Question 1

A 20 L vessel has milk and water in the ratio 3 : 2. Ten litres are removed and replaced with pure milk, and this process is repeated once more. Find the final milk : water ratio.

Accepted Answer

Introduction / Context: This is a successive replacement (removal and refill) problem. Removing a fixed volume of well-mixed solution takes out components in proportion to their current fractions. Replacing with pure milk changes the composition before the next removal. Given Data / Assumptions: Total volume stays 20 L. Initial ratio milk : water = 3 : 2 ⇒ milk = 12 L, water = 8 L. Remove 10 L of mixture each time; refill 10 L pure milk; repeat twice in total. Concept / Approach: At each removal, amounts removed are proportional to current composition. After removal, add 10 L milk to restore volume to 20 L, then repeat once more. Keep track of milk and water after each step. Step-by-Step Solution: Initial: milk 12, water 8. First removal (10 of 20): removes 6 milk and 4 water ⇒ remains milk 6, water 4. Add 10 L milk ⇒ milk 16, water 4. Second removal (10 of 20): composition milk 16/20, water 4/20 ⇒ removes 8 milk and 2 water ⇒ remains milk 8, water 2. Add 10 L milk ⇒ final milk 18, water 2 ⇒ ratio 18 : 2 = 9 : 1. Verification / Alternative check: Using fraction retention: milk fraction retained each removal = (1 − 10/20) = 1/2, but because we add back pure milk between removals, stepwise tracking as shown is more transparent here. Why Other Options Are Wrong: 5:3 and 4:1 do not match the increased milk dominance after two milk refills; 1:4 is impossible since milk is being added, not removed. Common Pitfalls: Forgetting that removal is proportional to current composition or failing to update before the second cycle. Always recalculate before each step. Final Answer: 9:1

Question 2

A 640 ml mixture has milk : water = 6 : 2. How many millilitres of water must be added so that the new mixture becomes half milk and half water?

Accepted Answer

Introduction / Context: We adjust the composition by adding water only. The milk quantity stays fixed, but total volume and water amount change. The objective is a 50% milk, 50% water mixture after addition. Given Data / Assumptions: Total initial volume = 640 ml. Milk : water = 6 : 2 ⇒ milk = 6/8 * 640 = 480 ml; water = 160 ml. Add x ml of water; milk remains 480 ml. Concept / Approach: For a half-and-half mixture, milk volume must equal water volume at the end. Set milk = water_final and solve for x (the added water). Step-by-Step Solution: Let added water = x ml. Final water = 160 + x; final milk = 480 (unchanged). Set equality: 480 = 160 + x ⇒ x = 320 ml. Therefore, add 320 ml of water. Verification / Alternative check: New total = 640 + 320 = 960 ml with milk = 480 ml and water = 480 ml, which is exactly 50% each. Why Other Options Are Wrong: 310, 330, 340 do not equalize milk and water; only 320 matches the requirement. Common Pitfalls: Trying to proportionally reduce milk, which does not happen here because only water is added. Keep milk constant and balance water to match it. Final Answer: 320 ml

Question 3

A shopkeeper buys 26 L of milk at Rs. 16/L and also purchases an inferior milk at Rs. 10/L. How many litres of the cheaper milk must be mixed so that selling at Rs. 14/L results in no loss?

Accepted Answer

Introduction / Context: We are to find a mixture whose cost price (weighted average) equals the selling price so that overall profit is zero. This requires solving for the unknown volume of cheaper milk that brings the average cost down to Rs. 14 per litre. Given Data / Assumptions: Cost1 = Rs. 16/L; Quantity1 = 26 L. Cost2 = Rs. 10/L; Quantity2 = x L. Target mixture price (no loss) = Rs. 14/L. Concept / Approach: Weighted average equation: (sum of cost) / (sum of litres) = target price. Solve linear equation for x. Alternatively, alligation could be used, but direct equation is straightforward here. Step-by-Step Solution: (26*16 + x*10) / (26 + x) = 14. 416 + 10x = 14(26 + x) = 364 + 14x. 416 − 364 = 14x − 10x ⇒ 52 = 4x ⇒ x = 13 L. Verification / Alternative check: Total litres = 39; total cost = 26*16 + 13*10 = 416 + 130 = 546. Average = 546/39 = 14, confirming no profit or loss at Rs. 14/L. Why Other Options Are Wrong: 12, 14, 16 do not satisfy the weighted average equation; only 13 produces exactly Rs. 14/L. Common Pitfalls: Forgetting to divide total cost by total litres or misplacing terms in the equation. Keep track of both totals carefully. Final Answer: 13 liter

Question 4

Two vessels contain milk-water mixtures in ratios 4:1 and 9:11, respectively. If they are combined in the ratio 3:2 by volume, find the milk : water ratio in the resulting mixture.

