Question 1

A merchant has 1000 kg of sugar. Part is sold at 8% profit and the rest at 18% profit. Overall gain is 14%. How many kg were sold at 18% profit?

Accepted Answer

Problem restatement 1000 kg are split across two profit rates (8% and 18%) to yield a 14% overall profit. Find the quantity at 18%. Given data Total quantity = 1000 kg. Let x kg be sold at 18%; then (1000 − x) kg at 8%. Overall profit% = 14%. Concept/Approach Use weighted average of percentages by weight. Step-by-step calculation (18x + 8(1000 − x)) / 1000 = 14 18x + 8000 − 8x = 14000 ⇒ 10x = 6000 ⇒ x = 600 kg Verification/Alternative Alligation check: distance from 14% to 8% is 6; from 18% to 14% is 4 ⇒ ratio at 8% : 18% = 4 : 6 = 2 : 3 ⇒ quantities 400 : 600 out of 1000 kg. Common pitfalls Using selling prices instead of profit percentages; here base cancels in weighted average. Final Answer 600 kg

Question 2

Type-1 rice costs Rs. 15/kg and Type-2 costs Rs. 20/kg. If mixed in the ratio 2 : 3, what is the price per kg of the mixture?

Accepted Answer

Problem restatement Find the weighted average price when Rs. 15/kg and Rs. 20/kg rice are mixed in the ratio 2 : 3. Given data Prices: 15, 20; ratio (Type-1 : Type-2) = 2 : 3. Concept/Approach Mixture price = (sum of price × quantity) / total quantity. Step-by-step calculation Mixture price = (2×15 + 3×20) ÷ (2 + 3) = (30 + 60) ÷ 5 = 90 ÷ 5 = 18 Verification/Alternative Alligation line: difference from 18 is 3 (for 15) and 2 (for 20) ⇒ ratio 2 : 3 matches given, confirming Rs. 18. Final Answer Rs. 18.00

Question 3

A vendor has two cans: Can-1 has 25% water (75% milk), Can-2 has 50% water (50% milk). How much from each should be mixed to get 12 litres with water : milk = 3 : 5?

Accepted Answer

Problem restatement Form 12 L mixture where water : milk = 3 : 5 (i.e., milk fraction 5/8). Can-1 has 75% milk; Can-2 has 50% milk. Find volumes x and y to draw from Can-1 and Can-2. Given data x + y = 12 (total litres). Milk in mix must be 12 × (5/8) = 7.5 L. Milk contributions: Can-1 → 0.75x; Can-2 → 0.50y. Concept/Approach Use two linear equations: volume balance and milk balance. Step-by-step calculation 0.75x + 0.50y = 7.5 x = 12 − y 0.75(12 − y) + 0.50y = 7.5 ⇒ 9 − 0.75y + 0.50y = 7.5 −0.25y = −1.5 ⇒ y = 6 x = 12 − 6 = 6 Verification/Alternative Milk = 0.75×6 + 0.5×6 = 4.5 + 3 = 7.5 L; Water = 12 − 7.5 = 4.5 L ⇒ ratio 4.5 : 7.5 = 3 : 5. Common pitfalls Targeting water fraction 3/8 but then equating to milk equation incorrectly; ensure milk = 5/8 of 12. Final Answer 6 litres from each can

Question 4

In what ratio should pulses costing Rs. 15/kg and Rs. 20/kg be mixed to get a mixture worth Rs. 16.50/kg?

Accepted Answer

Problem restatement Find the mixing ratio (cheaper : dearer) for a target mean price of Rs. 16.50 between Rs. 15 and Rs. 20. Concept/Approach Use alligation: Quantity ratio (cheaper : dearer) = (dearer − mean) : (mean − cheaper). Step-by-step calculation Ratio = (20 − 16.50) : (16.50 − 15) = 3.50 : 1.50 = 7 : 3 Verification/Alternative Weighted average check with 7 kg of Rs. 15 and 3 kg of Rs. 20: mean = (7×15 + 3×20)/10 = (105 + 60)/10 = 165/10 = 16.50. Final Answer 7 : 3

Question 5

Tea at Rs. 126/kg and Rs. 135/kg is mixed with a third variety in the ratio 1 : 1 : 2. If the mixture costs Rs. 153/kg, what is the price of the third variety?

