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A container contains a mixture of two liquids, A and B, in the proponion 7 : 5. If 9 liters of the m.ixmre is replaced by 9 liters of
liquid B, then the ratio of the two liquids becomes 7 : 9. How much of liquid A was there in the container initially?
21 liters
35 liters
40 liters
19 liters
Let’s break it down:
- The mixture is first in ratio A:B = 7:5.
- Let the total volume be x liters. So, A = (7/12)x and B = (5/12)x.
- You remove 9 liters of the mixture. That’s 9*(7/12) = 5.25 liters A and 9*(5/12) = 3.75 liters B.
- You add back 9 liters of B.
- After this, you have: A = (7/12)x – 5.25; B = (5/12)x – 3.75 + 9.
- New ratio: (A) : (B) = 7 : 9.
So, set up the equation:
[(7/12)x – 5.25] / [(5/12)x – 3.75 + 9] = 7/9
- Cross-multiplying: 9[(7/12)x – 5.25] = 7[(5/12)x + 5.25]
- Distribute: (21/4)x – 47.25 = (35/12)x + 36.75
- Get all x on one side: (21/4)x – (35/12)x = 36.75 + 47.25
- LCM of 4 and 12 is 12, so: (63/12)x – (35/12)x = 84
- (28/12)x = 84 ? (7/3)x = 84 ? x = 84 * 3 / 7 = 36
- Initial A = (7/12)x = (7/12)*36 = 21
So, here’s how the options play out:
1. 21 liters ? This is correct.
2. 35 liters (that’s if x?=?60, which doesn’t fit the ratios and process)
3. 40 liters (similarly, doesn’t match the equations)
4. 19 liters (again, doesn’t satisfy the ratio or process)
By: santosh ProfileResourcesReport error
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