Jar A has 3x litres of milk and Jar B has x liters of water. 25 liter each of milk and water was taken out from the respective jars and the resultant quantities from both the jars were mixed in Jar C. If the ratio of milk to water in Jar C was 19:3, what was the initial quantity of milk in Jar A? (in litres)

Options

A. 90
B. 60
C. 180
D. 120

Answer

To solve the problem step by step, we will follow the information given in the question and the video transcript.

### Step 1: Understand the quantities in the jars
- Jar A contains \( 3x \) liters of milk.
- Jar B contains \( x \) liters of water.

### Step 2: Determine the quantities after extraction
- From Jar A, 25 liters of milk is taken out. Therefore, the remaining milk in Jar A is:
\[ 3x - 25 \text{ liters} \]
- From Jar B, 25 liters of water is taken out. Therefore, the remaining water in Jar B is:
\[ x - 25 \text{ liters} \]

### Step 3: Mix the remaining quantities in Jar C
- The quantities mixed in Jar C are:
- Milk: \( 3x - 25 \) liters
- Water: \( x - 25 \) liters

### Step 4: Set up the ratio
According to the problem, the ratio of milk to water in Jar C is \( 19:3 \). This can be expressed as:
\[ \frac{3x - 25}{x - 25} = \frac{19}{3} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives us:
\[ 3(3x - 25) = 19(x - 25) \]

### Step 6: Expand both sides
Expanding both sides results in:
\[ 9x - 75 = 19x - 475 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 9x - 19x = -475 + 75 \]
\[ -10x = -400 \]

### Step 8: Solve for \( x \)
Dividing both sides by -10 gives:
\[ x = 40 \]

### Step 9: Find the initial quantity of milk in Jar A
Now, substituting \( x \) back into the expression for the initial quantity of milk in Jar A:
\[ 3x = 3 \times 40 = 120 \text{ liters} \]

### Conclusion
The initial quantity of milk in Jar A is **120 liters**.