Mixture A contains chocolate and milk in the ratio 4 ∶ 3 and mixture B contains chocolate and milk in the ratio 5 ∶ 2. A and B are taken in the ratio 5 ∶ 6 and mixed to form a new mixture. The percentage of chocolate in the new mixture is closest to:
65%
Solving the Mixture Problem: Finding Chocolate Percentage
This problem involves combining two different mixtures with specific ratios of chocolate and milk and then determining the percentage of chocolate in the resulting new mixture.
Let's break down the problem step-by-step to find the percentage of chocolate in the new mixture.
Understanding the Composition of Mixtures A and B
Mixture A contains chocolate and milk in the ratio 4:3.
- Total parts in Mixture A = $4 + 3 = 7$ parts.
- Fraction of chocolate in Mixture A = $\frac{4}{7}$.
- Fraction of milk in Mixture A = $\frac{3}{7}$.
Mixture B contains chocolate and milk in the ratio 5:2.
- Total parts in Mixture B = $5 + 2 = 7$ parts.
- Fraction of chocolate in Mixture B = $\frac{5}{7}$.
- Fraction of milk in Mixture B = $\frac{2}{7}$.
Combining Mixtures A and B
Mixture A and Mixture B are taken in the ratio 5:6 and mixed to form a new mixture.
Let's assume we take 5 units of Mixture A and 6 units of Mixture B. The total quantity of the new mixture will be $5 + 6 = 11$ units.
Calculating the Amount of Chocolate and Milk in the New Mixture
Now, we calculate the amount of chocolate and milk contributed by the taken quantities of Mixture A and Mixture B.
- Amount of chocolate from 5 units of Mixture A = $5 \times (\text{Fraction of chocolate in A}) = 5 \times \frac{4}{7} = \frac{20}{7}$ units.
- Amount of milk from 5 units of Mixture A = $5 \times (\text{Fraction of milk in A}) = 5 \times \frac{3}{7} = \frac{15}{7}$ units.
- Amount of chocolate from 6 units of Mixture B = $6 \times (\text{Fraction of chocolate in B}) = 6 \times \frac{5}{7} = \frac{30}{7}$ units.
- Amount of milk from 6 units of Mixture B = $6 \times (\text{Fraction of milk in B}) = 6 \times \frac{2}{7} = \frac{12}{7}$ units.
Total amount of chocolate in the new mixture = Amount of chocolate from A + Amount of chocolate from B
$$ \text{Total Chocolate} = \frac{20}{7} + \frac{30}{7} = \frac{20 + 30}{7} = \frac{50}{7} \text{ units} $$
Total amount of milk in the new mixture = Amount of milk from A + Amount of milk from B
$$ \text{Total Milk} = \frac{15}{7} + \frac{12}{7} = \frac{15 + 12}{7} = \frac{27}{7} \text{ units} $$
The total quantity of the new mixture is the sum of the total chocolate and total milk:
$$ \text{Total Mixture} = \frac{50}{7} + \frac{27}{7} = \frac{50 + 27}{7} = \frac{77}{7} = 11 \text{ units} $$
This matches our initial assumption of taking 5 units of A and 6 units of B (total 11 units).
Calculating the Percentage of Chocolate in the New Mixture
The percentage of chocolate in the new mixture is calculated as:
$$ \text{Percentage of Chocolate} = \left( \frac{\text{Total amount of Chocolate}}{\text{Total amount of New Mixture}} \right) \times 100\% $$
Substituting the values we found:
$$ \text{Percentage of Chocolate} = \left( \frac{\frac{50}{7}}{11} \right) \times 100\% $$
$$ \text{Percentage of Chocolate} = \left( \frac{50}{7 \times 11} \right) \times 100\% = \left( \frac{50}{77} \right) \times 100\% $$
Let's calculate the value:
$$ \frac{50}{77} \approx 0.64935 $$
$$ 0.64935 \times 100\% \approx 64.935\% $$
The percentage of chocolate in the new mixture is approximately 64.935%.
Finding the Closest Option
Comparing 64.935% with the given options:
- 35%
- 69%
- 31%
- 65%
The value 64.935% is closest to 65%.
| Mixture | Chocolate Ratio | Milk Ratio | Total Parts | Fraction of Chocolate | Fraction of Milk |
|---|---|---|---|---|---|
| A | 4 | 3 | 7 | $\frac{4}{7}$ | $\frac{3}{7}$ |
| B | 5 | 2 | 7 | $\frac{5}{7}$ | $\frac{2}{7}$ |
| Mixture | Ratio Taken | Assumed Quantity | Chocolate Amount | Milk Amount |
|---|---|---|---|---|
| A | 5 | 5 units | $5 \times \frac{4}{7} = \frac{20}{7}$ | $5 \times \frac{3}{7} = \frac{15}{7}$ |
| B | 6 | 6 units | $6 \times \frac{5}{7} = \frac{30}{7}$ | $6 \times \frac{2}{7} = \frac{12}{7}$ |
| New Mixture Component | Total Amount |
|---|---|
| Total Chocolate | $\frac{20}{7} + \frac{30}{7} = \frac{50}{7}$ |
| Total Milk | $\frac{15}{7} + \frac{12}{7} = \frac{27}{7}$ |
| Total New Mixture | $\frac{50}{7} + \frac{27}{7} = \frac{77}{7} = 11$ |
Conclusion
The percentage of chocolate in the new mixture is approximately 64.935%, which is closest to 65%.
Revision Table: Key Concepts in Mixture Problems
| Concept | Explanation | How it applies here |
|---|---|---|
| Ratio | A comparison of two quantities. e.g., a:b means a parts of one, b parts of another. | Used for component proportions (chocolate:milk) and mixture quantities (A:B). |
| Fraction of Component | Amount of a component divided by the total amount of the mixture. Calculated as $\frac{\text{Part}}{\text{Total Parts}}$. | Calculated for chocolate and milk in Mixtures A and B. |
| Combining Mixtures | When mixtures are combined, the total amount of each component (e.g., chocolate, milk) is the sum of the amounts from the individual mixtures. | We summed the amounts of chocolate from the taken quantities of A and B. |
| Percentage Calculation | $(\frac{\text{Part}}{\text{Whole}}) \times 100\%$. Used to express a fraction as a percentage. | Used to find the percentage of chocolate in the final combined mixture. |
Additional Information: Ratios and Proportions in Mixtures
Mixture problems often rely on understanding ratios and how quantities scale. A ratio like 4:3 for chocolate:milk means for every 4 units of chocolate, there are 3 units of milk. The total is 7 units. If you have a certain amount of this mixture, say $X$ units, the amount of chocolate will be $X \times \frac{4}{7}$ and the amount of milk will be $X \times \frac{3}{7}$.
When mixing two solutions, the total amount of solute (like chocolate) and the total amount of solvent (like milk, though here milk is also a component) are simply added together from the contributing solutions. The concentration (or percentage) in the final mixture depends on the total amount of the component and the total amount of the final mixture.
In this problem, even though Mixtures A and B have the same total number of parts in their internal ratios (7), they are mixed in a different external ratio (5:6). This is why the final percentage isn't simply the average of the percentages in A and B. We must account for the different quantities of A and B being mixed.
The method used here, calculating the total amount of the component (chocolate) and dividing by the total amount of the new mixture, is a standard approach for such problems.
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