Question 1

A dairy mixes cow's milk containing 10% fat with buffalo's milk containing 20% fat. After mixing, the resulting milk has fat equal to (120/7)%. In what ratio (cow's milk : buffalo's milk) were the two milks mixed to obtain this fat percentage?

Accepted Answer

Introduction:This is an alligation problem based on concentration (fat percentage). We mix a lower-fat milk (10%) with a higher-fat milk (20%) to obtain an intermediate fat percentage. The ratio is determined by how far the target lies from each concentration. The target here is (120/7)%, which is approximately 17.142857%. Given Data / Assumptions: Cow's milk fat = 10% Buffalo's milk fat = 20% Target fat = (120/7)% Let cow : buffalo = ? Concept / Approach:Using alligation:Ratio (cow : buffalo) = (buffalo% - target%) : (target% - cow%).Work exactly with fractions to avoid rounding errors: target = 120/7. Step-by-Step Solution:Cow% = 10, Buffalo% = 20, Target% = 120/7Compute (Buffalo - Target) = 20 - 120/7 = (140/7 - 120/7) = 20/7Compute (Target - Cow) = 120/7 - 10 = (120/7 - 70/7) = 50/7Required ratio (cow : buffalo) = (20/7) : (50/7)Cancel /7: ratio = 20 : 50Simplify ratio = 2 : 5 Verification / Alternative Check:Mix 2 parts cow and 5 parts buffalo.Fat contribution = 2*10 + 5*20 = 20 + 100 = 120.Total parts = 7.Fat% = 120/7, exactly the target. This confirms the ratio without any rounding. Why Other Options Are Wrong:1:5: gives fat% closer to 20% than required (too high).2:3 or 3:5: produces fat% lower than 120/7.2:1: produces fat% much closer to 10% than the target. Common Pitfalls:Rounding 120/7 too early and getting an inaccurate ratio.Reversing the alligation differences and flipping the ratio.Assuming equal volumes instead of calculating the correct ratio. Final Answer:2 : 5

Question 2

In a mixture of 35 litres, the ratio of milk to water is 4:1. If 7 litres of water is added to this mixture (and nothing is removed), what will be the new ratio of milk to water in the resulting mixture?

Accepted Answer

Introduction / Context:This question checks the core idea of mixture ratios. When you add only water to a milk-water mixture, the amount of milk stays exactly the same, while the amount of water increases. The ratio changes because the denominator side (water) changes, but the numerator side (milk) does not. The safest way to solve is to convert the ratio into actual litres, perform the addition, and then simplify the new ratio. Given Data / Assumptions: Total mixture = 35 litres Milk : Water = 4 : 1 Water added = 7 litres No milk removed, no mixture removed Concept / Approach:If milk:water = 4:1, total parts = 5. Milk is 4/5 of the mixture and water is 1/5 of the mixture. After adding water, milk remains constant but water becomes (original water + added water). Finally, simplify the ratio. Step-by-Step Solution: Total parts = 4 + 1 = 5 Milk = (4/5) * 35 = 28 litres Water = (1/5) * 35 = 7 litres New water = 7 + 7 = 14 litres New ratio = 28 : 14 Simplify by 14 => 2 : 1 Verification / Alternative check:Originally water was 7 litres. After adding 7 litres, water doubles to 14 litres, while milk stays 28 litres. So the ratio should become half of 4:1, which is 2:1. This quick logic matches the computed answer. Why Other Options Are Wrong: 4:1 is the original ratio and ignores the added water. 5:4 and 4:5 come from mixing up milk and water or simplifying incorrectly. 1:2 reverses the dominance even though milk is still greater than water here. Common Pitfalls:Adding 7 litres to the total but forgetting it is added specifically to water, not recomputing milk and water from parts, or not simplifying 28:14 correctly. Final Answer:The new ratio of milk to water is 2:1.

Question 3

The price of 1 kg of sugar is increased by 30%, and the increase amount is Rs 48. What is the new price (in Rs) of sugar per kg after applying this 30% increase?

Accepted Answer

Introduction / Context:This question tests percentage increase and the relationship between original value, percentage change, and change amount. If an item increases by 30%, then the increase is 30% of the original price, not 30% of the new price. Once the original price is found from the increase amount, the new price is obtained by adding the increase or multiplying by 1.30. The key skill here is translating the words into a clean equation. Given Data / Assumptions: Percentage increase = 30% Increase amount = Rs 48 Let original price = P (Rs per kg) New price = P + 48 Concept / Approach:Increase amount = (percentage increase) * (original price). So, 0.30 * P = 48. Solve for P. Then compute new price = P + 48 (or P * 1.30). Step-by-Step Solution: 0.30 * P = 48 P = 48 / 0.30 P = 160 New price = 160 + 48 = 208 Verification / Alternative check:30% of 160 is 0.30 * 160 = 48, exactly the given increase. Therefore the new price must be 160 + 48 = 208. Another check is 160 * 1.30 = 208, which confirms the same result. Why Other Options Are Wrong: 160 is the original price, not the new price. 216 would mean the increase was 56, not 48. 192 and 178 come from using the wrong base for 30% or arithmetic mistakes. Common Pitfalls:Applying 30% to the new price instead of the original, dividing 48 by 30 instead of 0.30, or forgetting to add the increase after finding the original price. Final Answer:The new price of sugar is Rs 208 per kg.

