Question 1

Two containers have milk and water in ratios 3:1 and 5:3 respectively. If equal quantities from each container are mixed together, what is the ratio of milk to water in the final mixture?

Accepted Answer

Step 1: Container 1 has milk:water = 3:1, so in 4 parts: milk = 3 parts, water = 1 part. Step 2: Container 2 has milk:water = 5:3, so in 8 parts: milk = 5 parts, water = 3 parts. Step 3: Let equal quantity = 4 litres from each (LCM of 4 and 8 = 8, so use 4 from each for simplicity). Step 4: From Container 1 (4L): milk = 3L, water = 1L. Step 5: From Container 2 (4L): milk = 2.5L, water = 1.5L. Step 6: Total milk = 3 + 2.5 = 5.5L, total water = 1 + 1.5 = 2.5L. Step 7: Ratio = 5.5:2.5 = 11:5.

Question 2

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg. In what ratio should they be mixed so that the cost of the mixture is ₹50 per kg?

Accepted Answer

Step 1: Use alligation method. Cost of cheaper rice = ₹40, cost of dearer rice = ₹60, cost of mixture = ₹50. Step 2: Difference between mixture cost and cheaper cost = 50 - 40 = 10. Step 3: Difference between dearer cost and mixture cost = 60 - 50 = 10. Step 4: Ratio of cheaper to dearer = 10:10 = 1:1.

Question 3

Three containers have milk in concentrations 20%, 30%, and 50% respectively. Equal quantities are taken from each and mixed. What is the milk concentration in the final mixture?

Accepted Answer

Step 1: Let equal quantity from each = 1 litre (or any unit). Step 2: Milk from container 1 = 1 × 20% = 0.2 litres. Step 3: Milk from container 2 = 1 × 30% = 0.3 litres. Step 4: Milk from container 3 = 1 × 50% = 0.5 litres. Step 5: Total milk = 0.2 + 0.3 + 0.5 = 1.0 litre. Step 6: Total mixture = 1 + 1 + 1 = 3 litres. Step 7: Concentration = (1.0/3) × 100% = 33.33% ≈ 33⅓%.

Question 4

Two containers have sugar solutions of concentrations 20% and 50% respectively. In what ratio should they be mixed to obtain a 30% solution?

Accepted Answer

Step 1: Use alligation method. Concentration 1 = 20%, Concentration 2 = 50%, Target = 30%. Step 2: Difference from target: |20 - 30| = 10, |50 - 30| = 20. Step 3: Ratio of quantities = (difference for higher concentration) : (difference for lower concentration) = 20 : 10 = 2 : 1.

Question 5

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. At what price per kg should he sell the mixture to gain 25% profit?

Accepted Answer

Step 1: Cost price of mixture per kg using alligation. Ratio = 3:2, so total parts = 5. Step 2: Cost = (3 × 40 + 2 × 60)/5 = (120 + 120)/5 = 240/5 = ₹48 per kg. Step 3: For 25% profit, selling price = 48 × 1.25 = ₹60 per kg.

Question 6

A container has 120 litres of a mixture of alcohol and water in the ratio 7:5. How much pure alcohol must be added to make the ratio 3:2?

Accepted Answer

Step 1: Original mixture: alcohol = 7/12 × 120 = 70 litres, water = 5/12 × 120 = 50 litres. Step 2: Let x litres of pure alcohol be added. New alcohol = 70 + x, water remains 50. Step 3: New ratio (70 + x):50 = 3:2. Step 4: 2(70 + x) = 3 × 50 → 140 + 2x = 150 → 2x = 10 → x = 5 litres.

Question 7

A chemist has two solutions with acid concentrations of 30% and 60%. He mixes them in a certain ratio and obtains 10 litres of 45% acid solution. How many litres of the 30% solution was used?

Accepted Answer

Step 1: Use alligation. Target concentration = 45%. Concentration 1 = 30%, Concentration 2 = 60%. Step 2: Differences: |60 - 45| = 15, |30 - 45| = 15. Step 3: Ratio = 15:15 = 1:1. Step 4: Equal parts of each solution. Step 5: From 10 litres total, 30% solution = 10/2 = 5 litres.

Question 8

Two alloys contain copper and tin in the ratios 3:2 and 5:4 respectively. In what ratio should these alloys be mixed to get a new alloy with copper and tin in the ratio 4:3?

Accepted Answer

Step 1: Alloy A has copper fraction = 3/5, tin fraction = 2/5. Alloy B has copper fraction = 5/9, tin fraction = 4/9. Step 2: Let alloys be mixed in ratio m:n. Step 3: Copper in mixture = (3m/5 + 5n/9)/(m+n), should equal 4/7. Step 4: (3m/5 + 5n/9)/(m+n) = 4/7. Step 5: 7(3m/5 + 5n/9) = 4(m+n) → 21m/5 + 35n/9 = 4m + 4n. Step 6: Multiply by 45: 189m + 175n = 180m + 180n → 9m = 5n → m:n = 5:9.

Question 9

A chemist mixes three solutions A, B, and C with acid concentrations 20%, 30%, and 50% respectively. If equal volumes of A and B are mixed, the resulting concentration is 25%. Now, if this mixture is combined with solution C in the ratio 2:1, what is the final acid concentration?

Accepted Answer

Step 1: Mix equal volumes of A (20%) and B (30%): concentration = (20 + 30)/2 = 25%. Step 2: This 25% mixture is combined with C (50%) in ratio 2:1. Step 3: Final concentration = (2 × 25 + 1 × 50)/(2 + 1) = (50 + 50)/3 = 100/3 ≈ 33.33%. Step 4: Expressed as fraction = 33⅓% or 100/3 %.

Question 10

A merchant has two types of tea costing ₹80 per kg and ₹120 per kg. He mixes them and sells the mixture at ₹117 per kg, making a profit of 30%. In what ratio should he mix the two types to achieve this profit margin?

Accepted Answer

Step 1: Selling price = ₹117 per kg, profit = 30%. Cost price = 117/1.30 = ₹90 per kg. Step 2: Let the ratio of ₹80 tea to ₹120 tea be x:y. Step 3: Weighted average cost = (80x + 120y)/(x + y) = 90. Step 4: 80x + 120y = 90x + 90y → 30y = 10x → x:y = 3:1.

SSC GD Constable Mixture & Alligation

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