Accepted Answer

Introduction / Context: When mixing mixtures, compute the absolute milk and water contributed from each vessel and then add them according to the given mixing ratio. Finally, express the totals as a simplified milk : water ratio. Given Data / Assumptions: Vessel A: milk : water = 4 : 1 ⇒ milk fraction = 4/5. Vessel B: milk : water = 9 : 11 ⇒ milk fraction = 9/20. Mixture proportions A : B = 3 : 2. Concept / Approach: Take notional volumes: 3 units from A and 2 units from B. Compute milk and water amounts using fractions, add them, and form the final ratio milk : water. Step-by-Step Solution: Milk from A = 3 * (4/5) = 12/5; water from A = 3 * (1/5) = 3/5. Milk from B = 2 * (9/20) = 18/20 = 9/10; water from B = 2 * (11/20) = 22/20 = 11/10. Total milk = 12/5 + 9/10 = 24/10 + 9/10 = 33/10. Total water = 3/5 + 11/10 = 6/10 + 11/10 = 17/10. Milk : Water = (33/10) : (17/10) = 33 : 17. Verification / Alternative check: The total milk fraction = 33/(33+17) = 33/50, which is consistent with intermediate calculations. Why Other Options Are Wrong: 34:16 and 16:34 are not supported by the computed fractions; 17:33 reverses the roles of milk and water. Common Pitfalls: Averaging ratios directly without converting to actual fractions and volumes. Always compute component amounts before forming the final ratio. Final Answer: 33 : 17

Question 5

A 6 L mixture contains 75% milk. How many litres of milk must be added to make the mixture 90% milk?

Accepted Answer

Introduction / Context: We increase the proportion of milk by adding pure milk. This leaves water quantity unchanged while increasing total volume and milk quantity. The target is a 90% milk concentration in the final mixture. Given Data / Assumptions: Initial volume = 6 L; milk = 75% of 6 = 4.5 L; water = 1.5 L. Add x L of milk; final volume = 6 + x; final milk = 4.5 + x. Target: milk fraction = 0.90. Concept / Approach: Set up the concentration equation (4.5 + x) / (6 + x) = 0.90 and solve for x. Water remains 1.5 L throughout because only milk is added. Step-by-Step Solution: (4.5 + x)/(6 + x) = 0.90. 4.5 + x = 0.9(6 + x) = 5.4 + 0.9x. x − 0.9x = 5.4 − 4.5 ⇒ 0.1x = 0.9 ⇒ x = 9 L. Verification / Alternative check: Final volume = 15 L; milk = 13.5 L; fraction = 13.5/15 = 0.9 = 90%, as required. Why Other Options Are Wrong: 8, 10, 12 litres do not satisfy the exact 90% equation; only 9 litres works. Common Pitfalls: Changing water amount by mistake. Only milk is added, so water remains at 1.5 L; adjust only the numerator and denominator accordingly. Final Answer: 9 liters

Question 6

In what ratio should water be added to spirit so that selling the resulting mixture at the cost price of spirit yields a 25% gain?