Question 4

Container A contains 54 litres of a milk-water mixture in the ratio 8:1 (milk:water). From container A, 18 litres of the mixture is taken out and poured into container B, in which initially the ratio of milk to water is 3:1. After this transfer, the…

Accepted Answer

Introduction / Context:This is a multi-step transfer mixture problem. The mixture taken from container A keeps the same milk-water ratio as in A, so you must first compute how much milk and how much water is inside the 18 litres transferred. Then, represent container B initial milk and water in terms of an unknown initial volume. After adding the transferred milk and water, use the given final condition (milk minus water equals 30 litres) to form an equation and solve for the initial volume in B. Given Data / Assumptions: Container A total = 54 litres, milk:water = 8:1 Transferred from A to B = 18 litres (same composition as A) Container B initial milk:water = 3:1 After transfer in B: (milk - water) = 30 litres Let initial volume in B = V litres Concept / Approach:Convert ratios into actual quantities. For A, use parts to get milk and water per litre. For the transferred 18 litres, scale down from the 54 litres or directly use the 8:1 ratio. For B, write initial milk and water as fractions of V based on 3:1. Add transferred amounts, then apply the difference condition to solve for V. Step-by-Step Solution: In A, total parts = 8 + 1 = 9 Milk in A = (8/9) * 54 = 48 litres Water in A = (1/9) * 54 = 6 litres 18 litres is 18/54 = 1/3 of A, so transferred milk = 48*(1/3)=16 and transferred water = 6*(1/3)=2 In B initially (3:1): milk = (3/4)V and water = (1/4)V After transfer: milk = (3/4)V + 16, water = (1/4)V + 2 Given milk - water = 30: ((3/4)V + 16) - ((1/4)V + 2) = 30 (1/2)V + 14 = 30 (1/2)V = 16 V = 32 litres Verification / Alternative check:If V=32, initial milk in B=24 and water=8. After adding 16 milk and 2 water, final milk=40 and final water=10. Difference is 40 - 10 = 30, which matches the condition. Why Other Options Are Wrong: 24, 28, 30 give smaller final differences when checked with the same transfer amounts. 36 gives a larger final difference than 30 after transfer. Common Pitfalls:Assuming the 18 litres transferred is pure milk or pure water, forgetting that the transferred mixture keeps the same ratio as A, or using the final condition as a ratio instead of a difference. Final Answer:The initial quantity of mixture in container B was 32 litres.

Question 5

In what ratio should sugar costing Rs 38 per kg and sugar costing Rs 30 per kg be mixed so that, on selling the mixture at Rs 35.2 per kg, there is a profit of 10%? (Assume profit percentage is calculated on the cost price of the mixture.)

Accepted Answer

Introduction / Context:This question combines two common aptitude ideas: profit percentage and alligation (mixing). The selling price is given, but the mixture must yield a 10% profit, so first you must find the required cost price of the mixture. Only after finding the correct mean cost price can you apply alligation between Rs 38/kg and Rs 30/kg to get the correct mixing ratio. Many wrong answers happen when people treat Rs 35.2 as the cost price directly. Given Data / Assumptions: Cost of sugar type 1 = Rs 38 per kg Cost of sugar type 2 = Rs 30 per kg Selling price of mixture = Rs 35.2 per kg Profit = 10% on cost price Concept / Approach:If profit is 10%, then SP = 1.10 * CP. So CP = SP / 1.10. Once the mean cost price (CP of mixture) is known, use alligation: Quantity ratio (dearer : cheaper) = (mean - cheaper) : (dearer - mean). Step-by-Step Solution: CP of mixture = 35.2 / 1.10 = 32 Now apply alligation with dearer=38, cheaper=30, mean=32 Required ratio (38 : 30) = (32 - 30) : (38 - 32) = 2 : 6 = 1 : 3 Verification / Alternative check:Mix 1 kg of Rs 38 with 3 kg of Rs 30. Total cost = 38 + 90 = 128 for 4 kg, so mean CP = 128/4 = 32. Selling at 35.2 gives profit = 35.2 - 32 = 3.2 per kg, which is 3.2/32 = 10%. Verified. Why Other Options Are Wrong: 13:7 and 9:4 use too much Rs 38 sugar, pushing CP above Rs 32. 3:7 and 2:5 do not produce the exact mean CP of Rs 32 when computed. Common Pitfalls:Using 10% of selling price instead of cost price, skipping the CP calculation and treating 35.2 as CP, or reversing the alligation differences incorrectly. Final Answer:The sugars should be mixed in the ratio 1:3 (Rs 38/kg : Rs 30/kg).

