Question 1

From a can full of milk, $10$ litres of milk is taken out and replaced with water and this operation is repeated twice. If the capacity of the can is $50$ litres, what is the ratio of water and milk in the can after this operation?

Accepted Answer

Concept & Successive Replacement: When a fixed quantity of a mixture is repeatedly removed and replaced with water, the amount of the original pure liquid remaining can be calculated using the successive dilution formula. Final Quantity = Initial Quantity × (1 - \frac{Replacement Volume}{Total Capacity})^n Where n is the *total* number of times the operation is performed. Step-by-Step Solution: • Given: - Initial quantity of milk (I) = 50 liters - Replacement volume (R) = 10 liters - The operation is done once, and then "repeated twice", meaning the total number of operations (n) is 1 + 2 = 3. • Calculation: 1. Calculate the final volume of milk left after 3 operations: Final Milk = 50 × (1 - (10/50))^3 Final Milk = 50 × (1 - (1/5))^3 Final Milk = 50 × ((4/5))^3 Final Milk = 50 × (64/125) = 2 × (64/5) = (128/5) = 25.6 liters 2. Calculate the final volume of water in the 50-liter can: Final Water = Total Capacity - Final Milk Final Water = 50 - 25.6 = 24.4 liters 3. Find the ratio of water to milk: Ratio = 24.4 : 25.6 = 244 : 256 4. Simplify the ratio by dividing by 4: 61 : 64 5. The ratio 61 : 64 is not present in the given options (2:3, 9:16, 16:9). Exam Strategy & Shortcut: Recognize the linguistic traps in dilution problems. Calculate the multiplier (1 - (10/50)) = (4/5). If the operation is done 3 times total, the final milk fraction is ((4/5))^3 = (64/125). Milk = 64 parts, Water = (125 - 64) = 61 parts. Ratio of Water to Milk = 61 : 64. None of the options match. Common Pitfall: A massive trap is misinterpreting the phrase "and this operation is repeated twice" as meaning the operation was performed a *total* of 2 times. If n=2, the final milk fraction is ((4/5))^2 = (16/25). The ratio of milk to total is 16:25, meaning the ratio of Water to Milk is 9:16, which perfectly matches option (b). However, strict English interpretation of "repeated twice" means 3 total operations. Final Answer: Therefore, the correct answer is None of these.

Question 2

In what ratio must a grocer mix two varieties of pulses costing ₹ 15 and ₹ 20 per kg respectively so as to get a mixture worth ₹ 16.50 per kg ?

Accepted Answer

Concept & Rule of Alligation: The Rule of Alligation allows us to find the ratio in which two ingredients at given prices must be mixed to produce a mixture at a desired mean price. \frac{Quantity of cheaper}{Quantity of dearer} = \frac{CP of dearer - Mean price}{Mean price - CP of cheaper} Step-by-Step Solution: 1. Identify the cost price (CP) of the two varieties: - CP of 1st kind (cheaper) = ₹ 15 - CP of 2nd kind (dearer) = ₹ 20 2. Identify the desired mean price of the mixture = ₹ 16.50 3. Apply the alligation cross-method: - Quantity of cheaper : Quantity of dearer = (20 - 16.50) : (16.50 - 15) - Quantity ratio = 3.50 : 1.50 4. Simplify the ratio: - 3.50 : 1.50 = 35 : 15 = 7 : 3 Exam Strategy & Shortcut: Write the values in a cross format: 15 20 16.50 3.50 1.50 The ratio is directly 3.50 : 1.50 which simplifies to 7 : 3. Common Pitfall: A common mistake is reversing the ratio by subtracting in the wrong order or matching the wrong quantity with the price. Always remember the cross-pattern maps the difference on the opposite side. Final Answer: Therefore, the correct answer is 7 : 3.

