Alligation

Answer: Option A
Solution
To solve this, we need to determine the ratio in which tea costing Rs. 192 per kg is to be mixed with tea costing Rs. 150 per kg such that the mixture gives a 20% profit when sold at Rs. 194.40 per kg.

Step 1: Calculate the cost price of the mixed tea

The selling price of the mixed tea is Rs. 194.40 per kg, and a profit of 20% is being made. So, let’s first find the cost price of the mixed tea.

Let the cost price of the mixed tea be \( C \).

Since the profit is 20%, the selling price (Rs. 194.40) is 120% of the cost price.

\[
194.40 = 1.2 \times C
\]

Solve for \( C \):

\[
C = \frac{194.40}{1.2} = 162
\]

So, the cost price of the mixed tea is Rs. 162 per kg.

Step 2: Apply the allegation method

Now, we need to use the allegation method to find the ratio in which the two types of tea should be mixed.

We know:
– The cost price of the first tea is Rs. 192 per kg.
– The cost price of the second tea is Rs. 150 per kg.
– The cost price of the mixture is Rs. 162 per kg.

The allegation method involves finding the differences between the cost prices and the mean cost price of the mixture.

– Difference between Rs. 192 and Rs. 162 = \( 192 – 162 = 30 \)
– Difference between Rs. 150 and Rs. 162 = \( 162 – 150 = 12 \)

Now, the ratio of the two teas is the inverse of these differences.

So, the ratio of tea costing Rs. 192 to tea costing Rs. 150 is:

\[
\text{Ratio} = 12 : 30 = 2 : 5
\]

Final Answer:
The tea costing Rs. 192 per kg should be mixed with tea costing Rs. 150 per kg in the ratio 2:5.

Answer: Option A
Solution:
Let’s solve the problem step by step.

Step 1: Determine the amount of milk and water in the initial mixture
The total volume of the mixture is 729 litres, and the ratio of milk to water is 7:2. This means for every 9 parts of the mixture, 7 parts are milk and 2 parts are water.

– The total parts in the mixture = \( 7 + 2 = 9 \) parts.
– The amount of milk in the mixture = \( \frac{7}{9} \times 729 \)
– The amount of water in the mixture = \( \frac{2}{9} \times 729 \)

Now, let’s calculate these values:

\[
\text{Amount of milk} = \frac{7}{9} \times 729 = 567 \text{ litres}
\] \[
\text{Amount of water} = \frac{2}{9} \times 729 = 162 \text{ litres}
\]

So, initially, the mixture contains:
– 567 litres of milk
– 162 litres of water

Step 2: Find the amount of water to be added to achieve the new ratio
We need the new ratio of milk to water to be 7:3. The amount of milk will remain the same (567 litres), but we need to adjust the amount of water.

Let the amount of water to be added be \( x \). After adding this amount, the new amount of water will be \( 162 + x \), and the total amount of the new mixture will be \( 729 + x \).

The new ratio of milk to water is 7:3, so we can set up the equation:

\[
\frac{567}{162 + x} = \frac{7}{3}
\]

Step 3: Solve the equation
Cross-multiply to solve for \( x \):

\[
3 \times 567 = 7 \times (162 + x)
\] \[
1701 = 1134 + 7x
\] \[
1701 – 1134 = 7x
\] \[
567 = 7x
\] \[
x = \frac{567}{7} = 81
\]

Final Answer:
To achieve the new mixture with a ratio of 7:3, 81 litres of water need to be added.

Answer: Option B
Solution:
To solve this, we will use the **allegation method** to find the ratio in which two solutions of different alcohol strengths must be mixed to get a mixture with a desired alcohol strength.

Given:
– The first solution has 30% alcohol strength.
– The second solution has 50% alcohol strength.
– The desired mixture has 45% alcohol strength.

Step 1: Use the allegation method

Let’s calculate the differences between the given strengths and the desired strength:

– Difference between 50% (second solution) and 45% (desired) = \( 50 – 45 = 5 \)
– Difference between 45% (desired) and 30% (first solution) = \( 45 – 30 = 15 \)

Now, the ratio in which the two solutions must be mixed is the inverse of these differences:

\[
\text{Ratio} = 5 : 15
\]

Simplifying the ratio:

\[
\text{Ratio} = 1 : 3
\]

Final Answer:
The two solutions must be mixed in a 1:3 ratio (1 part of the 30% alcohol solution and 3 parts of the 50% alcohol solution) to get a mixture with 45% alcohol strength.

Answer: Option B
Solution:
To solve this, let’s break it down step by step:

Given:
– The cost price of pure milk is the same as the selling price (we’ll consider the cost price as the price per kg of milk).
– The goal is to gain 10% profit.
– The initial quantity of pure milk is 50 kg.
– We need to mix water with this milk.

Step 1: Determine the total cost and selling price of the mixture

Let the cost price of 1 kg of pure milk be **C** (in terms of Rs., for simplicity).

– The total cost of 50 kg of pure milk = \( 50 \times C \).

To make a profit of 10%, the selling price should be 10% higher than the cost price.

\[
\text{Selling price of the mixture} = \text{Total cost} + 10\% \text{ of total cost}
\]

\[
\text{Selling price of the mixture} = 50C + 0.1 \times 50C = 50C \times 1.1 = 55C
\]

Now, let the quantity of water to be mixed with the 50 kg of pure milk be **x** kg.

Step 2: Find the selling price of the mixture

When water is added, the total quantity of the mixture becomes \( 50 + x \) kg.

Since water is free, the total cost of the mixture remains the same (i.e., 50 kg of milk costs \( 50C \)).

The selling price of the total mixture (which is now \( 50 + x \) kg) needs to be Rs. 55C, as calculated earlier.

\[
\text{Selling price of the mixture} = \text{Selling price per kg} \times \text{Total weight of the mixture}
\]

\[
55C = \text{Selling price per kg} \times (50 + x)
\]

Since the selling price per kg is equal to the cost price per kg (because we are selling at the cost price of pure milk), the selling price per kg is \( C \). So:

\[
55C = C \times (50 + x)
\]

Step 3: Solve for \( x \)

Cancel out \( C \) from both sides:

\[
55 = 50 + x
\]

Now, solve for \( x \):

\[
x = 55 – 50 = 5
\]

Final Answer:
To gain a 10% profit, 5 kg of water must be mixed with the 50 kg of pure milk.

Answer: Option C
Solution:
Let’s break it down step by step.

Given:
– A container is initially full of pure milk.
– 20% of the pure milk is replaced by water in each operation.
– This process is repeated three times.
– We need to find the remaining quantity of pure milk after the third operation.

Step 1: The process of replacing 20% of the milk
Each time we replace 20% of the pure milk with water, the amount of pure milk in the container decreases. Initially, the container has 100% pure milk. After the first operation, 20% of the pure milk is removed, and 20% of the container’s capacity is filled with water. This means that 80% of the milk remains after the first operation.

Step 2: Generalize the process
After each operation, 80% of the milk remains. So after \( n \) operations, the remaining quantity of pure milk will be:

\[
\text{Remaining pure milk after } n \text{ operations} = (0.8)^n \times \text{Initial quantity of pure milk}
\]

Step 3: Apply this formula to three operations
Let the initial quantity of pure milk be 100%. After 3 operations, the remaining quantity of pure milk is:

\[
\text{Remaining pure milk after 3 operations} = (0.8)^3 \times 100
\]

\[
(0.8)^3 = 0.512
\]

So, the remaining quantity of pure milk is:

\[
0.512 \times 100 = 51.2\%
\]

Final Answer:
After the third operation, 51.2% of the initial quantity of pure milk remains in the container.

Answer: Option A
Solution:
To solve this, we will use the **allegation method**, which helps determine the ratio in which two or more quantities (in this case, types of wheat with different costs) should be mixed to achieve a desired average.

Given:
– The price of wheat types: Rs. 1.27, Rs. 1.29, and Rs. 1.32 per kg.
– The desired selling price (average price) is Rs. 1.30 per kg.

Step 1: Set up the differences
Let the prices of the wheat types be \( P_1 = 1.27 \), \( P_2 = 1.29 \), and \( P_3 = 1.32 \), and the desired price is Rs. 1.30.

We will now calculate the differences between the given prices and the desired price:

– Difference between Rs. 1.27 and Rs. 1.30:
\( 1.30 – 1.27 = 0.03 \)

– Difference between Rs. 1.29 and Rs. 1.30:
\( 1.30 – 1.29 = 0.01 \)

– Difference between Rs. 1.32 and Rs. 1.30:
\( 1.32 – 1.30 = 0.02 \)

Step 2: Set up the ratio using these differences
The ratios of the quantities of each wheat type will be inversely proportional to these differences:

\[
\text{Ratio} = 0.01 : 0.02 : 0.03
\]

Step 3: Simplify the ratio
To simplify, multiply all parts by 100 to remove the decimals:

\[
\text{Ratio} = 1 : 2 : 3
\]

Final Answer:
The three types of wheat should be mixed in the ratio 1:2:3 to achieve the desired average price of Rs. 1.30 per kg.

Answer: Option C
Solution:
Let’s assume:
– The number of deers = \( x \)
– The number of ducks = \( y \)

Step 1: Set up the equations
From the problem, we know:
– The total number of heads is 180. Since both deers and ducks have one head each, the equation for heads is:
\[
x + y = 180
\] – The total number of legs is 448. Deers have 4 legs, and ducks have 2 legs, so the equation for legs is:
\[
4x + 2y = 448
\]

Step 2: Solve the system of equations
We have two equations:
1. \( x + y = 180 \)
2. \( 4x + 2y = 448 \)

Step 2.1: Solve for \( y \) in terms of \( x \) from the first equation
From \( x + y = 180 \), we can solve for \( y \):
\[
y = 180 – x
\]

Step 2.2: Substitute \( y = 180 – x \) into the second equation
Substitute \( y = 180 – x \) into the second equation \( 4x + 2y = 448 \):
\[
4x + 2(180 – x) = 448
\] Simplify the equation:
\[
4x + 360 – 2x = 448
\] \[
2x + 360 = 448
\] \[
2x = 448 – 360
\] \[
2x = 88
\] \[
x = \frac{88}{2} = 44
\]

Final Answer:
The number of deers in the zoo is 44.

Answer: Option A
Solution:
Let’s assume:
– The amount invested at 6% interest is Rs. 680.
– The amount invested at 10% interest is \( x \) (the unknown value we need to find).

Step 1: Set up the equation for average interest
The total interest earned from both investments is the sum of the interest from each investment.

1. Interest from the Rs. 680 invested at 6%:
\[
\text{Interest from 680} = 680 \times \frac{6}{100} = 680 \times 0.06 = 40.8
\]

2. Interest from the amount \( x \) invested at 10%:
\[
\text{Interest from } x = x \times \frac{10}{100} = x \times 0.10 = 0.1x
\]

The total interest is the sum of both interests:
\[
\text{Total interest} = 40.8 + 0.1x
\]

Step 2: Set up the equation for average interest
The total amount invested is \( 680 + x \), and the average interest rate is 7.5%. The total interest for the year is also equal to 7.5% of the total investment:
\[
\text{Total interest} = (680 + x) \times \frac{7.5}{100} = (680 + x) \times 0.075
\]

Step 3: Set up the equation and solve
We know the total interest is \( 40.8 + 0.1x \), so we can equate the two expressions for total interest:
\[
40.8 + 0.1x = (680 + x) \times 0.075
\] Now expand the right-hand side:
\[
40.8 + 0.1x = 0.075 \times 680 + 0.075 \times x
\] \[
40.8 + 0.1x = 51 + 0.075x
\] Now, subtract \( 0.075x \) from both sides:
\[
40.8 + 0.1x – 0.075x = 51
\] \[
40.8 + 0.025x = 51
\] Now, subtract 40.8 from both sides:
\[
0.025x = 51 – 40.8
\] \[
0.025x = 10.2
\] Solve for \( x \):
\[
x = \frac{10.2}{0.025} = 408
\]

Final Answer:
The amount invested at 10% interest is Rs. 408.

