Question 1

The average weight of boys in a class is 30 kg and the average weight of girls in the same class is 20kg. If the average weight of the whole class is 23.25 kg, what could be the possible strength of boys and girls respectively in the same class?

Accepted Answer

13 and 27

Question 2

Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.

Accepted Answer

2:3

Question 3

The ratio of water and alcohol in two different containers is 2:3 and 4:5. In what ratio we are required to mix the mixtures of two containers in order to get the new mixture in which the ratio of alcohol and water be 7:5?

Accepted Answer

5:3

Question 4

In the 75 litres of mixture of milk and water, the ratio of milk and water is 4:1. The quantity of water required to make the ratio of milk and water 3:1 is

Accepted Answer

5 litres

Question 5

From a container filled with milk, 6 litres of milk are drawn out and replaced with water. Again, 6 litres of the resulting mixture are drawn out and replaced with water. After these two operations, the ratio of milk to water in the container become…

Accepted Answer

Introduction: This problem involves repeated replacement in a mixture of milk and water, a classic concept in alligation and mixture. You remove some mixture and replace it with another liquid, which changes the ratio step by step. Understanding the formula for repeated replacement saves a lot of time in competitive exams. Given Data / Assumptions: The container is initially full of pure milk. First, 6 litres of milk are removed and replaced with 6 litres of water. Second, 6 litres of the resulting mixture are removed and replaced with 6 litres of water. After these two operations, the ratio of milk to water is 9 : 16. The total capacity of the container remains constant throughout. Concept / Approach: When a quantity x is removed from a container of volume V and replaced with another liquid, the fraction of the original liquid left after one operation is (1 - x / V). After n identical operations, the fraction left is (1 - x / V)^n. Here, milk is the original liquid, x is 6 litres, V is the capacity, and n = 2 operations. We will relate this fraction to the final milk fraction from the given ratio. Step-by-Step Solution: Step 1: Let the capacity of the container be V litres. Step 2: After two operations, the fraction of milk left is (1 - 6 / V)^2. Step 3: The final ratio of milk to water is 9 : 16. Step 4: So the fraction of milk in the final mixture is 9 / (9 + 16) = 9 / 25. Step 5: Equate the two expressions for the milk fraction: (1 - 6 / V)^2 = 9 / 25. Step 6: Take square roots on both sides: 1 - 6 / V = 3 / 5 (we discard the negative root since V must be greater than 6). Step 7: Solve for V: 6 / V = 1 - 3 / 5 = 2 / 5 V = 6 * 5 / 2 = 15 litres. Verification / Alternative check: If the capacity is 15 litres, after the first operation milk left is 15 - 6 = 9 litres. After the second operation, we remove 6 * (9 / 15) = 3.6 litres of milk, so milk left is 9 - 3.6 = 5.4 litres. Water then is 15 - 5.4 = 9.6 litres. The ratio 5.4 : 9.6 simplifies to 9 : 16, confirming that V = 15 litres is correct. Why Other Options Are Wrong: 16 litres: This capacity gives a different final ratio and does not match 9 : 16. 20 litres: The remaining fraction (1 - 6 / 20)^2 does not equal 9 / 25. 25 litres: With this capacity, milk remains too large a fraction after two operations. 30 litres: This also leads to a milk fraction much larger than 9 / 25. Common Pitfalls: Students often try to track milk and water litres stepwise without using the repeated replacement formula, which is error prone. Another mistake is taking the negative square root when solving, which would imply a capacity smaller than the amount removed, which is impossible. Also, sometimes the final ratio is misread as water to milk rather than milk to water. Final Answer: The total original quantity of the mixture in the container is 15 litres.

Question 6

In a pot there is a mixture of milk and water in the ratio 4 : 5. If an additional 8 litres of pure milk is poured into the pot, the pot becomes completely full and the ratio of milk to water changes to 6 : 5. What is the total capacity of the pot i…

Accepted Answer

Introduction: This mixture question deals with changing the ratio of milk to water in a container when extra milk is added. It uses the concept of ratios and basic algebra to relate the initial and final quantities and to determine the total capacity of the pot. Given Data / Assumptions: Initially the pot contains a mixture of milk and water in the ratio 4 : 5. Only milk and water are present in the mixture. 8 litres of pure milk are added to the pot. After adding the milk, the pot becomes full. The new ratio of milk to water becomes 6 : 5. We are asked to find the total capacity of the pot in litres. Concept / Approach: We use the idea of ratio representation. If a mixture has milk and water in the ratio 4 : 5, we can represent milk as 4k litres and water as 5k litres. When 8 litres of pure milk are added, only the milk quantity changes, while the amount of water remains the same. The final ratio becomes 6k1 : 5k1, but here we can directly set up an equation using the new ratio 6 : 5 and solve for k and for the pot capacity. Step-by-Step Solution: Step 1: Let the initial quantities be milk = 4k litres and water = 5k litres. Step 2: The initial total mixture is 4k + 5k = 9k litres. Step 3: 8 litres of pure milk are added, so the new milk quantity is 4k + 8 litres, water remains 5k litres. Step 4: The new ratio of milk to water is given as 6 : 5. Step 5: Set up the equation (4k + 8) / 5k = 6 / 5. Step 6: Cross multiply: 5 * (4k + 8) = 6 * 5k. Step 7: This gives 20k + 40 = 30k. Step 8: Rearrange: 30k - 20k = 40, so 10k = 40, hence k = 4. Step 9: Initial volume = 9k = 9 * 4 = 36 litres. Step 10: After addition, total capacity = 36 + 8 = 44 litres. Verification / Alternative check: With k = 4, initial milk = 16 litres, water = 20 litres. After adding 8 litres of milk, milk becomes 24 litres, water remains 20 litres. The ratio 24 : 20 simplifies to 6 : 5, confirming that the algebra is correct and that the total capacity of the pot is 44 litres. Why Other Options Are Wrong: 11 litres and 22 litres: These values are far less than the volume already present after adding 8 litres of milk, and they do not satisfy the ratio conditions. 33 litres: This would imply an initial volume too small to maintain the given ratios after adding 8 litres of milk. 36 litres: This is the initial volume before adding 8 litres, not the full pot capacity. The question specifically asks for the capacity when full, which is 36 + 8 = 44 litres. Common Pitfalls: A frequent error is to forget that only milk is added, so the water quantity remains unchanged. Another mistake is treating 8 litres as part of the ratio units directly, instead of using a variable k. Some learners also misinterpret the phrase that the pot becomes full and think they must change the total and the ratio in arbitrary ways rather than using a clear algebraic setup. Final Answer: The total capacity of the pot is 44 litres.

