Mixture and Alligation Questions in English

(B) Total volume of the mixture $= 60 \ L$.
The ratio of milk to water is $7:5$.
Sum of the ratio parts $= 7 + 5 = 12$.
Quantity of milk $= (7/12) \times 60 = 35 \ L$.
Quantity of water $= (5/12) \times 60 = 25 \ L$.
To make the ratio $1:1$,the quantity of water must be equal to the quantity of milk.
Required quantity of water $= 35 \ L$.
Water to be added $= 35 \ L - 25 \ L = 10 \ L$.

(A) The total volume of the mixture is $30$ $L$ and the ratio of milk to water is $8:7$.
Sum of the ratio terms $= 8 + 7 = 15$.
Quantity of milk $= (8/15) \times 30 = 16$ $L$.
Quantity of water $= (7/15) \times 30 = 14$ $L$.
Let $x$ $L$ of water be added to the mixture.
The new ratio of milk to water becomes $4:5$.
So,$16 / (14 + x) = 4 / 5$.
Cross-multiplying gives: $16 \times 5 = 4 \times (14 + x)$.
$80 = 56 + 4x$.
$4x = 80 - 56 = 24$.
$x = 6$ $L$.
Therefore,$6$ $L$ of water should be added.

(C) Using the method of alligation:
Cost price of first oil = $Rs. 62$ per $kg$
Cost price of second oil = $Rs. 72$ per $kg$
Mean price of the mixture = $Rs. 64.50$ per $kg$
Applying the alligation rule:
Ratio of first oil to second oil = (Mean price - Cost price of second oil) : (Cost price of first oil - Mean price)
Ratio = $(72 - 64.50) : (64.50 - 62)$
Ratio = $7.50 : 2.50$
Ratio = $3 : 1$
Thus,the oil must be mixed in the ratio $3:1$.

(B) Initial quantity of water in the $60 \, ltr$ mixture is $= \frac{20}{100} \times 60 = 12 \, ltr$.
Quantity of milk in the mixture $= 60 \, ltr - 12 \, ltr = 48 \, ltr$.
In the new mixture, water is $25 \%$, which implies milk is $100 \% - 25 \% = 75 \%$.
Since only water is added, the quantity of milk remains constant at $48 \, ltr$.
Let the total quantity of the new mixture be $x \, ltr$.
According to the condition, $75 \%$ of $x = 48 \, ltr$.
$\frac{75}{100} \times x = 48 \Rightarrow \frac{3}{4} \times x = 48$.
$x = 48 \times \frac{4}{3} = 64 \, ltr$.
The amount of water to be added is the difference between the new total quantity and the initial total quantity: $64 \, ltr - 60 \, ltr = 4 \, ltr$.

(C) By the rule of alligation:
Cheaper price $(Rs. 15)$ and Dearer price $(Rs. 20)$ are mixed to get a Mean price of $Rs. 16.50$.
Difference between Dearer price and Mean price $= 20 - 16.50 = 3.50$.
Difference between Mean price and Cheaper price $= 16.50 - 15 = 1.50$.
Required Ratio $= (\text{Difference between Dearer and Mean}) : (\text{Difference between Mean and Cheaper})$
Required Ratio $= 3.50 : 1.50 = 35 : 15 = 7 : 3$.

(C) Let the cost price $(CP)$ of $1$ litre of pure milk be $1$ unit.
Since the milkman sells the mixture at the cost price of pure milk,the selling price $(SP)$ of $1$ litre of the mixture is $1$ unit.
He gains $25 \%$,so the $CP$ of $1$ litre of the mixture is given by:
$CP = \frac{SP}{1 + \text{Gain} \%} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = \frac{100}{125} = \frac{4}{5}$ units.
Using the rule of alligation:
- $CP$ of water = $0$
- $CP$ of milk = $1$
- Mean price = $\frac{4}{5}$
Ratio of water to milk = $(1 - \frac{4}{5}) : (\frac{4}{5} - 0) = \frac{1}{5} : \frac{4}{5} = 1 : 4$.
This means in $5$ parts of the mixture,there is $1$ part of water.
Percentage of water = $(\frac{1}{1 + 4}) \times 100 \% = \frac{1}{5} \times 100 \% = 20 \%$.