Accepted Answer

Problem restatement Three teas priced at 126, 135, and x rupees per kg are mixed in the ratio 1 : 1 : 2 to yield Rs. 153/kg. Find x. Given data Weights: 1, 1, and 2 parts. Mean price of mixture = Rs. 153/kg. Concept/Approach Use weighted average: (sum of price × parts) / total parts = mixture price. Step-by-step calculation (126 + 135 + 2x) ÷ 4 = 153 261 + 2x = 612 ⇒ 2x = 351 ⇒ x = 175.5 Verification/Alternative Plug back: (126 + 135 + 2×175.5)/4 = (261 + 351)/4 = 612/4 = 153 (exact). Common pitfalls Using 1 : 1 : 1 or averaging the three prices without weighting. Final Answer Rs. 175.50 per kg

Question 6

A 40-litre container of milk has 4 litres removed and replaced by water, and this operation is repeated twice more. How much milk remains?

Accepted Answer

Problem restatement Successive replacement: each time remove 4 L of the current mixture from a 40 L container (initially pure milk) and refill with water. After 3 such operations, find remaining milk. Given data Initial milk = 40 L; each cycle removes 4 L and replaces with water. Number of cycles n = 3. Concept/Approach Use the standard successive-replacement formula: Remaining milk = Initial × (1 − removed/total)n. Step-by-step calculation Remaining milk = 40 × (1 − 4/40)3 = 40 × (36/40)3 = 40 × (0.9)3 = 40 × 0.729 = 29.16 L Verification/Alternative Iterative check: After 1st cycle milk = 40×0.9 = 36; 2nd = 36×0.9 = 32.4; 3rd = 32.4×0.9 = 29.16 L. Common pitfalls Subtracting 12 L instead of compounding proportions (replacement is proportional, not absolute). Final Answer 29.16 litres

Question 7

A milkman claims to sell at cost price but mixes water and gains 25%. What percentage of the mixture is water?

Accepted Answer

Problem restatement A vendor sells a milk–water mixture at the cost price of pure milk and still makes a 25% profit. Determine the percentage of water in the mixture. Given data Selling price per litre (SP) equals the cost price of pure milk per litre (assume Rs. 1 for convenience). Profit = 25%. Concept/Approach If SP = 1 and profit is 25%, then the effective cost price of the mixture must be 1 ÷ 1.25 = 0.8. Since only milk has cost and water is free, the fraction of milk in the mixture must equal 0.8. Hence water fraction = 1 − 0.8 = 0.2 = 20%. Step-by-step calculation SP = 1 (assumption)Profit% = 25% ⇒ Mixture CP = 1 ÷ 1.25 = 0.8Milk fraction = 0.8 ⇒ Water fraction = 1 − 0.8 = 0.2 = 20% Common pitfalls Using absolute rupee values unnecessarily or forgetting that water costs zero, so only the milk fraction contributes to cost. Final Answer 20%

Question 8

In what ratio should rice at Rs. 7.20/kg be mixed with rice at Rs. 5.70/kg to get a mixture worth Rs. 6.30/kg?

Accepted Answer

Problem restatement Mix two grades of rice at Rs. 7.20/kg and Rs. 5.70/kg to obtain a mixture at Rs. 6.30/kg. Find the mixing ratio. Given data Higher price (H) = 7.20 Lower price (L) = 5.70 Mean price (M) = 6.30 Concept/Approach Use the rule of alligation. Quantity ratio of H:L = (M − L) : (H − M). Step-by-step calculation H : L = (6.30 − 5.70) : (7.20 − 6.30)= 0.60 : 0.90 = 2 : 3 Common pitfalls Inverting the ratio. Remember H:L corresponds to (M − L):(H − M), not vice versa. Final Answer 2 : 3 (Rs. 7.20 : Rs. 5.70)

Question 9

How many kilograms of sugar at Rs. 9/kg must be mixed with 27 kg of sugar at Rs. 7/kg so that selling the mixture at Rs. 9.24/kg yields a 10% gain?