Answer: Option A
Solution:
Let’s assume:
– \( x \) kg of pulses are sold at 14% profit.
– \( 50 – x \) kg of pulses are sold at 6% loss.

Step 1: Calculate the total cost price
Since the total quantity of pulses is 50 kg, and let’s assume the cost price of 1 kg of pulses is \( C \), the total cost price is:
\[
\text{Total cost price} = 50 \times C = 50C
\]

Step 2: Calculate the total selling price
– The selling price of \( x \) kg of pulses sold at 14% profit is:
\[
\text{Selling price of } x \text{ kg} = x \times C \times \left( 1 + \frac{14}{100} \right) = x \times C \times 1.14
\] – The selling price of \( 50 – x \) kg of pulses sold at 6% loss is:
\[
\text{Selling price of } (50 – x) \text{ kg} = (50 – x) \times C \times \left( 1 – \frac{6}{100} \right) = (50 – x) \times C \times 0.94
\]

The total selling price is the sum of these two amounts:
\[
\text{Total selling price} = x \times C \times 1.14 + (50 – x) \times C \times 0.94
\]

Step 3: Set up the equation for the total loss
We are told that the overall loss is 4%. The total selling price should therefore be 96% of the total cost price (since a 4% loss means 96% of the cost price is recovered):
\[
\text{Total selling price} = 50C \times \left( 1 – \frac{4}{100} \right) = 50C \times 0.96
\]

Step 4: Set up the equation and solve
Now we can set the total selling price equal to \( 50C \times 0.96 \):
\[
x \times C \times 1.14 + (50 – x) \times C \times 0.94 = 50C \times 0.96
\] We can cancel \( C \) from both sides:
\[
x \times 1.14 + (50 – x) \times 0.94 = 50 \times 0.96
\] Simplify the equation:
\[
x \times 1.14 + 50 \times 0.94 – x \times 0.94 = 50 \times 0.96
\] \[
x \times (1.14 – 0.94) + 50 \times 0.94 = 50 \times 0.96
\] \[
x \times 0.20 + 47 = 48
\] Now, subtract 47 from both sides:
\[
x \times 0.20 = 1
\] Solve for \( x \):
\[
x = \frac{1}{0.20} = 5
\]

Final Answer:
The trader sells 5 kg of pulses at 14% profit and 45 kg (i.e., \( 50 – 5 \)) at 6% loss.

Answer: Option A
Solution:
Let’s assume:
– Ram’s weight is \( 4x \) kg.
– Shyam’s weight is \( 5x \) kg, based on the given ratio of their weights.

Step 1: Increase in Ram’s weight
Ram’s weight is increased by 10%. So, the new weight of Ram becomes:
\[
\text{New weight of Ram} = 4x \times (1 + 0.10) = 4x \times 1.10 = 4.4x
\]

Step 2: Total weight after increase
The total weight of Ram and Shyam after the increase is 82.8 kg. The total weight is the sum of the new weight of Ram and the increased weight of Shyam:
\[
\text{New total weight} = 4.4x + 5y
\] Where \( y \) is Shyam’s new weight, and it is increased by some percentage which we need to find.

We are also told that the total weight is increased by 15%. So the new total weight is 15% more than the original total weight:
\[
\text{Original total weight} = 4x + 5x = 9x
\] \[
\text{New total weight} = 9x \times (1 + 0.15) = 9x \times 1.15 = 10.35x
\]

We know the total weight after the increase is 82.8 kg, so:
\[
10.35x = 82.8
\] Solving for \( x \):
\[
x = \frac{82.8}{10.35} = 8
\]

Step 3: Find the new weight of Shyam
The original weight of Shyam is \( 5x = 5 \times 8 = 40 \) kg.

The new total weight of Shyam after the increase is \( 5y \), so:
\[
5y = 82.8 – 4.4x = 82.8 – 4.4 \times 8 = 82.8 – 35.2 = 47.6
\] Thus, the new weight of Shyam is:
\[
y = \frac{47.6}{5} = 9.52
\]

Step 4: Calculate the percentage increase in Shyam’s weight
The increase in Shyam’s weight is:
\[
\text{Increase in weight of Shyam} = 9.52 – 40 = 7.52
\] The percentage increase in Shyam’s weight is:
\[
\text{Percentage increase} = \frac{7.52}{40} \times 100 = 18.8\%
\]

Final Answer:
The weight of Shyam had to be increased by 18.8%.

Answer: Option A
Solution:
Let’s denote the original volume of the solution as \( V \).

The initial ratio of milk to water is 3:2, which means:
– Milk = \( \frac{3}{5}V \)
– Water = \( \frac{2}{5}V \)

Step 1: Increase the volume by 50%
The total volume of the solution is increased by 50%, so the new volume is:
\[
\text{New volume} = V + 0.50V = 1.5V
\] Since water is added to increase the volume, the amount of milk remains the same, and the water content increases. The amount of milk is still \( \frac{3}{5}V \), but the amount of water increases. The total amount of water becomes:
\[
\text{New amount of water} = \frac{2}{5}V + 0.50V = \frac{2}{5}V + \frac{5}{10}V = \frac{9}{10}V
\] So, after the volume is increased, we have:
– Milk = \( \frac{3}{5}V \)
– Water = \( \frac{9}{10}V \)

Step 2: Withdraw 30 L and replace with water
After withdrawing 30 L from the new solution and replacing it with water, the amount of milk and water in the solution changes. The milk-to-water ratio in the withdrawn 30 L is the same as the ratio in the new solution, i.e., the ratio of milk to water is \( \frac{3}{5} : \frac{9}{10} \). Simplifying this ratio:
\[
\frac{3}{5} : \frac{9}{10} = 3 \times 2 : 9 = 6 : 9 = 2 : 3
\] Thus, 30 L of the solution contains milk and water in the ratio of 2:3.

The amount of milk withdrawn is:
\[
\frac{2}{5} \times 30 = 12 \text{ L}
\] The amount of water withdrawn is:
\[
\frac{3}{5} \times 30 = 18 \text{ L}
\]

After withdrawing 30 L and replacing it with water, the amount of milk and water in the solution is:
– Milk = \( \frac{3}{5}V – 12 \)
– Water = \( \frac{9}{10}V – 18 + 30 \) (since 30 L of water is added)

So, the new amounts are:
– Milk = \( \frac{3}{5}V – 12 \)
– Water = \( \frac{9}{10}V + 12 \)

Step 3: Final ratio of milk to water
We are given that the final ratio of milk to water is 3:7. Therefore, we have the equation:
\[
\frac{\frac{3}{5}V – 12}{\frac{9}{10}V + 12} = \frac{3}{7}
\]

Step 4: Solve the equation
Cross-multiply to solve for \( V \):
\[
7 \left( \frac{3}{5}V – 12 \right) = 3 \left( \frac{9}{10}V + 12 \right)
\] Expand both sides:
\[
7 \times \frac{3}{5}V – 7 \times 12 = 3 \times \frac{9}{10}V + 3 \times 12
\] \[
\frac{21}{5}V – 84 = \frac{27}{10}V + 36
\] Multiply the entire equation by 10 to eliminate the denominators:
\[
42V – 840 = 27V + 360
\] Now, solve for \( V \):
\[
42V – 27V = 840 + 360
\] \[
15V = 1200
\] \[
V = \frac{1200}{15} = 80
\]

Final Answer:
The original volume of the solution is 80 L.

Answer: Option B
Solution:
Let the price per kilogram of the second quality of wheat be \( x \) (in Rs).

Step 1: Set up the equation based on the weighted average price

The two quantities of wheat are mixed in the ratio 8:7, and the mixture is worth Rs 10 per kg. To find the value of \( x \), we can use the concept of the weighted average price. The weighted average price of the mixture is given by the formula:
\[
\text{Weighted average price} = \frac{(8 \times 9.30) + (7 \times x)}{8 + 7}
\] The total weight of the mixture is \( 8 + 7 = 15 \) kg, and the total cost of the mixture is the sum of the costs of the two types of wheat. We are given that the weighted average price of the mixture is Rs 10 per kg, so:
\[
\frac{(8 \times 9.30) + (7 \times x)}{15} = 10
\]

Step 2: Solve for \( x \)

First, simplify the equation:
\[
\frac{(8 \times 9.30) + (7 \times x)}{15} = 10
\] \[
\frac{74.4 + 7x}{15} = 10
\] Multiply both sides by 15:
\[
74.4 + 7x = 150
\] Now, solve for \( x \):
\[
7x = 150 – 74.4
\] \[
7x = 75.6
\] \[
x = \frac{75.6}{7} = 10.80
\]

Final Answer:
The rate per kg of the second quality of wheat is Rs 10.80.

Answer: Option C
Solution:
We can solve this problem using a system of equations. Let’s define the following:

– Let \( r \) be the number of rabbits.
– Let \( p \) be the number of pigeons.

We know the following:

1. Each rabbit has 1 head and 4 legs.
2. Each pigeon has 1 head and 2 legs.

From the information given:

– The total number of heads is 200:
\[
r + p = 200
\]

– The total number of legs is 580:
\[
4r + 2p = 580
\]

Now, we solve this system of equations.

Step 1: Solve for \( r \) in terms of \( p \) from the first equation.
From \( r + p = 200 \), we can express \( r \) as:
\[
r = 200 – p
\]

Step 2: Substitute \( r = 200 – p \) into the second equation.
Substitute this into \( 4r + 2p = 580 \):
\[
4(200 – p) + 2p = 580
\] Simplifying:
\[
800 – 4p + 2p = 580
\] \[
800 – 2p = 580
\] \[
-2p = 580 – 800
\] \[
-2p = -220
\] \[
p = \frac{-220}{-2} = 110
\]

So, there are 110 pigeons. To find the number of rabbits, we substitute \( p = 110 \) into \( r + p = 200 \):
\[
r + 110 = 200
\] \[
r = 200 – 110 = 90
\]

Thus, there are 90 rabbits and 110 pigeons.

Answer: Option C
Solution:
Let’s solve this step by step.

Step 1: Define the total volume of the mixture
Let the total volume of the liquid in the vessel be \( V \).

– Initially, the mixture has 3 parts water and 5 parts syrup.
– The total number of parts is \( 3 + 5 = 8 \).
– Therefore, the amount of water in the mixture is:
\[
\frac{3}{8} \times V
\] – The amount of syrup in the mixture is:
\[
\frac{5}{8} \times V
\]

Step 2: How much mixture should be drawn off?
Let’s say we draw off \( x \) units of the mixture.

– When we draw off \( x \) units of the mixture, we are removing both water and syrup in the same proportion that they exist in the mixture.
– The amount of water removed is \( \frac{3}{8} \times x \).
– The amount of syrup removed is \( \frac{5}{8} \times x \).

Step 3: Replace the drawn-off mixture with water
After drawing off \( x \) units, we replace it with \( x \) units of pure water.

– After replacing, the new amount of water in the vessel becomes:
\[
\frac{3}{8} \times V – \frac{3}{8} \times x + x
\] This is the amount of water that remains plus the \( x \) units of water we added.

– The amount of syrup in the vessel remains the same as \( \frac{5}{8} \times V – \frac{5}{8} \times x \).

Step 4: Set up the equation for half water, half syrup
Now, we want the mixture to be half water and half syrup, meaning each should make up \( \frac{V}{2} \).