Question 6

In a mixture of 60 litres, the ratio of milk to water is 2:1. How many litres of water must be added to the mixture so that the new ratio of milk to water becomes 1:2?

Accepted Answer

Introduction / Context:This is a ratio-change problem where only water is added. The most important idea is that milk stays constant because nothing is removed and only water is poured in. So you first convert the initial ratio and total volume into actual milk and water quantities. Then you add x litres of water and impose the target ratio. This involves a simple equation but requires careful handling of the ratio meaning. Given Data / Assumptions: Total mixture = 60 litres Initial milk:water = 2:1 Only water is added Target milk:water = 1:2 Let added water = x litres Concept / Approach:From 2:1, total parts are 3. Milk = (2/3)*60 and water = (1/3)*60. After adding x, milk stays the same and water becomes (20 + x). For target 1:2, milk/water must equal 1/2. Step-by-Step Solution: Total parts = 2 + 1 = 3 Milk = (2/3) * 60 = 40 litres Water = (1/3) * 60 = 20 litres After adding x: water = 20 + x Target ratio 1:2 => 40 / (20 + x) = 1/2 Cross-multiply: 80 = 20 + x x = 60 litres Verification / Alternative check:After adding 60 litres, water becomes 80 litres while milk remains 40 litres. Ratio is 40:80 which simplifies to 1:2, exactly as required. This also makes sense because you need water to become double the milk quantity. Why Other Options Are Wrong: 42 and 56 do not make water reach 80 litres, so the ratio will not become 1:2. 77 makes water too large, producing a ratio smaller than 1:2. 50 makes water 70, giving ratio 40:70 = 4:7, not 1:2. Common Pitfalls:Changing milk when only water is added, using total volume after addition incorrectly, or flipping the target ratio as 2:1 instead of 1:2. Final Answer:60 litres of water must be added.

Question 7

The average age of 6 members of a family is 20 years. If the age of a servant is also included, the average age increases by 25%. What is the age (in years) of the servant?

Accepted Answer

Introduction / Context:This is an averages and totals question. Average is total sum divided by number of items (people). When a new person is included, both the total sum and the number of people change. Because the new average is given as a percentage increase of the old average, you can compute the new average directly. Then compute the new total sum and subtract the original total to isolate the servant's age. This approach is systematic and avoids guesswork. Given Data / Assumptions: Average age of 6 family members = 20 years Servant added, total people = 7 Average increases by 25% (relative increase) Let servant age = S years Concept / Approach:Original total = 6 * 20. New average = 20 * (1 + 25/100) = 25. New total = 7 * 25. Servant age is the difference between new total and original total. Step-by-Step Solution: Original total age = 6 * 20 = 120 New average = 20 * 1.25 = 25 New total age (7 people) = 7 * 25 = 175 Servant age S = 175 - 120 = 55 Verification / Alternative check:If the servant is 55, the new total is 120 + 55 = 175, and the new average is 175/7 = 25. Since 25 is 25% more than 20 (because 20 + 25% of 20 = 20 + 5 = 25), the statement is satisfied exactly. Why Other Options Are Wrong: 30 and 35 are too small; they cannot raise the average from 20 to 25. 50 gives total 170 and average 170/7 = 24.285..., not 25. 45 gives average 165/7 = 23.571..., also not 25. Common Pitfalls:Thinking “increase by 25%” means adding 25 years, forgetting the number of people becomes 7, or using 25% of the new average instead of 25% of the old average. Final Answer:The servant's age is 55 years.

Question 8

In a triangle XYZ, which one of the following conditions is always true for the side lengths (triangle inequality)? (Assume XY, YZ, and ZX are the lengths of the three sides.)

Accepted Answer

Introduction / Context:This question tests the triangle inequality theorem, one of the most fundamental rules for triangles. A triangle is possible only if any two sides together are longer than the third side. If this is not true, the shape cannot close into a triangle; the points would lie on a straight line (degenerate) or be impossible. The theorem applies to all triangles regardless of type (scalene, isosceles, equilateral) as long as side lengths are positive. Given Data / Assumptions: XYZ is a valid triangle XY, YZ, ZX are positive side lengths We need a condition that is always true Concept / Approach:Triangle inequality requires: XY + YZ > ZX YZ + ZX > XY ZX + XY > YZ Any condition using “<” is invalid for a real triangle, and “=” represents a degenerate case (not a proper triangle). Step-by-Step Solution: Pick the option that matches triangle inequality form “sum of two sides is greater than the third”. Option “XY + YZ > ZX” exactly matches the theorem. Therefore, it is always true for every valid triangle XYZ. Verification / Alternative check:Example triangle with sides 3, 4, 5: 3 + 4 = 7 > 5, so the condition holds. If you try a non-triangle like 2, 3, 5, then 2 + 3 = 5 which is not greater, and it becomes a straight line, confirming why “>” is necessary. Why Other Options Are Wrong: Any “<” option contradicts triangle inequality and implies a triangle cannot be formed. The “=” option describes a degenerate case where points are collinear, not a proper triangle. Common Pitfalls:Using “greater than or equal to” instead of strictly greater, confusing side rules with angle rules, or accidentally interpreting expressions without the plus sign correctly. Final Answer:The always true condition is XY + YZ > ZX.

