Let the amount of pure copper added be ‘x’ kg.

Amount of copper in 20 kg of alloy A = 2/ (2 + 3) × 20 = 8 kg

Amount of copper in 28 kg of alloy B = 3/ (3 + 4) × 28 = 12 kg

Total amount of copper in alloy C = 8 + 12 + x = 20 + x

Total amount of alloy C produced = 20 + 28 + x = 48 + x

⇒ 20 + x = 6/ (6 + 7) × (48 + x)

⇒ 20 + x = 6/13 × (48 + x)

⇒ 260 + 13x = 288 + 6x

⇒ 7x = 28

Quantity A:

⇒ Water in the mixture = 20% of 50 = 50 × 0.2 = 10 litres

⇒ Milk in the mixture = 50 – 10 = 40 litres

⇒ Water in 10 litres of the mixture = 20% of 10 = 10 × 0.2 = 2 litres

⇒ Milk in 10 litres of the mixture = 10 – 2 = 8 litres

After adding 10 litre of water,

⇒ Water in new mixture = 10 – 2 + 10 = 18 litres

⇒ Milk in the new mixture = 40 – 8 = 32 litres

⇒ Percentage of water in the new mixture = {18/(18 + 32)} × 100 = (18/50) × 100 = 36%

Quantity B:

Let the initial quantity of mixture is x litres and y litre is replaced;

According to the question;

0.4x + 0.2y = 0.35x

⇒ 0.2y = -0.05x

⇒ -4y = x

⇒ y/x = -¼

∴ Percentage of quantity replaced = -¼ = -25%(minus sign implies removal)

Sum of ratios = 4 + 2 + 3 = 9

Amount of batter that can be prepared using 20 cups of flour = (20/4) × 9 = 45 cups

Amount of batter that can be prepared using 8 cups of milk = (8/2) × 9 = 36 cups

∵ Milk is available only for 36 cups of batter, maximum possible batter that can be prepared is 36 cups

Amount of flour required for 36 cups of batter = (4/9) × 36 = 16 cups

Ratio of Chana dal and Masoor dal = 2 : 8 = 1 : 4

Total content removed from the jar = 20% of 10 kg

⇒ (20/100) × 10

⇒ 2 kg

Quantity of chana dal removed = 2 × (1/5)

⇒ 2/5 = 0.4 kg

Quantity of masoor dal removed = 2 × (4/5)

⇒ 8/5 = 1.6 kg

Quantity of chana dal remained = 2 - 0.4 = 1.6 kg      ----(1)

Quantity of masoor dal remained = 8 - 1.6 = 6.4 kg      ----(2)

We have to make their quantities in 4 ∶ 1 by adding chana dal

So, Quantity of chana dal should be = 6.4 × 4 = 25.6 kg      ----(3)

From equation (3) and (1), we get

Quantity of chana dal to be added = 25.6 - 1.6 = 24 kg

Accordingly, x = 24 kg

Now, ratio of chana dal and masoor dal is in 4 ∶ 1

We have to again make their ratio 1 ∶ 4 by adding masoor dal

So, Quantity of masoor dal should be = 25.6 × 4 = 102.4 kg      ----(4)

From equation (4) and (2), we get

Quantity of masoor dal to be added = 102.4 - 6.4 = 96 kg

Accordingly, y = 96 kg

Now, ratio of chana dal and masoor dal is 1 ∶ 4

We have to make their ratio 2 ∶ 5 by adding chana dal

So, Quantity of chana dal should be = (102.4/5) × 2

⇒ 20.48 × 2

⇒ 40.96 kg

Existing quantity of chana dal = 25.6 kg

Quantity of chana dal to be added = 40.96 - 25.6 = 15.36 kg

Accordingly, z = 15.36 kg

Here, we have been provided ratios instead of any cost. But we can still know the dearer and cheaper quantities through the amount of milk present in the mixture, considering that water doesn’t cost anything to the milkman.

Let the size of the vessels be 1 litre.