– The amount of water after replacement should be \( \frac{V}{2} \):
\[
\frac{3}{8} \times V – \frac{3}{8} \times x + x = \frac{V}{2}
\]

Step 5: Solve for \( x \)
Let’s simplify the equation:
\[
\frac{3}{8} \times V – \frac{3}{8} \times x + x = \frac{V}{2}
\] Multiply through by 8 to eliminate the fractions:
\[
3V – 3x + 8x = 4V
\] Simplifying:
\[
3V + 5x = 4V
\] \[
5x = 4V – 3V
\] \[
5x = V
\] \[
x = \frac{V}{5}
\]

Conclusion:
The amount of the mixture that must be drawn off and replaced with water is \( \frac{V}{5} \).

Answer: Option C
Solution:
We are given the following information:

– The first variety of tea costs Rs. 126 per kg.
– The second variety of tea costs Rs. 135 per kg.
– The third variety of tea costs \( x \) per kg (this is what we need to find).
– The ratio of mixing these teas is 1 : 1 : 2.
– The cost of the mixture is Rs. 153 per kg.

Let the total quantity of the mixture be \( 1 + 1 + 2 = 4 \) parts. The total cost of the mixture will be the weighted average of the costs of the three varieties, based on their respective proportions.

Step 1: Set up the equation for the cost of the mixture
The cost of the mixture is the sum of the individual costs of the three varieties, weighted by their proportions:

\[
\frac{1 \times 126 + 1 \times 135 + 2 \times x}{4} = 153
\]

Step 2: Simplify the equation
\[
\frac{126 + 135 + 2x}{4} = 153
\] \[
\frac{261 + 2x}{4} = 153
\] Multiply both sides by 4 to eliminate the denominator:
\[
261 + 2x = 612
\] Subtract 261 from both sides:
\[
2x = 612 – 261
\] \[
2x = 351
\] Divide both sides by 2:
\[
x = \frac{351}{2} = 175.5
\]

Conclusion:
The price of the third variety of tea is Rs. 175.50 per kg.

Answer: Option C
Solution:
We can solve this step by step using algebra.

Step 1: Define variables
Let the total volume of the mixture in the can be \( V \) litres.

– Initially, the ratio of liquid A to liquid B is 7:5. So, the initial amount of liquid A in the can is:
\[
A_{\text{initial}} = \frac{7}{12} \times V
\] And the initial amount of liquid B in the can is:
\[
B_{\text{initial}} = \frac{5}{12} \times V
\]

Step 2: After drawing off 9 litres and refilling with liquid B
When 9 litres of the mixture are drawn off, the amount of liquid A and liquid B removed will be in the same proportion as their initial amounts.

– The amount of liquid A removed is:
\[
A_{\text{removed}} = \frac{7}{12} \times 9 = \frac{63}{12} = 5.25 \text{ litres}
\] – The amount of liquid B removed is:
\[
B_{\text{removed}} = \frac{5}{12} \times 9 = \frac{45}{12} = 3.75 \text{ litres}
\]

After 9 litres are removed, the amounts of A and B left in the can are:

– Remaining liquid A:
\[
A_{\text{remaining}} = A_{\text{initial}} – A_{\text{removed}} = \frac{7}{12} \times V – 5.25
\] – Remaining liquid B:
\[
B_{\text{remaining}} = B_{\text{initial}} – B_{\text{removed}} = \frac{5}{12} \times V – 3.75
\]

Since the can is filled with 9 litres of liquid B after drawing off the mixture, the new amount of liquid B is:
\[
B_{\text{final}} = B_{\text{remaining}} + 9 = \frac{5}{12} \times V – 3.75 + 9 = \frac{5}{12} \times V + 5.25
\]

Step 3: Use the new ratio
After refilling, the ratio of liquid A to liquid B becomes 7:9. Therefore, we can set up the equation:
\[
\frac{A_{\text{remaining}}}{B_{\text{final}}} = \frac{7}{9}
\] Substitute the expressions for \( A_{\text{remaining}} \) and \( B_{\text{final}} \):
\[
\frac{\frac{7}{12} \times V – 5.25}{\frac{5}{12} \times V + 5.25} = \frac{7}{9}
\]

Step 4: Solve the equation
Now, we solve for \( V \):

Cross-multiply to eliminate the fraction:
\[
9 \times \left( \frac{7}{12} \times V – 5.25 \right) = 7 \times \left( \frac{5}{12} \times V + 5.25 \right)
\]

Distribute on both sides:
\[
9 \times \frac{7}{12} \times V – 9 \times 5.25 = 7 \times \frac{5}{12} \times V + 7 \times 5.25
\]

Simplifying the constants:
\[
\frac{63}{12} \times V – 47.25 = \frac{35}{12} \times V + 36.75
\]

Now, move the terms involving \( V \) to one side and the constant terms to the other side:
\[
\frac{63}{12} \times V – \frac{35}{12} \times V = 36.75 + 47.25
\] \[
\frac{28}{12} \times V = 84
\] Simplify:
\[
\frac{7}{3} \times V = 84
\] Multiply both sides by 3:
\[
7 \times V = 252
\] Finally, divide by 7:
\[
V = 36
\]

Step 5: Find the initial amount of liquid A
Now that we know the total volume of the mixture is 36 litres, the initial amount of liquid A is:
\[
A_{\text{initial}} = \frac{7}{12} \times 36 = 21 \text{ litres}
\]

Conclusion:
The initial amount of liquid A in the can was 21 litres.

Answer: Option B
Solution:
We need to find how much milk should be mixed from each of the two cans to get 12 litres of milk such that the ratio of water to milk is 3:5.

Step 1: Define the variables
Let the amount of milk taken from the first can be \( x \) litres, and the amount of milk taken from the second can be \( y \) litres. According to the problem:

– The total amount of milk should be 12 litres:
\[
x + y = 12
\] – The ratio of water to milk in the final mixture should be 3:5, which means the total amount of water should be \( \frac{3}{8} \times 12 = 4.5 \) litres, and the total amount of milk should be \( \frac{5}{8} \times 12 = 7.5 \) litres.

Step 2: Water and milk content in the cans
Now, let’s calculate the amount of water and milk in each can:

First can:
– The first can contains 25% water and 75% milk.
– The amount of water in \( x \) litres from the first can is:
\[
\text{Water from first can} = \frac{25}{100} \times x = 0.25x
\] – The amount of milk in \( x \) litres from the first can is:
\[
\text{Milk from first can} = \frac{75}{100} \times x = 0.75x
\]

Second can:
– The second can contains 50% water and 50% milk.
– The amount of water in \( y \) litres from the second can is:
\[
\text{Water from second can} = \frac{50}{100} \times y = 0.5y
\] – The amount of milk in \( y \) litres from the second can is:
\[
\text{Milk from second can} = \frac{50}{100} \times y = 0.5y
\]

Step 3: Set up the equations for water and milk
We want the total amount of water and the total amount of milk to satisfy the following:

– Total amount of water: \( 0.25x + 0.5y \)
– Total amount of milk: \( 0.75x + 0.5y \)

We know the total amount of water should be 4.5 litres, and the total amount of milk should be 7.5 litres. So, we have the system of equations:

\[
0.25x + 0.5y = 4.5 \quad \text{(Equation 1)}
\] \[
0.75x + 0.5y = 7.5 \quad \text{(Equation 2)}
\]

Step 4: Solve the system of equations
First, subtract Equation 1 from Equation 2 to eliminate \( y \):

\[
(0.75x + 0.5y) – (0.25x + 0.5y) = 7.5 – 4.5
\] \[
0.75x – 0.25x = 3
\] \[
0.5x = 3
\] \[
x = \frac{3}{0.5} = 6
\]

So, \( x = 6 \) litres. Now, substitute \( x = 6 \) into Equation 1 to find \( y \):

\[
0.25(6) + 0.5y = 4.5
\] \[
1.5 + 0.5y = 4.5
\] \[
0.5y = 4.5 – 1.5 = 3
\] \[
y = \frac{3}{0.5} = 6
\]

Conclusion:
The milk vendor should mix 6 litres from the first can and 6 litres from the second can to get 12 litres of mixture with a water to milk ratio of 3:5.

Answer: Option C
Solution:
To find the ratio in which the grocer should mix the two varieties of pulses, we can use the **algebraic method of allegation.

Step 1: Define the given costs
– Cost of the first variety of pulses = Rs. 15 per kg
– Cost of the second variety of pulses = Rs. 20 per kg
– Cost of the mixture = Rs. 16.50 per kg

Step 2: Set up the allegation
The allegation formula helps us to find the required ratio of the two quantities to be mixed. The formula is based on the difference in prices.

Let the required ratio of the two varieties be \( x : y \).

– The difference between the cost of the first variety and the cost of the mixture is:
\[
20 – 16.50 = 3.50
\] – The difference between the cost of the second variety and the cost of the mixture is:
\[
16.50 – 15 = 1.50
\]

Thus, the required ratio of the two varieties is \( 3.5 : 1.5 \).

Step 3: Simplify the ratio
To simplify, we can divide both numbers by 0.5:
\[
\frac{3.5}{0.5} : \frac{1.5}{0.5} = 7 : 3
\]

Conclusion:
The grocer must mix the two varieties of pulses in the ratio 7 : 3 to obtain a mixture worth Rs. 16.50 per kg.

Answer: Option C
Solution:
To solve this problem, let’s assume:

– The cost price of the milk per litre is \( C \).
– The milkman sells the milk at the same cost price \( C \), but he mixes it with water.

Step 1: Let the quantity of milk the milkman originally has be \( x \) litres.
– The cost of \( x \) litres of milk is \( C \times x \).

Step 2: The milkman mixes the milk with water and gains 25%.
– The selling price of the mixture is the same as the cost price of milk, so he sells \( x \) litres of milk at the price of \( C \times x \).
– If he gains 25%, then his profit is 25% of the cost price of the mixture.
– Therefore, the total selling price of the mixture is \( C \times x + 0.25 \times (C \times x) = 1.25 \times (C \times x) \).

Step 3: The total quantity of the mixture is now more than \( x \) litres because of the added water. Let the total volume of the mixture be \( y \) litres.
– Since the milkman sells the mixture at the same price, the amount of milk in the mixture is \( x \) litres, and the rest is water.
– The price at which the milkman sells the mixture is \( C \times y \), and this is equal to the selling price, which is \( 1.25 \times C \times x \).

Step 4: Set up the equation
We know that the total selling price of the mixture is \( 1.25 \times C \times x \), and the cost of \( y \) litres of mixture is \( C \times y \), so:
\[
C \times y = 1.25 \times C \times x
\] Cancel out \( C \) from both sides:
\[
y = 1.25 \times x
\]

Step 5: Find the percentage of water in the mixture
The amount of water in the mixture is:
\[
\text{Amount of water} = y – x = 1.25 \times x – x = 0.25 \times x
\] So, the percentage of water in the mixture is:
\[
\frac{0.25 \times x}{1.25 \times x} \times 100 = \frac{0.25}{1.25} \times 100 = 20\%
\]

Conclusion:
The percentage of water in the mixture is 20%.

Answer: Option D
Solution:
Let’s break down the process step by step to determine how much milk remains in the container after the process is repeated three times.

Initial Setup:
– The container initially contains 40 litres of milk.
– Each time 4 litres of milk is taken out and replaced with water.