Question 7

In a container there is a mixture of milk and water in which water makes up 26% of the mixture by volume. If 7 litres of this mixture are removed and replaced with 7 litres of pure milk, the percentage of milk in the mixture becomes 76%. What is the…

Accepted Answer

Introduction: This is a replacement type mixture question where a portion of a milk water mixture is removed and replaced with pure milk. The percentages before and after replacement are given, and we must find the original volume of the mixture. Such questions test your understanding of percentage composition and how replacement affects the concentration of components in a mixture. Given Data / Assumptions: Initially water is 26% of the mixture by volume. Therefore milk is 74% of the mixture initially. 7 litres of the original mixture are removed. These 7 litres are replaced by 7 litres of pure milk. After this, the percentage of milk in the mixture becomes 76%. The total volume of the mixture in the container remains constant. Concept / Approach: Let the initial volume of the mixture be V litres. Since water is 26%, milk is 74%. Removing 7 litres of mixture removes milk and water in the same 74 : 26 ratio. Then we add back pure milk, which changes only the milk content. We then set the final milk fraction equal to 76% and solve for V using a linear equation. Step-by-Step Solution: Step 1: Let the initial volume of the mixture be V litres. Step 2: Initial milk = 0.74 * V, initial water = 0.26 * V. Step 3: 7 litres of mixture are removed, so milk removed = 0.74 * 7 and water removed = 0.26 * 7. Step 4: Milk remaining after removal = 0.74 * V - 0.74 * 7. Step 5: Water remaining after removal = 0.26 * V - 0.26 * 7. Step 6: 7 litres of pure milk are added, so new milk = 0.74 * V - 0.74 * 7 + 7. Step 7: Total volume is still V litres, so final milk fraction = (0.74 * V - 0.74 * 7 + 7) / V. Step 8: This fraction is given as 76%, so set up the equation: (0.74 * V - 0.74 * 7 + 7) / V = 0.76. Step 9: Multiply both sides by V: 0.74 * V - 0.74 * 7 + 7 = 0.76 * V. Step 10: Rearrange terms: 7 - 0.74 * 7 = 0.76 * V - 0.74 * V = 0.02 * V. Step 11: Compute 7 - 0.74 * 7 = 7 * (1 - 0.74) = 7 * 0.26 = 1.82. Step 12: So 0.02 * V = 1.82, hence V = 1.82 / 0.02 = 91 litres. Verification / Alternative check: Initial milk = 0.74 * 91 = 67.34 litres, water = 23.66 litres. In 7 litres removed, milk removed = 0.74 * 7 = 5.18 litres, water removed = 1.82 litres. After removal, milk = 62.16 litres, water = 21.84 litres. Adding 7 litres of pure milk gives milk = 69.16 litres, water = 21.84 litres. Total volume remains 91 litres. Milk percentage = 69.16 / 91 which is approximately 0.76 or 76%, confirming the result. Why Other Options Are Wrong: 65 litres, 38 litres, and 104 litres do not satisfy the condition that the final milk percentage is exactly 76% after the described operation. Substituting these values for V in the equation leads to different final percentages. None of these is incorrect because 91 litres, which satisfies all conditions, is available among the options. Common Pitfalls: One common mistake is to think that simply adding 7 litres of milk increases the milk percentage in a straightforward manner without considering the removal of the mixed solution. Another error is to change the total volume instead of keeping it constant after replacement. Also, some learners misinterpret percentages as fixed amounts rather than proportions of the total volume. Final Answer: The original quantity of the mixture in the container is 91 litres.

Question 8

The ratio of water to milk in a container is 2 : 3. If 40 litres of the mixture are removed from the container and then 40 litres of pure milk are added in their place, the ratio of water to milk becomes 1 : 4. What was the initial quantity of the m…

Accepted Answer

Introduction: This question is a mixture and replacement problem involving changes in the ratio of water to milk when some mixture is removed and replaced by pure milk. It tests your ability to use ratios and algebra to track how the quantities of components change in a container. Given Data / Assumptions: Initially, the ratio of water to milk in a container is 2 : 3. A total of 40 litres of the mixture are removed. Then 40 litres of pure milk are added to the container. After these operations, the new ratio of water to milk becomes 1 : 4. The container volume remains constant throughout. We need to find the initial quantity of the mixture. Concept / Approach: We represent the initial amounts of water and milk using a common variable, based on the given ratio. Since the initial ratio is 2 : 3, we take water as 2k litres and milk as 3k litres. The total initial volume is 5k litres. When 40 litres of mixture are removed, water and milk are removed in the same ratio 2 : 3. Then 40 litres of pure milk are added, altering only the milk quantity. Using the final ratio, we set up an equation in k and solve for the initial total volume. Step-by-Step Solution: Step 1: Let initial water = 2k litres and initial milk = 3k litres. Step 2: Initial total mixture = 2k + 3k = 5k litres. Step 3: From 5k litres, 40 litres of mixture are removed. Step 4: Water removed = 40 * (2k / 5k) = 16 litres. Step 5: Milk removed = 40 * (3k / 5k) = 24 litres. Step 6: Water remaining = 2k - 16 litres. Step 7: Milk remaining before adding new milk = 3k - 24 litres. Step 8: Now add 40 litres of pure milk: new milk = 3k - 24 + 40 = 3k + 16 litres. Step 9: Final water = 2k - 16 litres, final milk = 3k + 16 litres. Step 10: The final ratio of water to milk is 1 : 4, so (2k - 16) / (3k + 16) = 1 / 4. Step 11: Cross multiply: 4 * (2k - 16) = 3k + 16. Step 12: This gives 8k - 64 = 3k + 16. Step 13: Rearrange: 8k - 3k = 16 + 64, so 5k = 80 and k = 16. Step 14: Initial total mixture = 5k = 5 * 16 = 80 litres. Verification / Alternative check: With k = 16, initial water = 32 litres, initial milk = 48 litres. Removing 40 litres in ratio 2 : 3 removes 16 litres of water and 24 litres of milk. Remaining water = 16 litres and remaining milk = 24 litres. After adding 40 litres of milk, final milk = 64 litres, final water = 16 litres. Thus final ratio of water to milk is 16 : 64 which simplifies to 1 : 4, confirming the result. Why Other Options Are Wrong: 75, 85, 90 and 100 litres do not satisfy the equation for the final ratio 1 : 4 when the same process is applied. Substituting these values produces different ratios that do not simplify to 1 : 4. Common Pitfalls: Students sometimes mistakenly subtract 40 litres only from milk or only from water instead of from the mixture as a whole. Another common error is forgetting that the removed mixture has the same composition as the original mixture. Also, some learners misread the final ratio as milk to water instead of water to milk, which leads to the wrong equation. Final Answer: The initial quantity of the mixture in the container is 80 litres.

