres
Milk in Vessel $B = 8$ litres
Water in Vessel $B = 22$ litres
Total mixture in Vessel $B = 8 + 22 = 30$ litres
After the second operation (transferring $6$ litres of mixture from $B$ to $A$):
The fraction of mixture taken out is $\frac{6}{30} = \frac{1}{5}$.
Amount of milk taken out from $B = \frac{1}{5} \times 8 = \frac{8}{5}$ litres.
Amount of water taken out from $B = \frac{1}{5} \times 22 = \frac{22}{5}$ litres.
Final quantity of milk in Vessel $A = 32 + \frac{8}{5} = \frac{160 + 8}{5} = \frac{168}{5}$ litres.
Final quantity of water in Vessel $B = 22 - \frac{22}{5} = \frac{110 - 22}{5} = \frac{88}{5}$ litres.
Required ratio = $\frac{168}{5} : \frac{88}{5} = 168 : 88 = 21 : 11$.

(A) Let the ratio in which the mixtures are mixed be $x:y$.
In vessel $A$,the fraction of milk is $\frac{4}{4+3} = \frac{4}{7}$.
In vessel $B$,the fraction of milk is $\frac{2}{2+3} = \frac{2}{5}$.
In the final mixture $C$,the ratio of milk to water is $1:1$,so the fraction of milk is $\frac{1}{1+1} = \frac{1}{2}$.
Using the rule of alligation:
Difference $1$ (between $B$ and $C$) $= \frac{1}{2} - \frac{2}{5} = \frac{5-4}{10} = \frac{1}{10}$.
Difference $2$ (between $A$ and $C$) $= \frac{4}{7} - \frac{1}{2} = \frac{8-7}{14} = \frac{1}{14}$.
The required ratio $x:y$ is the ratio of these differences:
$x:y = \frac{1}{10} : \frac{1}{14} = \frac{14}{10} = \frac{7}{5}$.
Thus,the liquids should be mixed in the ratio $7:5$.

(B) Using the rule of alligation:
We have two concentrations: $25 \%$ and $60 \%$.
The mean concentration required is $40 \%$.
Difference between $60$ and $40 = 20$.
Difference between $40$ and $25 = 15$.
The ratio of the first concentration to the second concentration is $20: 15$.
Simplifying the ratio $20: 15$ by dividing both terms by $5$,we get $4: 3$.
Therefore,the required ratio is $4: 3$.

(C) Total cost price of $30 \text{ kg}$ rice $= 30 \times 10 = ₹ 300$.
Total cost price of $35 \text{ kg}$ rice $= 35 \times 11 = ₹ 385$.
Total cost price of the mixture $= 300 + 385 = ₹ 685$.
Total quantity of the mixture $= 30 + 35 = 65 \text{ kg}$.
To gain a $30\%$ profit,the total selling price must be $₹ 685 \times (1 + 30/100) = 685 \times 1.3 = ₹ 890.5$.
Selling price per $\text{kg} = 890.5 / 65 = ₹ 13.7$.

(B) Let the total volume of each vessel be equal to the least common multiple of the sum of the parts $(13, 14, 15)$,which is $2730$.
For vessel $1$ ($6:7$,sum $= 13$): Water $= (6/13) \times 2730 = 1260$,Milk $= (7/13) \times 2730 = 1470$.
For vessel $2$ ($5:9$,sum $= 14$): Water $= (5/14) \times 2730 = 975$,Milk $= (9/14) \times 2730 = 1755$.
For vessel $3$ ($8:7$,sum $= 15$): Water $= (8/15) \times 2730 = 1456$,Milk $= (7/15) \times 2730 = 1274$.
Total Water $= 1260 + 975 + 1456 = 3691$.
Total Milk $= 1470 + 1755 + 1274 = 4499$.
The required ratio of water to milk is $3691:4499$.

(B) Initial mixture $= 40$ litres.
Ratio of alcohol to water $= 5:3$.
When $8$ litres of the mixture is removed,the remaining mixture $= 40 - 8 = 32$ litres.
In the remaining $32$ litres,the amount of alcohol $= \frac{5}{8} \times 32 = 20$ litres.
The amount of water $= \frac{3}{8} \times 32 = 12$ litres.
Now,$8$ litres of water is added to the mixture.
New amount of alcohol $= 20$ litres.
New amount of water $= 12 + 8 = 20$ litres.
Therefore,the required ratio of alcohol to water $= 20:20 = 1:1$.

(B) Using the method of alligation,we consider the fraction of milk in each mixture.
In the first vessel,the fraction of milk is $\frac{3}{3+2} = \frac{3}{5}$.
In the second vessel,the fraction of milk is $\frac{7}{7+3} = \frac{7}{10}$.
The desired fraction of milk in the final mixture is $\frac{2}{2+1} = \frac{2}{3}$.
Applying the alligation rule:
Ratio of vessel $I$ to vessel $II = \left| \frac{7}{10} - \frac{2}{3} \right| : \left| \frac{3}{5} - \frac{2}{3} \right|$
$= \left| \frac{21-20}{30} \right| : \left| \frac{9-10}{15} \right|$
$= \frac{1}{30} : \frac{1}{15}$
$= \frac{1}{30} : \frac{2}{30}$
$= 1:2$.