Accepted Answer

Problem restatement Two sugars (Rs. 9/kg and Rs. 7/kg) are mixed. The mixture is sold at Rs. 9.24/kg for a 10% profit. Find the required mass of Rs. 9/kg sugar to mix with 27 kg of Rs. 7/kg sugar. Given data SP (mixture) = Rs. 9.24/kg Profit% = 10% Quantity at Rs. 7/kg = 27 kg Quantity at Rs. 9/kg = x kg (unknown) Concept/Approach First find the mixture's cost price (CP): CP = SP / (1 + Profit). Then set the weighted average cost equal to this CP and solve for x. Step-by-step calculation Mixture CP = 9.24 / 1.10 = 8.40Weighted CP = (9x + 7 × 27) / (x + 27) = 8.409x + 189 = 8.40x + 226.80.60x = 37.8 ⇒ x = 63 kg Verification/Alternative Mixture mass = 90 kg; total CP = 8.4 × 90 = 756. Components CP = 63×9 + 27×7 = 567 + 189 = 756 (consistent). Common pitfalls Using SP instead of CP for the alligation/average, or forgetting to convert 10% profit into the divisor 1.10. Final Answer 63 kg

Question 10

In what ratio should teas costing Rs. 60/kg and Rs. 65/kg be mixed so that selling the mixture at Rs. 68.20/kg yields a 10% gain?

Accepted Answer

Problem restatement Two teas are priced at Rs. 60 and Rs. 65 per kg. The mixture is sold at Rs. 68.20/kg for a 10% profit. Find the ratio of quantities (Rs. 60 : Rs. 65). Given data SP (mixture) = 68.20 Profit% = 10% ⇒ Mixture CP = 68.20 / 1.10 = 62 H = 65, L = 60, M = 62 Concept/Approach By alligation, quantities of cheaper:dearer = (H − M) : (M − L). Step-by-step calculation Cheaper : Dearer = (65 − 62) : (62 − 60)= 3 : 2 ⇒ 3 : 2 (Rs. 60 : Rs. 65) Common pitfalls Applying alligation directly to SP without first converting to CP, which would give a wrong mean price. Final Answer 3 : 2 (Rs. 60 : Rs. 65)

Question 11

A whisky jar has 40% alcohol. A portion is replaced by a 19% alcohol whisky; the final alcohol becomes 26%. What fraction of the original whisky was replaced?

Accepted Answer

Problem restatement From a 40% alcohol solution, x litres are removed and replaced by a 19% solution, resulting in 26% alcohol overall. Find x as a fraction of the jar's capacity. Given data Initial alcohol fraction = 0.40 Added alcohol fraction = 0.19 Final alcohol fraction = 0.26 Concept/Approach Let the jar capacity be V. After removing x of the original mixture (all 40%) and adding x of 19%, set up an alcohol balance. Step-by-step calculation Final alcohol = 0.40(V − x) + 0.19x = 0.26V0.40V − 0.40x + 0.19x = 0.26V0.14V = 0.21x ⇒ x = (0.14/0.21)V = (2/3)V Common pitfalls Replacing with water (0%) by mistake, or using the replacement formula for repeated operations (here it is a single replacement with a nonzero-strength solution). Final Answer 2/3 of the jar's capacity

Question 12

A vessel contains a mixture with water:syrup = 3:5. How much of the mixture must be drawn off and replaced with water so the final mixture is half water and half syrup?

