In a mixture, milk and water are in ratio of 2 : 3. Some milk is added...
Let quantity of total mixture = k
Quantity of milk = 2k/5 and Quantity of water in it = k - 2k/5 = 3k/5
New ratio = 2 : 1
Let quantity of milk to be added further be x litres
Then, milk : water = (2k/5 + x) / (3k/5)
Now, = (2k/5 + x) / (3k/5) = 2/1
X = 4k/5
Quantity milk that was added as a percentage of initial mixture = (x/k) × 100 = 80%
∴ Option C is correct answer
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In a mixture, milk and water are in ratio of 2 : 3. Some milk is added...
To solve this problem, we can use the concept of ratios and proportions. Let's break down the given information and solve step by step.
Given:
- The initial ratio of milk and water in the mixture is 2:3.
- The ratio of milk and water after adding some milk becomes 2:1.
Let's assume the initial mixture contains 2x units of milk and 3x units of water.
Step 1: Finding the amount of milk and water in the mixture after adding some milk.
- After adding some milk, the ratio of milk and water becomes 2:1.
- This means the new mixture contains 2y units of milk and y units of water, where y is some positive number.
Step 2: Equating the amount of milk in the initial mixture to the amount of milk in the new mixture.
- The initial mixture contains 2x units of milk.
- After adding some milk, the new mixture contains 2y units of milk.
- Therefore, 2x = 2y.
Step 3: Finding the value of y.
- Dividing both sides of the equation by 2, we get x = y.
- This means the amount of milk added is equal to the amount of milk in the initial mixture.
Step 4: Finding the amount of milk added as a percentage of the initial mixture.
- The amount of milk added is y units.
- The initial mixture contains 2x units of milk.
- So, the percentage of milk added is (y/2x) * 100.
Step 5: Substituting the values of x and y.
- We know x = y.
- Therefore, the percentage of milk added is (y/2y) * 100 = 1/2 * 100 = 50%.
Therefore, the amount of milk added as a percentage of the initial mixture is 50%. Hence, the correct answer is option 'D' - 50%.