Question 3

The cost of Type 1 rice is ₹ 15 per kg and Type 2 rice is ₹ 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is

Accepted Answer

Concept & Weighted Average: When quantities and their respective prices are given, the price of the mixture is simply the weighted average of the individual prices. Mean Price = (n_1p_1 + n_2p_2/n_1 + n_2) Step-by-Step Solution: 1. Price of Type 1 (p_1) = ₹ 15 2. Price of Type 2 (p_2) = ₹ 20 3. Ratio of quantities (n_1 : n_2) = 2 : 3. We can assume quantities are 2 kg and 3 kg. 4. Calculate total cost: - Total cost = (15 × 2) + (20 × 3) = 30 + 60 = 90 5. Calculate total quantity: - Total quantity = 2 + 3 = 5 kg 6. Calculate mean price: - Mean price = Total cost / Total quantity = 90 / 5 = ₹ 18 Exam Strategy & Shortcut: Using assumed mean: Assume mean is 15. The deviation of Type 2 is +5. This +5 occurs 3 times (from the ratio), so total deviation is +15. Distribute this over the total parts (2+3=5): 15 / 5 = +3. Add to assumed mean: 15 + 3 = 18. Common Pitfall: Multiplying the prices by the wrong ratio parts, or adding the prices without weighting them by their respective quantities. Final Answer: Therefore, the correct answer is ₹ 18.

Question 4

In what ratio must a grocer mix two varieties of tea worth ₹ 60 a kg and ₹ 65 a kg so that by selling the mixture at ₹ 68.20 a kg he may gain 10%?

Accepted Answer

Concept & Alligation with Cost Price: The Rule of Alligation must ALWAYS be applied to the Cost Price (CP), not the Selling Price (SP). If SP and profit percentage are given, first calculate the CP of the mixture. CP = \frac{SP × 100}{100 + Profit %} Step-by-Step Solution: 1. Find the CP of the mixture: - Selling Price (SP) = ₹ 68.20 - Gain = 10% - Mean CP = (68.20 × 100/110) = (6820/110) = ₹ 62 2. Apply the alligation method using CPs: - CP of 1st variety = ₹ 60 - CP of 2nd variety = ₹ 65 - Mean CP = ₹ 62 3. Calculate the ratio: - Quantity of 1st : Quantity of 2nd = (65 - 62) : (62 - 60) - Ratio = 3 : 2 Exam Strategy & Shortcut: Quickly deduce the CP: 110% of CP = 68.20 \implies CP = 62. Then do mentally: difference from 65 is 3, difference from 60 is 2. Thus, the ratio is 3 : 2. Common Pitfall: Applying alligation directly to the Selling Price (68.20) alongside the Cost Prices (60 and 65). This is a fatal error and will yield an incorrect result. Always harmonize the units/types to Cost Price. Final Answer: Therefore, the correct answer is 3 : 2.

Question 5

A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is

Accepted Answer

Concept & Profit from Free Ingredient: When a mixture containing a free ingredient (water) and a pure ingredient (milk) is sold at the cost price of the pure ingredient, the profit percentage is equivalent to the ratio of the free ingredient to the pure ingredient. Ratio of Water : Milk = Profit Percentage : 100 Step-by-Step Solution: 1. Understand the source of profit: - The 25% gain is entirely due to the added water, since the milk is sold at its original cost price. 2. Determine the ratio of Water to Milk: - Water : Milk = 25 : 100 = 1 : 4 3. Calculate the total parts in the mixture: - Total mixture = 1 (water) + 4 (milk) = 5 parts 4. Find the percentage of water in the total mixture: - Percentage of water = ((1/5)) × 100% - Percentage of water = 20% Exam Strategy & Shortcut: If gain is P%, the proportion of water in the mixture is (P/100 + P). Here, (25/125) = (1/5) = 20%. This bypasses the ratio step entirely. Common Pitfall: A very common mistake is confusing the "percentage of water to milk" (25%) with the "percentage of water in the mixture" (20%). Always check what the denominator should be. Final Answer: Therefore, the correct answer is 20%.

Question 6

Two vessels A and B contain spirit and water mixed in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixtures be mixed to obtain a new mixture in vessel C containing spirit and water in the ratio 8 : 5 ?

