📑 Topics on this page (Jump to Topic):
📌 1. Rule of Alligation
If a cheaper item (price c) and a dearer item (price d) are mixed to get mean price m, then:
📘 Formulas:
- Cheaper quantity : dearer quantity = (d − m) : (m − c)
- The rule applies to any weighted average — percentages, speeds, averages, profit%
| c (cheap) | m (mean) | d (dear) |
|---|---|---|
| ₹40 | ₹50 | ₹60 |
| ratio → | (60−50) : (50−40) | = 1 : 1 |
📌 2. Removal & Replacement
📘 Formulas:
- From x litres, remove y litres and refill with water n times → pure liquid left = x(1 − y/x)n
⚠️ Exam Traps:
- Adding water does not change the quantity of milk — keep it fixed while solving.
- Always use per-unit prices; do not apply alligation to total prices.
📌 3. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. In what ratio must rice at ₹30/kg be mixed with rice at ₹50/kg to get a mixture worth ₹41/kg? Practice
Correct answer: (B)
Solution: Cheap : dear = (50 − 41) : (41 − 30) = 9 : 11 = 9 : 11.
Solution: Cheap : dear = (50 − 41) : (41 − 30) = 9 : 11 = 9 : 11.
Q2. 3 parts of tea at ₹40/kg are mixed with 5 parts at ₹55/kg. The price of the mixture per kg is: Practice
Correct answer: (C)
Solution: Mean = (40×3 + 55×5)/8 = ₹395/8.
Solution: Mean = (40×3 + 55×5)/8 = ₹395/8.
Q3. From a vessel of 60 litres of milk, 10 litres are removed and replaced with water. This is done 2 times in all. How much pure milk is now left? Practice
Correct answer: (D)
Solution: Milk = 60(1 − 10⁄60)2 = 125⁄3 L.
Solution: Milk = 60(1 − 10⁄60)2 = 125⁄3 L.
Q4. A 30-litre mixture contains 10% water. How much water must be added so that water becomes 20%? Practice
Correct answer: (A)
Solution: Milk = 27 L stays fixed. New total = 27 × 100⁄80 = 33.75 L → water added = 3.75 L.
Solution: Milk = 27 L stays fixed. New total = 27 × 100⁄80 = 33.75 L → water added = 3.75 L.
Q5. Two vessels contain milk and water in the ratios 3 : 1 and 1 : 3. If they are mixed in the ratio 2 : 3, milk : water = ? Practice
Correct answer: (B)
Solution: Milk = 3⁄4×2 + 1⁄4×3 = 9⁄4, water = 11⁄4 → 9 : 11.
Solution: Milk = 3⁄4×2 + 1⁄4×3 = 9⁄4, water = 11⁄4 → 9 : 11.
Q6. In a class, 10 boys average 50 marks and 30 girls average 80. The class average is: Practice
Correct answer: (C)
Solution: (50×10 + 80×30)/40 = 72.5. (Alligation also works with ratio 10:30)
Solution: (50×10 + 80×30)/40 = 72.5. (Alligation also works with ratio 10:30)
Q7. 1 parts of tea at ₹60/kg are mixed with 1 parts at ₹80/kg. The price of the mixture per kg is: Practice
Correct answer: (D)
Solution: Mean = (60×1 + 80×1)/2 = ₹70.
Solution: Mean = (60×1 + 80×1)/2 = ₹70.
Q8. From a vessel of 100 litres of milk, 20 litres are removed and replaced with water. This is done 2 times in all. How much pure milk is now left? Practice
Correct answer: (A)
Solution: Milk = 100(1 − 20⁄100)2 = 64 L.
Solution: Milk = 100(1 − 20⁄100)2 = 64 L.
Q9. A 40-litre mixture contains 10% water. How much water must be added so that water becomes 25%? Practice
Correct answer: (B)
Solution: Milk = 36 L stays fixed. New total = 36 × 100⁄75 = 48 L → water added = 8 L.
Solution: Milk = 36 L stays fixed. New total = 36 × 100⁄75 = 48 L → water added = 8 L.