Question 9

Three boxes of capacities 24 kg, 36 kg, and 84 kg are completely filled with three varieties of wheat A, B, and C respectively. All three boxes are emptied, the three types of wheat are thoroughly mixed to form a uniform mixture, and then the mixtur…

Accepted Answer

Introduction / Context:This question tests proportional distribution after uniform mixing. When different items are mixed thoroughly, every portion of the new mixture has the same composition as the whole. That means each box receives wheat A in proportion to how large that box is compared to the total amount of mixture. So you find the fraction of wheat A in the total mixture and then apply that same fraction to the capacity of the third box. Given Data / Assumptions: Box 1 capacity = 24 kg (all wheat A) Box 2 capacity = 36 kg (all wheat B) Box 3 capacity = 84 kg (all wheat C) After mixing, composition is uniform everywhere Mixture is redistributed back fully into the same capacities Concept / Approach:Total wheat quantity = 24 + 36 + 84. Total wheat A in the mixture is still 24 kg (mixing does not change amounts). Fraction of wheat A in the mixture = 24 / total. Wheat A in the 84 kg box = fraction * 84. Step-by-Step Solution: Total mixture = 24 + 36 + 84 = 144 kg Total wheat A = 24 kg Fraction of A in mixture = 24 / 144 = 1/6 Wheat A in 84 kg box = (1/6) * 84 = 14 kg Verification / Alternative check:Since the third box holds 84 out of 144 kg, it holds 84/144 = 7/12 of the total mixture. Wheat A total is 24, so A in third box should be 24 * (7/12) = 14. Same result, confirming correctness. Why Other Options Are Wrong: 10 and 12 underestimate the share of A in a large 84 kg box. 16 and 18 overestimate the share and would imply more than 1/6 fraction of A. Common Pitfalls:Assuming equal distribution of A into each box, forgetting to compute the total mixture, or using 24/84 instead of 24/144 for the fraction of A. Final Answer:The third (84 kg) box will contain 14 kg of wheat type A.

Question 10

In what ratio should a mixture having 20% milk strength be mixed with another mixture having 60% milk strength so that the new mixture becomes 25% milk strength? (Assume milk strength means the percentage of milk in the mixture.)

Accepted Answer

Introduction / Context:This is an alligation (percentage mixing) problem. You are mixing a weaker mixture (20% milk) with a stronger mixture (60% milk) to obtain a target mixture that is only slightly stronger than the weaker one (25% milk). Because the target is close to 20%, you will need much more of the 20% mixture than the 60% mixture. Alligation provides a fast and reliable method using only differences from the mean. Given Data / Assumptions: Mixture 1 milk percentage = 20% Mixture 2 milk percentage = 60% Target milk percentage = 25% Find ratio (20% mixture : 60% mixture) Concept / Approach:Alligation rule: Required ratio (weaker : stronger) = (stronger - mean) : (mean - weaker). Here weaker = 20, stronger = 60, mean = 25. Compute the two differences and simplify. Step-by-Step Solution: Weaker = 20, Stronger = 60, Mean = 25 Strong difference = 60 - 25 = 35 Weak difference = 25 - 20 = 5 Ratio (20% : 60%) = 35 : 5 Simplify = 7 : 1 Verification / Alternative check:Take 7 litres of 20% and 1 litre of 60%. Milk = 7*0.20 + 1*0.60 = 1.4 + 0.6 = 2.0 litres. Total = 8 litres. Milk% = 2.0/8 = 25%. Verified exactly. Why Other Options Are Wrong: 4:1 gives a stronger result than 25% because it uses relatively more 60% mixture. 5:2 and 9:2 shift the weighted average away from 25% when computed. 3:1 makes the mixture much stronger than 25%. Common Pitfalls:Reversing the ratio order, subtracting differences in the wrong direction, or confusing milk percentage with water percentage. Final Answer:The required ratio is 7:1 (20% mixture : 60% mixture).

Question 11

A mixture contains spirit and water in the ratio 3:2 (spirit:water). If the mixture contains 3 litres more spirit than water, what is the quantity of spirit (in litres) in the mixture?