Quantity of milk in 1 litre of mixture in first vessel = Quantity of cheaper = c = $\frac{Unitsofmilk}{Totalunits}=\frac{1}{4}$ 

Quantity of milk in 1 litre of mixture in second vessel = Quantity of dearer = d = $\frac{Unitsofmilk}{Totalunits}=\frac{3}{4}$

Desired quantity of milk in 1 litre of mixture in third vessel = Mean quantity = m = $\frac{2}{5}litre$

$\begin{array}{l}\Rightarrow Requiredrate=\frac{Quantityofcheaper}{Quantityofdearer}=\frac{Firstvessel}{Secondvessel}=\frac{d-m}{m-c}\\ \Rightarrow \frac{Firstvessel}{Secondvessel}=\frac{\frac{3}{4}-\frac{2}{5}}{\frac{2}{5}-\frac{1}{4}}\\ \Rightarrow \frac{3}{4}-\frac{2}{5}:\frac{2}{5}-\frac{1}{4}\\ \Rightarrow \frac{15-8}{20}:\frac{8-5}{20}\\ \Rightarrow \frac{7}{20}:\frac{3}{20}\end{array}$

⇒ 7 : 3

⇒ First vessel : Second vessel = 7 : 3

Let the milkman gets water free of cost. So, water is cheaper and milk is dearer product.

Let the cans in which the wife of the milkman had put the mixture initially had a volume of one litre.

Quantity of milk in first can = Quantity of cheaper = c = $\frac{Unitsofmilk}{Totalunits}=\frac{3}{8}litre$

Quantity of milk in second can = Quantity of dearer = d = $\frac{Unitsofmilk}{Totalunits}=\frac{3}{7}litre$

Desired quantity of milk in 1 litre of mixture in third can = Mean quantity = m = $\frac{1}{2}litre$

In both the cans the wife had created a mixture having milk less than half of the volume. So, in no condition, the milkman will be able to create a mixture having ratio of milk to water 1 ∶ 1.

Let T kg of first mixture is taken. In it, the weight of oil is twice that of water, means ratio of their weights is 2 : 1.

Weight of oil in first mixture = 2/(2 + 1) × T = 2T/3 kg

Weight of water in first mixture = 1/(2 + 1) × T = T/3 kg

Ratio of weights of oil and water in mixture $=\frac{\left(\frac{2T}{3}+2\right)}{\frac{T}{3}+2}=\frac{17}{9}$

$\Rightarrow \frac{18T}{3}+18=\frac{17T}{3}+34$ 

⇒ T/3 = 16

⇒ T = 48

∴ Weight of first mixture taken is 48 kg.

Both containers have equal capacity

Let us suppose that, each of them has a capacity of T liters.

Ratio of water and oil in first container = 3 : 1

Ratio of water and oil in second container = 4 : 3

Amount of water in first container = (3/(3+1)) × T = (3/4)T

Amount of oil in first container = (1/(3+1)) × T = (1/4)T

Amount of water in second container = (4/(4+3)) × T = (4/7)T

Amount of oil in second container = (3/(4+3)) × T = (3/4)T

After mixing

Total amount of water = (3/4)T + (4/7)T = (37/28)T

Total amount of Oil = (1/4)T + (3/7) T = (19/28)T

3 liters of liquid M and 5 liters of liquid N are mixed.

Total volume of mixture = (3+5) liters = 8 liters

After dividing into 4 equal parts,

Volume of each part = (8/4) liters = 2 liters

Volume of liquid M in each part = (3÷ (3+5)) ×2 = 0.75 liters.

Volume of liquid N in each part = (5÷ (3+5)) ×2 = 1.25 liters.

Suppose T liters of liquid M are added in one of these parts to make ratio of M and N to 6:5 in this part.

⇒ (0.75 + T)/1.25 = 6/5

⇒ 3.75 + 5T = 7.5

⇒ 5T = 3.75

⇒ T = 0.75

Now, all parts are mixed back together.

Volume of liquid M in mixture = (3 + 0.75) liters = 3.75 liters.

Volume of liquid N in mixture = 5 liters.

Given, volume of two jars is 3 and 5 liters respectively.

Volume of alcohol present in smaller jar = 25% of volume of smaller jar

⇒ volume of alcohol present in smaller jar = 0.25 × 3 = 0.75 litres

Volume of alcohol present in the bigger jar = 75 % of volume of bigger jar

⇒ Volume of alcohol present in the bigger jar = 0.75 × 5 = 3.75 litres

Now, both the jars are emptied in a 9 litre cask, which is then filled up with water.