Q10. Two vessels contain milk and water in the ratios 2 : 3 and 4 : 1. If they are mixed in the ratio 2 : 3, milk : water = ? Practice
Correct answer: (C)
Solution: Milk = 2⁄5×2 + 4⁄5×3 = 16⁄5, water = 9⁄5 → 16 : 9.
Solution: Milk = 2⁄5×2 + 4⁄5×3 = 16⁄5, water = 9⁄5 → 16 : 9.
Q11. In what ratio should pulses at ₹24/kg and ₹44/kg be mixed so that selling the mixture at ₹42/kg gives 25% profit? Practice
Correct answer: (D)
Solution: CP of mixture = 42 × 100⁄125 = ₹168⁄5; ratio = (44 − 168⁄5) : (168⁄5 − 24) = 13 : 12.
Solution: CP of mixture = 42 × 100⁄125 = ₹168⁄5; ratio = (44 − 168⁄5) : (168⁄5 − 24) = 13 : 12.
Q12. In a class, 30 boys average 50 marks and 20 girls average 80. The class average is: Practice
Correct answer: (A)
Solution: (50×30 + 80×20)/50 = 62. (Alligation also works with ratio 30:20)
Solution: (50×30 + 80×20)/50 = 62. (Alligation also works with ratio 30:20)
Q13. In what ratio must rice at ₹45/kg be mixed with rice at ₹75/kg to get a mixture worth ₹64/kg? Practice
Correct answer: (B)
Solution: Cheap : dear = (75 − 64) : (64 − 45) = 11 : 19 = 11 : 19.
Solution: Cheap : dear = (75 − 64) : (64 − 45) = 11 : 19 = 11 : 19.
Q14. 3 parts of tea at ₹60/kg are mixed with 5 parts at ₹80/kg. The price of the mixture per kg is: Practice
Correct answer: (C)
Solution: Mean = (60×3 + 80×5)/8 = ₹145/2.
Solution: Mean = (60×3 + 80×5)/8 = ₹145/2.
Q15. From a vessel of 50 litres of milk, 4 litres are removed and replaced with water. This is done 3 times in all. How much pure milk is now left? Practice
Correct answer: (D)
Solution: Milk = 50(1 − 4⁄50)3 = 38.9344 L.
Solution: Milk = 50(1 − 4⁄50)3 = 38.9344 L.
Q16. Two vessels contain milk and water in the ratios 3 : 2 and 1 : 2. If they are mixed in the ratio 2 : 3, milk : water = ? Practice
Correct answer: (A)
Solution: Milk = 3⁄5×2 + 1⁄3×3 = 11⁄5, water = 14⁄5 → 11 : 14.
Solution: Milk = 3⁄5×2 + 1⁄3×3 = 11⁄5, water = 14⁄5 → 11 : 14.
Q17. In a class, 10 boys average 40 marks and 30 girls average 60. The class average is: Practice
Correct answer: (B)
Solution: (40×10 + 60×30)/40 = 55. (Alligation also works with ratio 10:30)
Solution: (40×10 + 60×30)/40 = 55. (Alligation also works with ratio 10:30)
Q18. In what ratio must rice at ₹28/kg be mixed with rice at ₹40/kg to get a mixture worth ₹32/kg? Practice
Correct answer: (C)
Solution: Cheap : dear = (40 − 32) : (32 − 28) = 8 : 4 = 2 : 1.
Solution: Cheap : dear = (40 − 32) : (32 − 28) = 8 : 4 = 2 : 1.
Q19. 1 parts of tea at ₹20/kg are mixed with 4 parts at ₹30/kg. The price of the mixture per kg is: Practice
Correct answer: (D)
Solution: Mean = (20×1 + 30×4)/5 = ₹28.
Solution: Mean = (20×1 + 30×4)/5 = ₹28.
Q20. From a vessel of 40 litres of milk, 10 litres are removed and replaced with water. This is done 3 times in all. How much pure milk is now left? Practice
Correct answer: (A)
Solution: Milk = 40(1 − 10⁄40)3 = 16.875 L.
Solution: Milk = 40(1 − 10⁄40)3 = 16.875 L.