Question 9

From a container full of pure wine, a thief steals 15 litres of wine and replaces it with 15 litres of water. He repeats this process two more times, each time removing 15 litres of the current mixture and replacing it with 15 litres of water. After…

Accepted Answer

Introduction: This is a classic repeated replacement mixture problem, where a fixed quantity of liquid is removed and replaced with another liquid several times. The question provides the final ratio of wine to water and asks for the initial volume of wine. Understanding the replacement formula for mixtures is crucial in solving this efficiently. Given Data / Assumptions: The container initially contains only pure wine. In each operation, 15 litres of the mixture are removed. These 15 litres are then replaced with 15 litres of water. This process is repeated three times in total. After three operations, the ratio of wine to water is 343 : 169. The total volume of the liquid in the container remains constant. Concept / Approach: Let V be the initial volume of wine in the container. In each operation, a fraction 15 / V of the current mixture is removed and replaced by water. After one operation, the fraction of wine left is (1 - 15 / V). After n repeated operations, the fraction of wine remaining is (1 - 15 / V)^n. Here n = 3. This fraction must equal the final wine fraction obtained from the ratio 343 : 169. Step-by-Step Solution: Step 1: Let the initial volume of wine be V litres. Step 2: After each replacement, the fraction of wine remaining is (1 - 15 / V). Step 3: After three such operations, the fraction of wine remaining is (1 - 15 / V)^3. Step 4: The final ratio of wine to water is 343 : 169, so total parts = 343 + 169 = 512. Step 5: The fraction of wine finally is 343 / 512. Step 6: Set up the equation: (1 - 15 / V)^3 = 343 / 512. Step 7: Take cube roots: 1 - 15 / V = 7 / 8 (we discard the negative root by logic). Step 8: Solve for V: 15 / V = 1 - 7 / 8 = 1 / 8. Step 9: Therefore V = 15 * 8 = 120 litres. Verification / Alternative check: Start with 120 litres of wine. After first operation, wine left = 120 * (1 - 15 / 120) = 120 * (105 / 120) = 105 litres. After second operation, wine left = 105 * (105 / 120) = 105^2 / 120. After third operation, wine left = (105^3) / (120^2). The fraction of wine is (105 / 120)^3. Simplify 105 / 120 to 7 / 8. So final fraction = (7 / 8)^3 = 343 / 512, matching the given ratio, confirming that V = 120 litres is correct. Why Other Options Are Wrong: 75 litres, 100 litres, and 150 litres produce different fractions of wine after three operations and do not yield the ratio 343 : 169. 90 litres similarly does not satisfy the equation (1 - 15 / V)^3 = 343 / 512. Common Pitfalls: A common mistake is to subtract 15 litres of wine each time directly, forgetting that after the first replacement the container holds a mixture, not pure wine. Another frequent error is to ignore the exponential nature of repeated replacement and simply multiply or subtract linearly. It is also easy to miscalculate the final ratio by treating 343 : 169 as a difference rather than a ratio. Final Answer: The initial amount of wine in the container was 120 litres.

Question 10

A vessel contains 20 litres of a mixture of milk and water in the ratio 3 : 2. First, 10 litres of this mixture are removed and replaced with 10 litres of pure milk. This process of removing 10 litres of the mixture and replacing it with 10 litres o…

Accepted Answer

Introduction: This problem is about successive replacement in a milk water mixture, where mixture is removed and replaced with pure milk twice. It tests your understanding of how the composition of a mixture changes step by step and how to compute the final ratio after repeated operations. Given Data / Assumptions: A vessel initially contains 20 litres of a mixture of milk and water. The initial ratio of milk to water is 3 : 2. First, 10 litres of this mixture are removed. These 10 litres are replaced with 10 litres of pure milk. The same operation is repeated a second time. We must find the final ratio of milk to water. Concept / Approach: We can track the exact quantities of milk and water after each step rather than relying on a general formula. Starting from the initial ratio, we compute the amounts removed in the same ratio, then update the mixture after adding pure milk. Repeating the process once more gives us the final amounts, from which we can obtain the final ratio of milk to water. Step-by-Step Solution: Step 1: Initial mixture = 20 litres with milk : water = 3 : 2. Step 2: Total parts = 3 + 2 = 5. So milk = 20 * 3 / 5 = 12 litres, water = 20 * 2 / 5 = 8 litres. Step 3: First operation: 10 litres of mixture are removed. Step 4: Milk removed = 10 * (12 / 20) = 6 litres, water removed = 10 * (8 / 20) = 4 litres. Step 5: After removal, milk = 12 - 6 = 6 litres, water = 8 - 4 = 4 litres. Step 6: Add 10 litres of pure milk. New milk = 6 + 10 = 16 litres, water remains 4 litres. Step 7: Second operation: again 10 litres of current mixture are removed. Step 8: Current total = 20 litres, milk = 16 litres, water = 4 litres. Step 9: Milk removed = 10 * (16 / 20) = 8 litres, water removed = 10 * (4 / 20) = 2 litres. Step 10: After removal, milk = 16 - 8 = 8 litres, water = 4 - 2 = 2 litres. Step 11: Add 10 litres of pure milk again, so final milk = 8 + 10 = 18 litres. Step 12: Final water remains 2 litres, so final ratio milk : water = 18 : 2 = 9 : 1. Verification / Alternative check: We can also reason that each replacement operation increases the proportion of milk since only milk is added and mixture is removed. After the first replacement, milk is 16 litres and water 4 litres, ratio 4 : 1. After the second replacement, milk increases further relative to water, ending at 18 : 2 which simplifies to 9 : 1. Both the stepwise computation and this reasoning are consistent. Why Other Options Are Wrong: 5 : 3 and 1 : 4 do not reflect the fact that milk proportion is rising due to addition of pure milk each time. 4 : 1 is the ratio after the first replacement only, not after the second replacement. 7 : 3 does not match the exact calculated final amounts of 18 litres milk and 2 litres water. Common Pitfalls: A typical error is to treat each 10 litre removal as affecting only water or only milk, rather than both in the existing ratio. Some students also mistakenly adjust the total volume after replacements, forgetting it stays at 20 litres. Another mistake is to stop after the first replacement instead of repeating the process as described. Final Answer: The ratio of milk to water in the final mixture is 9 : 1.