Accepted Answer

Concept & Alligation with Fractions: When mixing two compound mixtures to get a third, apply the Rule of Alligation to the fractional concentration of a single component (e.g., just the spirit). \frac{Quantity of A}{Quantity of B} = \frac{Fraction in B - Mean Fraction}{Mean Fraction - Fraction in A} Step-by-Step Solution: 1. Identify the fraction of spirit in each vessel: - Vessel A: Spirit = 5, Water = 2 \implies Fraction of Spirit = (5/5 + 2) = (5/7) - Vessel B: Spirit = 7, Water = 6 \implies Fraction of Spirit = (7/7 + 6) = (7/13) - Vessel C (Target): Spirit = 8, Water = 5 \implies Fraction of Spirit = (8/8 + 5) = (8/13) 2. Set up the alligation cross for the spirit fractions: - Difference for A = \left|(8/13) - (7/13)\right| = (1/13) - Difference for B = \left|(5/7) - (8/13)\right| = (65 - 56/91) = (9/91) 3. The required ratio of A : B corresponds to the cross differences: - Ratio = (1/13) : (9/91) 4. Simplify the ratio: - Multiply by the LCM of denominators (91): - 91 × ((1/13)) : 91 × ((9/91)) - 7 : 9 Exam Strategy & Shortcut: To avoid fraction subtraction, equalize all denominators first. Fractions are (5/7), (7/13), and (8/13). LCM of 7 and 13 is 91. - A = (65/91) - B = (49/91) - Mean = (56/91) Apply alligation on numerators: 65 and 49 with mean 56. Ratio = (56 - 49) : (65 - 56) = 7 : 9. Common Pitfall: Mixing up the ratios of spirit and water when determining the fraction. Ensure you divide the part by the *total sum of parts* (e.g., 5 / (5+2)), not by the other component. Final Answer: Therefore, the correct answer is 7 : 9.

Question 7

Two vessels A and B contain milk and water mixed in the ratio 8 : 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing $69 \frac{3}{13}\%$ milk, is

Accepted Answer

Concept & Concentration Alligation: Convert mixed fraction percentages into simple fractions, then apply the Rule of Alligation using the fraction of milk in each mixture. Step-by-Step Solution: 1. Convert the target milk percentage to a fraction: - Target milk = 69 (3/13)% = ((69 × 13) + 3/13)% = (897 + 3/13)% = (900/13)% - As a fraction: (900/13) × (1/100) = (9/13) 2. Identify the fraction of milk in vessels A and B: - Vessel A (8 : 5) \implies Fraction of milk = (8/8 + 5) = (8/13) - Vessel B (5 : 2) \implies Fraction of milk = (5/5 + 2) = (5/7) 3. Apply alligation on the milk fractions: - Difference 1 (for A) = (5/7) - (9/13) = (65 - 63/91) = (2/91) - Difference 2 (for B) = (9/13) - (8/13) = (1/13) 4. Determine the ratio of A : B: - Ratio = (2/91) : (1/13) - Multiply by 91: 2 : ((91/13)) = 2 : 7 Exam Strategy & Shortcut: Notice the denominators: 13 and 7. Scale all fractions to a common denominator of 91. - Target: (9/13) = (63/91) - A: (8/13) = (56/91) - B: (5/7) = (65/91) Alligate numerators: (65 - 63) : (63 - 56) = 2 : 7. Common Pitfall: A calculation error when converting the complex mixed fraction 69 (3/13)% into a standard improper fraction. Always double-check the multiplication step: 69 × 13. Final Answer: Therefore, the correct answer is 2 : 7.

Question 8

A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5 ?