(B) Using the rule of alligation:
Cost price of first type of rice = $Rs. 7.20$ per $kg$
Cost price of second type of rice = $Rs. 5.70$ per $kg$
Mean price of the mixture = $Rs. 6.30$ per $kg$
By the rule of alligation,the ratio of the quantities is:
(Mean price - Cost price of second type) : (Cost price of first type - Mean price)
$= (6.30 - 5.70) : (7.20 - 6.30)$
$= 0.60 : 0.90$
$= 6 : 9$
$= 2 : 3$
Thus,the required ratio is $2:3$.

(B) In a business partnership, the profit is distributed in the same ratio as the capital invested by the partners.
Given investment of Aditya $= 45000$.
Given investment of Sanjay $= 30000$.
Required Ratio of profit $= \text{Investment of Aditya} : \text{Investment of Sanjay}$
$= 45000 : 30000$
$= 45 : 30$
$= 3 : 2$
Therefore, the ratio between their profit is $3:2$.

(A) Using the rule of alligation:
Cost price of first type of sugar = $Rs. 2.00$ per $kg$
Cost price of second type of sugar = $Rs. 3.50$ per $kg$
Mean price of the mixture = $Rs. 2.50$ per $kg$
By the rule of alligation:
(Quantity of first sugar) / (Quantity of second sugar) = (Price of second sugar - Mean price) / (Mean price - Price of first sugar)
Ratio = $(3.50 - 2.50) : (2.50 - 2.00)$
Ratio = $1.00 : 0.50$
Ratio = $100 : 50 = 2 : 1$
Therefore,the required ratio is $2:1$.

(C) Initial amount of solution $= 600 \, gm$.
Initial amount of sugar $= 40 \% \text{ of } 600 = \frac{40}{100} \times 600 = 240 \, gm$.
Initial amount of water $= 600 - 240 = 360 \, gm$.
Let the amount of sugar to be added be $x \, gm$.
After adding sugar,the new amount of sugar $= 240 + x \, gm$.
The new total amount of solution $= 600 + x \, gm$.
We want the new concentration of sugar to be $50 \%$,which means the ratio of sugar to the total solution is $\frac{1}{2}$.
$\frac{240 + x}{600 + x} = \frac{50}{100} = \frac{1}{2}$.
$2(240 + x) = 600 + x$.
$480 + 2x = 600 + x$.
$2x - x = 600 - 480$.
$x = 120 \, gm$.

(D) Using the rule of alligation:
Density of Gold $= 21$
Density of Copper $= 11$
Mean density of mixture $= 17$
By applying the alligation rule:
(Gold) : (Copper) = $(17 - 11) : (21 - 17)$
(Gold) : (Copper) = $6 : 4$
Simplifying the ratio,we get $3 : 2$.

(D) Using the method of alligation:
Profit on first part = $12 \%$
Loss on second part = $-8 \%$
Overall profit = $11 \%$
Applying alligation:
Ratio of books sold at $12 \%$ profit to books sold at $8 \%$ loss = $(11 - (-8)) : (12 - 11) = 19 : 1$
Total parts = $19 + 1 = 20$
Number of books sold at $12 \%$ profit = $\frac{19}{20} \times 100 = 95$

(B) The ratio of milk to water in the three tubs is $5:2$,$4:3$,and $3:1$.
Fraction of milk in the first tub $= \frac{5}{5+2} = \frac{5}{7}$.
Fraction of milk in the second tub $= \frac{4}{4+3} = \frac{4}{7}$.
Fraction of milk in the third tub $= \frac{3}{3+1} = \frac{3}{4}$.
Total milk $= \frac{5}{7} + \frac{4}{7} + \frac{3}{4} = \frac{20 + 16 + 21}{28} = \frac{57}{28}$.
Fraction of water in the first tub $= \frac{2}{7}$.
Fraction of water in the second tub $= \frac{3}{7}$.
Fraction of water in the third tub $= \frac{1}{4}$.
Total water $= \frac{2}{7} + \frac{3}{7} + \frac{1}{4} = \frac{8 + 12 + 7}{28} = \frac{27}{28}$.
Ratio of milk to water $= \frac{57}{28} : \frac{27}{28} = 57 : 27 = 19 : 9$.