Accepted Answer

Introduction / Context: Water is free; spirit has cost. If the mixture is sold at the cost price of spirit, the only source of profit is the added water. A 25% gain means the water amount is one-quarter of the spirit content in the mixture. Given Data / Assumptions: Selling price equals cost price of spirit. Profit% = (water quantity / spirit quantity) * 100. Desired profit = 25%. Concept / Approach: Set water/spirit = 25/100 = 1/4. Then express the mixing ratio (water : spirit) accordingly. This directly gives the answer without additional algebra. Step-by-Step Solution: Let spirit = 4 units (paid component). To gain 25%, water must be 1 unit so that profit = 1/4 of spirit cost. Therefore, ratio water : spirit = 1 : 4. Verification / Alternative check: Sell 5 units (4 paid + 1 free) at the cost of 4 units. Revenue equals the cost of 4 units; the extra 1 unit creates a 25% gain on the cost outlay of spirit. Why Other Options Are Wrong: 4 : 1 and others imply far more water or the wrong proportion, leading to different profit percentages. Common Pitfalls: Confusing profit% with water% in the mixture. The 25% gain relates to water compared to spirit, not to the total mixture. Final Answer: 1 : 4

Question 7

A shopkeeper has 50 kg of rice. Part is sold at 20% profit and the rest at 40% profit. If the overall gain is 25%, find the quantities sold at 20% and at 40% profit.

Accepted Answer

Introduction / Context: We are given two profit rates on parts of a single stock and an overall profit rate. This is a weighted average situation. Let x be the amount sold at 20% profit; the remainder is sold at 40% profit. Use the average profit equation to find x. Given Data / Assumptions: Total quantity = 50 kg. Profit rates = 20% on x kg; 40% on (50 − x) kg. Overall profit = 25% on the whole. Concept / Approach: Average profit% = (20*x + 40*(50 − x)) / 50. Set equal to 25 and solve for x. The quantities are x at 20% and (50 − x) at 40%. Step-by-Step Solution: (20x + 40(50 − x)) / 50 = 25. 20x + 2000 − 40x = 1250 ⇒ −20x = −750 ⇒ x = 37.5 kg. Hence, at 20%: 37.5 kg; at 40%: 12.5 kg. Verification / Alternative check: Weighted profit = (20*37.5 + 40*12.5)/50 = (750 + 500)/50 = 1250/50 = 25% overall, as required. Why Other Options Are Wrong: 12.5 & 37.5 reverse the roles; other pairs do not satisfy the weighted average equation at all. Common Pitfalls: Averaging 20 and 40 directly to 30% and assuming equal split—this ignores the required 25% overall and the unknown split. Final Answer: 37.5 kgs and 12.5 kgs

Question 8

An organisation has 400 employees (officers and workers). The overall average monthly salary is Rs. 3000; officers average Rs. 10,000 and workers Rs. 2000. Find the numbers of officers and workers.

Accepted Answer

Introduction / Context: This is a weighted average and mixture problem: the overall average salary is formed from two groups with known averages. Using total salary, we can determine how many belong to each category (officers vs. workers). Given Data / Assumptions: Total employees = 400. Overall average = Rs. 3000 per month. Officer average = Rs. 10,000; worker average = Rs. 2000. Concept / Approach: Total salary = overall average * total employees. Let O be officers and W be workers with O + W = 400. Then 10000*O + 2000*W = total salary. Solve for O, then W = 400 − O. Step-by-Step Solution: Total salary = 3000 * 400 = Rs. 1,200,000. Equation: 10000O + 2000W = 1,200,000; also O + W = 400. Substitute W = 400 − O: 10000O + 2000(400 − O) = 1,200,000. 10000O + 800,000 − 2000O = 1,200,000 ⇒ 8000O = 400,000 ⇒ O = 50; W = 350. Verification / Alternative check: Check total salary: 50*10000 + 350*2000 = 500,000 + 700,000 = 1,200,000, which matches the overall figure. Why Other Options Are Wrong: 300,100 or 250,150 or 310,90 do not satisfy both the headcount and salary equations simultaneously. Common Pitfalls: Misapplying averages (e.g., averaging 10,000 and 2,000 directly) or forgetting to multiply by counts to get total salary. Final Answer: 50 , 350

Question 9

Alloys mixed: 4 kg of an alloy with 1/4 iron (rest tin) is blended with 6 kg of another alloy with 2/3 iron (rest tin). Find the iron : tin ratio in the final mixture.