Step 1: After the first replacement
– The amount of milk taken out is 4 litres.
– The amount of milk remaining after the first replacement is:
\[
40 – 4 = 36 \text{ litres of milk}
\] – The total volume of the container remains 40 litres, but now 4 litres of water has been added. So, the proportion of milk left in the container after the first replacement is:
\[
\frac{36}{40} = 0.9 \quad \text{(or 90% of the original milk)}
\]

Step 2: After the second replacement
– 4 litres of the mixture (which now contains both milk and water) is taken out. Since the remaining milk is 90% of the original amount, 90% of the 4 litres taken out will be milk.
– The amount of milk taken out is:
\[
0.9 \times 4 = 3.6 \text{ litres of milk}
\] – After the second replacement, the amount of milk remaining in the container is:
\[
36 – 3.6 = 32.4 \text{ litres of milk}
\] – The total volume remains 40 litres, so the proportion of milk left in the container after the second replacement is:
\[
\frac{32.4}{40} = 0.81 \quad \text{(or 81% of the original milk)}
\]

Step 3: After the third replacement
– Again, 4 litres of the mixture is taken out. The proportion of milk in the container after the second replacement is 81%, so 81% of the 4 litres taken out will be milk.
– The amount of milk taken out is:
\[
0.81 \times 4 = 3.24 \text{ litres of milk}
\] – After the third replacement, the amount of milk remaining in the container is:
\[
32.4 – 3.24 = 29.16 \text{ litres of milk}
\]

Conclusion:
After the process is repeated three times, 29.16 litres of milk remains in the container.

Answer: Option B
Solution:
Let’s break down the problem and solve it step by step.

Step 1: Define the variables
– Let the total volume of whisky in the jar be \( V \) litres.
– Let the amount of whisky replaced be \( x \) litres.

Step 2: Initial amount of alcohol
Initially, the whisky contains 40% alcohol. Therefore, the amount of alcohol in the jar before any replacement is:
\[
\text{Alcohol initially} = 0.40 \times V
\]

Step 3: After removing \( x \) litres of whisky
When \( x \) litres of whisky is removed, the amount of alcohol removed is:
\[
\text{Alcohol removed} = 0.40 \times x
\] Thus, the remaining alcohol in the jar after removal is:
\[
\text{Remaining alcohol} = 0.40 \times V – 0.40 \times x
\]

Step 4: Adding the new whisky
Next, \( x \) litres of a new whisky containing 19% alcohol is added. The amount of alcohol in the new whisky added is:
\[
\text{Alcohol added} = 0.19 \times x
\]

Step 5: Total amount of alcohol after replacement
After the replacement, the total amount of alcohol in the jar becomes:
\[
\text{Total alcohol} = \left( 0.40 \times V – 0.40 \times x \right) + 0.19 \times x
\] This simplifies to:
\[
\text{Total alcohol} = 0.40 \times V – 0.40 \times x + 0.19 \times x
\] \[
\text{Total alcohol} = 0.40 \times V – 0.21 \times x
\]

Step 6: New percentage of alcohol
After the replacement, the new percentage of alcohol in the jar is given as 26%. The total amount of alcohol after replacement is 26% of the total volume of the jar, which is \( V \). Therefore, we have the equation:
\[
\text{Total alcohol} = 0.26 \times V
\]

Step 7: Set up the equation
Now, equate the two expressions for the total amount of alcohol:
\[
0.40 \times V – 0.21 \times x = 0.26 \times V
\]

Step 8: Solve for \( x \)
Rearrange the equation to isolate \( x \):
\[
0.40 \times V – 0.26 \times V = 0.21 \times x
\] \[
0.14 \times V = 0.21 \times x
\] Now, solve for \( x \):
\[
x = \frac{0.14 \times V}{0.21} = \frac{2}{3} \times V
\]

Conclusion:
The quantity of whisky replaced is \( \frac{2}{3} \) of the total volume of the jar, or two-thirds of the jar’s volume.

Answer: Option A
Solution:
To solve this, we can apply the concept of **allegation**. The idea is to mix water (which costs nothing) with milk (which has a certain cost) to gain a profit when selling the mixture at the cost price of the milk.

We are given that the milkman gains a profit of \( 16 \frac{2}{3} \) or \( \frac{50}{3} \)% on selling the mixture at the cost price of the milk.

Step 1: Convert the profit percentage to a decimal
The percentage profit is \( 16 \frac{2}{3} \)%, which can be written as:
\[
16 \frac{2}{3} \% = \frac{50}{3} \% = \frac{50}{3} \times \frac{1}{100} = \frac{1}{6}
\] Thus, the milkman gains \( \frac{1}{6} \) profit on the cost price.

Step 2: Use allegation to determine the ratio
– The cost price of the milk is \( C \) (per litre), and water costs nothing.
– Since the milkman gains \( \frac{1}{6} \) profit by selling at cost price, the total cost of the mixture must be less than the selling price.

In the mixture:
– The cost of milk is \( C \), and the cost of water is \( 0 \).

To gain a profit of \( \frac{1}{6} \), the amount of milk in the mixture should cost \( \frac{5}{6} \) of the cost price of the milk (since \( \frac{5}{6} \) is the part that is not profit).

Using the allegation rule, we want the amount of milk to correspond to a cost of \( \frac{5}{6} \) of the total cost, and the amount of water to correspond to a difference of \( \frac{1}{6} \).

So, we set up the following:

– The cost of milk = \( C \)
– The cost of water = \( 0 \)
– The desired cost ratio of milk to water = \( \frac{5}{6} : \frac{1}{6} \)

This simplifies to:
\[
\text{Ratio of milk to water} = 5 : 1
\]

Conclusion:
The ratio in which water must be mixed with milk is 5 : 1.

Answer: Option B
Solution:
To solve this, we will use the **allegation method**.

Step 1: Define the variables
– The cost of the first variety of rice = Rs. 7.20 per kg
– The cost of the second variety of rice = Rs. 5.70 per kg
– The desired cost of the mixture = Rs. 6.30 per kg

Step 2: Set up the allegation
The allegation method helps us determine the ratio in which the two varieties should be mixed. The formula involves finding the differences between the cost of each variety and the desired cost of the mixture.

– The difference between the cost of the first variety and the cost of the mixture:
\[
7.20 – 6.30 = 0.90
\]

– The difference between the cost of the second variety and the cost of the mixture:
\[
6.30 – 5.70 = 0.60
\]

Step 3: Find the ratio
Now, the ratio of the two rice varieties to be mixed is the inverse of these differences:
\[
\text{Ratio} = 0.60 : 0.90
\]

To simplify, divide both sides by 0.30:
\[
\frac{0.60}{0.30} : \frac{0.90}{0.30} = 2 : 3
\]

Conclusion:
The ratio in which the two varieties of rice should be mixed is 2 : 3.

Answer: Option A
Solution:
Let’s solve this step-by-step using the **allegation method**.

Step 1: Define the variables
– Cost of the first variety of tea = Rs. 60 per kg
– Cost of the second variety of tea = Rs. 65 per kg
– Selling price of the mixture = Rs. 68.20 per kg
– Desired gain = 10%

Step 2: Calculate the cost price of the mixture
The grocer wants to gain 10% profit by selling the mixture at Rs. 68.20 per kg. To find the cost price of the mixture, use the formula for selling price with profit:
\[
\text{Selling Price} = \text{Cost Price} + 10\% \text{ of Cost Price}
\] Let the cost price of the mixture be \( C \). Since the grocer sells at Rs. 68.20, the equation is:
\[
68.20 = C + 0.10C = 1.10C
\] Solve for \( C \):
\[
C = \frac{68.20}{1.10} = 62 \text{ per kg}
\]

So, the cost price of the mixture is Rs. 62 per kg.

Step 3: Use the allegation method
Now, we apply the allegation method to determine the ratio in which the two varieties of tea should be mixed to obtain a mixture costing Rs. 62 per kg.

– The first variety costs Rs. 60 per kg.
– The second variety costs Rs. 65 per kg.
– The desired cost price of the mixture is Rs. 62 per kg.

Allegation calculation:
– The difference between the cost of the second variety and the desired cost price:
\[
65 – 62 = 3
\]

– The difference between the cost of the first variety and the desired cost price:
\[
62 – 60 = 2
\]

Thus, the ratio of the two varieties of tea is:
\[
\text{Ratio} = 2 : 3
\]

Conclusion:
The grocer should mix the two varieties of tea in the ratio 2 : 3 to achieve the desired cost price of Rs. 62 per kg and gain a 10% profit.

Answer: Option A
Solution:
To find the price per kg of the mixed variety of rice, we can use the **weighted average formula** based on the given ratio of mixing.

Step 1: Define the given variables
– Cost of Type 1 rice = Rs. 15 per kg
– Cost of Type 2 rice = Rs. 20 per kg
– The ratio of mixing Type 1 to Type 2 rice is 2 : 3.

Step 2: Apply the weighted average formula
The price per kg of the mixture can be calculated using the following formula for a weighted average:
\[
\text{Price of mixture} = \frac{(C_1 \times w_1) + (C_2 \times w_2)}{w_1 + w_2}
\] Where:
– \( C_1 \) and \( C_2 \) are the costs of Type 1 and Type 2 rice, respectively.
– \( w_1 \) and \( w_2 \) are the respective weights (or amounts) of Type 1 and Type 2 rice in the mixture.

Here, \( C_1 = 15 \), \( C_2 = 20 \), \( w_1 = 2 \), and \( w_2 = 3 \).

Step 3: Calculate the price of the mixture
Substitute the values into the formula:
\[
\text{Price of mixture} = \frac{(15 \times 2) + (20 \times 3)}{2 + 3}
\] \[
\text{Price of mixture} = \frac{30 + 60}{5}
\] \[
\text{Price of mixture} = \frac{90}{5} = 18
\]

Conclusion:
The price per kg of the mixed variety of rice is Rs. 18.

Answer: Option B
Solution:
Let’s solve this problem step by step.

Step 1: Define the variables
Let the total capacity of the cask be \( x \) litres, and initially, it is filled with wine.

Each time, 8 litres of the mixture is drawn out and replaced by water. This operation is repeated 4 times. After the operation, the ratio of wine to water in the cask is given as 16:65.

Step 2: The process of removing and replacing wine
Each time we draw out 8 litres of the mixture, the amount of wine that is removed is proportional to the amount of wine left in the cask.

1. Initially, the cask is full of wine, so it contains \( x \) litres of wine.
2. After the first operation, 8 litres are drawn and replaced by water. The remaining wine will be \( x – 8 \) litres, but this is \( \frac{x – 8}{x} \) of the original amount of wine.
3. After the second operation, 8 litres of the new mixture (which contains both wine and water) is drawn and replaced with water. The remaining wine is now \( \frac{x – 8}{x} \times \frac{x – 8}{x} \), or \( \left( \frac{x – 8}{x} \right)^2 \) of the original wine.

So, after \( n \) operations, the amount of wine left in the cask is given by:
\[
\text{Amount of wine after } n \text{ operations} = x \left( \frac{x – 8}{x} \right)^n
\]

Step 3: Use the ratio after 4 operations
After 4 operations, we are told that the ratio of the amount of wine left to the amount of water is 16:65. This means that the amount of wine left is \( \frac{16}{16 + 65} = \frac{16}{81} \) of the total amount in the cask.

Thus, the amount of wine left after 4 operations is:
\[
\frac{16}{81} \times x
\]

We also know that after 4 operations, the remaining wine is \( x \left( \frac{x – 8}{x} \right)^4 \). Therefore, we can set up the equation:
\[
x \left( \frac{x – 8}{x} \right)^4 = \frac{16}{81} \times x
\]

Step 4: Solve the equation
First, cancel \( x \) from both sides of the equation:
\[
\left( \frac{x – 8}{x} \right)^4 = \frac{16}{81}
\]

Take the fourth root of both sides:
\[
\frac{x – 8}{x} = \left( \frac{16}{81} \right)^{1/4}
\] \[
\frac{x – 8}{x} = \frac{2}{3}
\]

Step 5: Solve for \( x \)
Now, solve for \( x \):
\[
\frac{x – 8}{x} = \frac{2}{3}
\] Multiply both sides by \( x \):
\[
x – 8 = \frac{2}{3}x
\] Multiply both sides by 3 to eliminate the fraction:
\[
3(x – 8) = 2x
\] Expand the equation:
\[
3x – 24 = 2x
\] Solve for \( x \):
\[
3x – 2x = 24
\] \[
x = 24
\]

Conclusion:
The cask originally held 24 litres of wine.