(D) Using the rule of alligation:
Cost price of first sugar = $Rs. 17/kg$
Cost price of second sugar = $Rs. 29/kg$
Mean price of mixture = $Rs. 20/kg$
Difference between mean price and cost price of second sugar = $|29 - 20| = 9$
Difference between mean price and cost price of first sugar = $|17 - 20| = 3$
The ratio of the first sugar to the second sugar is $9:3$.
Simplifying the ratio: $9:3 = 3:1$.
Therefore,the required proportion is $3:1$.

(D) Using the rule of alligation:
Cost price of first pulse $= Rs. 70/kg$
Cost price of second pulse $= Rs. 45/kg$
Mean price of mixture $= Rs. 60/kg$
Difference between mean price and second price $= 60 - 45 = 15$
Difference between first price and mean price $= 70 - 60 = 10$
Required ratio of first pulse to second pulse $= 15 : 10 = 3 : 2$.

(B) Using the method of alligation:
Profit on first part = $10 \%$
Loss on second part = $-5 \%$
Overall gain = $7 \%$
Applying the alligation rule:
Ratio of quantities = $|-5 - 7| : |10 - 7| = |-12| : |3| = 12 : 3 = 4 : 1$
The total quantity is $50 \ kg$.
Quantity sold at $10 \%$ profit = $\frac{4}{4+1} \times 50 = \frac{4}{5} \times 50 = 40 \ kg$.

(C) Let the cost price of the first cow be $C_1$ and the second cow be $C_2$.
Given $C_1 + C_2 = 2700$.
Since there is neither gain nor loss on the whole,the loss on the first cow equals the gain on the second cow.
$6\% \text{ of } C_1 = 7.5\% \text{ of } C_2$.
$\frac{C_1}{C_2} = \frac{7.5}{6} = \frac{75}{60} = \frac{5}{4}$.
Thus,the ratio of the cost prices is $C_1 : C_2 = 5 : 4$.
The cost price of the $2^{nd}$ cow is $C_2 = \frac{4}{5+4} \times 2700 = \frac{4}{9} \times 2700 = 4 \times 300 = Rs. 1200$.

(D) Let the investment of Nutan be $x$.
Then,Sonal's investment $= x + 5,000$.
Aditya's investment $= (x + 5,000) + 4,000 = x + 9,000$.
The total investment is $x + (x + 5,000) + (x + 9,000) = 50,000$.
$3x + 14,000 = 50,000$.
$3x = 36,000 \Rightarrow x = 12,000$.
Nutan's investment $= 12,000$.
Sonal's investment $= 12,000 + 5,000 = 17,000$.
Aditya's investment $= 12,000 + 9,000 = 21,000$.
The ratio of their investments is $21,000 : 17,000 : 12,000 = 21 : 17 : 12$.
The sum of the ratio parts is $21 + 17 + 12 = 50$.
Aditya's share of the profit $= \frac{21}{50} \times 70,000 = 21 \times 1,400 = 29,400$.

(A) Using the rule of alligation:
Concentration of first solution = $15 \%$
Concentration of second solution = $40 \%$
Mean concentration = $30 \%$
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 = $10 : 15$
Simplifying the ratio by dividing by $5$,we get $2 : 3$.
Therefore,the required ratio is $2 : 3$.

(A) Cost Price $(CP)$ of $30 \text{ kg}$ of wheat $= 30 \times 9.50 = Rs. 285$.
Cost Price $(CP)$ of $40 \text{ kg}$ of wheat $= 40 \times 8.50 = Rs. 340$.
Total Cost Price $(CP)$ of $70 \text{ kg}$ of mixture $= 285 + 340 = Rs. 625$.
Selling Price $(SP)$ of $70 \text{ kg}$ of mixture $= 70 \times 8.90 = Rs. 623$.
Since $CP > SP$,there is a loss.
Loss $= CP - SP = 625 - 623 = Rs. 2$.

(D) Let the price of milk be $108$ paise per litre and the price of water be $0$ paise per litre.
The mean price of the mixture is $90$ paise per litre.
Using the rule of alligation:
Milk (Price: $108$) : Water (Price: $0$) = $(90 - 0) : (108 - 90) = 90 : 18 = 5 : 1$.
Given that the quantity of water is $16$ litres.
Let the quantity of milk be $x$ litres.
So,$x / 16 = 5 / 1$.
$x = 16 \times 5 = 80$ litres.
Thus,there are $80$ litres of milk in the mixture.