Question 11

A tin contains a mixture of two liquids A and B in the proportion 4 : 1 by volume. If 45 litres of this mixture are removed and replaced by 45 litres of pure liquid B, the ratio of liquid A to liquid B in the tin becomes 2 : 5. How many litres of li…

Accepted Answer

Introduction: This is an alligation and mixture problem involving two liquids A and B in a tin. A portion of the mixture is removed and replaced with pure liquid B, changing the ratio of A to B. The question asks for the initial amount of B and the capacity of the tin, but the multiple choice answers point to the total capacity. We will find the capacity first and then the amount of B initially. Given Data / Assumptions: Initially, the ratio of liquid A to liquid B in the tin is 4 : 1. 45 litres of the mixture are removed. Those 45 litres are replaced with 45 litres of pure B. After this operation, the ratio of A to B becomes 2 : 5. The total volume in the tin remains constant throughout. We must determine the capacity of the tin and the initial amount of B. Concept / Approach: Let the total capacity of the tin be V litres. Initially A and B are 4V / 5 and V / 5 litres respectively. Removing 45 litres removes A and B in the same 4 : 1 ratio. Then adding 45 litres of pure B changes only B. Using the final ratio 2 : 5, we form an equation in V and solve. Once we know V, we can find the initial amount of B as V / 5. Step-by-Step Solution: Step 1: Let the capacity of the tin be V litres. Step 2: Initial quantities: A = 4V / 5 litres, B = V / 5 litres. Step 3: 45 litres of mixture are removed in the 4 : 1 ratio. Step 4: A removed = 45 * (4 / 5) = 36 litres, B removed = 45 * (1 / 5) = 9 litres. Step 5: Remaining A = 4V / 5 - 36, remaining B = V / 5 - 9. Step 6: 45 litres of pure B are added, so new B = V / 5 - 9 + 45 = V / 5 + 36. Step 7: New ratio of A to B is 2 : 5, so set up the equation: (4V / 5 - 36) / (V / 5 + 36) = 2 / 5. Step 8: Cross multiply: 5 * (4V / 5 - 36) = 2 * (V / 5 + 36). Step 9: Simplify left side: 5 * 4V / 5 = 4V, and 5 * (-36) = -180, so left side = 4V - 180. Step 10: Right side: 2 * (V / 5 + 36) = 2V / 5 + 72. Step 11: Equation: 4V - 180 = 2V / 5 + 72. Step 12: Multiply all terms by 5 to clear the denominator: 20V - 900 = 2V + 360. Step 13: Rearrange: 20V - 2V = 360 + 900, so 18V = 1260. Step 14: Hence V = 1260 / 18 = 70 litres. Step 15: Initial liquid B = V / 5 = 70 / 5 = 14 litres. Verification / Alternative check: Initially A = 56 litres, B = 14 litres. Removing 45 litres in ratio 4 : 1 removes 36 litres A and 9 litres B. Remaining A = 20 litres, remaining B = 5 litres. Adding 45 litres of B gives final A = 20 litres, B = 50 litres. The final ratio A : B = 20 : 50 simplifies to 2 : 5, confirming that a capacity of 70 litres is correct. Why Other Options Are Wrong: 50, 58, 62 and 65 litres all fail to satisfy the ratio equation when substituted for V. They lead to different final A : B ratios and do not give 2 : 5. Common Pitfalls: Some students mistakenly remove 45 litres only from A or only from B instead of from the mixture. Others forget that the mixture removed has the same composition as the mixture in the tin at that moment. It is also easy to misinterpret the question and try to match options to the initial amount of B directly without checking the ratio condition after replacement. Final Answer: The capacity of the tin is 70 litres, and initially it contained 14 litres of liquid B.

Question 12

In a mixture of milk and water, the proportion of water by weight is 75%. If to 60 g of this mixture an additional 15 g of water are added, what will be the new percentage of water in the resulting mixture? (All weights are in grams.)

Accepted Answer

Introduction: This is a straightforward percentage and mixture problem involving milk and water. You are given the initial percentage of water in a mixture by weight, then an additional amount of water is added. The goal is to calculate the new percentage of water in the final mixture. This type of question frequently appears in aptitude and bank exams. Given Data / Assumptions: The mixture consists only of milk and water. Initially, water makes up 75% of the mixture by weight. The initial amount of mixture considered is 60 g. An extra 15 g of water is added to the 60 g of mixture. We need to find the new percentage of water in the final mixture. Concept / Approach: Percentages can be directly converted to actual weights when we know the total amount of mixture. Here, water is 75% of 60 g. Once we know the initial water and milk weights, we can add the 15 g of water, compute the new total weight and the new water weight, and then express water as a percentage of the new total weight using the formula: percentage of water = (weight of water / total weight of mixture) * 100. Step-by-Step Solution: Step 1: Initial total mixture = 60 g. Step 2: Water is 75% of 60 g, so water weight = 0.75 * 60 = 45 g. Step 3: Milk weight initially = 60 - 45 = 15 g. Step 4: Now an additional 15 g of water is added to the mixture. Step 5: New water weight = 45 + 15 = 60 g. Step 6: Milk weight stays the same at 15 g. Step 7: New total mixture weight = 60 g (water) + 15 g (milk) = 75 g. Step 8: New percentage of water = (60 / 75) * 100. Step 9: 60 / 75 = 4 / 5, so water percentage = (4 / 5) * 100 = 80%. Verification / Alternative check: You can reason that initially water is 45 g out of 60 g, so there is already a high proportion of water. Adding more pure water increases the water percentage. Since we add 15 g water, the total increases by one quarter (from 60 g to 75 g), but water increases from 45 g to 60 g, which is four fifths of the new total. That gives 80%, confirming our calculation. Why Other Options Are Wrong: 70% and 75%: These are lower than the original 75% or equal to it, but adding water must increase the water percentage, not decrease or keep it the same. 62% and 65%: These are far below the original percentage and do not match the simple calculation of 60 / 75 * 100. Common Pitfalls: A common mistake is to add 15 g to both milk and water or to think of percentage changes directly without recalculating absolute weights. Another error is to forget that the total weight also changes when extra water is added, so you must divide the new water weight by the new total weight to get the correct percentage. Final Answer: The new percentage of water in the mixture is 80%.