Answer: Option C
Solution:
Let’s solve this problem step by step.

Step 1: Define the variables
– Let \( x \) be the amount of sugar sold at 18% profit (in kg).
– The remaining amount of sugar, \( 1000 – x \), is sold at 8% profit.

Step 2: Calculate the total profit
The total profit is the sum of the profit from the sugar sold at 18% profit and the sugar sold at 8% profit.

– Profit from the sugar sold at 18% profit = \( 18\% \) of \( x \) = \( 0.18 \times x \)
– Profit from the sugar sold at 8% profit = \( 8\% \) of \( 1000 – x \) = \( 0.08 \times (1000 – x) \)

Step 3: Total profit from the entire 1000 kg of sugar
The merchant gains a total of 14% profit on the whole 1000 kg of sugar. So, the total profit from selling all the sugar is:
\[
\text{Total profit} = 14\% \text{ of } 1000 = 0.14 \times 1000 = 140
\]

Step 4: Set up the equation
Now, set up the equation for the total profit:
\[
0.18 \times x + 0.08 \times (1000 – x) = 140
\]

Step 5: Solve the equation
Expand the equation:
\[
0.18 \times x + 0.08 \times 1000 – 0.08 \times x = 140
\] \[
0.18x + 80 – 0.08x = 140
\] Combine like terms:
\[
(0.18x – 0.08x) + 80 = 140
\] \[
0.10x + 80 = 140
\] Subtract 80 from both sides:
\[
0.10x = 60
\] Solve for \( x \):
\[
x = \frac{60}{0.10} = 600
\]

Conclusion:
The quantity of sugar sold at 18% profit is 600 kg.

Answer: Option A
Solution:
Let’s solve this step by step.

Step 1: Define the given quantities
– The total volume of the mixture is 30 litres.
– The initial ratio of milk to water is 7:3. This means that:
– Milk in the mixture = \( \frac{7}{10} \times 30 = 21 \) litres
– Water in the mixture = \( \frac{3}{10} \times 30 = 9 \) litres

We need to add water to the mixture such that the resultant mixture has 40% water.

Step 2: Define the variable
Let \( x \) be the number of litres of water to be added to the mixture.

After adding \( x \) litres of water:
– The total amount of water in the mixture becomes \( 9 + x \) litres.
– The total volume of the mixture becomes \( 30 + x \) litres.

Step 3: Set up the equation
We are told that the resultant mixture should have 40% water. Therefore, the amount of water in the new mixture should be 40% of the total volume, i.e.,
\[
\text{Amount of water} = 40\% \times (30 + x)
\] \[
9 + x = 0.40 \times (30 + x)
\]

Step 4: Solve the equation
Now, solve for \( x \):
\[
9 + x = 0.40 \times (30 + x)
\] Expand the equation:
\[
9 + x = 12 + 0.40x
\] Bring like terms involving \( x \) to one side:
\[
9 – 12 = 0.40x – x
\] \[
-3 = -0.60x
\] Solve for \( x \):
\[
x = \frac{-3}{-0.60} = 5
\]

Conclusion:
The amount of water that should be added is 5 litres.

Answer: Option D
Solution:
Let’s solve this step by step.

Step 1: Define the variables
Let the quantities of the mixtures taken from vessels A and B be \( x \) litres and \( y \) litres, respectively.

– Vessel A contains spirit and water in the ratio of 5:2. This means in \( x \) litres from vessel A:
– Amount of spirit = \( \frac{5}{7} \times x \)
– Amount of water = \( \frac{2}{7} \times x \)

– Vessel B contains spirit and water in the ratio of 7:6. This means in \( y \) litres from vessel B:
– Amount of spirit = \( \frac{7}{13} \times y \)
– Amount of water = \( \frac{6}{13} \times y \)

We want to mix these in such a way that the resulting mixture in vessel C has spirit and water in the ratio 8:5.

Step 2: Set up the equations
In the new mixture, the total amount of spirit and the total amount of water must have the ratio 8:5. Thus, the ratio of spirit to water in vessel C is:
\[
\frac{\text{Total spirit}}{\text{Total water}} = \frac{8}{5}
\]

– The total amount of spirit in the new mixture:
\[
\text{Spirit from A} + \text{Spirit from B} = \frac{5}{7} \times x + \frac{7}{13} \times y
\]

– The total amount of water in the new mixture:
\[
\text{Water from A} + \text{Water from B} = \frac{2}{7} \times x + \frac{6}{13} \times y
\]

So, we have the equation:
\[
\frac{\frac{5}{7}x + \frac{7}{13}y}{\frac{2}{7}x + \frac{6}{13}y} = \frac{8}{5}
\]

Step 3: Simplify the equation
To eliminate the fractions, multiply both the numerator and the denominator of the left-hand side by 91 (LCM of 7 and 13):
\[
\frac{\frac{5}{7}x + \frac{7}{13}y}{\frac{2}{7}x + \frac{6}{13}y} = \frac{8}{5}
\] Multiplying by 91:
\[
\frac{65x + 49y}{26x + 42y} = \frac{8}{5}
\]

Step 4: Cross-multiply to solve for the ratio of \( x \) and \( y \)
Cross-multiply:
\[
5(65x + 49y) = 8(26x + 42y)
\] Simplify both sides:
\[
325x + 245y = 208x + 336y
\] Now, bring the terms involving \( x \) and \( y \) to one side:
\[
325x – 208x = 336y – 245y
\] \[
117x = 91y
\] So, the ratio of \( x \) to \( y \) is:
\[
\frac{x}{y} = \frac{91}{117}
\] Simplify the ratio:
\[
\frac{x}{y} = \frac{13}{17}
\]

Conclusion:
The quantities of the mixtures to be mixed from vessels A and B should be in the ratio 13 : 17.

Answer: Option A
Solution:
We are given a mixture of 30 litres containing milk and water in the ratio 7:3, and we need to add water to this mixture such that the resultant mixture contains 40% water.
Step 1: Initial amount of water and milk
In the original mixture of 30 litres:

– The total ratio of milk and water is 7:3.
– The total parts in the ratio = \(7 + 3 = 10\) parts.

– The amount of water in the original mixture:
\[
\text{Water} = \frac{3}{10} \times 30 = 9 \text{ litres}.
\]

– The amount of milk in the original mixture:
\[
\text{Milk} = \frac{7}{10} \times 30 = 21 \text{ litres}.
\]

Step 2: Adding water
Let \( x \) be the amount of water to be added to the mixture. After adding \( x \) litres of water, the new total volume of the mixture will be:
\[
\text{New total volume} = 30 + x \text{ litres}.
\]

The new amount of water in the mixture will be:
\[
\text{New amount of water} = 9 + x \text{ litres}.
\]

Step 3: New concentration of water
We want the resultant mixture to have 40% water. So, the amount of water should be 40% of the new total volume:
\[
\text{New amount of water} = 0.40 \times (30 + x).
\]

We now equate the two expressions for the amount of water:
\[
9 + x = 0.40 \times (30 + x).
\]

Step 4: Solve for \( x \)
Expand the equation:
\[
9 + x = 0.40 \times 30 + 0.40 \times x,
\] \[
9 + x = 12 + 0.40x.
\]

Now, move the terms involving \( x \) to one side and constant terms to the other side:
\[
x – 0.40x = 12 – 9,
\] \[
0.60x = 3.
\]

Solve for \( x \):
\[
x = \frac{3}{0.60} = 5 \text{ litres}.
\]

Conclusion:
You need to add 5 litres of water to the mixture.

Answer: Option D
Solution:
We need to find the ratio in which the mixtures from vessels A and B should be combined to obtain a new mixture in vessel C with a spirit-to-water ratio of 8:5.

Step 1: Let the quantities of mixtures from vessels A and B
Let the quantities of mixtures taken from vessels A and B be \( x \) litres and \( y \) litres, respectively.

Step 2: Quantities of spirit and water in each mixture

– Mixture A (Ratio 5:2 of spirit to water):
– The total ratio of spirit to water is 5:2, so out of every 7 parts of the mixture:
– Spirit = \(\frac{5}{7}\) of the total mixture.
– Water = \(\frac{2}{7}\) of the total mixture.
– In \( x \) litres of mixture A:
– Spirit from A = \( \frac{5}{7}x \) litres.
– Water from A = \( \frac{2}{7}x \) litres.

– **Mixture B** (Ratio 7:6 of spirit to water):
– The total ratio of spirit to water is 7:6, so out of every 13 parts of the mixture:
– Spirit = \(\frac{7}{13}\) of the total mixture.
– Water = \(\frac{6}{13}\) of the total mixture.
– In \( y \) litres of mixture B:
– Spirit from B = \( \frac{7}{13}y \) litres.
– Water from B = \( \frac{6}{13}y \) litres.
Step 3: Final mixture in vessel C

We want the final mixture in vessel C to have a spirit-to-water ratio of 8:5. The total amount of spirit and water in the final mixture will be the sum of the spirit and water from A and B.

– Total spirit in C = Spirit from A + Spirit from B = \( \frac{5}{7}x + \frac{7}{13}y \).
– Total water in C = Water from A + Water from B = \( \frac{2}{7}x + \frac{6}{13}y \).

The ratio of spirit to water in the final mixture should be 8:5, which gives the equation:
\[
\frac{\frac{5}{7}x + \frac{7}{13}y}{\frac{2}{7}x + \frac{6}{13}y} = \frac{8}{5}.
\]

Step 4: Solve the equation

Now, we cross-multiply to simplify the equation:
\[
5 \left( \frac{5}{7}x + \frac{7}{13}y \right) = 8 \left( \frac{2}{7}x + \frac{6}{13}y \right).
\]

Distribute on both sides:
\[
\frac{25}{7}x + \frac{35}{13}y = \frac{16}{7}x + \frac{48}{13}y.
\]

Multiply the entire equation by 91 (the least common multiple of 7 and 13) to eliminate the denominators:
\[
91 \left( \frac{25}{7}x + \frac{35}{13}y \right) = 91 \left( \frac{16}{7}x + \frac{48}{13}y \right),
\] \[
325x + 245y = 208x + 336y.
\]

Now, move terms involving \( x \) and \( y \) to opposite sides:
\[
325x – 208x = 336y – 245y,
\] \[
117x = 91y.
\]

Finally, solve for the ratio of \( x \) to \( y \):
\[
\frac{x}{y} = \frac{91}{117}.
\]

Simplify the ratio by dividing both the numerator and the denominator by their greatest common divisor, which is 13:
\[
\frac{x}{y} = \frac{7}{9}.
\]

Conclusion:
The mixtures from vessels A and B should be mixed in the ratio \( 7:9 \) to obtain the desired mixture in vessel C.

Answer: Option C
Solution:
We are given that the initial ratio of milk and water in the mixture is 4:3, and when 6 litres of water is added, the new ratio becomes 8:7. We need to find the quantity of milk in the original mixture.

Step 1: Let the quantity of milk and water in the original mixture be
Let the quantity of milk be \( 4x \) litres and the quantity of water be \( 3x \) litres (since the ratio of milk to water is 4:3).

Step 2: After adding 6 litres of water
When 6 litres of water is added, the new quantity of water becomes:
\[
\text{New quantity of water} = 3x + 6 \text{ litres}.
\] The quantity of milk remains the same, which is \( 4x \) litres.

