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Amount of first alloy = 24 kg

∴ Amount of A = 5/12 × 24 = 10 kg

⇒ Amount of B = (24 - 10) = 14 kg

Amount of second alloy = 7 kg

∴ Amount of A = 4 kg

⇒ Amount of B = 3 kg

Let the amount of METAL B mixed be X kg

∴ TOTAL amount of A = 10 + 4 = 14 kg

⇒ Total amount of B = 14 + 3 + x = (17 + x) kg

⇒ Total weight = A + B = 14 + 17 + x = (x + 31) kg

Percentage of A in mixture = 40%

∴ [14/(x + 31)] × 100 = 40

⇒ (14 × 100)/40 = x + 31

⇒ 35 = x + 31

⇒ x = 4 kg

From the data given in the problem, we can conclude the solution from the table below

⇒ Required concentration = (16.75/40) × 100%

⇒ 41.875%

Let the equal QUANTITIES of each alloy which are melted be a

Given,

Alloy A metals are in the ratio 7 : 3

∴ AMOUNT of gold = 7a/10

∴ Amount of copper = 3a/10

Given,

Alloy B metals are in the ratio 7 : 13

∴ Amount of gold = 7a/20

∴ Amount of copper = 13a/20

∴ Ratio of gold and copper in alloy C $(= \;\FRAC{{\frac{{7a}}{{10}}\; + \;\frac{{7a}}{{20}}}}{{\frac{{3a}}{{10}}\; + \;\frac{{13a}}{{20}}}})$ 

⇒ Ratio of gold and copper in alloy C = 21a/19a = 21 : 19

LET the weights of two alloys mixed in RATIO 2 : 3 be 2T and 3T, respectively.

Weight of Silver in first alloy = (2/(3 + 2 + 3)) × 2T = 0.5T

Weight of Silver in SECOND alloy = (1/(4 + 1 + 5)) × 3T = 0.3T

Amount of Silver in final mixture = 0.5T + 0.3T = 0.8T

⇒ In 5T KG of mixture, there is 0.8T of Silver.

Ratio of milk and water in these mixtures are 7 ? 13, 4 ? 1, 1 ? 3 and 11 ? 9

∴ Percentage of milk in these mixtures will be 35%, 80%, 25% and 55%

Quantity of water in the jar D is 22 LITRES less than the quantity of milk;

Since quantity of milk in jar D is 55% and water is 45%;

∴ 0.55x – 0.45x = 22

⇒ x = 220 litres

∴ Quantity of water in the jar D = 0.45 × 220 = 99 litres

The quantity of milk in jar B is 100 litres;

TOTAL quantity in jar B = 100/0.8 = 125 litres

∴ Quantity of water in jar B = 125 – 100 = 25 litres

If total quantity of water in jar B, C and D is 229 litres;

∴ Quantity of water in jar C = 229 – 99 – 25 = 105 litres

Total quantity in jar C = 105/0.75 = 140 litres

And quantity of milk in jar C = 140 – 105 = 35 litres

Let the total quantity in jar A = y litres

According to the question;

After mixing jar C with jar A, quantity of milk;

∴ 0.35y + 35 = (y + 140) × 0.28

⇒ 0.07y = 4.2

⇒ y = 60

∴ Total quantity in jar A = 60 litres

∴ Quantity of water in jar A = 60 × 0.65 = 39 litres

Let AMOUNT of mixture TAKEN be T kg.

Amount of sugar in T kg of mixture = 4T/9 kg

Amount of salt in T kg of mixture = 5T/9 kg

Now, 4.5 kg of salt is added so that ratio of sugar and salt becomes 2:3.

⇒ (4T/9)/(4.5 + (5T/9)) = 2/3

⇒ 12T = 81 + 10T

⇒ 2T = 81

⇒ T = 40.5 kg

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)

Percentage of milk in these mixtures will be 35%, 80%, 25% and 55%.

Percentage of milk in jar C and D are 25% and 55% and that of water are 75% and 45%;

Let TOTAL QUANTITIES in two jars be C and D;

According to the question;

0.75C – 0.55D = 21.6

0.25C + 0.45D = 189.6

Solving both the equations;

C = 240 litres and D = 288 litres

Let z litres water is mixed in jar D to get a new mixture containing 36% milk and 64% water;

∴ 288 × 0.45 + z = (288 + z) × 0.64

⇒ 0.36z = 288 × 0.19

⇒ 0.36z = 54.72

⇒ z = 152 litres

∴ 152 litres water should be mixed in jar D to get a new mixture containing 36% milk

Let the INITIAL quantity of jar B is p litres and amount of milk added in jar A is Q litres;

Then initial quantity in jar A = 2p/3 litres (33.33% LESS than jar B)

After adding 36 litres water in jar B, mixture X with 50% water is prepared;

∴ 0.2p + 36 = (p + 36) × 0.5

⇒ 0.3p = 18

⇒ p = 60 litres

∴ Initial quantity of jar B is 60 litres

Now when q litres milk is added to jar A, mixture Y with 52% water (48% milk) is prepared;

∴ 0.35 × 2p/3 + q = (q + 2p/3) × 0.48

⇒ 0.52q = 0.96p/3 – 0.7p/3

Putting p = 60 litres;

⇒ 0.52q = 5.2

⇒ q = 10

∴ Amount of milk added to jar A is 10 litres

∴ Required Ratio = 60 ? 10 = 6 ? 1

Amount of milk left in the container after four times extraction of 4 liters of milk

⇒ [40{1 – (4/40)}⁴],

⇒ [40{1 – 0.1}⁴],

⇒ [40 {0.9}⁴],

⇒ [40 × 0.6561],

⇒ 26.244 liters

⇒ Amount of WATER in the container = 40 – 26.244 = 13.756 litres

Now, 10 liters of the mixture is also added to make the quantity of mixture 50 liters

⇒ Quantity of milk to be added = 3/5 of 10 = 30/5 = 6 litres

⇒ Quantity of water to be added = 2/5 of 10 = 20/5 = 4 litres

⇒ TOTAL quantity of milk in mixture = 26.244 + 6 = 32.244 litres

⇒ Total quantity of water in mixture = 13.756 + 4 = 17.756 litres

Let us assume that the quantity of water is 1 liter

We need the quantity of milk as 1.5 liters to make the ratio of milk to water 3 ? 2

If 1 = 17.756 litres. Then, 1.5 = 17.756 × 1.5 = 26.634 litres

It means that we need 26.634 liters of milk in the mixture.

⇒ Excess quantity of milk = 32.244 – 26.634 = 5.61 litres

⇒ Percentage of excess quantity of milk = (5.61 × 100)/32.244 = 17.4%

Volume of three types of pulse are in proportion 5, 4 and 3

Weight of EQUAL volume are in proportion 6, 5 and 4

⇒ Ratio of weights of three types of pulse = (5 × 6) : (4 × 5) : (3 × 4)

⇒ 30 : 20 : 12 = 15 : 10 : 6

Let the weight of first type be 15x, second type be 10x and third type be 6x

⇒ 15x + 10x + 6x = 31x

⇒ x = 248/31 = 8