Question 13

In a 48 litre mixture, the ratio of milk to water is 5 : 3. How many litres of water must be added to this mixture so that the ratio of milk to water becomes 3 : 5 in the final mixture?

Accepted Answer

Introduction: This question involves changing the ratio of milk to water in a mixture by adding water only. Initially, milk is in higher proportion, and we want water to become dominant, so more water must be added. The problem tests skills in ratio handling and algebraic manipulation in mixture situations. Given Data / Assumptions: Total initial mixture volume = 48 litres. Initial ratio of milk to water = 5 : 3. Only water is added, milk quantity remains unchanged. After adding water, the ratio of milk to water becomes 3 : 5. We must find how many litres of water are added. Concept / Approach: We first determine the initial volumes of milk and water from the given ratio. Then we let x be the amount of water added. Milk remains constant, while water increases by x. Using the final ratio, we create an equation and solve for x. This is a standard application of ratio algebra in mixtures. Step-by-Step Solution: Step 1: Initial total mixture = 48 litres with milk : water = 5 : 3. Step 2: Total parts = 5 + 3 = 8 parts. Step 3: Milk initially = 48 * 5 / 8 = 30 litres. Step 4: Water initially = 48 * 3 / 8 = 18 litres. Step 5: Let x litres of water be added to the mixture. Step 6: Milk after addition = 30 litres (unchanged). Step 7: Water after addition = 18 + x litres. Step 8: The final ratio of milk to water is 3 : 5, so: 30 / (18 + x) = 3 / 5. Step 9: Cross multiply: 30 * 5 = 3 * (18 + x). Step 10: This gives 150 = 54 + 3x. Step 11: Subtract 54 from both sides: 96 = 3x. Step 12: So x = 96 / 3 = 32 litres. Verification / Alternative check: After adding 32 litres of water, water amount becomes 18 + 32 = 50 litres. New total volume = 30 litres milk + 50 litres water = 80 litres. New ratio of milk to water = 30 : 50, which simplifies to 3 : 5 as required. Thus the calculation is consistent and correct. Why Other Options Are Wrong: 24 litres or 16 litres: These give water amounts of 42 litres or 34 litres respectively, leading to ratios that are not equal to 3 : 5. 8 litres and 20 litres also produce final ratios that do not simplify to 3 : 5 when combined with the unchanged 30 litres of milk. Common Pitfalls: Common errors include changing the milk quantity along with water when only water is added, or attempting to scale the ratio directly without accounting for the absolute volumes. Another mistake is cross multiplying incorrectly when setting up the ratio equation, which can lead to incorrect values of x. Final Answer: The amount of water that must be added is 32 litres.

Question 14

A jar is completely filled with pure honey. A person repeatedly performs the following operation: he draws out 20% of the honey from the jar and replaces it with an equal quantity of sugar solution. He repeats this process four times. After the four…

Accepted Answer

Introduction: This question is another example of repeated replacement in a mixture, this time involving pure honey and sugar solution. A fixed percentage of the mixture is removed and replaced multiple times. The problem provides the amount of honey left after several such operations and asks for the initial amount of honey in the jar. Given Data / Assumptions: Initially the jar contains only pure honey. Each time, 20% of the contents are removed. The removed portion is replaced with an equal amount of sugar solution. This operation is repeated 4 times. After the fourth operation, 512 g of honey remain in the jar. The total volume of liquid in the jar remains constant. We must find the initial quantity of honey in kilograms. Concept / Approach: If a fraction f of a mixture is removed and replaced each time, the fraction of the original component left after one operation is (1 - f). After n operations, the fraction of the original component left is (1 - f)^n. Here, f = 20% = 0.20, so (1 - f) = 0.80. There are n = 4 operations. Let the initial amount of honey be H grams. Then H * (0.80)^4 must equal 512 grams. Step-by-Step Solution: Step 1: Let the initial amount of honey be H grams. Step 2: Each time, 20% is removed, so 80% of honey remains after one operation. Step 3: Fraction of honey remaining after one operation = 0.80. Step 4: After 4 operations, fraction of honey remaining = (0.80)^4. Step 5: Amount of honey left after 4 operations = H * (0.80)^4. Step 6: We are told that this amount equals 512 g, so: H * (0.80)^4 = 512. Step 7: Compute (0.80)^4 = 0.4096. Step 8: So H * 0.4096 = 512. Step 9: Therefore H = 512 / 0.4096 = 1250 grams. Step 10: Convert grams to kilograms: 1250 g = 1.25 kg. Verification / Alternative check: We can verify by working forward. Starting with 1250 g of honey, after first operation honey left = 1250 * 0.80 = 1000 g. After second operation = 1000 * 0.80 = 800 g. After third operation = 800 * 0.80 = 640 g. After fourth operation = 640 * 0.80 = 512 g, which matches the given condition, confirming that the initial quantity is indeed 1.25 kg. Why Other Options Are Wrong: 1 kg and 1.5 kg: These values, when multiplied by (0.80)^4, produce amounts of honey left that are not equal to 512 g. None of these is incorrect because we have found a valid option that matches the condition, namely 1.25 kg. 2 kg also does not satisfy the equation H * 0.4096 = 512. Common Pitfalls: Learners sometimes forget that the replacement is of the mixture, not just honey, and attempt to subtract a fixed amount of honey each time. Another error is to ignore the exponential nature of repeated percentage reduction, mistakenly using linear subtraction. Some also confuse grams and kilograms, leading to unit mistakes in the final answer. Final Answer: The initial amount of honey in the jar was 1.25 kg.