Step 3: Set up the new ratio
The new ratio of milk to water is given as 8:7. So, we can set up the following equation:
\[
\frac{4x}{3x + 6} = \frac{8}{7}.
\]

Step 4: Solve for \( x \)
Now, we cross-multiply to solve for \( x \):
\[
7 \times 4x = 8 \times (3x + 6),
\] \[
28x = 8(3x + 6),
\] \[
28x = 24x + 48.
\]

Now, subtract \( 24x \) from both sides:
\[
4x = 48.
\]

Solve for \( x \):
\[
x = 12.
\]

Step 5: Find the quantity of milk
Since the quantity of milk in the original mixture is \( 4x \), we substitute \( x = 12 \) into this expression:
\[
\text{Quantity of milk} = 4 \times 12 = 48 \text{ litres}.
\]
Conclusion:
The quantity of milk in the original mixture is 48 litres.

Answer: Option B
Solution:
We are given that 35 kg of type A sandal powder is mixed with a certain amount of type B sandal powder, and the mixture is sold at a profit. We need to find the amount of type B sandal powder in the mixture.

Step 1: Let the amount of type B sandal powder be \( x \) kg.

Step 2: Cost price of the mixture
– The cost of 35 kg of type A sandal powder is:
\[
\text{Cost of type A powder} = 35 \times 614 = 21490 \, \text{Rs}.
\] – The cost of \( x \) kg of type B sandal powder is:
\[
\text{Cost of type B powder} = x \times 695 \, \text{Rs}.
\]

Thus, the total cost price of the mixture is:
\[
\text{Total cost of the mixture} = 21490 + 695x \, \text{Rs}.
\]

Step 3: Selling price of the mixture
– The total weight of the mixture is \( 35 + x \) kg.
– The mixture is sold at Rs. 767 per kg, so the total selling price is:
\[
\text{Selling price of the mixture} = (35 + x) \times 767.
\]

Step 4: Profit and selling price relation
We are told that an 18% profit was earned. The relationship between the cost price and selling price with profit is given by:
\[
\text{Selling price} = \text{Cost price} \times (1 + \frac{\text{Profit percentage}}{100}).
\] Thus:
\[
\text{Selling price} = \text{Cost price} \times 1.18.
\] Substituting the expressions for cost price and selling price:
\[
(35 + x) \times 767 = (21490 + 695x) \times 1.18.
\]

Step 5: Solve the equation
First, expand both sides:
\[
(35 + x) \times 767 = 35 \times 767 + x \times 767 = 26845 + 767x.
\] On the right-hand side:
\[
(21490 + 695x) \times 1.18 = 21490 \times 1.18 + 695x \times 1.18 = 25357.2 + 821.1x.
\]

Equating both sides:
\[
26845 + 767x = 25357.2 + 821.1x.
\]

Now, subtract \( 25357.2 \) and \( 767x \) from both sides:
\[
26845 – 25357.2 = 821.1x – 767x,
\] \[
1487.8 = 54.1x.
\]

Solve for \( x \):
\[
x = \frac{1487.8}{54.1} \approx 27.5 \, \text{kg}.
\]

Conclusion:
The amount of type B sandal powder in the mixture is approximately 27.5 kg.

Answer: Option A
Solution:
We are given two vessels A and B, each containing milk and water in different ratios, and we need to find the ratio in which these two mixtures should be mixed to get a new mixture containing \( 68\frac{3}{13} \)% milk.

Step 1: Convert the percentage of milk in the final mixture to a fraction
First, let’s convert \( 68\frac{3}{13} \)% to an improper fraction. We have:
\[
68\frac{3}{13} = \frac{68 \times 13 + 3}{13} = \frac{884 + 3}{13} = \frac{887}{13}.
\] So, the percentage of milk in the final mixture is \( \frac{887}{13} \)% or \( \frac{887}{1300} \) as a fraction of the total mixture.

Step 2: Find the quantity of milk in each mixture

– Vessel A: The ratio of milk to water in vessel A is 8:5. So, out of every 13 parts of the mixture, 8 parts are milk. Therefore, the fraction of milk in vessel A is:
\[
\frac{8}{13}.
\]

– **Vessel B**: The ratio of milk to water in vessel B is 5:2. So, out of every 7 parts of the mixture, 5 parts are milk. Therefore, the fraction of milk in vessel B is:
\[
\frac{5}{7}.
\]

Step 3: Set up the equation for the final mixture

Let \( x \) be the amount of mixture taken from vessel A and \( y \) be the amount of mixture taken from vessel B. The total amount of milk in the final mixture will be the sum of the milk from A and the milk from B.

The total amount of milk in the final mixture is:
\[
\text{Milk from A} = \frac{8}{13}x, \quad \text{Milk from B} = \frac{5}{7}y.
\]

The total amount of the final mixture is \( x + y \), and the fraction of milk in the final mixture is given as \( \frac{887}{1300} \). Therefore, we can set up the equation:
\[
\frac{\frac{8}{13}x + \frac{5}{7}y}{x + y} = \frac{887}{1300}.
\]

Step 4: Solve the equation

Now, cross-multiply to solve for \( x \) and \( y \):
\[
1300 \left( \frac{8}{13}x + \frac{5}{7}y \right) = 887(x + y).
\]

Expand both sides:
\[
1300 \times \frac{8}{13}x + 1300 \times \frac{5}{7}y = 887x + 887y,
\] \[
800x + 928.57y = 887x + 887y.
\]

Now, move the terms involving \( x \) and \( y \) to opposite sides:
\[
800x – 887x = 887y – 928.57y,
\] \[
-87x = -41.57y.
\]

Divide both sides by \( -87 \):
\[
\frac{x}{y} = \frac{41.57}{87} \approx \frac{2}{4} = 2:1.
\]

Conclusion:
The mixtures from vessels A and B should be mixed in the ratio \( 2:1 \) to get a mixture containing \( 68\frac{3}{13} \)% milk.

Answer: Option C
Solution:
We are given a 20-litre mixture containing milk and water in the ratio 3:1. We need to find the amount of milk to be added to the mixture to change the ratio to 4:1.

Step 1: Amount of milk and water in the original mixture
The total mixture is 20 litres, and the ratio of milk to water is 3:1. Therefore, the total number of parts in the mixture is:
\[
3 + 1 = 4 \text{ parts}.
\]

– The amount of milk in the original mixture is:
\[
\text{Milk} = \frac{3}{4} \times 20 = 15 \text{ litres}.
\]

– The amount of water in the original mixture is:
\[
\text{Water} = \frac{1}{4} \times 20 = 5 \text{ litres}.
\]

Step 2: Amount of milk after adding milk
Let \( x \) be the amount of milk to be added to the mixture. After adding \( x \) litres of milk, the new amount of milk in the mixture will be:
\[
\text{New amount of milk} = 15 + x.
\] The amount of water remains the same, which is 5 litres.

Step 3: New ratio of milk to water
We want the new mixture to have a milk-to-water ratio of 4:1. This means:
\[
\frac{15 + x}{5} = \frac{4}{1}.
\]

Step 4: Solve for \( x \)
Now, solve the equation:
\[
15 + x = 4 \times 5,
\] \[
15 + x = 20,
\] \[
x = 20 – 15 = 5.
\]

Conclusion:
The amount of milk to be added to the mixture is 5 litres.

Answer: Option A
Solution:
We are given that milk costs Rs. 12 per litre and we need to mix it with water (which costs Rs. 0 per litre) to obtain a mixture worth Rs. 8 per litre. We need to find the ratio of water to milk for the desired mixture.

Step 1: Let the amount of milk and water be
Let the amount of milk be \( x \) litres and the amount of water be \( y \) litres.

Step 2: Cost of milk and water
– The cost of \( x \) litres of milk is \( 12x \) Rs (since milk costs Rs. 12 per litre).
– The cost of \( y \) litres of water is \( 0 \times y = 0 \) Rs (since water is free).

The total cost of the mixture is:
\[
\text{Total cost} = 12x + 0 = 12x \, \text{Rs}.
\]

The total amount of the mixture is \( x + y \) litres, and we want the mixture to be worth Rs. 8 per litre. Therefore, the total cost of the mixture should be:
\[
\text{Cost of mixture} = 8(x + y) \, \text{Rs}.
\]

Step 3: Set up the equation
We equate the total cost of the mixture to the cost of the mixture with the desired price per litre:
\[
12x = 8(x + y).
\]

Step 4: Solve for the ratio of water to milk
Now, expand and solve for \( y \):
\[
12x = 8x + 8y,
\] \[
12x – 8x = 8y,
\] \[
4x = 8y,
\] \[
\frac{x}{y} = 2.
\]

Thus, the ratio of milk to water is 2:1, meaning the amount of water should be half the amount of milk.

Conclusion:
The ratio of water to milk is 1:2.

Answer: Option D
Solution:
To solve this, let’s assume the milkman mixes \( x \) litres of water with milk and sells the mixture at the cost price of milk, gaining 25% profit.

Step 1: Let the cost price of 1 litre of milk be \( C \).

When the milkman mixes water with milk, the cost of water is zero (since water is free). The milkman sells the mixture at the cost price of milk \( C \), but gains 25% profit.

Step 2: Gain of 25% means
The milkman sells the mixture at a price that is 25% more than the cost price of the mixture. Therefore, the cost price of the mixture and the selling price are related as:
\[
\text{Selling Price} = \text{Cost Price} \times (1 + 0.25) = \text{Cost Price} \times 1.25.
\]

Since the milkman sells the mixture at the cost price of milk \( C \), we have:
\[
C = \text{Cost Price of the Mixture} \times 1.25.
\]

Step 3: Find the cost price of the mixture
Let the amount of milk in the mixture be \( m \) litres and the amount of water added be \( w \) litres. The total amount of the mixture is \( m + w \) litres.

– The cost of milk in the mixture is \( m \times C \) (since the cost of milk is \( C \) per litre).
– The cost of water is 0 (since water is free).

Thus, the total cost price of the mixture is:
\[
\text{Cost Price of the Mixture} = m \times C.
\]

The total volume of the mixture is \( m + w \), so the cost price per litre of the mixture is:
\[
\frac{m \times C}{m + w}.
\]

Step 4: Set up the equation
We know that the milkman sells the mixture at the cost price of milk, which is \( C \). Therefore:
\[
C = \frac{m \times C}{m + w} \times 1.25.
\]

Simplifying the equation:
\[
\frac{m}{m + w} = \frac{4}{5}.
\]

Step 5: Solve for the ratio of water to milk
We now solve for the ratio of water to milk. From the equation:
\[
\frac{m}{m + w} = \frac{4}{5},
\] cross-multiply:
\[
5m = 4(m + w),
\] \[
5m = 4m + 4w,
\] \[
m = 4w.
\]

Thus, the amount of milk is 4 times the amount of water, so the ratio of water to milk is:
\[
\frac{w}{m} = \frac{1}{4}.
\]

Conclusion:
The ratio of water to milk is 1:4.

Answer: Option D
Solution:
We are given that a vessel contains a mixture of grape, pineapple, and banana juices in the ratio 4:6:5. After 15 litres of the mixture is taken out, 8 litres of grape juice and 2 litres of pineapple juice are added to the vessel. The resultant quantity of grape juice is 10 litres less than the resultant quantity of pineapple juice. We need to find the initial quantity of the mixture in the vessel.