(B) Let the cost price of the chemical be $Rs. 25$ per $litre$ and the cost price of water be $Rs. 0$ per $litre$.
Given,selling price of the mixture $= Rs. 20$ per $litre$ and profit $= 25\%$.
Cost price $(CP)$ of the mixture $= \frac{100}{100 + \text{Profit}\%} \times \text{Selling Price} = \frac{100}{125} \times 20 = 16$.
Using the rule of alligation:
Chemical $(25)$ : Water $(0)$ = Mixture $(16)$
Difference between chemical and mixture $= 25 - 16 = 9$
Difference between water and mixture $= 16 - 0 = 16$
Therefore,the ratio of chemical to water is $16:9$.

(C) Initial quantity of mixture $= 40$ $litres$.
Percentage of water $= 10 \%$,so quantity of water $= 0.10 \times 40 = 4$ $litres$.
Quantity of milk $= 40 - 4 = 36$ $litres$.
Let $x$ $litres$ of water be added to the mixture.
New quantity of water $= 4 + x$ $litres$.
New total quantity of mixture $= 40 + x$ $litres$.
According to the problem,the new percentage of water is $20 \%$,so:
$\frac{4 + x}{40 + x} = \frac{20}{100} = \frac{1}{5}$.
$5(4 + x) = 40 + x$.
$20 + 5x = 40 + x$.
$4x = 20$.
$x = 5$ $litres$.

(B) Quantity of zinc in the first metal $= \frac{1}{3} \times 2 = \frac{2}{3} \, kg$.
Quantity of copper in the first metal $= 2 - \frac{2}{3} = \frac{4}{3} \, kg$.
Quantity of zinc in the second metal $= \frac{1}{4} \times 3 = \frac{3}{4} \, kg$.
Quantity of copper in the second metal $= 3 - \frac{3}{4} = \frac{9}{4} \, kg$.
Total quantity of zinc in the mixture $= \frac{2}{3} + \frac{3}{4} = \frac{8 + 9}{12} = \frac{17}{12} \, kg$.
Total quantity of copper in the mixture $= \frac{4}{3} + \frac{9}{4} = \frac{16 + 27}{12} = \frac{43}{12} \, kg$.
Therefore,the required ratio of zinc to copper $= \frac{17}{12} : \frac{43}{12} = 17 : 43$.

(B) Initial weight of alloy $= 50 \, g$.
Weight of gold in the alloy $= 80 \% \text{ of } 50 \, g = 0.80 \times 50 = 40 \, g$.
Weight of silver in the alloy $= 50 - 40 = 10 \, g$.
Let the quantity of gold to be added be $x \, g$.
After adding $x \, g$ of gold,the new weight of gold $= 40 + x \, g$.
The total weight of the alloy becomes $50 + x \, g$.
According to the problem,the new percentage of gold is $95 \%$.
Therefore,the percentage of silver becomes $100 \% - 95 \% = 5 \%$.
Since the weight of silver remains constant at $10 \, g$,we can write:
$5 \% \text{ of } (50 + x) = 10 \, g$
$0.05 \times (50 + x) = 10$
$50 + x = \frac{10}{0.05} = 200$
$x = 200 - 50 = 150 \, g$.
Thus,$150 \, g$ of gold must be added.

(C) Let the number of $5$-paise coins be $x$ and the number of $10$-paise coins be $y$.
Given the total number of coins is $80$,so $x + y = 80$ ... $(i)$.
The total value of the coins is $Rs. 6.40$,which is $640$ paise.
Therefore,$5x + 10y = 640$ ... $(ii)$.
Divide equation $(ii)$ by $5$,we get $x + 2y = 128$ ... $(iii)$.
Subtract equation $(i)$ from equation $(iii)$:
$(x + 2y) - (x + y) = 128 - 80$
$y = 48$.
Substitute $y = 48$ into equation $(i)$:
$x + 48 = 80$
$x = 80 - 48 = 32$.
Thus,there are $32$ coins of $5$-paise.