Accepted Answer

Introduction / Context: When mixing alloys, compute the actual masses of each component (iron and tin) from each alloy, then add them to obtain totals and simplify to a ratio. This problem uses simple fractions of iron in each alloy. Given Data / Assumptions: Alloy 1: 4 kg with iron fraction 1/4 ⇒ iron 1 kg, tin 3 kg. Alloy 2: 6 kg with iron fraction 2/3 ⇒ iron 4 kg, tin 2 kg. Concept / Approach: Add iron masses and tin masses separately: iron_total = iron1 + iron2; tin_total = tin1 + tin2. Express as iron : tin and reduce if possible. Step-by-Step Solution: Iron_total = 1 + 4 = 5 kg. Tin_total = 3 + 2 = 5 kg. Therefore, iron : tin = 5 : 5 = 1 : 1. Verification / Alternative check: The totals sum to 10 kg, matching the combined mass. Equal iron and tin indicate a balanced mixture. Why Other Options Are Wrong: 2 : 1, 1 : 2, 3 : 2 do not match the computed component totals for this specific mixing. Common Pitfalls: Confusing “fraction of iron” with “fraction of tin” or failing to multiply by the mass of each alloy before adding. Final Answer: 1 : 1

Question 10

In a courtyard there are chickens and goats. Heads counted = 100; legs counted = 320. Find the number of chickens and goats.

Accepted Answer

Introduction / Context: This is a standard heads-and-legs puzzle. With two types of animals, each with a known number of legs (2 for chickens, 4 for goats), we can set up two linear equations to find the exact counts given total heads and total legs. Given Data / Assumptions: Chickens (2 legs), goats (4 legs). Total heads = 100. Total legs = 320. Concept / Approach: Let c = chickens, g = goats. Then c + g = 100 and 2c + 4g = 320. These two linear equations uniquely determine c and g. Step-by-Step Solution: c + g = 100 … (1) 2c + 4g = 320 ⇒ divide by 2 ⇒ c + 2g = 160 … (2) Subtract (1) from (2): g = 60 ⇒ c = 100 − 60 = 40. Verification / Alternative check: Legs check: 40 chickens * 2 = 80 legs; 60 goats * 4 = 240 legs; total = 320, matching the data. Why Other Options Are Wrong: 20,50 totals only 70 heads; 30,70 yields 320 legs but 100 heads? (30+70=100) but legs 30*2 + 70*4 = 60 + 280 = 340 (too high). 50,50 gives 300 legs (too low). Common Pitfalls: Solving with trial instead of equations can lead to errors. Using simultaneous equations ensures correctness. Final Answer: 40 , 60

Question 11

A container is initially full of milk. One-third is removed and replaced with water. This removal-and-replacement by the same quantity of water is repeated 4 times. If 16 L of milk remain after the 4th operation, find the container’s capacity.

Accepted Answer

Introduction / Context: This is a repeated dilution problem. Each cycle removes one-third of the current mixture and then replaces that amount with water. The milk fraction shrinks by the same retention factor every time. After several iterations, the amount of milk left is related to the initial capacity by a power of the retention fraction. Given Data / Assumptions: Each operation removes 1/3 of the mixture and refills with water. Number of operations = 4. Milk remaining after 4th operation = 16 L. Let capacity be V litres. Concept / Approach: After removing 1/3, the milk retained is (1 − 1/3) = 2/3 of what it was before the removal. Repeating this 4 times means milk becomes V * (2/3)^4. Set this equal to the final milk and solve for V. Step-by-Step Solution: Milk after n operations = V * (2/3)^n. For n = 4: V * (2/3)^4 = V * (16/81) = 16. Thus V * 16/81 = 16 ⇒ V = 16 * 81 / 16 = 81 L. Verification / Alternative check: Retention factor check: After each step milk fraction multiplies by 2/3. Four steps give (2/3)^4 = 16/81 of original, consistent with the result. Why Other Options Are Wrong: 76, 82, 85 L do not satisfy V*(16/81) = 16 when substituted. Common Pitfalls: Forgetting to raise the retention fraction to the power of the number of operations, or using 1/3 instead of 2/3 for retention. Final Answer: 81 litres

Question 12

Two types of rice are available: Type I costs Rs. 15 per kg and Type II costs Rs. 20 per kg. If the two types are mixed in the ratio 2 : 3 (by weight), what is the price per kilogram of the mixed variety sold at cost?

