Solving Milk and Water Ratio Problems

This problem involves understanding how ratios change when quantities of components are added to a mixture. We start with a milk and water solution and make two additions, changing the ratio each time. We are given a relationship between the amounts added and need to find the initial quantity of the solution.

Initial Setup of the Milk and Water Solution

Let the initial quantity of the solution be \(S\) litres.

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

This means that for every 1 part milk, there are 4 parts water, making a total of \(1 + 4 = 5\) parts.

• Initial quantity of milk = \(\frac{1}{5} \times S\)
• Initial quantity of water = \(\frac{4}{5} \times S\)

First Change: Adding M litres of Milk

M litres of milk are added to the solution.

The quantity of water remains the same, but the quantity of milk increases.

• New quantity of milk = Initial milk + M = \(\frac{1}{5}S + M\)
• Quantity of water = Initial water = \(\frac{4}{5}S\)

After adding M litres of milk, the ratio of milk to water becomes 1 : 3.

We can write this as an equation:

\(\frac{\text{New quantity of milk}}{\text{Quantity of water}} = \frac{1}{3}\)

\(\frac{\frac{1}{5}S + M}{\frac{4}{5}S} = \frac{1}{3}\)

Now, we cross-multiply to solve for S in terms of M:

\(3 \times (\frac{1}{5}S + M) = 1 \times (\frac{4}{5}S)\)

\(\frac{3}{5}S + 3M = \frac{4}{5}S\)

Subtract \(\frac{3}{5}S\) from both sides:

\(3M = \frac{4}{5}S - \frac{3}{5}S\)

\(3M = \frac{1}{5}S\)

Multiply both sides by 5:

\(15M = S\)

So, we have our first relationship: \(S = 15M\).

Second Change: Adding N litres of Water

N litres of water are added to the solution from the state after adding M litres of milk.

The quantity of milk remains the same as after the first addition, but the quantity of water increases.

• Quantity of milk = \(\frac{1}{5}S + M\)
• New quantity of water = Water after first addition + N = \(\frac{4}{5}S + N\)

After adding N litres of water, the ratio of milk to water becomes 2 : 7.

We set up the new ratio equation:

\(\frac{\text{Quantity of milk}}{\text{New quantity of water}} = \frac{2}{7}\)

\(\frac{\frac{1}{5}S + M}{\frac{4}{5}S + N} = \frac{2}{7}\)

Cross-multiply:

\(7 \times (\frac{1}{5}S + M) = 2 \times (\frac{4}{5}S + N)\)

\(\frac{7}{5}S + 7M = \frac{8}{5}S + 2N\)

Rearrange the terms to get S on one side and M and N on the other:

\(7M - 2N = \frac{8}{5}S - \frac{7}{5}S\)

\(7M - 2N = \frac{1}{5}S\)

Multiply both sides by 5:

\(35M - 10N = S\)

So, we have our second relationship: \(S = 35M - 10N\).

Using the Given Condition N - M = 4

We have a system of equations:

1. \(S = 15M\)
2. \(S = 35M - 10N\)
3. \(N - M = 4\)

From equation 3, we can express N in terms of M: \(N = M + 4\).

Now, substitute \(S\) from equation 1 into equation 2:

\(15M = 35M - 10N\)

Subtract 15M from both sides:

\(0 = 20M - 10N\)

Divide by 10:

\(0 = 2M - N\)

This gives us a simple relationship between M and N: \(N = 2M\).

Now we have two expressions for N:

• \(N = M + 4\)
• \(N = 2M\)

Set them equal to each other to find the value of M:

\(M + 4 = 2M\)

Subtract M from both sides:

\(4 = M\)

So, M = 4 litres.

Now find N using either equation. Using \(N = M + 4\):

\(N = 4 + 4\)

\(N = 8\)

So, N = 8 litres.

Calculating the Initial Quantity of Solution

We found the relationship \(S = 15M\). Now substitute the value of M = 4 into this equation:

\(S = 15 \times 4\)

\(S = 60\)

The initial quantity of the solution is 60 litres.

Verification

Let's verify the values of S, M, and N with the ratios.

Initial solution: \(S = 60\) litres.

• Initial milk = \(\frac{1}{5} \times 60 = 12\) litres
• Initial water = \(\frac{4}{5} \times 60 = 48\) litres

Initial ratio: 12 : 48 = 1 : 4 (Correct)

Add M = 4 litres of milk:

• Milk = 12 + 4 = 16 litres
• Water = 48 litres

New ratio: 16 : 48 = 1 : 3 (Correct)

Add N = 8 litres of water:

• Milk = 16 litres
• Water = 48 + 8 = 56 litres

New ratio: 16 : 56 = Divide both by 8: 2 : 7 (Correct)

Also, check the condition N - M = 4:

\(8 - 4 = 4\) (Correct)

All conditions are satisfied, so the initial quantity of solution is indeed 60 litres.

Summary of Steps

1. Set up expressions for initial milk and water quantities based on the total initial solution \(S\) and the initial ratio 1:4.
2. Formulate an equation based on the first addition (M litres of milk) and the resulting ratio 1:3. Simplify this to find a relationship between S and M.
3. Formulate a second equation based on the second addition (N litres of water) and the resulting ratio 2:7. This involves the quantities after the first addition. Simplify this to find a relationship between S, M, and N.
4. Use the given condition \(N - M = 4\) along with the equations from step 2 and 3 to solve for the values of M and N.
5. Substitute the value of M back into the equation relating S and M to find the initial quantity of the solution, S.
6. Verify the answer by checking if all conditions and ratios are satisfied with the calculated values.

The initial quantity of the solution is 60 litres.

Ratio and Mixture Problem Revision

Understanding ratio and mixture problems requires careful tracking of quantities as changes occur. Each step where a component is added or removed alters the overall quantity and the relative proportions of the components.

• Ratio: A ratio compares two or more quantities. A ratio \(a:b\) means the first quantity is \(a\) parts and the second quantity is \(b\) parts. The total number of parts is \(a+b\). If the total mixture is \(T\), the first quantity is \(\frac{a}{a+b}T\) and the second is \(\frac{b}{a+b}T\).
• Mixture Problems: These problems involve combining different substances or changing the composition of a mixture. The key is to keep track of the absolute quantities of each component, not just the ratio, especially when amounts are added or removed.
• Setting up Equations: Each piece of information about the ratio at a specific stage of the mixture can be translated into an algebraic equation. Often, you'll end up with a system of equations that you need to solve simultaneously.
• Solving Simultaneous Equations: Techniques like substitution or elimination are used to find the values of unknown variables. In this problem, we used substitution by expressing S in terms of M and then using the given condition \(N - M = 4\).

Additional Information on Ratio and Proportion

Ratio and proportion are fundamental concepts in mathematics used to compare quantities. A ratio is a comparison of two quantities by division. A proportion is an equation stating that two ratios are equal.

When dealing with mixture problems, it's important to distinguish between adding a component and adding a mixture. Adding a pure component (like just milk or just water) changes the quantity of only that component, while adding another mixture changes the quantities of all components present in the added mixture based on its ratio.

Always ensure that when you set up ratio equations, you are comparing the correct quantities at the specific stage described in the problem.

Practice with different types of mixture problems (adding components, removing components, mixing two different mixtures) to build confidence.