Accepted Answer

Concept & Target Ratio Alligation: When given the target ratio of components, first convert that target ratio into a percentage or fraction, then use alligation to find the volume ratio of the two source mixtures. Target Percentage = \frac{Part}{Total} × 100 Step-by-Step Solution: 1. Find the target percentage of water in the final mixture: - Target ratio of water to milk = 3 : 5 - Total parts = 3 + 5 = 8 - Target % of water = ((3/8)) × 100 = 37.5% 2. Identify the percentage of water in the two cans: - Can 1 = 25% water - Can 2 = 50% water 3. Apply the Rule of Alligation using water percentages: - Can 1 proportion = 50% - 37.5% = 12.5% - Can 2 proportion = 37.5% - 25% = 12.5% - Ratio of Can 1 to Can 2 = 12.5 : 12.5 = 1 : 1 4. Calculate the volume needed from each can: - Total required volume = 12 litres. - Since they are mixed in a 1 : 1 ratio, divide the total volume equally. - Volume from Can 1 = 12 × ((1/2)) = 6 litres. - Volume from Can 2 = 12 × ((1/2)) = 6 litres. Exam Strategy & Shortcut: Observe the percentages: 25% and 50%. The target is exactly halfway between them (37.5%). Therefore, they must be mixed in exactly equal parts. 12 / 2 = 6 litres each. Common Pitfall: Using the target ratio numbers (3 and 5) directly in the alligation instead of converting them into an overall fraction or percentage ((3/8)). Final Answer: Therefore, the correct answer is 6 litres, 6 litres.

Question 9

One quality of wheat at ₹ 9.30 per kg is mixed with another quality at a certain rate in the ratio 8 : 7. If the mixture so formed be worth ₹ 10 per kg, what is the rate per kg of the second quality of wheat ?

Accepted Answer

Concept & Weighted Average: The mean value of a mixture is the weighted average of the prices of its components based on their mixing ratio. Mean Price = (n_1p_1 + n_2p_2/n_1 + n_2) Step-by-Step Solution: 1. Define the given variables: - p_1 (price of first quality) = ₹ 9.30 - n_1 (ratio proportion of first quality) = 8 - n_2 (ratio proportion of second quality) = 7 - Mean price = ₹ 10 2. Let the price of the second quality be x. 3. Set up the weighted average equation: - ((8 × 9.30) + (7 × x)/8 + 7) = 10 4. Solve for x: - (74.40 + 7x/15) = 10 - 74.40 + 7x = 150 - 7x = 150 - 74.40 - 7x = 75.60 - x = (75.60/7) = 10.80 Exam Strategy & Shortcut: Use deviation from the mean: The mean is 10. The first type is 9.30, which is -0.70 below the mean. For 8 parts, total deficit = 8 × -0.70 = -5.60. This deficit must be balanced by the second type. So, the 7 parts of the second type must have a total surplus of +5.60. Surplus per part = (5.60/7) = +0.80. Therefore, the price of the second type = 10 + 0.80 = ₹ 10.80. Common Pitfall: Miscalculating the multiplication of decimals (8 × 9.30) leading to an incorrect constant in the algebraic equation. Final Answer: Therefore, the correct answer is ₹ 10.80.

Question 10

A merchant has $1000$ kg of sugar, part of which he sells at $8\%$ profit and the rest at $18\%$ profit. He gains $14\%$ on the whole. The quantity sold at $18\%$ profit is

Accepted Answer

Concept & Rule of Alligation: The rule of alligation is a quick way to find the ratio in which two ingredients or components are mixed to produce a mixture of a desired average value. \frac{Quantity of Cheaper}{Quantity of Dearer} = \frac{Dearer Rate - Mean Rate}{Mean Rate - Cheaper Rate} Step-by-Step Solution: • Given: Total quantity = 1000 kg. Profit on first part = 8%, Profit on second part = 18%, Mean profit = 14%. • By the rule of alligation: • Ratio of quantity at 8% to quantity at 18% = (18 - 14) : (14 - 8) = 4 : 6 = 2 : 3. • Total ratio parts = 2 + 3 = 5. • The quantity sold at 18% profit corresponds to 3 parts out of 5. • Quantity sold at 18% profit = (3/5) × 1000 = 600 kg. Exam Strategy & Shortcut: Use the alligation cross method visually: 8% (Part 1) ---- 18% (Part 2) \ / 14% (Mean) / \ (18 - 14) ---- (14 - 8) 4 : 6 Ratio = 2:3. Quick mental math: (3/5) of 1000 is 600. Common Pitfall: Mixing up the ratios! A common mistake is assigning the calculated ratio 4 to the 18% profit instead of the 8% profit. Remember, the difference crosses over diagonally. Final Answer: Therefore, the correct answer is 600 kg.