Accepted Answer

Introduction / Context:This is a direct weighted-average (alligation) problem. When two qualities are mixed in a given ratio, the mean price equals the weighted average of the individual prices according to their quantities. Given Data / Assumptions: Type I price = Rs. 15 per kg. Type II price = Rs. 20 per kg. Mixing ratio (Type I : Type II) = 2 : 3 by weight. No wastage or additional costs; mixture is sold at cost. Concept / Approach:For prices p1 and p2 mixed in weights w1 and w2, the mixture price p is p = (w1*p1 + w2*p2) / (w1 + w2). Alligation gives the same result here. Step-by-Step Solution:Let weights be 2 kg (Type I) and 3 kg (Type II).Total cost = 2*15 + 3*20 = 30 + 60 = Rs. 90.Total weight = 2 + 3 = 5 kg.Mixture price per kg = 90 / 5 = Rs. 18. Verification / Alternative check:By alligation: differences are 20 − 18 = 2 and 18 − 15 = 3, giving ratio 3 : 2 (Type I : Type II), which matches the given 2 : 3 after swapping sides, confirming consistency. Why Other Options Are Wrong:19.50, 19, and 18.50 are higher than the true weighted average. 17.50 is too low and inconsistent with the ratio favoring the costlier type. Common Pitfalls:Taking a simple average (15 and 20) to get 17.5 without using the ratio, or reversing the 2 : 3 weights. Weighted average must honor the given quantities. Final Answer:18

Question 13

A milkman sells a mixture at the cost price of pure milk. In what ratio should water be mixed with milk to earn a gain of 16 2/3% (i.e., one-sixth) on selling the mixture at the cost price of milk?

Accepted Answer

Introduction / Context:When a seller mixes free water with milk and sells the mixture at the cost price of milk, the observed profit arises from the water content. We must determine the water-to-milk ratio that yields a 16 2/3% (one-sixth) gain. Given Data / Assumptions: Selling price per liter of mixture = cost price per liter of pure milk. Profit% on cost = 16 2/3% = 1/6. Water is free; milk costs 1 (normalized) per liter for ease of proportion. Concept / Approach:If milk fraction in the mixture is m, then revenue per liter = 1, cost per liter = m*1 + (1 − m)*0 = m. Profit% = (revenue − cost)/cost * 100 = (1 − m)/m * 100. Set this to 16 2/3%. Step-by-Step Solution:(1 − m)/m = 1/61/m − 1 = 1/6 ⇒ 1/m = 7/6 ⇒ m = 6/7Water fraction = 1 − m = 1/7Hence water : milk = (1/7) : (6/7) = 1 : 6 Verification / Alternative check:With milk 6 liters and water 1 liter (7 liters total), revenue at “1 per liter” is 7, cost is 6, profit = 1, profit% = 1/6 = 16 2/3%. Why Other Options Are Wrong:Ratios like 4 : 3 or 6 : 1 would produce very different profits; only 1 : 6 gives exactly one-sixth profit at cost price. Common Pitfalls:Confusing water : milk with milk : water, or using profit on selling price instead of profit on cost price. Final Answer:1 : 6

Question 14

A 729 ml mixture contains milk and water in the ratio 7 : 2. How much additional water must be added so that the new mixture has milk and water in the ratio 7 : 3?

Accepted Answer

Introduction / Context:The mixture is changed by adding only water. Since no milk is added or removed, the quantity of milk remains constant while water increases until the desired ratio is reached. Given Data / Assumptions: Total initial mixture = 729 ml. Initial milk : water = 7 : 2. Final milk : water = 7 : 3 (milk unchanged). No evaporation or losses. Concept / Approach:Compute the initial milk and water. Because milk stays constant, compute the water required to make water = (3/7) of milk (as per the target ratio). Step-by-Step Solution:Initial milk = (7/9)*729 = 567 mlInitial water = (2/9)*729 = 162 mlTarget water for 7 : 3 = (3/7)*milk = (3/7)*567 = 243 mlAdditional water to add = 243 − 162 = 81 ml Verification / Alternative check:New totals: milk = 567 ml, water = 243 ml; ratio 567 : 243 = 7 : 3 (divide by 81). Why Other Options Are Wrong:60, 72, and 96 ml do not produce water = 243 ml. Only 81 ml gets the exact 7 : 3 ratio. Common Pitfalls:Altering both milk and water or using 7 : 3 as milk : total instead of milk : water. Keep milk fixed and adjust water only. Final Answer:81 ml

Question 15

A 60-liter mixture has milk and water in the ratio 2 : 1 (milk = 40 L, water = 20 L). How much additional water must be added so that the ratio becomes 1 : 2 (milk : water)?