Accepted Answer

Introduction / Context:This is a ratio-and-difference question. When two quantities are in a ratio, you can represent them as multiples of a common factor. The difference between the two quantities then becomes the difference in parts times that common factor. Since the difference is provided as a real number (3 litres), you can solve for the factor quickly and then compute the required component (spirit). This method is faster and clearer than trial and error. Given Data / Assumptions: Spirit : Water = 3 : 2 Spirit is 3 litres more than water Let common factor = k litres per part Concept / Approach:If spirit:water = 3:2, then spirit = 3k and water = 2k. The difference is 3k - 2k = k. Since the difference equals 3 litres, k = 3. Then spirit = 3k. Step-by-Step Solution: Let spirit = 3k and water = 2k Difference = 3k - 2k = k Given difference = 3 litres, so k = 3 Spirit = 3k = 3 * 3 = 9 litres Verification / Alternative check:If spirit is 9 litres, then water must be 6 litres to maintain the 3:2 ratio. The difference 9 - 6 = 3 litres, which matches the condition. Both ratio and difference are satisfied, so the solution is correct. Why Other Options Are Wrong: 8 and 10 do not allow water to be exactly 3 litres less while keeping ratio 3:2. 12 would pair with water 9 giving ratio 4:3, not 3:2. 15 would pair with water 12 giving ratio 5:4, not 3:2. Common Pitfalls:Treating 3 litres as total mixture, not as the difference, or using 3/2 directly as a fraction without forming part variables. Final Answer:The quantity of spirit is 9 litres.

Question 12

Two bottles contain acid and water in the ratios 2:3 and 1:2 respectively (acid:water). These two mixtures are mixed together in the ratio 1:3 (first mixture : second mixture). What is the ratio of acid to water in the final mixture?

Accepted Answer

Introduction / Context:This question tests compound mixture calculation. Each bottle has its own internal composition (acid fraction and water fraction). When you mix the two bottles in a given external ratio, you must scale the acid and water contributions accordingly and then add them. The final ratio is based on total acid and total water, not on any single bottle ratio. The safest method is to assume convenient volumes (like 1 litre and 3 litres) matching the mixing ratio. Given Data / Assumptions: Mixture 1 acid:water = 2:3 Mixture 2 acid:water = 1:2 Mixed in ratio (Mixture 1 : Mixture 2) = 1:3 Assume 1 litre of mixture 1 and 3 litres of mixture 2 Concept / Approach:Convert each ratio to fractions. For 2:3, total parts 5, acid fraction = 2/5, water fraction = 3/5. For 1:2, total parts 3, acid fraction = 1/3, water fraction = 2/3. Multiply fractions by assumed litres, add totals, and simplify the final ratio. Step-by-Step Solution: Take 1 litre of mixture 1 and 3 litres of mixture 2 Mixture 1: acid = (2/5)*1 = 2/5, water = (3/5)*1 = 3/5 Mixture 2: acid = (1/3)*3 = 1, water = (2/3)*3 = 2 Total acid = 2/5 + 1 = 7/5 Total water = 3/5 + 2 = 13/5 Final ratio acid:water = (7/5):(13/5) = 7:13 Verification / Alternative check:You can also convert both totals to the same denominator and confirm the denominator cancels. The result 7:13 remains stable no matter what base litres you assume, as long as you keep the 1:3 mixing ratio. Why Other Options Are Wrong: 1:3 ignores internal compositions and wrongly assumes mixture 2 dominates as if it were pure water. 23:37 and 11:57 come from incorrect fraction conversion (like using 2/3 instead of 2/5). 5:9 does not match the computed acid and water totals. Common Pitfalls:Using 2:3 as acid fraction 2/3 instead of 2/5, forgetting to multiply mixture 2 contributions by 3, or reversing the external mixing ratio 1:3 as 3:1. Final Answer:The ratio of acid to water in the final mixture is 7:13.

Question 13

A jar contains a blend of fruit juice and water in the ratio 5:x (fruit juice:water). When 1 litre of water is added to 4 litres of this blend, the ratio of fruit juice to water becomes 1:1. What is the value of x?

Accepted Answer

Introduction / Context:This is a ratio reconstruction problem. The jar has a fixed internal ratio of fruit juice to water (5:x). When you take 4 litres of the blend, the quantities of juice and water in that 4 litres are proportional to the 5:x ratio. Then you add 1 litre of water, which changes only the water amount. The final ratio becomes 1:1, meaning juice and water quantities are equal. Setting up this equality gives an equation for x. Given Data / Assumptions: Initial ratio (juice:water) = 5:x Blend taken = 4 litres Added water = 1 litre Final ratio juice:water = 1:1 (equal quantities) Concept / Approach:In ratio 5:x, total parts = (5 + x). In 4 litres: Juice = 4 * 5/(5+x) Water = 4 * x/(5+x) After adding 1 litre water, water becomes 4x/(5+x) + 1. For 1:1, set juice = water and solve for x. Step-by-Step Solution: Juice in 4 litres = 4 * 5/(5 + x) = 20/(5 + x) Water in 4 litres = 4 * x/(5 + x) = 4x/(5 + x) After adding 1 litre water: water = 4x/(5 + x) + 1 Since final ratio is 1:1, juice = water: 20/(5 + x) = 4x/(5 + x) + 1 Multiply by (5 + x): 20 = 4x + (5 + x) 20 = 5x + 5 5x = 15 x = 3 Verification / Alternative check:If x=3, ratio is 5:3. In 4 litres, juice = 4*(5/8)=2.5 and water = 4*(3/8)=1.5. Add 1 litre water gives water = 2.5, equal to juice, so the ratio becomes 1:1. Verified. Why Other Options Are Wrong: x=1 or 2 makes water too small; even after adding 1 litre, juice remains greater than water. x=4 or 5 makes water too large relative to juice so equality after addition does not hold. Common Pitfalls:Forgetting that 4 litres is a portion of the blend (not pure juice), treating 5:x as 5/x instead of parts, or assuming 1:1 means total volume doubles (it does not). Final Answer:The value of x is 3.

