DEBUG_INFO: title=Mixture & Alligation, type=object, isArray=, length=5
Visualizing Balance.
- Concept: If 10kg is at 20Rs and 20kg is at 50Rs.
- Balance: Weight Ratio 1:2. The balance point shifts towards 50. Mean = 40.
Example:
Q: Avg of Class A (20 students) is 60. Class B (30 students) is 70.
Solution: Ratio 2:3. Gap 10. Break Gap in 3:2 inverse. 3/5 of 10 is 6. 60+6=66. To find Ratio when Mean is given.
- Rule: Diagonally subtract. тИгDтИТMтИг and тИгMтИТCтИг.
- Important: All units must be same (e.g., all CP or all SP).
Example:
Q: Rice @ 30 mixed with Rice @ 40 to get 33.
Solution: 30 ... 40
..33..
Diff: 7 : 3. Ratio 7:3.
Repeated dilution logic.
- Formula: Rem=Initial├Ч(1тИТCapacityOutтАЛ)n.
- Use for pure liquid left after n operations.
Example:
Q: 100L Milk. 10L replaced with water. 2 times.
Solution: Rem=100(1тИТ10/100)2=100(0.9)2=81L. If mixture sold at Cost Price of Milk yields Profit.
- Hack: Profit % is exactly the quantity of Water added (if Water is free).
- Rule: If 20% Profit, Ratio Water:Milk = 20:100 = 1:5.
Example:
Q: Sold at CP, gained 25%. Find Water:Milk.
Solution: Gain 25%. Values: Water=25, Milk=100. Ratio 1:4.
When only water is added, Milk remains constant.
- Method: Equate the constant part in both ratios.
- If Milk:Water changes 4:1тЖТ4:3. Milk is same (4). Water increased by 2 units.
Example:
Q: Mix 4:1. Add 10L water, becomes 4:3.
Solution: Milk constant. Water gap 3тИТ1=2 units. 2u=10LтЖТ1u=5L.