(C) Let the initial quantity of liquid $A$ be $7x$ litres and liquid $B$ be $5x$ litres.
Total initial quantity $= 7x + 5x = 12x$ litres.
When $9$ litres of the 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 $A = 7x - \frac{21}{4}$.
Remaining quantity of $B = 5x - \frac{15}{4} + 9 = 5x + \frac{21}{4}$.
According to the problem,the new ratio is $7:9$:
$\frac{7x - 21/4}{5x + 21/4} = \frac{7}{9}$.
$9(7x - 5.25) = 7(5x + 5.25)$.
$63x - 47.25 = 35x + 36.75$.
$28x = 84$.
$x = 3$.
Initial quantity of $A = 7x = 7 \times 3 = 21$ litres.

(D) Let the quantity of acid in the original mixture be $x$ litres and the quantity of water be $3x$ litres.
According to the problem,when $5$ litres of acid is added,the new ratio of acid to water becomes $1:2$.
So,$\frac{x+5}{3x} = \frac{1}{2}$.
Cross-multiplying,we get $2(x+5) = 3x$.
$2x + 10 = 3x$.
$3x - 2x = 10$,which gives $x = 10$.
The original quantity of the mixture was $x + 3x = 4x = 4(10) = 40$ litres.
After adding $5$ litres of acid,the quantity of the new mixture is $40 + 5 = 45$ litres.

(A) Let the total volume of the mixture be $V$. Initially,the amount of acid is $0.8V$ and water is $0.2V$.
Let $x$ be the fraction of the mixture removed and replaced by water.
After removing $xV$ of the mixture,the remaining acid is $0.8V(1-x)$.
After adding $xV$ of water,the new amount of water is $0.2V(1-x) + xV$.
The new ratio of acid to water is given as $4:3$,so:
$\frac{0.8V(1-x)}{0.2V(1-x) + xV} = \frac{4}{3}$
$\frac{0.8(1-x)}{0.2 - 0.2x + x} = \frac{4}{3}$
$\frac{0.8(1-x)}{0.2 + 0.8x} = \frac{4}{3}$
$2.4(1-x) = 4(0.2 + 0.8x)$
$2.4 - 2.4x = 0.8 + 3.2x$
$1.6 = 5.6x$
$x = \frac{1.6}{5.6} = \frac{16}{56} = \frac{2}{7}$.
Wait,re-evaluating the ratio: $80\%$ acid means $4:1$ ratio. If we replace part $x$ with water,the acid content becomes $0.8(1-x)$. The total volume remains $V$. The new acid amount is $0.8(1-x)V$. Since the ratio is $4:3$,acid is $4/7$ of the total volume.
$0.8(1-x) = 4/7$
$4/5(1-x) = 4/7$
$1-x = 5/7$
$x = 1 - 5/7 = 2/7$.

(D) Initially,the mixture contains $80 \%$ acid and $20 \%$ water,so the ratio of Acid : Water $= 80:20 = 4:1$.
Let the total volume of the mixture be $1$ unit. Let $x$ be the part of the mixture removed and replaced by water.
After removing $x$ part of the mixture,the remaining mixture is $(1-x)$.
The amount of acid remaining is $4(1-x)/5$ and the amount of water remaining is $(1-x)/5$.
After adding $x$ amount of water,the new amount of water is $(1-x)/5 + x$.
According to the problem,the new ratio of acid to water is $4:3$:
$\frac{4(1-x)/5}{(1-x)/5 + x} = \frac{4}{3}$
$\frac{4(1-x)}{1-x+5x} = \frac{4}{3}$
$\frac{1-x}{1+4x} = \frac{1}{3}$
$3 - 3x = 1 + 4x$
$7x = 2$
$x = \frac{2}{7}$

(C) Let the weight of alloy $A$ be $1 \text{ kg}$.
$\therefore$ Gold in $A = \frac{7}{9} \text{ kg}$ and copper in $A = \frac{2}{9} \text{ kg}$.
In $1 \text{ kg}$ of alloy $B$,Gold $= \frac{7}{18} \text{ kg}$ and copper $= \frac{11}{18} \text{ kg}$.
Since equal quantities are mixed,the total gold in alloy $C = \frac{7}{9} + \frac{7}{18} = \frac{14+7}{18} = \frac{21}{18} \text{ kg}$.
The total copper in alloy $C = \frac{2}{9} + \frac{11}{18} = \frac{4+11}{18} = \frac{15}{18} \text{ kg}$.
$\therefore$ Required ratio of gold to copper in $C = \frac{21}{18} : \frac{15}{18} = 21 : 15 = 7 : 5$.

(C) Let the ratio of the first mixture to the second mixture be $x:y$.
In the first bottle,the fraction of acid is $\frac{2}{2+5} = \frac{2}{7}$.
In the second bottle,the fraction of acid is $\frac{7}{7+3} = \frac{7}{10}$.
In the final mixture,the fraction of acid is $\frac{2}{2+3} = \frac{2}{5}$.
Using the rule of alligation:
Difference between the second mixture and the final mixture: $\frac{7}{10} - \frac{2}{5} = \frac{7-4}{10} = \frac{3}{10}$.
Difference between the final mixture and the first mixture: $\frac{2}{5} - \frac{2}{7} = \frac{14-10}{35} = \frac{4}{35}$.
The ratio $x:y$ is $\frac{3}{10} : \frac{4}{35}$.
Multiplying both sides by $70$,we get $x:y = (\frac{3}{10} \times 70) : (\frac{4}{35} \times 70) = 21 : 8$.

(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: $0.600 < 0.625 < 0.636 < 0.666$.
The minimum value is $0.6$,which corresponds to the third container.