(D) Using the rule of alligation:
Price of first tea = $Rs. 4/kg$
Price of second tea = $Rs. 10/kg$
Mean price of mixture = $Rs. 6.50/kg$
Applying alligation:
(Price of second tea - Mean price) : (Mean price - Price of first tea)
$= (10 - 6.50) : (6.50 - 4)$
$= 3.50 : 2.50$
$= 35 : 25 = 7 : 5$
Let the quantity of tea at $Rs. 4/kg$ be $x \, kg$.
Given,quantity of tea at $Rs. 10/kg = 15 \, kg$.
Ratio of quantities = $x : 15 = 7 : 5$
$x / 15 = 7 / 5$
$x = (7 \times 15) / 5$
$x = 7 \times 3 = 21 \, kg$.
Therefore,$21 \, kg$ of tea should be added.

(B) Using the rule of alligation:
$1$. The density of gold relative to water is $19$.
$2$. The density of copper relative to water is $9$.
$3$. The desired density of the mixture relative to water is $15$.
Applying the alligation method:
- Difference between copper and mixture $= |9 - 15| = 6$
- Difference between gold and mixture $= |19 - 15| = 4$
Therefore,the ratio of gold to copper is $6:4$.
Simplifying the ratio $6:4$ gives $3:2$.
Thus,the metals should be mixed in the ratio $3:2$.

(D) Initial quantity of milk $(x)$ = $81$ $\text{litres}$.
Amount replaced in each operation $(y)$ = $\frac{1}{3} \times 81 = 27$ $\text{litres}$.
Number of operations $(n)$ = $2$.
Using the formula for the quantity of milk left after $n$ operations: $Milk_{left} = x(1 - \frac{y}{x})^n$.
$Milk_{left} = 81(1 - \frac{27}{81})^2 = 81(1 - \frac{1}{3})^2 = 81 \times (\frac{2}{3})^2$.
$Milk_{left} = 81 \times \frac{4}{9} = 9 \times 4 = 36$ $\text{litres}$.
Quantity of water in the mixture = Total volume - Quantity of milk = $81 - 36 = 45$ $\text{litres}$.
Required ratio of milk to water = $36 : 45$.
Dividing both by $9$, we get $4 : 5$.

(D) Using the method of alligation:
Profit percentage $= 12 \%$,Loss percentage $= -8 \%$,Overall profit percentage $= 11 \%$.
Ratio of pens sold at profit to pens sold at loss:
$= (11 - (-8)) : (12 - 11)$
$= (11 + 8) : 1$
$= 19 : 1$
Total parts $= 19 + 1 = 20$.
Number of pens sold at $12 \%$ profit $= \frac{19}{20} \times 60 = 19 \times 3 = 57$.

(B) Let the cost price $(CP)$ of $1$ litre of milk be $Rs. 1$.
Since the mixture is sold at the cost price of milk,the selling price $(SP)$ of $1$ litre of the mixture is $Rs. 1$.
Given that the gain is $16 \%$,we can find the $CP$ of $1$ litre of the mixture using the formula: $CP = SP \times \frac{100}{100 + \text{gain} \%}$.
$CP = 1 \times \frac{100}{100 + 16} = \frac{100}{116} = \frac{25}{29}$.
Using the rule of alligation:
Cost of water = $0$
Cost of milk = $1$
Mean cost of mixture = $\frac{25}{29}$
Ratio of water to milk = $(\text{Cost of milk} - \text{Mean cost}) : (\text{Mean cost} - \text{Cost of water})$
Ratio = $(1 - \frac{25}{29}) : (\frac{25}{29} - 0) = \frac{4}{29} : \frac{25}{29} = 4 : 25$.

(A) First,calculate the Cost Price $(CP)$ of the mixture.
Given Selling Price $(SP)$ = $Rs. 68.20$ per $kg$ and Profit = $10\%$.
$CP = \frac{100}{100 + \text{Profit}\%} \times SP$
$CP = \frac{100}{110} \times 68.20 = \frac{10}{11} \times 68.20 = Rs. 62$ per $kg$.
Now,apply the rule of Alligation:
Cost of first variety = $60$
Cost of second variety = $65$
Mean price = $62$
Difference between second variety and mean price = $|65 - 62| = 3$
Difference between mean price and first variety = $|62 - 60| = 2$
Therefore,the required ratio is $3:2$.