Question 15

A milkman pays Rs 6.40 per litre to buy pure milk. He then adds some amount of water, which is free, and sells the resulting mixture at Rs 8 per litre, thereby making a profit of 37.5% on his cost price. What is the ratio of water to milk in the mix…

Accepted Answer

Introduction: This question combines profit and loss with mixtures. The milkman buys milk at a certain cost price, mixes in free water, and sells the mixture at the original cost price of milk, yet earns a profit. We must compute the proportion of water and milk in the final mixture based on the percentage profit. Given Data / Assumptions: Cost price of pure milk = Rs 6.40 per litre. Water is assumed to be free of cost. The mixture is sold at Rs 8 per litre. The milkman claims to sell at cost, but effectively earns a profit of 37.5%. We must find the ratio of water to milk in the mixture supplied to customers. Concept / Approach: Let m be the cost of pure milk per litre, here m = 6.40. Let the fraction of milk in 1 litre of the mixture be k, so the fraction of water is 1 - k. The cost price of 1 litre of mixture is then m * k, since only milk costs money. Selling price per litre of mixture is given as Rs 8. Profit percentage is 37.5%, that is 3 / 8 in fractional form. Use the relation: SP = CP * (1 + profit rate). We equate SP to 8, CP to m * k, and profit rate to 3 / 8, and solve for k. Step-by-Step Solution: Step 1: Cost price of milk m = Rs 6.40 per litre. Step 2: Let fraction of milk in the mixture be k, so CP of 1 litre of mixture = 6.40 * k. Step 3: Profit percentage = 37.5% = 3 / 8. Step 4: Selling price SP = CP * (1 + 3 / 8) = CP * (11 / 8). Step 5: Given SP = Rs 8, so 8 = (6.40 * k) * 11 / 8. Step 6: Multiply both sides by 8: 64 = 6.40 * k * 11. Step 7: 6.40 = 32 / 5, so 64 = (32 / 5) * k * 11. Step 8: Simplify: 64 = (352 / 5) * k, so 64 * 5 = 352 * k. Step 9: 320 = 352 * k, hence k = 320 / 352 = 10 / 11. Step 10: Fraction of milk in mixture = 10 / 11, so fraction of water = 1 / 11. Step 11: Ratio of water to milk = (1 / 11) : (10 / 11) = 1 : 10. Verification / Alternative check: Assume 11 litres of mixture: milk = 10 litres, water = 1 litre. Cost of milk = 10 * 6.40 = Rs 64. Selling price at Rs 8 per litre for 11 litres = 11 * 8 = Rs 88. Profit = 88 - 64 = Rs 24. Profit percentage = 24 / 64 * 100 = 37.5%, matching the given profit. Thus the ratio 1 : 10 is correct for water to milk. Why Other Options Are Wrong: 10 : 1 and 11 : 9 suggest more water than milk or comparable quantities, which would drastically change the cost and profit structure and not yield 37.5% profit. 9 : 11 and 1 : 4 do not match the calculated fractions obtained from the profit condition when you test them with assumed volumes and cost calculations. Common Pitfalls: Many learners mistakenly assume the selling price is equal to the cost price of pure milk, ignoring the given profit percentage. Others try to handle this as a simple alligation without incorporating profit, which leads to incorrect ratios. It is also easy to invert the ratio and give milk to water instead of water to milk, so careful reading is important. Final Answer: The ratio of water to milk in the mixture is 1 : 10.

Question 16

A certain liquid mixture in a container contains 25% milk and the remaining 75% is water. What percentage of this mixture must be removed and replaced with an equal quantity of pure milk so that, in the resulting mixture, the quantities of milk and …

Accepted Answer

Introduction: This problem deals with replacing part of a mixture by pure milk in order to change the percentage composition. Initially, milk is in the minority. By removing some mixture and adding pure milk, we want the final mixture to have equal amounts of milk and water. This is a classic replacement percentage question in mixture and alligation. Given Data / Assumptions: Initial mixture contains 25% milk and 75% water. A certain percentage of this mixture is removed. The same percentage volume is replaced with pure milk. After this operation, milk and water in the mixture are equal in quantity. We must determine the percentage of the mixture that is removed and replaced. Concept / Approach: For simplicity, assume the total initial mixture is 1 unit (or 100%). Then milk is 0.25 units and water is 0.75 units. Let the fraction of mixture removed be f. The removed portion contains milk and water in the same 25% to 75% ratio. After removing that portion, we add back the same fraction f, but now as pure milk. We then set the final milk amount equal to the final water amount and solve for f. Step-by-Step Solution: Step 1: Assume total mixture = 1 unit. Step 2: Initial milk = 0.25 units, initial water = 0.75 units. Step 3: Let fraction of mixture removed be f. Step 4: Milk removed = 0.25 * f, water removed = 0.75 * f. Step 5: After removal, milk = 0.25 - 0.25 * f, water = 0.75 - 0.75 * f. Step 6: The removed fraction f is replaced by pure milk. Step 7: New milk amount = (0.25 - 0.25 * f) + f = 0.25 + 0.75 * f. Step 8: New water amount = 0.75 - 0.75 * f. Step 9: In the final mixture, milk and water must be equal, so: 0.25 + 0.75 * f = 0.75 - 0.75 * f. Step 10: Rearrange: 0.75 * f + 0.75 * f = 0.75 - 0.25. Step 11: 1.5 * f = 0.50, so f = 0.50 / 1.5 = 1 / 3. Step 12: Convert f to percentage: f = 1 / 3 is approximately 33.33%. Verification / Alternative check: Suppose there are 100 litres initially. Then milk = 25 litres, water = 75 litres. Remove 33.33 litres of mixture (approximately 1 / 3 of the total). Milk removed = 25 * 1 / 3 ≈ 8.33 litres, water removed = 75 * 1 / 3 ≈ 25 litres. Remaining milk ≈ 16.67 litres, remaining water ≈ 50 litres. Add back 33.33 litres of pure milk, giving final milk ≈ 50 litres and water ≈ 50 litres, which are equal as required. Why Other Options Are Wrong: 31.8%, 31%, and 29.85% do not give exactly equal quantities of milk and water when used in the above calculation. 40% overshoots, resulting in more milk than water in the final mixture rather than equal amounts. Common Pitfalls: A common mistake is to remove a fixed volume in litres without relating it proportionally to the total, which makes the percentage calculation inconsistent. Another frequent error is to equate percentages directly instead of actual quantities, or to forget that the total mixture remains the same after replacement. Careful algebra and assuming a simple total like 1 or 100 makes the algebra more manageable. Final Answer: The percentage of the mixture that must be removed and replaced with milk is 33.33%.