Accepted Answer

Problem restatement From an initial ratio water:syrup = 3:5 (total 8 parts), remove some mixture and replace it with water such that final water:syrup becomes 1:1. Given data Initial water fraction = 3/8 Initial syrup fraction = 5/8 Final syrup fraction required = 1/2 Concept/Approach Let the vessel capacity be V and the amount drawn off be x. Syrup remaining after removal is (5/8)(V − x). This must equal V/2. Step-by-step calculation (5/8)(V − x) = V/25(V − x) = 4V ⇒ 5V − 5x = 4Vx = V/5 ⇒ Draw off 1/5 of the vessel and replace with water Verification/Alternative After replacement: syrup = (5/8)(4V/5) = V/2; water = V − V/2 = V/2 (as required). Common pitfalls Equating water amounts instead of syrup amounts; either works if set up consistently, but syrup is simpler here. Final Answer 1/5 of the vessel

Question 13

In what ratio should water be mixed with milk to gain 16½% (i.e., 16.67%) by selling the mixture at the cost price of milk?

Accepted Answer

Problem restatement A vendor adds water (free) to milk and sells the mixture at the milk's cost price. What water:milk ratio yields a 16½% profit? Concept/Approach Assume 1 litre of milk costs Rs. 1. Add w litres of free water; total litres sold = 1 + w at Rs. 1 per litre. Step-by-step calculation Cost = 1, Revenue = (1 + w) × 1Profit = (1 + w − 1) = w ⇒ Profit% = w × 100%Given 16½% = 1/6 ⇒ w = 1/6Water : Milk = (1/6) : 1 = 1 : 6 Common pitfalls Using rupee arithmetic instead of the simpler litre-based approach. Final Answer 1 : 6

Question 14

A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is refilled with B, the ratio of A to B becomes 7 : 9. How many litres of liquid A were in the can initially?

Accepted Answer

Problem restatement From a can containing a well-mixed liquid A and B in the ratio 7 : 5, 9 L are drawn out and replaced by pure B. The new ratio becomes 7 : 9. Find the original litres of A. Given data Initial ratio A : B = 7 : 5 (i.e., A = (7/12) of total, B = (5/12) of total) Drawn off and replaced volume = 9 L (drawn from the mixture, refilled with pure B) Final ratio A : B = 7 : 9 Concept/Approach When a well-mixed quantity is withdrawn, each component reduces proportionally. Let the can's capacity be T litres. After removing 9 L, the remaining amounts of A and B are each multiplied by (T − 9)/T. Then 9 L of pure B is added. Set up the final ratio and solve for T, then compute initial A = (7/12)T. Step-by-step calculation Initial A = (7/12)T, Initial B = (5/12)TAfter removing 9 L (proportionally):Arem = (7/12)T − (7/12)×9 = (7/12)(T − 9)Brem = (5/12)T − (5/12)×9 = (5/12)(T − 9)After adding 9 L of pure B: Bnew = (5/12)(T − 9) + 9Final ratio given: Arem : Bnew = 7 : 9⇒ 9 × (7/12)(T − 9) = 7 × (5/12)(T − 9) + 9Multiply by 12: 63(T − 9) = 35(T − 9) + 75628(T − 9) = 756 ⇒ T − 9 = 27 ⇒ T = 36Initial A = (7/12) × 36 = 21 litres Verification/Alternative With T = 36: After removal, remaining total = 27 L.Arem = (7/12)×27 = 15.75 L; Brem = (5/12)×27 = 11.25 L.After adding 9 L of B: Bnew = 11.25 + 9 = 20.25 L.Ratio = 15.75 : 20.25 = (divide by 2.25) = 7 : 9 — matches perfectly. Common pitfalls Forgetting that the 9 L removed is a mixture (so both A and B reduce proportionally). Adding 9 L to both A and B (only B increases because the refill is pure B). Setting up the ratio as totals rather than component-wise after each operation. Final Answer 21 litres

Question 15

From a full cask, 8 litres are drawn and replaced with water. This is repeated 4 times in total. The final wine:water ratio is 16:65. How many litres of wine did the cask originally hold?

