Question 1

An alloy is made by mixing metal A costing Rs 2000 per kg and metal B costing Rs 400 per kg in the ratio A:B = 3:1. What is the total cost (in Rs) of 8 kilograms of this alloy?

Accepted Answer

Introduction / Context: This is a weighted average cost problem based on mixing ratio. If metals are mixed in ratio 3:1, then out of every 4 equal parts of alloy, 3 parts come from metal A and 1 part comes from metal B. You can either compute the mean cost per kg first and multiply by 8 kg, or directly compute how many kg of each metal are present in 8 kg and then calculate the total cost. Both methods should give the same result. Given Data / Assumptions: Cost of metal A = Rs 2000 per kg Cost of metal B = Rs 400 per kg Mixing ratio A:B = 3:1 Total alloy quantity required = 8 kg Concept / Approach: Method 1 (average cost): Mean cost per kg = (3*2000 + 1*400) / (3+1). Total cost for 8 kg = (mean cost per kg) * 8. Method 2 (component quantities): A quantity = (3/4)*8 and B quantity = (1/4)*8, then compute total cost directly. Step-by-Step Solution: Total parts = 3 + 1 = 4 Cost of 4 kg equivalent mixture = 3*2000 + 1*400 = 6000 + 400 = 6400 Mean cost per kg = 6400 / 4 = 1600 Total cost for 8 kg = 1600 * 8 = 12800 Verification / Alternative check: Component method: A = (3/4)*8 = 6 kg and B = (1/4)*8 = 2 kg. Cost = 6*2000 + 2*400 = 12000 + 800 = 12800. This matches the average-cost method, confirming correctness. Why Other Options Are Wrong: 1600 is the cost per kg, not the total for 8 kg. 6400 is the cost corresponding to 4 kg in the 3:1 ratio, not 8 kg. 9800 and 11200 come from incorrect weighting or wrong component quantities. Common Pitfalls: Forgetting to divide by total parts when finding mean cost, multiplying the 4 kg cost by 8 instead of scaling correctly, or confusing per-kg cost with total cost. Final Answer: The cost of 8 kg of the alloy is Rs 12800.

Question 2

In what ratio should tea costing Rs 300 per kg be mixed with tea costing Rs 200 per kg so that the final mixture costs exactly Rs 225 per kg?

Accepted Answer

Introduction: This question tests the alligation (weighted average) concept used in mixtures. When two items with different costs are mixed, the mean price of the mixture depends on how much of each item is used. Alligation gives a quick way to find the required mixing ratio without solving lengthy equations. Here, we are mixing two teas costing Rs 300/kg and Rs 200/kg to obtain a mixture costing Rs 225/kg. We must find the ratio in which the Rs 300/kg tea and the Rs 200/kg tea are combined. Given Data / Assumptions: • Cost of dearer tea = Rs 300 per kg• Cost of cheaper tea = Rs 200 per kg• Mean (mixture) cost = Rs 225 per kg Concept / Approach: By alligation, the required quantities are in the inverse ratio of the differences from the mean: quantity of dearer : quantity of cheaper = (mean - cheaper) : (dearer - mean). This works because the extra cost paid for the dearer item must be balanced by the savings from the cheaper item to land exactly at the mean cost. Step-by-Step Solution: Step 1: Compute differences from the mean cost.mean - cheaper = 225 - 200 = 25dearer - mean = 300 - 225 = 75Step 2: Apply alligation ratio.Quantity of Rs 300 tea : Quantity of Rs 200 tea = 25 : 75Step 3: Simplify the ratio by dividing by 25.25 : 75 = 1 : 3 Verification / Alternative check: If we take 1 kg of Rs 300 tea and 3 kg of Rs 200 tea, total cost = 1*300 + 3*200 = 300 + 600 = 900. Total quantity = 4 kg. Average cost = 900/4 = 225 per kg, which matches the required mean. So the ratio is correct. Why Other Options Are Wrong: 3 : 1 gives a mean much closer to Rs 300, so it exceeds Rs 225.2 : 5 and 5 : 2 produce different weighted averages, not Rs 225.4 : 1 is also too heavy on the Rs 300 tea, so the mean becomes higher than Rs 225. Common Pitfalls: • Reversing the ratio (writing 3 : 1 instead of 1 : 3).• Subtracting in the wrong order while computing differences.• Forgetting that quantities are inversely proportional to the price differences from the mean. Final Answer: Tea costing Rs 300/kg should be mixed with tea costing Rs 200/kg in the ratio 1 : 3.