Question 11

$8$ litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is $16 : 65$. How much wine did the cask hold original…

Accepted Answer

Concept & Reverse Dilution Logic: We use the standard replacement formula, but in reverse, by linking it to the final ratio of the mixture components. \frac{Final Wine}{Initial Capacity} = (1 - \frac{Replaced Volume}{Initial Capacity})^n Step-by-Step Solution: • Given: Replaced volume = 8 L. Total operations n = 1 + 3 = 4. Final ratio of Wine : Water = 16 : 65. • Let the original capacity of the cask be x litres. • The ratio of Wine to Total Mixture (which equals the original capacity x) is (16/16 + 65) = (16/81). • Using the formula: \frac{Final Wine}{x} = (1 - (8/x))^4 • Substitute the known ratio: (16/81) = (1 - (8/x))^4 • Take the fourth root of both sides: • ((2/3))^4 = (1 - (8/x))^4 • (2/3) = 1 - (8/x) • Solve for x: • (8/x) = 1 - (2/3) = (1/3) • x = 8 × 3 = 24 litres. Exam Strategy & Shortcut: Whenever a process is done 4 times and you see numbers like 16 and 81 (total), recognize them immediately as 2^4 and 3^4. The ratio of remaining wine to total capacity is ((2/3))^4. The fraction removed is 1 - (2/3) = (1/3). Since (1/3) of the volume is 8L, the total volume is 24L. Common Pitfall: Using the ratio of Wine to Water ((16/65)) in the formula instead of Wine to Total Mixture ((16/81)). The formula gives the ratio of remaining pure liquid to the *total* initial volume. Final Answer: Therefore, the correct answer is 24 litres.

Question 12

A vessel contains a mixture of Grape, Pineapple and Banana juices in the respective ratio of $4 : 6 : 5$. $15$ litres of this mixture is taken out and $8$ litres of grape juice and $2$ litres of pineapple juice is added to the vessel. If the resulta…

Accepted Answer

Concept & Ratio Equations: When a specific amount of mixture is removed, it is drawn out in the exact same ratio as the original mixture. Tracking the algebraic changes to each component allows us to solve for the original total. *(Note: The question contains a redundant phrase error, but the mathematical logic relies on: "grape juice is 10 litres less than... pineapple juice")*. Step-by-Step Solution: • Given: Initial ratio G : P : B = 4 : 6 : 5. Total initial parts = 15. Let total initial quantity be 15x. • 15 L of mixture is removed. It leaves in the ratio 4:6:5. • Grape removed = (4/15) × 15 = 4 L. • Pineapple removed = (6/15) × 15 = 6 L. • Initial component quantities: Grape = 4x, Pineapple = 6x. • Quantities after removal: • Grape = 4x - 4 • Pineapple = 6x - 6 • Quantities after addition (8L Grape, 2L Pineapple): • Final Grape = 4x - 4 + 8 = 4x + 4 • Final Pineapple = 6x - 6 + 2 = 6x - 4 • The problem states Final Grape is 10 L less than Final Pineapple: • (4x + 4) = (6x - 4) - 10 • 4x + 4 = 6x - 14 • 18 = 2x \Rightarrow x = 9 • Initial quantity = 15x = 15 × 9 = 135 L. Exam Strategy & Shortcut: You can bypass calculating the removed fractions by treating the remaining mixture as the baseline. Let the remaining mixture components be 4k, 6k, 5k. Final Grape = 4k + 8. Final Pineapple = 6k + 2. 4k + 8 = (6k + 2) - 10 \Rightarrow 4k + 8 = 6k - 8 \Rightarrow 2k = 16 \Rightarrow k = 8. Total remaining mixture = 15k = 15 × 8 = 120 L. Initial total = Remaining + Removed = 120 + 15 = 135 L. This avoids fractional subtraction early on! Common Pitfall: Getting confused by the typo in the original text ("resultant quantity of grape juice is 10 litres less than the resultant quantity of grape juice is 10 litres less than..."). Ignore the repetition and extract the core mathematical equation: G = P - 10. Final Answer: Therefore, the correct answer is 135.

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