Question 17

A container initially holds 50 litres of pure milk. First, 8 litres of milk are removed from the container and replaced with 8 litres of water. This process of removing 8 litres of the current mixture and replacing it with 8 litres of water is repea…

Accepted Answer

Introduction: This question is a classic repeated replacement problem where pure milk in a container is gradually replaced by water in several steps. Each time a fixed amount of the current mixture is removed and replaced with water. We must find the quantity of milk remaining after three such operations. Given Data / Assumptions: Initial volume in the container = 50 litres, all milk. In each step, 8 litres of the current mixture are removed. The 8 litres removed are replaced with 8 litres of water. This process is repeated 3 times in total. The total volume in the container remains 50 litres at all times. We must compute the approximate volume of milk left after these 3 steps. Concept / Approach: If V is the total volume and x litres are removed and replaced each time, the fraction of the original liquid remaining after one operation is (1 - x / V). After n identical operations, the fraction remaining is (1 - x / V)^n. Here V = 50, x = 8, and n = 3. Multiplying this fraction by 50 litres gives the milk left after three operations. Step-by-Step Solution: Step 1: Total volume V = 50 litres, removed each time x = 8 litres. Step 2: Fraction of milk remaining after one operation = 1 - x / V = 1 - 8 / 50. Step 3: 8 / 50 = 0.16, so fraction remaining each time = 1 - 0.16 = 0.84. Step 4: After 3 operations, fraction of milk remaining = (0.84)^3. Step 5: Compute (0.84)^3 = 0.84 * 0.84 * 0.84 ≈ 0.5927. Step 6: Milk remaining = 50 * 0.5927 ≈ 29.635 litres. Step 7: Rounding to two decimal places, milk remaining ≈ 29.63 litres. Verification / Alternative check: We can also approximate stepwise. After first operation, milk = 50 * 0.84 = 42 litres. After second operation, milk = 42 * 0.84 ≈ 35.28 litres. After third operation, milk = 35.28 * 0.84 ≈ 29.63 litres. This matches the value obtained using the direct exponential formula and confirms the result is correct. Why Other Options Are Wrong: 24.52, 28.21, and 25.14 litres are too low compared to the calculated value and do not match the repeated 16% removal pattern correctly. 30.00 litres is close but slightly higher than the precise result; it is an overestimate and does not match the exact calculation of 29.63 litres. Common Pitfalls: Students sometimes subtract 8 litres of milk each time, ignoring that after the first step the liquid is a mixture, not pure milk. Another mistake is to forget that the removal each time is of the current mixture in the same proportion, so a constant fraction, not a constant amount, of milk is removed. Using linear subtraction instead of an exponential fraction is a frequent source of error. Final Answer: The container now contains approximately 29.63 litres of milk.

Question 18

A 4 kg sample of an alloy contains 1/5 copper and the rest zinc. Another 5 kg sample of a different alloy contains 1/6 copper and the rest zinc. If the two alloys are melted together to form a single mixture, what is the ratio of copper to zinc in t…

Accepted Answer

Introduction: This alloy problem involves combining two different metal mixtures, each containing copper and zinc in different proportions. After mixing, we must determine the overall ratio of copper to zinc in the final alloy. This is a direct application of weighted averages and ratio calculations based on given fractions and masses. Given Data / Assumptions: First alloy has a total mass of 4 kg. In the first alloy, copper is 1/5 of the mass and the rest is zinc. Second alloy has a total mass of 5 kg. In the second alloy, copper is 1/6 of the mass and the rest is zinc. The two alloys are completely melted and mixed to form one homogeneous alloy. We must find the ratio of copper to zinc in this final mixture. Concept / Approach: We first compute the absolute masses of copper and zinc in each alloy using the given fractions. Then we add the copper masses to get total copper and add the zinc masses to get total zinc. Finally, we express these as a simplified ratio copper : zinc. Fractions are helpful to keep the calculations exact. Step-by-Step Solution: Step 1: In the first alloy of 4 kg, copper mass = 1/5 of 4 kg = 4/5 kg. Step 2: Zinc mass in the first alloy = 4 - 4/5 = 20/5 - 4/5 = 16/5 kg. Step 3: In the second alloy of 5 kg, copper mass = 1/6 of 5 kg = 5/6 kg. Step 4: Zinc mass in the second alloy = 5 - 5/6 = 30/6 - 5/6 = 25/6 kg. Step 5: Total copper in the final mixture = 4/5 + 5/6. Step 6: Find a common denominator: 4/5 = 24/30, 5/6 = 25/30, so total copper = (24 + 25) / 30 = 49/30 kg. Step 7: Total zinc in the final mixture = 16/5 + 25/6. Step 8: Convert: 16/5 = 96/30, 25/6 = 125/30, so total zinc = (96 + 125) / 30 = 221/30 kg. Step 9: Ratio of copper to zinc = (49/30) : (221/30). Step 10: The denominators cancel, giving copper : zinc = 49 : 221. Verification / Alternative check: We can check that the total mass is preserved. Total initial mass = 4 kg + 5 kg = 9 kg. Final copper + final zinc = 49/30 + 221/30 = 270/30 = 9 kg, which matches the initial total. This confirms that no arithmetic errors have been made and that the ratio 49 : 221 is correct. Why Other Options Are Wrong: 39 : 231, 94 : 181, and 1 : 5 do not result from the exact calculations of copper and zinc masses. When you convert these ratios back to fractions of a 9 kg mixture, they do not match the computed values of 49/30 kg and 221/30 kg. None of these is incorrect because we have a valid matching option, 49 : 221. Common Pitfalls: Some learners mistakenly average the given fractions 1/5 and 1/6 directly without weighting by the masses 4 kg and 5 kg. Others subtract copper percentage incorrectly, leading to wrong zinc amounts. It is also easy to make mistakes when converting fractions to a common denominator, so care is required in these arithmetic steps. Final Answer: The ratio of copper to zinc in the resulting alloy is 49 : 221.