Accepted Answer

Introduction / Context:Only water is added to change the composition. Milk remains constant while water increases until the desired ratio is reached. This is a straightforward ratio adjustment. Given Data / Assumptions: Total initial = 60 L; milk : water = 2 : 1 ⇒ milk = 40 L, water = 20 L. Add x liters of water; milk remains 40 L. Target milk : water = 1 : 2. Concept / Approach:Set up the ratio equation with constant milk and unknown water after addition, then solve for x. Step-by-Step Solution:After adding x, water = 20 + xRequire 40 : (20 + x) = 1 : 2Cross-multiply: 2*40 = 20 + x ⇒ 80 = 20 + x ⇒ x = 60 Verification / Alternative check:New amounts: milk = 40 L; water = 80 L; ratio 40 : 80 = 1 : 2 as required. Why Other Options Are Wrong:20, 30, 40, and 50 do not yield water = 80 L. Only 60 L addition achieves the target ratio. Common Pitfalls:Changing milk as well or forming milk : total ratio instead of milk : water. Final Answer:60 litres

Question 16

A merchant has 50 kg of sugar. Part is sold at 8% profit and the remainder at 18% profit. If his overall gain is 14%, how many kilograms were sold at 18% profit?

Accepted Answer

Introduction / Context:This is a weighted-average profit problem. The overall profit percentage is an average of the two profit rates, weighted by quantities sold at those rates. Given Data / Assumptions: Total sugar = 50 kg. Profit rates: 8% on x kg; 18% on (50 − x) kg. Overall profit = 14%. Concept / Approach:Use the equation for weighted average of percentages: (8*x + 18*(50 − x)) / 50 = 14. Solve for x and then compute the 18% quantity. Step-by-Step Solution:(8x + 18(50 − x)) / 50 = 148x + 900 − 18x = 700 ⇒ −10x = −200 ⇒ x = 20So 18% portion = 50 − x = 30 kg Verification / Alternative check:Alligation check around 14 between 8 and 18 gives distances 6 and 4 → ratio 6 : 4 = 3 : 2 (8% : 18%). For total 50, 18% share = 2/(3+2)*50 = 20? Careful: 3 parts at 8% and 2 parts at 18% means 8%-quantity is larger; solving directly confirmed 18% = 30 kg (consistent with algebra). Why Other Options Are Wrong:20, 15, 35, 25 do not satisfy the weighted-average 14% across 50 kg. Common Pitfalls:Confusing which portion corresponds to which distance in alligation; algebra avoids this mix-up. Final Answer:30 kg

Question 17

In what ratio must wheat at Rs. 1.60 per kg be mixed with wheat at Rs. 1.45 per kg so that the mixture is worth Rs. 1.54 per kg (no wastage)?

Accepted Answer

Introduction / Context:Alligation is the quickest way to find the mixing ratio for a target mean price between two known prices. The quantities are inversely proportional to the differences from the mean. Given Data / Assumptions: Dearer wheat: Rs. 1.60/kg. Cheaper wheat: Rs. 1.45/kg. Mixture price: Rs. 1.54/kg. Concept / Approach:For prices P1 (dearer) and P2 (cheaper) yielding mean M, the required ratio P1 : P2 = (M − P2) : (P1 − M). Step-by-Step Solution:Differences: M − P2 = 1.54 − 1.45 = 0.09P1 − M = 1.60 − 1.54 = 0.06Hence 1.60 : 1.45 quantities = 0.09 : 0.06 = 3 : 2 Verification / Alternative check:Weighted average with 3 kg of 1.60 and 2 kg of 1.45: price = (3*1.60 + 2*1.45)/5 = (4.80 + 2.90)/5 = 7.70/5 = 1.54. Why Other Options Are Wrong:2 : 3 or 4 : 3 give means other than 1.54; only 3 : 2 balances the differences correctly. Common Pitfalls:Using direct differences instead of inverse differences, which reverses the intended ratio. Final Answer:3 : 2

Question 18

An adulterated milk sample weighs 66 kg with milk : water = 5 : 1. How much water must be added to make the ratio 5 : 3 (milk : water)?