Question 3

The ratio of milk and water in three samples is 1 : 3, 3 : 5, and 11 : 5 (milk : water). Equal quantities of all three samples are mixed together. What will be the final ratio of milk and water in the new mixture?

Accepted Answer

Introduction: This question checks mixture averaging using ratios. When we mix equal quantities from different samples, we can compute the final composition by converting each ratio into fractions (milk fraction and water fraction), then averaging those fractions because each sample contributes the same total volume. The key is to avoid directly averaging the ratios (that would be wrong). Instead, convert each sample into milk and water fractions first, then combine. Given Data / Assumptions: • Sample 1 milk : water = 1 : 3• Sample 2 milk : water = 3 : 5• Sample 3 milk : water = 11 : 5• Equal quantities of all three samples are mixed Concept / Approach: If each sample contributes the same total quantity (say 1 unit each), then total milk in final mixture = sum of milk fractions from each sample, and total water = sum of water fractions. Finally, convert total milk : total water into simplest integer ratio. Step-by-Step Solution: Step 1: Convert each ratio to a milk fraction.For 1 : 3, total parts = 1+3 = 4, milk fraction = 1/4, water fraction = 3/4For 3 : 5, total parts = 8, milk fraction = 3/8, water fraction = 5/8For 11 : 5, total parts = 16, milk fraction = 11/16, water fraction = 5/16Step 2: Assume 1 unit of each sample is taken.Total milk = 1/4 + 3/8 + 11/16Convert to common denominator 16: 1/4 = 4/16, 3/8 = 6/16, so total milk = (4+6+11)/16 = 21/16Total mixture quantity = 3 units, so milk fraction in mixture = (21/16) / 3 = 21/48 = 7/16Step 3: Water fraction = 1 - 7/16 = 9/16Therefore milk : water = 7/16 : 9/16 = 7 : 9 Verification / Alternative check: You can also add total water fractions directly: 3/4 + 5/8 + 5/16 = 12/16 + 10/16 + 5/16 = 27/16. Dividing by 3 gives water fraction = (27/16)/3 = 27/48 = 9/16, matching our result. So the ratio 7 : 9 is consistent. Why Other Options Are Wrong: 5 : 7 and 9 : 11 suggest a different milk fraction (around 0.41), but actual milk fraction is 7/16 = 0.4375.9 : 7 reverses the dominance; water is actually more than milk here.15 : 13 is close to 1 : 1, which is not possible given two samples are water-heavy. Common Pitfalls: • Averaging ratios directly instead of converting to fractions.• Forgetting that samples are mixed in equal quantities.• Making arithmetic errors while using common denominators. Final Answer: The ratio of milk to water in the new mixture is 7 : 9.

Question 4

A container initially has 960 litres of pure milk. If 48 litres of milk is removed and replaced with 48 litres of water, and this replacement process is repeated two more times (total 3 replacements), what will be the amount of pure milk left in the…

Accepted Answer

Introduction: This is a classic repeated replacement (dilution) problem. When a fixed volume is removed from a mixture and replaced with another liquid, the remaining amount of the original liquid decreases by a constant factor each time. The same logic applies whether we talk about milk, alcohol, or any solution. Here, the container starts with pure milk, and each operation removes some milk and replaces it with water. After multiple operations, we compute how much pure milk remains. Given Data / Assumptions: • Initial volume = 960 litres (pure milk)• Each time removed = 48 litres• Each time replaced with = 48 litres water• Number of replacements = 3 Concept / Approach: After each replacement, the fraction of the original liquid that remains is (1 - removed/total). After n identical operations, remaining original liquid = initial * (1 - removed/total)^n. This works because each removal takes away the same fraction of whatever milk remains at that moment. Step-by-Step Solution: Step 1: Compute the fraction remaining after one replacement.removed/total = 48/960 = 0.05remaining fraction = 1 - 0.05 = 0.95Step 2: Apply the repeated replacement formula for n = 3.Milk left = 960 * (0.95)^3Step 3: Calculate (0.95)^3.0.95^2 = 0.90250.95^3 = 0.9025 * 0.95 = 0.857375Step 4: Multiply by 960.Milk left = 960 * 0.857375 = 823.08 litres (approx) Verification / Alternative check: Because each operation removes 5% of the mixture, it removes 5% of the milk present at that time. So milk decreases by a factor of 0.95 each step. Applying the factor three times (0.95 * 0.95 * 0.95) confirms the final factor 0.857375. Multiplying by 960 gives the same result, so the calculation is consistent. Why Other Options Are Wrong: 901.54 ltr is too high; it would correspond to fewer replacements or a smaller removed fraction.821.54 ltr and 719.64 ltr are too low, implying more dilution than actually performed.857.38 ltr matches 960*0.8931 approximately, not the correct 0.95^3 factor. Common Pitfalls: • Subtracting 48 litres three times directly from 960, which is incorrect because removed liquid is partly water after the first step.• Using n = 2 instead of n = 3.• Forgetting to use the fraction removed (48/960) rather than absolute litres in the formula. Final Answer: The amount of pure milk left after three replacements is 823.08 litres.