Step 1: Define the initial quantity of the mixture
Let the initial quantity of the mixture in the vessel be \( x \) litres. The ratio of grape juice, pineapple juice, and banana juice is 4:6:5, so the initial amounts of each juice are:

– Grape juice = \( \frac{4}{15} \times x \)
– Pineapple juice = \( \frac{6}{15} \times x \)
– Banana juice = \( \frac{5}{15} \times x \)

Step 2: Quantity of juices removed
When 15 litres of the mixture is taken out, the proportion of grape juice, pineapple juice, and banana juice removed will be the same as their respective ratios. Therefore, the quantities removed are:

– Grape juice removed = \( \frac{4}{15} \times 15 = 4 \) litres
– Pineapple juice removed = \( \frac{6}{15} \times 15 = 6 \) litres
– Banana juice removed = \( \frac{5}{15} \times 15 = 5 \) litres

Step 3: New quantities of juices in the vessel
After 15 litres are removed, the new amounts of juice in the vessel will be:

– Grape juice remaining = \( \frac{4}{15} \times x – 4 \)
– Pineapple juice remaining = \( \frac{6}{15} \times x – 6 \)
– Banana juice remaining = \( \frac{5}{15} \times x – 5 \)

Now, 8 litres of grape juice and 2 litres of pineapple juice are added. So, the new quantities of juice are:

– Grape juice = \( \frac{4}{15} \times x – 4 + 8 = \frac{4}{15} \times x + 4 \)
– Pineapple juice = \( \frac{6}{15} \times x – 6 + 2 = \frac{6}{15} \times x – 4 \)

Step 4: Set up the equation based on the condition
We are told that the resultant quantity of grape juice is 10 litres less than the resultant quantity of pineapple juice. Therefore, we can set up the equation:
\[
\frac{4}{15} \times x + 4 = \frac{6}{15} \times x – 4 – 10.
\] Simplify the right-hand side:
\[
\frac{4}{15} \times x + 4 = \frac{6}{15} \times x – 14.
\] Now, move all the terms involving \( x \) to one side and constant terms to the other side:
\[
\frac{4}{15} \times x – \frac{6}{15} \times x = -14 – 4,
\] \[
-\frac{2}{15} \times x = -18.
\] Multiply both sides by \( -15/2 \):
\[
x = \frac{-18 \times -15}{2} = \frac{270}{2} = 135.
\]

Conclusion:
The initial quantity of the mixture in the vessel was 135 litres.

Answer: Option A
Solution:
We are given two types of tea:

– Tea A costs Rs. 62 per kg.
– Tea B costs Rs. 72 per kg.

We need to find the ratio in which these two teas should be mixed to obtain a mixture worth Rs. 64.50 per kg.

Step 1: Let the amounts of tea A and tea B be
Let the amount of tea A to be mixed be \( x \) kg, and the amount of tea B to be mixed be \( y \) kg.

Step 2: Total cost of the mixture
– The cost of \( x \) kg of tea A is \( 62x \) Rs.
– The cost of \( y \) kg of tea B is \( 72y \) Rs.

The total cost of the mixture is:
\[
\text{Total cost} = 62x + 72y.
\]

The total weight of the mixture is \( x + y \) kg.

Step 3: Cost of the mixture per kg
The cost of the mixture is given as Rs. 64.50 per kg. So, the total cost of the mixture is also:
\[
\text{Total cost} = 64.50(x + y).
\]

Step 4: Set up the equation
Now, we equate the total cost from both expressions:
\[
62x + 72y = 64.50(x + y).
\]

Step 5: Solve the equation
First, expand the right-hand side:
\[
62x + 72y = 64.50x + 64.50y.
\]

Now, collect the terms involving \( x \) and \( y \):
\[
62x – 64.50x = 64.50y – 72y,
\] \[
-2.5x = -7.5y.
\]

Divide both sides by -2.5:
\[
x = 3y.
\]

Step 6: Ratio of tea A to tea B
Thus, the ratio of tea A to tea B is:
\[
\frac{x}{y} = \frac{3}{1}.
\]

Conclusion:
The two teas must be mixed in the ratio 3:1.

Answer: Option A
Solution:
We are given that a vessel contains 60 litres of milk. 12 litres of milk is taken out and replaced with water twice. We need to find the final ratio of milk and water in the resultant mixture.

Step 1: First step of removing milk and adding water
Initially, the vessel contains 60 litres of milk and 0 litres of water.

1. **First removal**: 12 litres of milk is removed. The amount of milk remaining after the first removal is:
\[
60 – 12 = 48 \text{ litres of milk}.
\] The amount of water is now 12 litres (since 12 litres of milk was replaced by 12 litres of water).

Step 2: Second step of removing milk and adding water
Now, 12 litres of the new mixture is taken out, and it is replaced with 12 litres of water.

– The new mixture now consists of 48 litres of milk and 12 litres of water.
– The total mixture after the second removal is still 60 litres.
– Since the mixture now consists of 48 litres of milk and 12 litres of water, the proportion of milk in the mixture is:
\[
\frac{48}{60} = \frac{4}{5}.
\] So, when 12 litres of this mixture is taken out, the amount of milk removed will be:
\[
12 \times \frac{4}{5} = 9.6 \text{ litres of milk}.
\] The amount of water removed will be:
\[
12 – 9.6 = 2.4 \text{ litres of water}.
\]

After removing 9.6 litres of milk and 2.4 litres of water, the amount of milk left in the vessel is:
\[
48 – 9.6 = 38.4 \text{ litres of milk}.
\] The amount of water left in the vessel is:
\[
12 – 2.4 = 9.6 \text{ litres of water}.
\]

Now, 12 litres of water is added to the vessel, so the amount of water becomes:
\[
9.6 + 12 = 21.6 \text{ litres of water}.
\]

Step 3: Final ratio of milk to water
After both steps, the vessel contains:
– 38.4 litres of milk
– 21.6 litres of water

The ratio of milk to water in the resultant mixture is:
\[
\frac{38.4}{21.6} = \frac{384}{216} = \frac{16}{9}.
\]

Conclusion:
The final ratio of milk to water in the resultant mixture is 16:9.

Answer: Option D
Solution:
We are given a mixture of 40 litres containing 10% water, and we need to find how much water must be added to make the water content 20% in the new mixture.

Step 1: Determine the initial amount of water in the mixture
The mixture contains 40 litres, and 10% of it is water. Therefore, the amount of water in the mixture is:
\[
\text{Amount of water} = 10\% \times 40 = \frac{10}{100} \times 40 = 4 \text{ litres of water}.
\]

Step 2: Determine the amount of milk in the mixture
Since the total mixture is 40 litres and the amount of water is 4 litres, the amount of milk in the mixture is:
\[
\text{Amount of milk} = 40 – 4 = 36 \text{ litres of milk}.
\]

Step 3: Let \( x \) litres of water be added
Let \( x \) litres of water be added to the mixture. After adding \( x \) litres of water, the total amount of water in the mixture will be:
\[
4 + x \text{ litres of water}.
\] The total volume of the mixture will now be:
\[
40 + x \text{ litres}.
\]

Step 4: Set up the equation for 20% water content in the new mixture
We are told that the new mixture must contain 20% water. Therefore, the amount of water in the new mixture should be 20% of the total mixture. This gives us the equation:
\[
\frac{4 + x}{40 + x} = 20\% = \frac{20}{100} = \frac{1}{5}.
\]

Step 5: Solve the equation
Now, solve for \( x \):
\[
\frac{4 + x}{40 + x} = \frac{1}{5}.
\] Cross-multiply:
\[
5(4 + x) = 1(40 + x),
\] \[
20 + 5x = 40 + x.
\] Move terms involving \( x \) to one side and constant terms to the other side:
\[
5x – x = 40 – 20,
\] \[
4x = 20,
\] \[
x = \frac{20}{4} = 5.
\]

Conclusion:
The amount of water that must be added to the mixture is 5 litres.

Answer: Option C
Solution:
We are given a 2 kg mixture of copper and aluminium, where 30% of the mixture is copper. We need to find how much aluminium powder should be added to the mixture so that the quantity of copper becomes 20%.

Step 1: Determine the amount of copper in the initial mixture
The total weight of the mixture is 2 kg, and 30% of it is copper. Therefore, the amount of copper in the mixture is:
\[
\text{Amount of copper} = 30\% \times 2 = \frac{30}{100} \times 2 = 0.6 \, \text{kg of copper}.
\]

Step 2: Determine the amount of aluminium in the initial mixture
The remaining part of the mixture is aluminium. Therefore, the amount of aluminium in the mixture is:
\[
\text{Amount of aluminium} = 2 – 0.6 = 1.4 \, \text{kg of aluminium}.
\]

Step 3: Let \( x \) be the amount of aluminium to be added
Let \( x \) kg of aluminium powder be added to the mixture. After adding \( x \) kg of aluminium, the new total weight of the mixture becomes:
\[
\text{New total weight} = 2 + x \, \text{kg}.
\]

The amount of copper remains unchanged at 0.6 kg.

Step 4: Set up the equation for 20% copper in the new mixture
We want the quantity of copper to become 20% of the total new mixture. Therefore, we have the equation:
\[
\frac{0.6}{2 + x} = 20\% = \frac{20}{100} = \frac{1}{5}.
\]

Step 5: Solve for \( x \)
Now, solve for \( x \):
\[
\frac{0.6}{2 + x} = \frac{1}{5}.
\] Cross-multiply:
\[
0.6 \times 5 = 1 \times (2 + x),
\] \[
3 = 2 + x.
\] Subtract 2 from both sides:
\[
x = 1.
\]

Conclusion:
The amount of aluminium powder that should be added is 1 kg.

Answer: Option B
Solution:
We are given that the initial sugar solution contains 3 litres of solution, and 60% of it is sugar. We need to find the percentage of sugar in the new solution after adding 1 litre of water.

Step 1: Calculate the amount of sugar in the initial solution
The total volume of the solution is 3 litres, and 60% of it is sugar. Therefore, the amount of sugar in the initial solution is:
\[
\text{Amount of sugar} = 60\% \times 3 = \frac{60}{100} \times 3 = 1.8 \text{ litres of sugar}.
\]

Step 2: Add 1 litre of water to the solution
When 1 litre of water is added, the total volume of the solution becomes:
\[
\text{New total volume} = 3 + 1 = 4 \text{ litres}.
\] Since water has no sugar, the amount of sugar in the solution remains the same at 1.8 litres.

Step 3: Calculate the percentage of sugar in the new solution
The new percentage of sugar in the solution is the ratio of the amount of sugar to the total volume, multiplied by 100:
\[
\text{Percentage of sugar} = \frac{\text{Amount of sugar}}{\text{Total volume}} \times 100 = \frac{1.8}{4} \times 100 = 45\%.
\]

Conclusion:
The percentage of sugar in the new solution is 45%.

Answer: Option D
Solution:
To solve this, we need to apply the concept of **algebraic mixture problems** where the cost of the mixture is determined by the weighted average of the costs of the individual types of tea.

Step 1: Find the cost price of the mixture
The given selling price of the mixture is Rs. 320/kg. The profit is 20%. To find the cost price (C.P.) of the mixture, we use the formula:

\[
\text{Selling Price} = \text{Cost Price} + \text{Profit}
\]

Let the cost price be \(C.P.\). The profit is 20% of \(C.P.\), so:

\[
320 = C.P. + 0.20 \times C.P.
\]

This simplifies to:

\[
320 = 1.20 \times C.P.
\]

Now, solving for \(C.P.\):

\[
C.P. = \frac{320}{1.20} = 266.67 \, \text{Rs/kg}
\]

Step 2: Set up the equation for the mixture
Let the ratio in which the two types of tea (costing Rs. 180/kg and Rs. 280/kg) are mixed be \(x : y\), where \(x\) is the amount of the first tea and \(y\) is the amount of the second tea.