Accepted Answer

Introduction / Context:Adding only water increases the water quantity while milk remains constant. Set up the ratio with fixed milk to compute the extra water required for the new ratio. Given Data / Assumptions: Total = 66 kg; milk : water = 5 : 1 initially. Initial milk = 55 kg; initial water = 11 kg. Target milk : water = 5 : 3. Concept / Approach:Let y be water added. Then water becomes 11 + y. Milk remains 55. Use 55 : (11 + y) = 5 : 3. Step-by-Step Solution:55 : (11 + y) = 5 : 33*55 = 5*(11 + y)165 = 55 + 5y ⇒ 110 = 5y ⇒ y = 22 kg Verification / Alternative check:New water = 33 kg; milk : water = 55 : 33 = 5 : 3; total becomes 88 kg, consistent. Why Other Options Are Wrong:24.750, 16.500, and 20 do not lead to a water total of 33 kg; only 22 kg satisfies the ratio condition. Common Pitfalls:Accidentally changing milk or targeting milk : total instead of milk : water. Final Answer:22 kg

Question 19

Two vessels contain milk : water as follows — Vessel A: 5 : 3 and Vessel B: 2 : 3. In what ratio should mixtures from A and B be combined to obtain a final mixture that is exactly half milk and half water?

Accepted Answer

Introduction / Context:We have two mixtures with different milk fractions. To achieve a 1 : 1 milk-to-water target, use alligation on milk fractions to find the mixing ratio of the two sources. Given Data / Assumptions: Vessel A milk fraction = 5/8 = 0.625. Vessel B milk fraction = 2/5 = 0.4. Target milk fraction = 1/2 = 0.5. Concept / Approach:Alligation formula for fractions f1 (A) and f2 (B) to reach m: required A : B = (f2 − m) : (m − f1), ensuring f1 > m > f2 or vice versa with correct ordering. Step-by-Step Solution:A : B = (0.4 − 0.5) : (0.5 − 0.625) in magnitudeUse absolute differences with proper sides: A : B = (m − f2) : (f1 − m) = (0.5 − 0.4) : (0.625 − 0.5)= 0.1 : 0.125 = 4 : 5 Verification / Alternative check:With 4 parts of A and 5 parts of B: milk fraction = (4*0.625 + 5*0.4) / 9 = (2.5 + 2) / 9 = 4.5 / 9 = 0.5. Why Other Options Are Wrong:Other ratios give milk fraction different from 0.5. Only 4 : 5 balances the fractions. Common Pitfalls:Reversing the alligation differences and obtaining 5 : 4 instead of 4 : 5. Final Answer:4 : 5

Question 20

Water (cost Rs. 0) is mixed with wine priced at Rs. 22.50 per liter to reduce the price to Rs. 18 per liter. In what proportion should water and wine be mixed (water : wine)?

Accepted Answer

Introduction / Context:Water has zero cost. Mixing water with wine reduces the average price. Alligation directly gives the water-to-wine ratio to hit the target price of Rs. 18 per liter. Given Data / Assumptions: Cheaper (water) = Rs. 0. Dearer (wine) = Rs. 22.50. Mean (mixture) = Rs. 18. Concept / Approach:By alligation: quantities are proportional to the differences from the mean. Water : Wine = (22.50 − 18) : (18 − 0). Step-by-Step Solution:Water : Wine = 4.5 : 18 = 1 : 4 Verification / Alternative check:Take 1 L water and 4 L wine → total 5 L, total cost = 4*22.5 = 90; price per liter = 90/5 = 18. Why Other Options Are Wrong:Other ratios produce means higher or lower than 18; only 1 : 4 matches the target. Common Pitfalls:Mixing up water : wine with wine : water, or using direct instead of inverse differences. Final Answer:1 : 4

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