Quantitative Aptitude — Mixture & Alligation

Let the cost of the cheaper pulse be C1 = Rs. 60/kg. Let the cost of the dearer pulse be C2 = Rs. 80/kg. Let the mean cost of the mixture be M = Rs. 65/kg. Using the Alligation Rule: Quantity of cheaper pulse / Quantity of dearer pulse = (C2 - M) / (M - C1) Ratio = (80 - 65) / (65 - 60) Ratio = 15 / 5 Ratio = 3 / 1 So, the pulses must be mixed in the ratio 3:1.

For Vessel 1: Milk proportion = 3/(3+1) = 3/4 Water proportion = 1/(3+1) = 1/4 For Vessel 2: Milk proportion = 5/(5+3) = 5/8 Water proportion = 3/(5+3) = 3/8 Since the contents are mixed in the ratio 1:1, we can assume equal quantities from each vessel. To simplify calculations, let's assume we take 8 units from each vessel (LCM of denominators 4 and 8). From Vessel 1 (8 units): Quantity of Milk = (3/4) * 8 = 6 units Quantity of Water = (1/4) * 8 = 2 units From Vessel 2 (8 units): Quantity of Milk = (5/8) * 8 = 5 units Quantity of Water = (3/8) * 8 = 3 units In the new mixture: Total Milk = 6 + 5 = 11 units Total Water = 2 + 3 = 5 units The ratio of milk to water in the new mixture is 11:5.

Initial quantity of milk (Q) = 40 litres. Quantity of milk taken out and replaced (x) = 4 litres. Number of times the process is repeated (n) = 1 (first time) + 1 (once more) = 2 times. Formula for final quantity of milk after 'n' operations: Final Quantity = Q * (1 - x/Q)^n Final Quantity = 40 * (1 - 4/40)^2 Final Quantity = 40 * (1 - 1/10)^2 Final Quantity = 40 * (9/10)^2 Final Quantity = 40 * (81/100) Final Quantity = (40 * 81) / 100 Final Quantity = 3240 / 100 Final Quantity = 32.4 litres.

Selling Price (SP) of the mixture = Rs. 154 per kg. Profit percentage = 10%. First, calculate the Cost Price (CP) of the mixture: CP = SP / (1 + Profit/100) CP = 154 / (1 + 10/100) CP = 154 / (1 + 0.10) CP = 154 / 1.1 CP = 140 Rs/kg. Now, use the Alligation Rule with the cost prices: Cost of Tea 1 (C1) = Rs. 120/kg Cost of Tea 2 (C2) = Rs. 180/kg Mean Cost (M) = Rs. 140/kg Ratio of Tea 1 : Tea 2 = (C2 - M) : (M - C1) Ratio = (180 - 140) : (140 - 120) Ratio = 40 : 20 Ratio = 2 : 1. So, the two types of tea were mixed in the ratio 2:1.

Initial total mixture = 60 litres. Initial percentage of water = 10%. Initial quantity of water = 10% of 60 = (10/100) * 60 = 6 litres. Initial quantity of milk = 60 - 6 = 54 litres. Let 'x' litres of water be added to the mixture. New total mixture quantity = (60 + x) litres. New quantity of water = (6 + x) litres. Quantity of milk remains constant = 54 litres. In the new mixture, water is 25%, which means milk is (100 - 25)% = 75%. So, 54 litres of milk represents 75% of the new total mixture. 0.75 * (60 + x) = 54 (3/4) * (60 + x) = 54 60 + x = 54 * (4/3) 60 + x = 18 * 4 60 + x = 72 x = 72 - 60 x = 12 litres. Therefore, 12 litres of water should be added.

Let the initial total quantity of the mixture be Q litres. Initial ratio of milk to water = 4:1. Initial proportion of milk = 4/(4+1) = 4/5. When 12 litres of the mixture is taken out, the quantity of milk removed is (4/5) * 12 litres. Remaining quantity of milk = (4/5)Q - (4/5)*12 = (4/5)(Q - 12) litres. When 12 litres of water is added, the total volume of the mixture is restored to Q litres. The quantity of milk remains the same as in the previous step because only water was added. In the new mixture, the ratio of milk to water is 7:3. New proportion of milk = 7/(7+3) = 7/10. So, the quantity of milk in the new mixture is (7/10)Q. Equating the milk quantities: (4/5)(Q - 12) = (7/10)Q Multiply both sides by 10 to clear denominators: 8(Q - 12) = 7Q 8Q - 96 = 7Q 8Q - 7Q = 96 Q = 96 litres. The initial quantity of the mixture was 96 litres.

Let the quantities of the three varieties of rice be 2x, 3x, and 5x kg respectively. Cost of 1st variety = Rs. 10/kg Cost of 2nd variety = Rs. 15/kg Cost of 3rd variety = Rs. 20/kg Total quantity of the mixture = 2x + 3x + 5x = 10x kg. Total cost of the mixture: Cost = (Quantity1 * Price1) + (Quantity2 * Price2) + (Quantity3 * Price3) Cost = (2x * 10) + (3x * 15) + (5x * 20) Cost = 20x + 45x + 100x Cost = 165x Rs. Average cost of the mixture per kg = Total Cost / Total Quantity Average Cost = 165x / 10x Average Cost = 16.50 Rs/kg.

Percentage of sugar in 1st solution (P1) = 20%. Percentage of sugar in 2nd solution (P2) = 50%. Desired mean percentage of sugar in the mixture (M) = 30%. Using the Alligation Rule: Ratio of 1st solution : 2nd solution = (P2 - M) : (M - P1) Ratio = (50 - 30) : (30 - 20) Ratio = 20 : 10 Ratio = 2 : 1. So, the two solutions should be mixed in the ratio 2:1. Given that 20 litres of the first solution is used. Let the quantity of the second solution be Y litres. (Quantity of 1st solution) / (Quantity of 2nd solution) = 2/1 20 / Y = 2 / 1 2Y = 20 Y = 10 litres. Therefore, 10 litres of the second solution is used.

Initial quantity of milk (Q) = 100 litres. Quantity of milk taken out and replaced (x) = 10 litres. Total number of times the process is repeated (n) = 1 (first time) + 2 (two more times) = 3 times. Formula for the final quantity of milk after 'n' operations: Final Quantity = Q * (1 - x/Q)^n Final Quantity = 100 * (1 - 10/100)^3 Final Quantity = 100 * (1 - 1/10)^3 Final Quantity = 100 * (9/10)^3 Final Quantity = 100 * (729/1000) Final Quantity = 729 / 10 Final Quantity = 72.9 litres. So, 72.9 litres of milk is now in the container.

From Vessel A: Ratio of Milk : Water = 4:1 Total quantity drawn = 10 litres Quantity of Milk from A = (4/5) * 10 = 8 litres Quantity of Water from A = (1/5) * 10 = 2 litres From Vessel B: Ratio of Milk : Water = 7:3 Total quantity drawn = 10 litres Quantity of Milk from B = (7/10) * 10 = 7 litres Quantity of Water from B = (3/10) * 10 = 3 litres When contents are mixed in Vessel C: Total Milk in C = Milk from A + Milk from B = 8 + 7 = 15 litres Total Water in C = Water from A + Water from B = 2 + 3 = 5 litres Ratio of Milk to Water in Vessel C = 15 : 5 Simplifying the ratio by dividing by 5: 3 : 1.