Question 14

Jar A contains 36 litres of a milk-water mixture in the ratio 5:4 (milk:water). Jar B contains 20 litres of a milk-water mixture (composition unknown). Jar B is completely emptied into Jar A. After mixing, the ratio of milk to water in Jar A becomes…

Accepted Answer

Introduction / Context:This is a mixture combination problem where one mixture is known and the other mixture is unknown. The key strategy is to convert Jar A ratio into exact litres of milk and water, then represent Jar B milk and water as variables that sum to 20 litres. After combining, the final ratio is given (5:3), which creates an equation relating total milk and total water. Solving that equation gives the water in Jar B. Given Data / Assumptions: Jar A volume = 36 litres, milk:water = 5:4 Jar B volume = 20 litres, composition unknown Jar B is fully poured into Jar A Final milk:water ratio = 5:3 Let water in Jar B = w and milk in Jar B = m m + w = 20 Concept / Approach:Compute milk and water in Jar A from ratio 5:4. Then write total milk and total water after mixing as (milkA + m) and (waterA + w). Use the final ratio equation (total milk)/(total water) = 5/3. Substitute m = 20 - w and solve for w. Step-by-Step Solution: Jar A total parts = 5 + 4 = 9 Milk in A = (5/9) * 36 = 20 litres Water in A = (4/9) * 36 = 16 litres After adding Jar B: total milk = 20 + m, total water = 16 + w Final ratio 5:3 => (20 + m)/(16 + w) = 5/3 m + w = 20 => m = 20 - w (20 + 20 - w)/(16 + w) = 5/3 (40 - w)/(16 + w) = 5/3 3*(40 - w) = 5*(16 + w) 120 - 3w = 80 + 5w 40 = 8w w = 5 litres Verification / Alternative check:If w=5, then m=15. Totals become milk=20+15=35 and water=16+5=21. Ratio 35:21 simplifies to 5:3, matching the final condition exactly. Why Other Options Are Wrong: 2 or 3 litres makes Jar B too milk-heavy, increasing the final ratio above 5:3. 7 or 8 litres adds too much water, reducing the final ratio below 5:3. Common Pitfalls:Mixing up milk and water quantities from 5:4, forgetting that Jar B is completely poured, or applying the final ratio to Jar B only instead of to the combined totals. Final Answer:Jar B initially contained 5 litres of water.

Question 15

The ratio of milk to water in a mixture is 5:4. If 2 litres of water is added to the mixture, the ratio becomes 10:9. Find the new amount of water (in litres) in the mixture after adding the 2 litres of water.

Accepted Answer

Introduction / Context:This is a ratio update problem where only water is added. Since milk is not removed or added, the milk quantity stays constant. The final ratio is given after the water addition, so you can represent the original mixture using a variable k based on the 5:4 ratio. Then increase only the water part by 2 litres and enforce the final ratio 10:9. Solving the resulting equation gives the original scale and therefore the new water quantity. Given Data / Assumptions: Initial milk:water = 5:4 Water added = 2 litres Final milk:water = 10:9 Let initial milk = 5k and initial water = 4k Concept / Approach:Milk remains 5k. Water becomes 4k + 2 after addition. Apply final ratio: 5k / (4k + 2) = 10/9. Solve for k, then compute the new water amount (4k + 2). Step-by-Step Solution: Initial milk = 5k, initial water = 4k After adding 2 litres water: water = 4k + 2 Final ratio 10:9 => 5k/(4k + 2) = 10/9 Cross-multiply: 9*5k = 10*(4k + 2) 45k = 40k + 20 5k = 20 k = 4 New water amount = 4k + 2 = 16 + 2 = 18 litres Verification / Alternative check:Milk = 5k = 20 litres. Water new = 18 litres. Ratio 20:18 simplifies to 10:9, exactly matching the condition, so the computed new water amount is correct. Why Other Options Are Wrong: 16 litres would imply ratio 20:16 = 5:4, meaning no effective change after addition. 12, 14, 17 do not yield 10:9 when milk remains fixed and only water increases by 2. Common Pitfalls:Adding 2 litres to milk by mistake, setting up the equation as (milk + 2)/water, or simplifying ratios incorrectly after solving for k. Final Answer:The new amount of water in the mixture is 18 litres.

Question 16

A vessel, full of water, weighs 24 kg. When the same vessel is 1/4 full of water, it weighs 9 kg. Find the weight (in kg) of the empty vessel (the vessel without any water).