(B) Let the quantity of milk taken from the first can be $x$ litres and from the second can be $(12 - x)$ litres.
In the first can,water is $25\%$,so milk is $75\% = \frac{3}{4}$.
In the second can,water is $50\%$,so milk is $50\% = \frac{1}{2}$.
The final mixture of $12$ litres has a water to milk ratio of $3:5$,which means milk is $\frac{5}{3+5} = \frac{5}{8}$ of the total volume.
Total milk in the final mixture $= 12 \times \frac{5}{8} = 7.5$ litres.
Now,set up the equation based on the milk content:
$\frac{3}{4}x + \frac{1}{2}(12 - x) = 7.5$
Multiply by $4$ to clear the denominators:
$3x + 2(12 - x) = 30$
$3x + 24 - 2x = 30$
$x = 6$ litres.
Therefore,the quantity from the first can is $6$ litres and from the second can is $12 - 6 = 6$ litres.

(D) Let the cost price of the mixture be $CP$. Given that the selling price is $Rs. 9.24$ per $kg$ with a gain of $10\%$.
$CP = \frac{100}{110} \times 9.24 = \frac{10}{11} \times 9.24 = 8.4$.
Using the rule of alligation:
Cost of first type of sugar = $9$
Cost of second type of sugar = $7$
Mean price = $8.4$
Difference between mean price and second cost = $8.4 - 7 = 1.4$
Difference between first cost and mean price = $9 - 8.4 = 0.6$
Ratio of quantities = $1.4 : 0.6 = 14 : 6 = 7 : 3$.
Let the quantity of the first type of sugar be $x\, kg$.
Then,$\frac{x}{27} = \frac{7}{3}$.
$x = \frac{7}{3} \times 27 = 7 \times 9 = 63\, kg$.

(D) The formula for the quantity of milk remaining after $n$ operations is given by:
$\text{Remaining Milk} = x \left(1 - \frac{y}{x}\right)^n$
Where:
$x = 50 \, \text{litres}$ (initial quantity of milk)
$y = 5 \, \text{litres}$ (quantity replaced in each operation)
$n = 3$ (total number of operations,as the process was repeated two more times after the first)
Substituting the values:
$\text{Remaining Milk} = 50 \left(1 - \frac{5}{50}\right)^3$
$= 50 \left(1 - \frac{1}{10}\right)^3$
$= 50 \left(\frac{9}{10}\right)^3$
$= 50 \times \frac{729}{1000}$
$= \frac{729}{20} = 36.45 \, \text{litres}$

(D) Total volume of the mixture $= 30 \ L$.
The ratio of water to milk is $3:7$.
Sum of ratio parts $= 3 + 7 = 10$.
Quantity of water $= (3/10) \times 30 = 9 \ L$.
Quantity of milk $= (7/10) \times 30 = 21 \ L$.
Let the quantity of water added be $x \ L$.
New quantity of water $= 9 + x$.
New ratio of milk to water $= 1:2$.
Therefore,$\frac{\text{Milk}}{\text{Water}} = \frac{21}{9 + x} = \frac{1}{2}$.
Cross-multiplying,we get: $21 \times 2 = 9 + x$.
$42 = 9 + x$.
$x = 42 - 9 = 33 \ L$.

(A) Let the price per $kg$ of the mixture be $x$.
Using the weighted average formula:
Price of mixture $= \frac{(\text{Quantity}_1 \times \text{Price}_1) + (\text{Quantity}_2 \times \text{Price}_2)}{\text{Total Quantity}}$
Price of mixture $= \frac{(2 \times 15) + (3 \times 20)}{2 + 3}$
Price of mixture $= \frac{30 + 60}{5}$
Price of mixture $= \frac{90}{5} = 18$
Thus,the price per $kg$ of the mixed variety is $Rs. 18$.

(C) Using the method of alligation:
Profit percentage of first part = $8 \%$
Profit percentage of second part = $18 \%$
Mean profit percentage = $14 \%$
Difference between second part and mean = $18 - 14 = 4$
Difference between mean and first part = $14 - 8 = 6$
Ratio of quantities = $4 : 6 = 2 : 3$
Total parts = $2 + 3 = 5$
Quantity sold at $18 \%$ profit = $\frac{3}{5} \times 1000 = 600 \ kg$.

