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 any quantity, milk fraction = 3/4, water fraction = 1/4. Step 2: Container 2 has milk:water = 5:3, so milk fraction = 5/8, water fraction = 3/8. Step 3: Let equal quantity = L litres from each. Step 4: Total milk = L(3/4) + L(5/8) = L(6/8 + 5/8) = L(11/8). Step 5: Total water = L(1/4) + L(3/8) = L(2/8 + 3/8) = L(5/8). Step 6: Ratio of milk to water = (11/8):(5/8) = 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

A container has 120 litres of a solution with alcohol and water in ratio 7:5. How much pure alcohol must be added to make the ratio 3:1?

Accepted Answer

Step 1: Original alcohol = 120 × (7/12) = 70 litres, water = 120 × (5/12) = 50 litres. Step 2: Let x litres of pure alcohol be added. Step 3: New alcohol = 70 + x, water remains 50 litres. Step 4: New ratio is 3:1, so (70 + x)/50 = 3/1. Step 5: 70 + x = 150 → x = 80 litres.

Question 4

Three containers have solutions with salt concentrations of 20%, 30%, and 50% respectively. If equal volumes are mixed from all three, what is the salt concentration in the final mixture?

Accepted Answer

Step 1: Let equal volume from each = V litres. Step 2: Salt from container 1 = V × 0.20 = 0.20V. Step 3: Salt from container 2 = V × 0.30 = 0.30V. Step 4: Salt from container 3 = V × 0.50 = 0.50V. Step 5: Total salt = 0.20V + 0.30V + 0.50V = 1.00V. Step 6: Total volume = V + V + V = 3V. Step 7: Concentration = (1.00V)/(3V) × 100% = (1.00/3) × 100% = 33.33% ≈ 33⅓%.

Question 5

A chemist mixes two solutions of acid with concentrations 30% and 50%. If 40 litres of the first solution and 60 litres of the second solution are mixed, what is the concentration of acid in the resulting mixture?

Accepted Answer

Step 1: Acid from first solution = 40 × 0.30 = 12 litres. Step 2: Acid from second solution = 60 × 0.50 = 30 litres. Step 3: Total acid = 12 + 30 = 42 litres. Step 4: Total mixture = 40 + 60 = 100 litres. Step 5: Concentration = (42/100) × 100% = 42%.

Question 6

A merchant has two types of rice costing ₹40 per kg and ₹60 per kg. He mixes them and sells the mixture at ₹55 per kg, making a profit of 10%. In what ratio should he mix the two types of rice?

Accepted Answer

Step 1: Selling price = ₹55 per kg, profit = 10%, so cost price = 55/1.10 = ₹50 per kg. Step 2: Using alligation method: Rice A costs ₹40, Rice B costs ₹60, mixture costs ₹50. Step 3: Difference for A = 50 - 40 = 10. Difference for B = 60 - 50 = 10. Step 4: Ratio = 10:10 = 1:1.

Question 7

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) = 4/7. Step 4: Cross-multiply: 7(3m/5 + 5n/9) = 4(m + n). Step 5: 21m/5 + 35n/9 = 4m + 4n. Step 6: Multiply by 45: 189m + 175n = 180m + 180n → 9m = 5n → m:n = 5:9.

Question 8

A merchant mixes two types of rice costing Rs. 40 per kg and Rs. 60 per kg in a certain ratio. The cost price of the mixture is Rs. 50 per kg. Later, he adds 20 kg of the cheaper rice to 80 kg of the mixture. The new cost price per kg becomes Rs. 48. Find the original ratio of the two types of rice in the mixture.

Accepted Answer

Step 1: Let the ratio of rice at Rs. 40 and Rs. 60 be x:y. Using alligation: (60 - 50)/(50 - 40) = x/y → 10/10 = x/y → x:y = 1:1. Step 2: Verify: In 1:1 ratio, cost = (1×40 + 1×60)/2 = 50 ✓. Step 3: Original mixture of 80 kg has 40 kg of each type. Step 4: After adding 20 kg of cheaper rice: total = 100 kg, cheaper rice = 40 + 20 = 60 kg, costlier rice = 40 kg. Step 5: New cost = (60×40 + 40×60)/100 = (2400 + 2400)/100 = 4800/100 = 48 ✓. Step 6: Original ratio = 1:1.

Question 9

A chemist has three solutions with acid concentrations of 20%, 30%, and 50% respectively. He mixes them in the ratio 2:3:5 to create a final solution. If the final solution has a total volume of 100 litres, what is the concentration of acid in the final mixture?

Accepted Answer

Step 1: Ratio of solutions = 2:3:5, total parts = 10. Step 2: Volume of 20% solution = (2/10) × 100 = 20L, volume of 30% solution = (3/10) × 100 = 30L, volume of 50% solution = (5/10) × 100 = 50L. Step 3: Acid from 20% solution = 20 × 0.20 = 4L. Step 4: Acid from 30% solution = 30 × 0.30 = 9L. Step 5: Acid from 50% solution = 50 × 0.50 = 25L. Step 6: Total acid = 4 + 9 + 25 = 38L. Step 7: Concentration in final mixture = (38/100) × 100% = 38%.

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