The cost price of the mixture is the weighted average of the individual costs:

\[
\frac{180x + 280y}{x + y} = 266.67
\]

Step 3: Solve the equation
We now solve for the ratio \(x : y\). Multiply both sides by \(x + y\):

\[
180x + 280y = 266.67(x + y)
\]

Expanding both sides:

\[
180x + 280y = 266.67x + 266.67y
\]

Now, bring the terms involving \(x\) and \(y\) to one side:

\[
180x – 266.67x = 266.67y – 280y
\]

Simplify the terms:

\[
-86.67x = -13.33y
\]

Now, divide both sides by -13.33:

\[
\frac{x}{y} = \frac{13.33}{86.67} = \frac{1}{6.5}
\]

So, the ratio of the two types of tea to be mixed is:

\[
x : y = 1 : 6.5
\]

To express this ratio as whole numbers, multiply both terms by 2:

\[
x : y = 2 : 13
\]

Final Answer:
The two types of tea should be mixed in the ratio 2:13 to get the desired mixture.

Answer: Option C
Solution:
To find the ratio of gold and copper in the third alloy \( C \), we need to calculate the amount of gold and copper in alloys \( A \) and \( B \), then combine them and find the ratio in alloy \( C \).

Step 1: Gold and Copper in Alloy A
In alloy \( A \), the ratio of gold to copper is 5:3. This means for every 8 parts of the alloy, 5 parts are gold and 3 parts are copper.

Let the total quantity of alloy \( A \) be \( x \) kg.

– Amount of gold in \( A \) = \( \frac{5}{8} \times x \)
– Amount of copper in \( A \) = \( \frac{3}{8} \times x \)

Step 2: Gold and Copper in Alloy B
In alloy \( B \), the ratio of gold to copper is 5:11. This means for every 16 parts of the alloy, 5 parts are gold and 11 parts are copper.

Let the total quantity of alloy \( B \) be \( x \) kg (since equal quantities of alloys \( A \) and \( B \) are melted).

– Amount of gold in \( B \) = \( \frac{5}{16} \times x \)
– Amount of copper in \( B \) = \( \frac{11}{16} \times x \)

Step 3: Total Gold and Copper in Alloy C
When equal quantities of alloys \( A \) and \( B \) are melted, the total amount of gold and copper in alloy \( C \) is the sum of the gold and copper from both alloys.

– Total amount of gold in \( C \) = \( \frac{5}{8} \times x + \frac{5}{16} \times x \)
– Total amount of copper in \( C \) = \( \frac{3}{8} \times x + \frac{11}{16} \times x \)

Step 4: Simplifying the Gold and Copper Expressions
Let’s simplify the expressions for gold and copper.

Total Gold in \( C \):
\[
\frac{5}{8} \times x + \frac{5}{16} \times x = \frac{10}{16} \times x + \frac{5}{16} \times x = \frac{15}{16} \times x
\]

Total Copper in \( C \):
\[
\frac{3}{8} \times x + \frac{11}{16} \times x = \frac{6}{16} \times x + \frac{11}{16} \times x = \frac{17}{16} \times x
\]

Step 5: Ratio of Gold to Copper in Alloy C
Now, we can find the ratio of gold to copper in alloy \( C \):

\[
\text{Ratio of gold to copper} = \frac{\frac{15}{16} \times x}{\frac{17}{16} \times x} = \frac{15}{17}
\]

Final Answer:
The ratio of gold to copper in alloy \( C \) is 15:17.

Answer: Option C
Solution:
To solve this problem, we need to calculate the cost price of the mixture, the total revenue from selling the mixture, and then find the profit percentage.

Step 1: Cost of the Mixture
We are given two blends of a commodity:

– First blend costs Rs. 35 per kg.
– Second blend costs Rs. 40 per kg.

These two blends are mixed in the ratio of 2:3 by weight. Let the total weight of the mixture be \( 5x \) kg (since the ratio is 2:3, the total parts are 2 + 3 = 5).

– Weight of the first blend = \( 2x \) kg
– Weight of the second blend = \( 3x \) kg

The cost of the first blend (Rs. 35 per kg) for \( 2x \) kg is:

\[
\text{Cost of first blend} = 35 \times 2x = 70x
\]

The cost of the second blend (Rs. 40 per kg) for \( 3x \) kg is:

\[
\text{Cost of second blend} = 40 \times 3x = 120x
\]

Thus, the total cost of the mixture is:

\[
\text{Total cost} = 70x + 120x = 190x
\]

The total weight of the mixture is \( 5x \) kg, so the cost price per kg of the mixture is:

\[
\text{Cost price per kg} = \frac{190x}{5x} = 38 \, \text{Rs/kg}
\]

Step 2: Revenue from Selling the Mixture
We are told that one-fifth of the mixture is sold at Rs. 46 per kg and the remaining four-fifths are sold at Rs. 55 per kg.

– One-fifth of the total mixture is \( \frac{5x}{5} = x \) kg.
– The remaining four-fifths of the total mixture is \( \frac{4 \times 5x}{5} = 4x \) kg.

Now, let’s calculate the revenue:

– Revenue from selling \( x \) kg at Rs. 46 per kg:

\[
\text{Revenue from first part} = 46 \times x = 46x
\]

– Revenue from selling \( 4x \) kg at Rs. 55 per kg:

\[
\text{Revenue from second part} = 55 \times 4x = 220x
\]

Thus, the total revenue is:

\[
\text{Total revenue} = 46x + 220x = 266x
\]

Step 3: Profit Calculation
The profit is the difference between the total revenue and the total cost:

\[
\text{Profit} = \text{Total revenue} – \text{Total cost} = 266x – 190x = 76x
\]

Step 4: Profit Percentage
Finally, the profit percentage is given by:

\[
\text{Profit percentage} = \frac{\text{Profit}}{\text{Total cost}} \times 100 = \frac{76x}{190x} \times 100 = \frac{76}{190} \times 100
\]

\[
\text{Profit percentage} = 40\%
\]

Final Answer:
The profit percentage is 40%.

Answer: Option D
Solution:
Let’s denote:

– The number of boys who appeared in the exam as \( b \).
– The number of girls who appeared in the exam as \( g \).

From the given information, we have the following equations:

1. Total number of students = 12500, so:
\[
b + g = 12500
\]

2. 50% of the boys cleared the exam, so the number of boys who cleared the exam is \( 0.50b \).

3. 70% of the girls cleared the exam, so the number of girls who cleared the exam is \( 0.70g \).

4. The total number of students who cleared the exam is 60% of the total number of students, which is \( 0.60 \times 12500 = 7500 \).

Thus, the total number of students who cleared the exam is the sum of the boys and girls who cleared:

\[
0.50b + 0.70g = 7500
\]

Step 1: Solve the system of equations

We now have the system of two equations:

1. \( b + g = 12500 \)
2. \( 0.50b + 0.70g = 7500 \)

Step 1.1: Express \( b \) in terms of \( g \)

From the first equation:

\[
b = 12500 – g
\]

Step 1.2: Substitute \( b \) in the second equation

Substitute \( b = 12500 – g \) into the second equation:

\[
0.50(12500 – g) + 0.70g = 7500
\]

Now, simplify the equation:

\[
6250 – 0.50g + 0.70g = 7500
\]

\[
6250 + 0.20g = 7500
\]

Step 1.3: Solve for \( g \)

Subtract 6250 from both sides:

\[
0.20g = 1250
\]

Now, divide by 0.20:

\[
g = \frac{1250}{0.20} = 6250
\]

Final Answer:
The number of girls who appeared in the exam is 6250.

Answer: Option A
To find the amount of tin in the new alloy, we need to calculate the amount of tin in each of the original alloys and then combine them.

Step 1: Tin in Alloy A
Alloy A has lead and tin in the ratio 3:2. This means for every 5 parts of the alloy, 2 parts are tin.

– The total weight of alloy A is 60 kg.
– Amount of tin in alloy A = \( \frac{2}{5} \times 60 = 24 \) kg.

Step 2: Tin in Alloy B
Alloy B has tin and copper in the ratio 1:4. This means for every 5 parts of the alloy, 1 part is tin.

– The total weight of alloy B is 100 kg.
– Amount of tin in alloy B = \( \frac{1}{5} \times 100 = 20 \) kg.

Step 3: Total Amount of Tin in the New Alloy
When the two alloys are mixed, the total amount of tin in the new alloy is the sum of the tin from both alloys:

\[
\text{Total tin} = 24 \, \text{kg (from alloy A)} + 20 \, \text{kg (from alloy B)} = 44 \, \text{kg}.
\]

Final Answer:
The amount of tin in the new alloy is 44 kg.

Answer: Option D
Solution:
To solve this, let’s first determine the initial amounts of milk and water in the mixture and then figure out how much water needs to be added to achieve the desired ratio.

Step 1: Initial Amounts of Milk and Water
The total mixture is 80 litres, and the ratio of milk to water is 7:3. This means the mixture can be divided into 7 + 3 = 10 parts.

– Amount of milk in the mixture = \( \frac{7}{10} \times 80 = 56 \) litres.
– Amount of water in the mixture = \( \frac{3}{10} \times 80 = 24 \) litres.

Step 2: Desired Ratio of Milk to Water
We want to change the ratio of milk to water to 2:1. Let \( x \) litres of water be added to the mixture to achieve this new ratio. The amount of milk will remain the same (56 litres), but the amount of water will increase to \( 24 + x \) litres.

The new ratio of milk to water should be 2:1. Therefore, we can set up the following equation:

\[
\frac{\text{Amount of milk}}{\text{Amount of water}} = \frac{2}{1}
\]

Substitute the known values:

\[
\frac{56}{24 + x} = 2
\]

Step 3: Solve for \( x \)
To solve for \( x \), multiply both sides of the equation by \( 24 + x \):

\[
56 = 2(24 + x)
\]

Simplify:

\[
56 = 48 + 2x
\]

Subtract 48 from both sides:

\[
8 = 2x
\]

Now, divide by 2:

\[
x = 4
\]

Final Answer:
To make the ratio of milk to water 2:1, 4 litres of water should be added.

Answer: Option C
Solution:
To solve this, let’s first determine how much sugar is present in the initial solution and then figure out how much more sugar needs to be added to achieve the desired concentration.

Step 1: Initial Amount of Sugar
We are given:

– The total weight of the solution = 300 kg.
– The concentration of sugar = 40%.

The amount of sugar in the solution is:

\[
\text{Sugar in the solution} = 40\% \text{ of } 300 = \frac{40}{100} \times 300 = 120 \, \text{kg}.
\]

Step 2: Amount of Sugar to be Added
Let \( x \) kg of sugar be added to the solution. After adding \( x \) kg of sugar, the total amount of sugar in the solution will be:

\[
\text{New amount of sugar} = 120 + x \, \text{kg}.
\]

The total weight of the solution will also increase. After adding \( x \) kg of sugar, the total weight of the solution becomes:

\[
\text{New total weight of solution} = 300 + x \, \text{kg}.
\]

We want the new concentration of sugar to be 50%. This means:

\[
\frac{\text{New amount of sugar}}{\text{New total weight of solution}} = 50\%
\]

Substitute the expressions for the new amount of sugar and the new total weight of the solution:

\[
\frac{120 + x}{300 + x} = \frac{50}{100} = 0.50
\]

Step 3: Solve for \( x \)
Now, we solve for \( x \):

\[
120 + x = 0.50 \times (300 + x)
\]

Simplify the equation:

\[
120 + x = 150 + 0.50x
\]

Now, subtract \( 0.50x \) from both sides:

\[
120 + 0.50x = 150
\]

Subtract 120 from both sides:

\[
0.50x = 30
\]

Now, divide by 0.50:

\[
x = \frac{30}{0.50} = 60
\]

Final Answer:
To make the solution 50% sugar, 60 kg of sugar should be added.