(C) Using the rule of alligation:
Boys pass percentage = $85 \%$
Girls pass percentage = $70 \%$
Total pass percentage = $75 \%$
Difference for boys = $|75 - 70| = 5$
Difference for girls = $|85 - 75| = 10$
Ratio of boys to girls = $5 : 10 = 1 : 2$
Total students = $900$
Number of girls = $\frac{2}{1 + 2} \times 900 = \frac{2}{3} \times 900 = 600$

(D) Initial amount of solution $= 400 \, gm$.
Amount of salt $= 40 \% \text{ of } 400 = \frac{40}{100} \times 400 = 160 \, gm$.
Amount of water $= 400 - 160 = 240 \, gm$.
Let $x \, gm$ of salt be added.
New amount of salt $= 160 + x$.
New total amount of solution $= 400 + x$.
We want the new concentration of salt to be $50 \%$,so $\frac{160 + x}{400 + x} = \frac{50}{100} = \frac{1}{2}$.
$2(160 + x) = 400 + x$.
$320 + 2x = 400 + x$.
$x = 400 - 320 = 80 \, gm$.

(D) Let the price per dozen of the mixture be $x$.
Using the weighted average method:
Price of mixture $= \frac{(3 \times 12) + (5 \times 10)}{3 + 5}$
Price of mixture $= \frac{36 + 50}{8}$
Price of mixture $= \frac{86}{8}$
Price of mixture $= 10.75$
Thus,the price per dozen of the mixture is $Rs. 10.75$.

(A) Let the mean price of the mixture be $x$ per dozen.
Using the method of alligation:
Ratio of quantities = $\frac{\text{Price}_2 - x}{x - \text{Price}_1} = \frac{3}{5}$
$\frac{5.40 - x}{x - 4.20} = \frac{3}{5}$
$5(5.40 - x) = 3(x - 4.20)$
$27.00 - 5x = 3x - 12.60$
$8x = 39.60$
$x = \frac{39.60}{8} = 4.95$
Thus,the price per dozen of the mixture is $Rs. 4.95$.

(B) Initial mixture volume $= 48 \text{ litre}$.
Water content $= 10 \% \text{ of } 48 = 4.8 \text{ litre}$.
Milk content $= 48 - 4.8 = 43.2 \text{ litre}$.
Let the amount of water to be added be $x \text{ litre}$.
In the new mixture,the amount of milk remains constant at $43.2 \text{ litre}$,but this now represents $(100 - 20) \% = 80 \%$ of the total new volume.
So,$80 \% \text{ of } (48 + x) = 43.2$.
$0.8 \times (48 + x) = 43.2$.
$48 + x = \frac{43.2}{0.8} = 54$.
$x = 54 - 48 = 6 \text{ litre}$.

(B) Let the quantity of milk be $M$ and water be $W$. Total quantity $M + W = 160 \text{ litres}$.
By the rule of alligation:
Milk $(70 \%)$ and Water $(30 \%)$ are mixed to get a final mixture of $55 \%$.
The differences are:
$|55 - 30| = 25$
$|70 - 55| = 15$
The ratio of milk to water is $25 : 15 = 5 : 3$.
Total parts $= 5 + 3 = 8$.
Quantity of milk $= \frac{5}{8} \times 160 = 100 \text{ litres}$.
Quantity of water $= \frac{3}{8} \times 160 = 60 \text{ litres}$.

(C) Initial mixture $= 80 \text{ litre}$.
Water content $= 10 \% \text{ of } 80 = 8 \text{ litre}$.
Milk content $= 80 - 8 = 72 \text{ litre}$.
Let the amount of water to be added be $x \text{ litre}$.
New total mixture $= 80 + x \text{ litre}$.
New water content $= 8 + x \text{ litre}$.
According to the problem,the new water content is $20 \%$ of the new mixture:
$8 + x = 0.20 \times (80 + x)$
$8 + x = 16 + 0.2x$
$x - 0.2x = 16 - 8$
$0.8x = 8$
$x = \frac{8}{0.8} = 10 \text{ litre}$.