Question 19

A milkman claims to sell milk at its original cost price per litre, yet he still makes a profit of 30% because he mixes some amount of free water with the milk. Assuming he sells the adulterated mixture at the same price per litre that he paid for t…

Accepted Answer

Introduction: This question combines profit concept with mixture adulteration. The milkman adds free water to the milk but sells the mixture at the original cost price of pure milk. Despite this, he earns a 30% profit. From this information, we must determine what percentage of the mixture is actually milk. Given Data / Assumptions: The milkman buys pure milk at some cost price per litre, say Rs M. He mixes water, which is free, with the milk. He sells the mixture at Rs M per litre, the same as the cost price of pure milk. He earns a profit of 30% on his cost price. We must find the percentage of milk in the mixture. Concept / Approach: Let the fraction of milk in 1 litre of the mixture be k, so the fraction of water is (1 - k). Because water is free, the cost price of 1 litre of the mixture is only due to the milk and equals M * k. The selling price per litre is M. Profit percentage is defined as: profit% = (SP - CP) / CP * 100. Given the profit is 30%, SP = CP * 1.3. We equate the selling price M to 1.3 times the mixture cost M * k and solve for k. Step-by-Step Solution: Step 1: Let cost price of pure milk per litre be M rupees. Step 2: Let fraction of milk in the mixture be k, so CP of 1 litre of mixture = M * k. Step 3: Selling price of mixture per litre = M rupees. Step 4: Profit percentage = 30%, so SP = CP * (1 + 0.30) = CP * 1.3. Step 5: Thus M = 1.3 * (M * k). Step 6: Divide both sides by M (M is not zero): 1 = 1.3 * k. Step 7: So k = 1 / 1.3 = 10 / 13. Step 8: Fraction of milk in mixture is 10 / 13, so percentage of milk = (10 / 13) * 100. Step 9: (10 / 13) * 100 ≈ 76.92%. Verification / Alternative check: Assume the milkman sells 13 litres of mixture. Then milk in the mixture = 10 litres, water = 3 litres. Cost price of milk = 10 * M = 10M. Selling price of mixture = 13 * M = 13M. Profit = 13M - 10M = 3M. Profit percentage = 3M / 10M * 100 = 30%, which matches the condition given in the question and confirms that 10 / 13 or 76.92% of the mixture is milk. Why Other Options Are Wrong: 71.02%, 63.22%, and 86.42% do not correspond to any simple fraction that gives exactly 30% profit given the structure of costs and selling price. Testing them by assuming a convenient volume of mixture would show that the profit percentage is not 30% for those compositions. 60% is too low and would yield a much higher profit percentage because the cost price per litre would be only 60% of M while the selling price remains M. Common Pitfalls: Many students confuse the direction of the profit calculation and set SP = CP / 1.3 instead of SP = CP * 1.3. Others wrongly treat M as a specific number instead of a variable, or mistakenly add water cost. Some learners also misinterpret the question and attempt to find the percentage of water rather than milk, leading to an inverted answer. Final Answer: The percentage of milk in the mixture is 76.92%.

Question 20

How many litres of oil costing Rs 40 per litre should be mixed with 240 litres of a second variety of oil costing Rs 60 per litre so as to obtain a mixture whose cost price is Rs 52 per litre?

Accepted Answer

Introduction: This is an alligation and mixture problem on cost price. Two types of oil with different cost prices are mixed to obtain a mixture with an intermediate cost price. We are given the quantity of one type and must find how much of the other type is needed so that the final mixture has a specified cost per litre. Given Data / Assumptions: First variety of oil costs Rs 40 per litre. Second variety of oil costs Rs 60 per litre. There are 240 litres of the second variety of oil. The required mixture must have a cost price of Rs 52 per litre. We must find how many litres of the first variety should be added. Concept / Approach: We can use the alligation rule or directly set up a weighted average equation. Let x be the litres of first oil (Rs 40 per litre). Then total cost of the mixture is 40x + 60 * 240. Total volume is x + 240 litres. The average cost per litre (which is the cost price of the mixture) equals Rs 52, so: (40x + 60 * 240) / (x + 240) = 52. We solve this equation for x to find the required quantity of the first oil. Step-by-Step Solution: Step 1: Let x = litres of oil costing Rs 40 per litre. Step 2: Total cost of first oil = 40x rupees. Step 3: Total cost of second oil = 60 * 240 = 14400 rupees. Step 4: Total volume of mixture = x + 240 litres. Step 5: Required average cost per litre = 52 rupees. Step 6: Set up equation: (40x + 14400) / (x + 240) = 52. Step 7: Cross multiply: 40x + 14400 = 52 * (x + 240). Step 8: Expand right side: 40x + 14400 = 52x + 12480. Step 9: Rearrange: 14400 - 12480 = 52x - 40x. Step 10: 1920 = 12x, so x = 1920 / 12 = 160. Verification / Alternative check: If we mix 160 litres of Rs 40 oil with 240 litres of Rs 60 oil, total volume = 160 + 240 = 400 litres. Total cost = 160 * 40 + 240 * 60 = 6400 + 14400 = 20800 rupees. Average cost per litre = 20800 / 400 = 52 rupees, which matches the target price. This confirms that 160 litres is correct. Why Other Options Are Wrong: 170, 150, 140 and 200 litres, when substituted as x, give mixture cost prices different from Rs 52. Each leads to a weighted average price that is either too low or too high compared to 52 rupees per litre. Common Pitfalls: A common mistake is to misapply the alligation rule by reversing the differences or mixing up which cost goes where. Another is to ignore the given fixed quantity of the second oil and treat both quantities as unknown. Some students also attempt to use percentage change concepts instead of the simple weighted average formula, which can complicate the problem unnecessarily. Final Answer: The quantity of Rs 40 per litre oil to be mixed is 160 litres.

Alligation or Mixture Questions