Accepted Answer

Problem restatement Repeatedly remove 8 L of wine and replace with water, 4 times total. Final ratio wine:water = 16:65. Find the initial capacity. Concept/Approach For repeated replacement, the fraction of wine remaining after n operations is (1 − d/V)n, where d is the amount removed each time and V is the initial volume. The final fraction equals wine/total. Step-by-step calculation Let V = initial volumeWine fraction after 4 operations = (1 − 8/V)4 = 16/(16 + 65) = 16/81(1 − 8/V) = (16/81)1/4 = 2/38/V = 1/3 ⇒ V = 24 litres Verification/Alternative (2/3)4 = 16/81, consistent with the given final ratio. Common pitfalls Using the single-replacement balance instead of the repeated replacement formula, or miscounting the number of operations. Final Answer 24 litres

Question 16

Mixture adjustment (milk–water): A 45-litre mixture has milk and water in the ratio 4 : 1 (i.e., milk = 36 L, water = 9 L). How many litres of water must be added so that the final ratio becomes 3 : 2 (milk : water)?

Accepted Answer

Introduction / Context: This is a standard mixture proportioning problem. When only water is added, the amount of milk remains unchanged. We use the target ratio to determine the exact quantity of water to add. Given Data / Assumptions: Total initial mixture = 45 L with milk : water = 4 : 1 → milk = 36 L; water = 9 L. Only water is added; milk remains 36 L. Final desired ratio (milk : water) = 3 : 2. Concept / Approach: Let x be the litres of water added. The final water becomes 9 + x. Since milk stays at 36 L, impose 36 : (9 + x) = 3 : 2 and solve for x. Step-by-Step Solution: 36 / (9 + x) = 3 / 2.Cross-multiply: 72 = 27 + 3x ⇒ 3x = 45 ⇒ x = 15. Verification / Alternative check: After adding 15 L water, amounts are milk = 36 L, water = 24 L. Ratio 36 : 24 simplifies to 3 : 2, confirming the result. Why Other Options Are Wrong: 24 L would overshoot; 1.5 L and 72 L are far from the needed adjustment. Common Pitfalls: Changing the milk amount (it remains constant when only water is added). Confusing milk : total with milk : water. Final Answer: 15 litres

Question 17

Weighted-average ambiguity check (boys vs. girls): In a class of 30 students, the average weight of boys is 20 kg and the average weight of girls is 25 kg. What fraction of the class are boys?

Accepted Answer

Introduction / Context: This question probes understanding of weighted averages. Knowing subgroup averages (boys vs. girls) is not enough to determine subgroup sizes unless the overall average (or equivalent information) is also provided. Given Data / Assumptions: Total students = 30. Average weight of boys = 20 kg. Average weight of girls = 25 kg. Overall average of the entire class is not given. Concept / Approach: Let the number of boys be b and girls be g with b + g = 30. The overall average A must satisfy 20*b + 25*g = A*(b + g) = 30A. Without A (or an equivalent constraint), infinitely many (b, g) pairs can satisfy different overall averages, making the fraction of boys indeterminate. Step-by-Step Solution: 20b + 25g = 30A.But g = 30 − b, so 20b + 25(30 − b) = 30A.This reduces to an equation containing A and b. Without A, b cannot be uniquely determined. Verification / Alternative check: Pick two different overall averages between 20 and 25 (say 22 and 24). Each will lead to different b values that still respect the subgroup averages and total count, demonstrating non-uniqueness of the answer. Why Other Options Are Wrong: 4/5, 5/6, and 3/4 assert specific fractions without any overall average or other binding information to fix b uniquely. Common Pitfalls: Assuming the overall average is the arithmetic mean of 20 and 25; it is not unless subgroup sizes are equal. Forgetting that weighted averages depend on counts as well as subgroup means. Final Answer: Data insufficient

Question 18

Mixing water with milk but selling at cost price: In what ratio should water be mixed with milk so that selling the resulting mixture at the original cost price of milk yields a profit of 50/3 %?

