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

In a mixture of 45 liters, the ratio of milk and water is 4:1. How much water must be added to make the mixture ratio 3:2?

Accepted Answer

Let us assume the ratio factor is x. According to question, Quantity of milk = 4x liters and Quantity of Water = x liters. According to question, 4x + x = 45 ⇒ 5x = 45 ⇒ x = 9 Quantity of milk = 4x = 9 x 4 = 36 liters Quantity of water = x = 9 liters Let y liters of water be added to make the ratio 3:2 Then, 36/(9 + y) = 3/2 ⇒ 72 = 27 + 3y ⇒ y = 15 liters If you add 15 liters of water, the ratio will become 3:2.

Question 17

In a class of 30 students , the average weight of boys is 20 kg and the average weight of the girls is 25 kg . The fraction of boys out of the total students of the class is

Accepted Answer

Let us assume the number of boys = B and number of girls = G. According to question, B + G = 30 Lets us assume total weight of boys = W1 and total weight of girls = W2 average weight of boys = total weight of boys/number of boys total weight of boys/number of boys = 20 W1/B = 20 W1 = 20B average weight of girls = total weight of girls/number of girls 25 = W2/G W2 = 25G Data is not sufficient to solve the equation. since we do not know either the average weight of the whole class or the ratio of no. of boys to girls.

Question 18

In what ratio should milk and water be mixed so that after selling the mixture at the cost price a profit of 50/3 % is made?

Accepted Answer

Method 1 to solve the equation. Let us assume Cost Price (C.P) of 1 liter milk be Y rupees. According to question, Selling Price (S.P) of 1 liter of mixture Y rupees. (Selling price should be same as Cost price) Profit = 50/3 % Let us assume Cost price of 1 Liter of Mixture of Water and Milk = C Cost price of 1 Liter of Mixture of Water and Milk = Selling price of Mixture - Profit C = Y - C x 50/3% ⇒ C = Y - C x 50/3x100 ⇒ C = Y - C x 1/3x2 ⇒ C = Y - C/6 ⇒ Y = C + C/6 ⇒ Y = (6C + C)/6 ⇒ Y = (6C + C)/6 ⇒ Y = 7C/6 ⇒ C = 6Y/7 Since in Y rupees we will get 1 liter milk. Hence in 1 rupees we will get 1/Y liter milk. Hence in 6Y/7 rupees we will get 1 x 6Y/7 x Y liter milk. Hence in 6Y/7 rupees we will get 6/7 liter milk. We need 1 liter mixture of milk and water for sold on the same price, we need to mix the water. So water quantity in mixture = 1 - 6/7 = 1/7 Ratio of Milk and water in mixture = quantity of Milk in Mixture/quantity of Water in Mixture Ratio of Milk and water in mixture = 6/7 / 1/7 Ratio of Milk and water in mixture = 6 x 7 / 7 Ratio of Milk and water in mixture = 6 / 1 Ratio of Milk and water in mixture = 6 : 1 Method 2 to solve the question. Let the original amount of milk be 1 liter and the cost price is 1 rupees per liter. Cost price of milk = 1 rupees. Selling price of Mixture = 1 rupees. When the milk is mix with x liters of water milk remaining is 1- x. Let us assume the cost price of milk in mixture is Y and given that the Sold Price is 1. we should use the Profit formula in algebra, Y + Y x 50/3 % = 1 Y + Y x 50/3 x 100 = 1 Y + Y x /3 x 2 = 1 7Y/6 = 1 Y = 6/7 Y = 6/7 is the cost price of Mixture for 1 liter. Since water is free so there is not cost of water in mixture. So price of milk is 6/7 rs in mixture. Since as per question, In 1 rupees we will get 1 liter of milk. hence in 6/7 rupees we will get 1 x 6/7 = 6/7 liter of milk. Now we have to make 1 liter of mixture with water and milk. quantity of Water + Quantity of Milk = 1 liter quantity of Water + 6/7 = 1 quantity of Water = 1 - 6/7 = (7 - 6)/7 quantity of Water = 1/7 Ratio of milk and Water = Quantity of Milk / quantity of Water Ratio of milk and Water = (6/7) / (1/7) Ratio of milk and Water = (6/7) x (7/1) Ratio of milk and Water = 6 x 7/ 7 x 1 Ratio of milk and Water = 6 :1

Question 19

Two solution of 90% and 97% purity and mixed resulting in 21 liters of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?

Accepted Answer

Method 1 to solve the equation. Let us assume the number of liters of the 90% purity solution = A and the number of liters of the 97% purity solution = B. According to question, Since there are 21 liters of the solution, A + B = 21 ...................... (1) Since after mixing the two solutions the new mixture has 94% purity, Concentrate of A + Concentrate of B = Concentrate of (A + B) A x 90% + B x 94% = (A+ B) x 97% ⇒ A x 90/100 + B x 97/100 = (A + B) x 94/100 ⇒ 90A + 97B = (A + B) x 94 ⇒ 90A + 97B = 94A + 94B ⇒ 94A + 94B - 90A- 97B = 0 ⇒ 4A - 3B = 0 ........................(2) Multiply the 3 with Equation (1) and add with Equation (2), 3A + 3B + 4A - 3B = 63 + 0 ⇒ 7A = 63 ⇒ A = 63/7 = 9 Put the value of A in Equation (1) , we will get 9 + B = 21 B = 21 - 9 B = 12 The first solution would be A = 9 liters. Method 2 to solve the equation. Hit and trail method. 94% is closer to 97% but barely meaning the mixtures will not be equal parts but will be slightly more of the higher purity. Quickly eliminate A and B. Out of the others 9 is the easy choice. If the other choices were closer to half this wouldn't work.

Question 20

In what proportion must tea at Rs. 62 per kg be mixed with tea at Rs. 72 per kg in order to obtain the mixture worth Rs. 65 per kg?

Accepted Answer

Using Alligation rule, (Quantity of cheaper tea) / (Quantity of dearer tea) = (d - m) / (m - c) = 7/3 Therefore, they must be mixed in the ratio of 7 : 3.

Alligation or Mixture Questions