Accepted Answer

Introduction / Context:This problem tests forming and solving linear equations using weight compositions. The total weight of a vessel with water equals the weight of the empty vessel plus the weight of the water inside it. When the vessel is only 1/4 full, the water weight becomes 1/4 of the full-water weight, but the vessel weight remains unchanged. By writing two equations (full and quarter-full) and subtracting, you can eliminate the vessel weight and solve for the water weight, then get the empty vessel weight. Given Data / Assumptions: Weight when full = 24 kg Weight when 1/4 full = 9 kg Let empty vessel weight = E kg Let full water weight = W kg Concept / Approach:Set equations: Full: E + W = 24 Quarter-full: E + (W/4) = 9 Subtract second from first to isolate W, then compute E = 24 - W. Step-by-Step Solution: Equation (1): E + W = 24 Equation (2): E + W/4 = 9 Subtract (2) from (1): W - W/4 = 24 - 9 (3/4)W = 15 W = 15 * (4/3) = 20 E = 24 - 20 = 4 Verification / Alternative check:If empty vessel is 4 kg and full water is 20 kg, then at 1/4 full the water weighs 5 kg. Total becomes 4 + 5 = 9 kg, matching the given condition, so the result is consistent. Why Other Options Are Wrong: 3 kg would imply full water weight 21 kg and quarter-full total 3 + 5.25 = 8.25, not 9. 5 kg would imply full water weight 19 and quarter-full total 5 + 4.75 = 9.75, not 9. 6 kg or 8 kg similarly fail one of the two conditions when checked. Common Pitfalls:Assuming 1/4 full means total weight is 1/4 of 24 (incorrect), forgetting that vessel weight is constant, or not handling the subtraction step correctly. Final Answer:The empty vessel weighs 4 kg.

Question 17

Three bottles of equal capacity contain milk-water mixtures in the ratios 2:5, 3:4, and 4:5 respectively (milk:water). All three bottles are emptied into a large container. What will be the ratio of milk to water in the final mixture in the large co…

Accepted Answer

Introduction / Context:This question tests combining mixtures with equal volumes. Because all three bottles have equal capacity, you can assume each bottle contains 1 litre (or any same unit volume). Then compute the milk fraction and water fraction for each bottle based on its ratio. Add total milk from all bottles and add total water from all bottles. Finally, express the totals as a simplified ratio. The most common mistake is trying to add ratios directly, which is not valid. Given Data / Assumptions: Bottle 1 milk:water = 2:5 Bottle 2 milk:water = 3:4 Bottle 3 milk:water = 4:5 All bottles have equal capacity Assume each bottle has 1 litre for calculation Concept / Approach:For ratio a:b, milk fraction = a/(a+b) and water fraction = b/(a+b). Compute fractions for each bottle, sum milk and sum water. Convert to integer ratio by using a common denominator and then simplify if possible. Step-by-Step Solution: Bottle 1 (2:5): milk = 2/7, water = 5/7 Bottle 2 (3:4): milk = 3/7, water = 4/7 Bottle 3 (4:5): milk = 4/9, water = 5/9 Total milk = 2/7 + 3/7 + 4/9 = 5/7 + 4/9 Use denominator 63: 5/7 = 45/63 and 4/9 = 28/63 Total milk = (45 + 28)/63 = 73/63 Total water = 5/7 + 4/7 + 5/9 = 9/7 + 5/9 Convert to /63: 9/7 = 81/63 and 5/9 = 35/63 Total water = (81 + 35)/63 = 116/63 Final ratio milk:water = (73/63):(116/63) = 73:116 Verification / Alternative check:Since the denominator 63 cancels out, only the numerators matter. Also, 73 is prime relative to 116 (no common factor), so the ratio is already in simplest form. Why Other Options Are Wrong: 73:113 and 73:106 come from incorrect water totals or wrong fraction conversion for 4:5. 73:189 is far too water-heavy and indicates a major arithmetic error. 59:94 does not match the exact computed totals. Common Pitfalls:Adding ratios directly (2+3+4):(5+4+5), forgetting equal capacity assumption, or using 4/5 instead of 4/9 for the third bottle milk fraction. Final Answer:The final ratio of milk to water is 73:116.

Question 18

In a mixture of milk and water, milk constitutes 19% of the total mixture. If the quantity of milk in the mixture is 437 millilitres, what is the total quantity (in millilitres) of the mixture?

Accepted Answer

Introduction / Context:This question checks percentage-to-total conversion. If milk is 19% of the mixture, then the milk quantity equals 19/100 of the total mixture quantity. Given the milk amount, you can find the total by dividing by 19% (or multiplying by 100/19). Precision matters here because a small arithmetic slip can move you to a nearby option. The cleanest approach is to set up an equation and solve directly. Given Data / Assumptions: Milk percentage in mixture = 19% Milk quantity = 437 millilitres Total mixture quantity = T millilitres Concept / Approach:Milk = (19/100) * T. So T = Milk * (100/19). This uses the idea that percentage is a fraction of the total. Once T is computed, it represents the total mixture (milk + water) in millilitres. Step-by-Step Solution: 437 = (19/100) * T T = 437 * (100/19) T = 43700 / 19 19 * 2300 = 43700 So T = 2300 millilitres Verification / Alternative check:Check: 19% of 2300 = (19/100)*2300 = 19*23 = 437 ml. This matches the given milk quantity exactly, so the total mixture value is correct. Why Other Options Are Wrong: 2200 gives milk = 0.19*2200 = 418 ml, not 437 ml. 2400 gives milk = 456 ml, too high. 2350 gives milk = 446.5 ml, not 437 ml. 2100 gives milk = 399 ml, too low. Common Pitfalls:Using 81% instead of 19%, dividing 437 by 19 without multiplying by 100, or rounding too early instead of checking exact multiplication. Final Answer:The total quantity of the mixture is 2300 millilitres.