Question 5

Three bottles contain mixtures of syrup and water in the ratios 2:3, 3:4, and 7:5 (syrup : water). If 10 litres from the first bottle and 21 litres from the second bottle are taken, how many litres should be taken from the third bottle so that the f…

Accepted Answer

Introduction: This problem tests advanced mixture combination: we are taking specified amounts from different sources, each with its own syrup-to-water ratio, and then choosing an unknown amount from a third source to achieve a target final ratio. The correct method is to compute how much syrup and how much water are contributed by each selected quantity, add them, and then enforce the final target ratio (here 1:1, meaning equal syrup and water). Given Data / Assumptions: • Bottle 1 ratio (syrup:water) = 2:3, quantity taken = 10 L• Bottle 2 ratio (syrup:water) = 3:4, quantity taken = 21 L• Bottle 3 ratio (syrup:water) = 7:5, quantity taken = x L• Final desired ratio (syrup:water) = 1:1 Concept / Approach: Convert each ratio into syrup fraction and water fraction. Multiply by the litres taken to get actual litres of syrup and water. Add contributions, then set total syrup = total water (because 1:1 means equal). Solve for x. Step-by-Step Solution: Step 1: Bottle 1 (2:3) total parts = 5.Syrup fraction = 2/5, water fraction = 3/5From 10 L: syrup = 10*(2/5)=4 L, water = 10*(3/5)=6 LStep 2: Bottle 2 (3:4) total parts = 7.Syrup fraction = 3/7, water fraction = 4/7From 21 L: syrup = 21*(3/7)=9 L, water = 21*(4/7)=12 LStep 3: Combine totals so far.Total syrup = 4 + 9 = 13 LTotal water = 6 + 12 = 18 LStep 4: Bottle 3 (7:5) total parts = 12.From x L: syrup = x*(7/12), water = x*(5/12)Step 5: Enforce final ratio 1:1, so total syrup = total water.13 + (7x/12) = 18 + (5x/12)(7x - 5x)/12 = 52x/12 = 5 => x/6 = 5 => x = 30 Verification / Alternative check: If x = 30, then syrup added = 30*(7/12)=17.5 and water added = 30*(5/12)=12.5. New totals: syrup = 13+17.5=30.5 and water = 18+12.5=30.5. They are equal, so the final mixture ratio is exactly 1:1, confirming the answer. Why Other Options Are Wrong: 20 L and 25 L do not add enough syrup (since bottle 3 is syrup-heavy), so water stays higher than syrup.35 L adds too much syrup, making syrup exceed water.15 L is far too small to balance the initial excess water (18 vs 13). Common Pitfalls: • Treating ratios as litres directly instead of converting to fractions.• Forgetting that 1:1 means total syrup equals total water, not total volume equals something.• Making fraction errors with 7/12 and 5/12. Final Answer: 30 Litres should be taken from the third bottle.

Question 6

In a colored picture, the overall ratio of blue to yellow color used is 4:3. If in the upper half of the picture the ratio of blue to yellow is 2:3, then what must be the ratio of blue to yellow in the lower half of the picture?