(D) Let the initial amount of liquid $P$ be $4x$ and liquid $Q$ be $x$.
When $10 \, L$ of the mixture is removed,the amount of $P$ removed is $\frac{4}{5} \times 10 = 8 \, L$ and the amount of $Q$ removed is $\frac{1}{5} \times 10 = 2 \, L$.
After removing the mixture,the remaining amount of $P$ is $4x - 8$ and the remaining amount of $Q$ is $x - 2$.
Then,$10 \, L$ of liquid $Q$ is added,so the new amount of $Q$ is $x - 2 + 10 = x + 8$.
The new ratio is given as $2:3$,so $\frac{4x - 8}{x + 8} = \frac{2}{3}$.
Cross-multiplying gives $3(4x - 8) = 2(x + 8)$,which simplifies to $12x - 24 = 2x + 16$.
$10x = 40$,so $x = 4$.
The initial quantity of liquid $P$ was $4x = 4 \times 4 = 16 \, L$.

(D) Initial quantity of milk $(x)$ = $81$ litres.
Amount replaced in each operation $(y)$ = $\frac{1}{3}$ of $81 = 27$ litres.
Number of operations $(n)$ = $2$.
The quantity of milk remaining after $n$ operations is given by the formula: $Milk_{final} = x \left(1 - \frac{y}{x}\right)^n$.
$Milk_{final} = 81 \left(1 - \frac{27}{81}\right)^2 = 81 \left(1 - \frac{1}{3}\right)^2 = 81 \times \left(\frac{2}{3}\right)^2$.
$Milk_{final} = 81 \times \frac{4}{9} = 9 \times 4 = 36$ litres.
Quantity of water in the final mixture = Total volume - Quantity of milk = $81 - 36 = 45$ litres.
Ratio of milk to water = $36 : 45$.
Dividing both by $9$,we get $4 : 5$.

(D) Total profit $= Rs. 1020$.
Aditya's share $= Rs. 495$.
Manish's share $= 1020 - 495 = Rs. 525$.
The ratio of their profit shares is equal to the ratio of their investments multiplied by time.
Let Aditya invest for $x$ months. Then Manish invests for $(36 - x)$ months.
Ratio of profit $= \frac{300 \times x}{500 \times (36 - x)} = \frac{495}{525}$.
Simplifying the fraction $\frac{495}{525}$ by dividing by $15$,we get $\frac{33}{35}$.
So,$\frac{3x}{5(36 - x)} = \frac{33}{35}$.
$\frac{x}{36 - x} = \frac{33}{35} \times \frac{5}{3} = \frac{11}{7}$.
$7x = 11(36 - x)$.
$7x = 396 - 11x$.
$18x = 396$.
$x = \frac{396}{18} = 22$.
Thus,Aditya kept his money for $22$ months.

(C) Let the initial quantity of liquid $A$ be $3x$ and liquid $B$ be $2x.$ The total initial quantity is $5x.$
When $5$ litres of the mixture is removed,the amount of $A$ removed is $\frac{3}{5} \times 5 = 3$ litres,and the amount of $B$ removed is $\frac{2}{5} \times 5 = 2$ litres.
Remaining quantity of $A = 3x - 3$.
Remaining quantity of $B = 2x - 2$.
After adding $5$ litres of liquid $B$,the new quantity of $B = 2x - 2 + 5 = 2x + 3$.
The new ratio is given as $2 : 3$,so $\frac{3x - 3}{2x + 3} = \frac{2}{3}$.
Cross-multiplying gives: $3(3x - 3) = 2(2x + 3)$.
$9x - 9 = 4x + 6$.
$5x = 15$,which means $x = 3$.
The initial quantity of liquid $A = 3x = 3 \times 3 = 9$ litres.

(C) Let the cost price of $1 \ kg$ of tea be $100$ units.
Total cost price of $40 \ kg$ of tea $= 40 \times 100 = 4000$ units.
He sells a part at $5 \%$ loss (i.e.,$95$ units per $kg$) and the rest at cost price (i.e.,$100$ units per $kg$).
Overall loss is $3 \%$,so the total selling price is $4000 \times 0.97 = 3880$ units.
Let $x$ be the quantity sold at $5 \%$ loss and $(40 - x)$ be the quantity sold at cost price.
$95x + 100(40 - x) = 3880$
$95x + 4000 - 100x = 3880$
$-5x = 3880 - 4000$
$-5x = -120$
$x = 24 \ kg$ (quantity sold at $5 \%$ loss).
Quantity sold at cost price $= 40 - 24 = 16 \ kg$.