Accepted Answer

Introduction / Context: This is a classic adulteration/mixture scenario. If a milk seller adds free water and still sells at the milk’s cost price, all the revenue from the water portion becomes profit. The profit percentage directly relates to the water-to-milk ratio. Given Data / Assumptions: Let milk quantity be m and added water be w. Selling price per unit equals milk’s cost price per unit. Profit % = (50/3)% = 16.666…%. Concept / Approach: Because water is free, profit comes from selling w units at milk’s unit cost. Hence Profit% = (w / m) * 100. Set (w / m) * 100 = 50/3 and solve for w : m, then invert if needed to express milk : water. Step-by-Step Solution: (w / m) * 100 = 50 / 3 ⇒ w / m = 1 / 6.Therefore, milk : water = m : w = 6 : 1. Verification / Alternative check: If milk = 6 L and water = 1 L, total = 7 L. Revenue at milk’s unit cost equals 7 units of cost while cost is only 6 units → profit = 1/6 of cost → 16.666…%, as required. Why Other Options Are Wrong: 1:2, 2:3, and 2:5 do not produce a 16.666…% profit when selling at cost price. Common Pitfalls: Treating profit% as w / (m + w) instead of w / m (since selling at cost price of milk). Final Answer: 6:1

Question 19

Purity mixing via alligation: Two solutions, one of 90% purity and another of 97% purity, are mixed to get 21 litres of a 94% pure mixture. How many litres of the 90% solution were used?

Accepted Answer

Introduction / Context: We are mixing two purity solutions to reach a target purity. This can be handled either by a direct equation on pure content or by the alligation rule. Both give the same result quickly. Given Data / Assumptions: Total mixture = 21 L. Purities: 90% and 97%; target = 94%. Concept / Approach: Let x be the volume from the 90% solution. Then (21 − x) comes from the 97% solution. Equate total pure content to the pure content in the final mixture and solve for x. Step-by-Step Solution: 0.90x + 0.97(21 − x) = 0.94 * 21.0.90x + 20.37 − 0.97x = 19.74 ⇒ −0.07x = −0.63 ⇒ x = 9. Verification / Alternative check: Alligation differences: (97 − 94) : (94 − 90) = 3 : 4. Therefore, volumes are in ratio 3 (from 90%) to 4 (from 97%). For 21 L total, 3 + 4 = 7 parts → each part 3 L → 90% share = 3 parts = 9 L, matching the equation method. Why Other Options Are Wrong: 12 L, 15 L, and 6 L do not satisfy the purity balance for a 21 L mixture at 94%. Common Pitfalls: Swapping the alligation differences, which inverts the component volumes. Final Answer: 9 litres

Question 20

Tea-price alligation: In what proportion should teas priced at ₹ 62/kg and ₹ 72/kg be mixed to obtain a blend worth ₹ 65/kg?

Accepted Answer

Introduction / Context: Alligation is a quick method to find the ratio of two ingredients to achieve a desired mean price or concentration. We apply it directly to per-kilogram prices here. Given Data / Assumptions: Cheaper tea = ₹ 62/kg; dearer tea = ₹ 72/kg. Target blend price = ₹ 65/kg. Concept / Approach: Alligation ratio (cheaper : dearer) = (dearer − mean) : (mean − cheaper) = (72 − 65) : (65 − 62) = 7 : 3. Mix in this ratio by weight. Step-by-Step Solution: Compute differences: 72 − 65 = 7 and 65 − 62 = 3.Therefore, ratio cheaper : dearer = 7 : 3. Verification / Alternative check: Pick 7 kg of ₹ 62 and 3 kg of ₹ 72. Total cost = 7*62 + 3*72 = 434 + 216 = 650 for 10 kg → price per kg = ₹ 65, confirming the ratio. Why Other Options Are Wrong: 4 : 6 reduces to 2 : 3 (not matching the computed 7 : 3). 2 : 3 and 4 : 7 do not yield ₹ 65/kg when tested. Common Pitfalls: Reversing the differences leading to the inverse ratio. Final Answer: 7 : 3

Alligation or Mixture Questions