Question 19

In a mixture of 25 litres, the ratio of milk to water is 4:1. If another 3 litres of water is added to the mixture (without removing anything), what will be the new ratio of milk to water?

Accepted Answer

Introduction / Context:This is a straightforward mixture ratio update problem. The mixture initially has milk and water in a known ratio. When you add water only, the milk quantity remains the same and only the water quantity changes. To avoid confusion, convert the ratio into actual litres using the total volume, then add the extra water, and finally simplify the new ratio. This method is reliable for all “add water” ratio questions. Given Data / Assumptions: Total mixture = 25 litres Initial milk:water = 4:1 Added water = 3 litres Nothing is removed Concept / Approach:Total parts = 4 + 1 = 5. Milk = (4/5)*25 and water = (1/5)*25. After adding 3 litres of water, water becomes (5 + 3). Then form the ratio milk:water and simplify. Step-by-Step Solution: Total parts = 5 Milk = (4/5) * 25 = 20 litres Water = (1/5) * 25 = 5 litres New water = 5 + 3 = 8 litres New ratio = 20 : 8 Simplify by 4 => 5 : 2 Verification / Alternative check:The ratio must become smaller than 4:1 because water increased. 5:2 is 2.5:1, which is smaller than 4:1, so the direction of change is consistent with the operation performed. Why Other Options Are Wrong: 5:1 implies too little water (would correspond to 20:4, not 20:8). 5:3 and 5:4 do not match 20:8 after simplification. 4:3 is not equal to 20:8 and often appears from swapping numbers incorrectly. Common Pitfalls:Adding 3 to total and forgetting it is specifically added to water, not simplifying 20:8 properly, or using ratio parts without converting to litres correctly. Final Answer:The new ratio of milk to water is 5:2.

Question 20

Two types of brass contain Copper and Zinc in the ratios 8:3 and 15:7 respectively (Copper:Zinc). If these two types of brass are melted and mixed in the ratio 5:2 (first type : second type), what will be the ratio of Copper to Zinc in the new brass…

Accepted Answer

Introduction / Context:This is an alloy composition (mixture) problem. Each brass type has a fixed copper-to-zinc ratio, which means each type has a fixed fraction of copper and a fixed fraction of zinc. When you mix them in a given amount ratio, you must compute copper and zinc contributed by each portion and then add the totals. Finally, convert the resulting totals into the simplest copper:zinc ratio. This is similar to mixing solutions by concentration. Given Data / Assumptions: Type 1 brass: Copper:Zinc = 8:3 Type 2 brass: Copper:Zinc = 15:7 Mixing ratio (Type 1 : Type 2) = 5:2 Assume 5 units of Type 1 and 2 units of Type 2 Concept / Approach:Convert ratios to fractions. Type 1 total parts = 11, so copper fraction = 8/11 and zinc fraction = 3/11. Type 2 total parts = 22, so copper fraction = 15/22 and zinc fraction = 7/22. Multiply each fraction by its mixed amount (5 or 2), then add copper totals and zinc totals, and simplify. Step-by-Step Solution: Type 1 (amount 5): copper = 5*(8/11) = 40/11, zinc = 5*(3/11) = 15/11 Type 2 (amount 2): copper = 2*(15/22) = 30/22 = 15/11, zinc = 2*(7/22) = 14/22 = 7/11 Total copper = 40/11 + 15/11 = 55/11 = 5 Total zinc = 15/11 + 7/11 = 22/11 = 2 Copper:Zinc = 5:2 Verification / Alternative check:The totals became clean integers after combining, which strongly indicates correct fraction handling and common denominators. Since 5 and 2 share no common factor, the ratio is already simplified. Why Other Options Are Wrong: 3:2 and 3:4 do not match the computed totals and imply incorrect contribution calculations. 2:3 reverses dominance and would mean zinc exceeds copper, which is impossible here because both brass types have more copper than zinc. 7:5 suggests too much zinc relative to copper compared to both input alloys. Common Pitfalls:Using 8/3 or 15/7 directly as copper fraction, forgetting to convert to total parts (8+3 and 15+7), or forgetting to scale each type by the mixing ratio 5:2. Final Answer:The ratio of Copper to Zinc in the new alloy is 5:2.

Alligation or Mixture Questions

Questions

Practice smarter