Volume of alcohol present in the cask = sum of volume alcohol present in the two jars

⇒ Volume of alcohol present in the cask = 0.75 + 3.75 = 4.5 litres

% of alcohol present in the cask = ( volume of alcohol present / volume of cask ) × 100 %

⇒ % of alcohol present in the cask = (4.5/9) × 100 % = 50 %

For the first alloy,

The ratio by weight of X:Y in the first alloy is 6:5

⇒ In fraction the amount of X in the alloy = 6/(6 + 5) = 6/11

Similarly

⇒ In fraction the amount of Y in the alloy = 5/(6 + 5) = 5/11

Hence, In 11 kg of the first alloy,

∴Amount of X = 11 × 6/11 kg = 6kg

∴The amount of Y = 11 × 5/11 kg = 5kg

 For the second alloy,

The ratio by weight of X : Y in the second alloy is 7:13

⇒ In fraction the amount of X in the alloy = 7/(7 + 13) = 7/20

Similarly,

⇒ In fraction the amount of Y in the alloy = 13/(7 + 13) = 13/20

In 20 kg of the second alloy,

∴The amount of X = 20 × 7/20 kg = 7kg

∴The amount of Y = 20 × 13/20 kg = 13kg

∴Total amount of X in the two alloys = 6 kg + 7 kg = 13kg

∴Total amount of Y in the two alloys = 5kg + 13 kg = 18 kg

 Suppose we have to add "a" kg of X, in order to make the percentage of X = 75%

So now,

∴ Total amount of X in the two alloys = 6 kg + 7 kg + a kg = (13 + a) kg

∴ Total amount of Y in the two alloys = 5kg + 13 kg = 18 kg

According to Question the Ratio of X and Y required is 75 : 25 = 3:1

So,

$\frac{13+a}{18}=\frac{3}{1}$

⇒ 13 + a = 54

⇒ a = 41

Amount of compound A in 60 liters mixture = $\frac{5}{12}\times 60=25\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}$

Amount of compound B in 60 liters mixture = $\frac{7}{12}\times 60=35\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}$

Amount of mixture replaced = $\frac{60}{2}=30\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}$

Amount of compound A in 30 liters mixture = $\frac{5}{12}\times 30=12.5\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}$

Amount of compound B in 60 liters mixture = $\frac{7}{12}\times 30=17.5\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}$

∴ Final amount of compound B in the mixture = 35 − 17.5 = 17.5 litres

⇒ Presence of milk = 1/3 : 2/5 : 2/3 : 3/5

⇒ Quantity of milk from container 1 = 1/3 × 2 = 2/3 liter

⇒ Quantity of milk from container 2 = 2/5 × 3 = 6/5 liter

⇒ Quantity of milk from container 3 = 2/3 × 1 = 2/3 liter

⇒ Quantity of milk from container 4 = 3/5 × 4 = 12/5 liter

⇒ So, total milk in new container = 2/3 + 6/5 + 2/3 + 12/5 = 74/15 liter

⇒ Total water in new container = 10 – 74/15 = 76/15 liter

Let the volume of container be ‘Y’ Lit.

From given information,

5Y/6 – x = 2Y/3

⇒ Y/6 = x        ….(1)

And then 47 litres is added.

⇒ 5Y/6 – x + 47 = 6Y/7

⇒ (–Y/42) + 47 = x

⇒ 47 = [(1/42) + (1/6)]Y

⇒ 4Y/21 = 47

⇒ Y = 246.75 Lit

⇒ x = 246.75/6 = 41.125 Lit

30 litres of milk-water mixture contains,

⇒ Milk = 8/15 × 30 = 16 litres

⇒ Water = 7/15 × 30 = 14 litres

20 litres of milk-water mixture contains,

⇒ Milk = 7/10 × 20 = 14 litres

⇒ Water = 3/10 × 20 = 6 litres

Hence, new mixture contains,

⇒ Milk = 16 + 14 = 30 litres

⇒ Water = 14 + 6 = 20 litres