Accepted Answer

Introduction: This is a ratio-combination question where an overall ratio is formed by combining two equal halves (upper and lower) that have different internal ratios. The key idea is that the picture has two halves of equal size, so the total amount of color used in the upper half equals the total amount used in the lower half. Using the given overall blue:yellow ratio and the upper-half ratio, we can compute the actual blue and yellow amounts in each half and then determine the required ratio in the lower half. Given Data / Assumptions: • Overall ratio (blue:yellow) = 4:3• Upper half ratio (blue:yellow) = 2:3• Upper half and lower half are equal in size, so total color used in each half is equal Concept / Approach: Let total color used in the whole picture be a convenient multiple of (4+3)=7 parts. Then each half uses half of the total color. Use the upper-half ratio to split its half into blue and yellow. Subtract from total blue and yellow to get lower-half blue and yellow, then form the lower-half ratio. Step-by-Step Solution: Step 1: Assume total color used in the picture = 7k parts.Overall blue = 4k, overall yellow = 3kStep 2: Since upper and lower halves are equal, each half uses (7k)/2 = 3.5k total color.Step 3: Upper half ratio is 2:3 (total 5 parts).Upper blue = (2/5)*3.5k = 1.4kUpper yellow = (3/5)*3.5k = 2.1kStep 4: Compute lower half amounts by subtraction.Lower blue = 4k - 1.4k = 2.6kLower yellow = 3k - 2.1k = 0.9kStep 5: Form the ratio in the lower half.Lower blue : Lower yellow = 2.6k : 0.9k = 26 : 9 Verification / Alternative check: If lower half ratio is 26:9, then lower-half blue fraction is 26/35 and yellow fraction is 9/35. Combining with upper half (2:3) while keeping equal total amounts in each half will reproduce the overall 4:3 ratio. This supports that the computed ratio is logically consistent with the given totals. Why Other Options Are Wrong: 1 : 1 and 2 : 1 make the lower half too balanced or too blue-heavy; then the overall would not remain 4:3.9 : 26 reverses the result, making the lower half yellow-heavy, contradicting the subtraction outcome.13 : 5 is close but not exact; it would not match the exact overall ratio when combined with the upper half. Common Pitfalls: • Ignoring that halves are equal and assuming unknown unequal quantities.• Averaging ratios directly instead of converting to actual amounts first.• Making mistakes while subtracting upper-half amounts from the overall totals. Final Answer: The ratio of blue to yellow in the lower half is 26 : 9.

Question 7

An alloy contains copper and tin in the ratio 3 : 2. If 250 g of copper is added to this alloy, the quantity of copper becomes exactly double the quantity of tin. What is the amount of tin (in grams) present in the original alloy?

Accepted Answer

Introduction: This question tests ratio-based composition in alloys and how the composition changes when one component is added. An alloy is a mixture of metals, and the ratio tells us their relative quantities. Here, copper and tin are in the ratio 3:2 initially. When 250 g of copper is added, the new condition says copper becomes double tin. We must find the original tin quantity. The key is to represent original quantities using a common multiplier and then apply the new relationship after the addition. Given Data / Assumptions: • Initial copper : tin = 3 : 2• Copper added = 250 g• After addition: copper = 2 * tin Concept / Approach: If copper:tin = 3:2, let copper = 3x and tin = 2x for some x. After adding 250 g copper, new copper amount becomes 3x + 250, while tin remains 2x. Use the condition that new copper equals double the tin to form an equation and solve for x. Then compute tin = 2x. Step-by-Step Solution: Step 1: Represent original alloy quantities.Copper = 3x, Tin = 2xStep 2: Add 250 g copper.New copper = 3x + 250Tin stays the same = 2xStep 3: Apply the condition “copper becomes double tin”.3x + 250 = 2*(2x)3x + 250 = 4xStep 4: Solve for x.x = 250Step 5: Find tin in the original alloy.Tin = 2x = 2*250 = 500 g Verification / Alternative check: Original copper = 3x = 750 g and tin = 500 g. After adding 250 g copper, new copper = 1000 g while tin remains 500 g. Indeed, 1000 is exactly double 500, so the condition is satisfied, confirming the calculation. Why Other Options Are Wrong: 250 g would make tin too small; copper could not become exactly double after adding only 250 g.750 g and 1000 g are too large and would break the required doubling relationship.625 g is not consistent with a clean integer ratio 3:2 and the exact doubling condition. Common Pitfalls: • Adding 250 g to the total alloy instead of only to copper.• Misreading “copper becomes double tin” as “tin becomes double copper”.• Forgetting that tin amount does not change during the addition. Final Answer: The amount of tin in the original alloy is 500 g.

Question 8

A drum contains 80 litres of ethanol initially. If 20 litres of the liquid is removed and replaced with water, and then again 20 litres of the resulting mixture is removed and replaced with water, how many litres of water are present in the drum now?

Accepted Answer

Introduction: This is a two-step replacement (dilution) problem. The drum starts with pure ethanol. When we remove 20 litres and replace with water, the ethanol concentration decreases. In the second replacement, we remove a mixture (not pure ethanol), so we remove some ethanol and some water, and then add water again. The goal is to calculate how much water accumulates in the drum after two such operations. The safest approach is to track ethanol and water amounts after each step. Given Data / Assumptions: • Initial total volume = 80 litres• Initially: ethanol = 80 L, water = 0 L• Each operation: remove 20 L and replace with 20 L water• Number of operations = 2 Concept / Approach: After each removal, the removed portion has the same composition as the drum at that moment. So we compute the ethanol and water removed based on current fractions, subtract them, and then add water replacement. After two operations, read the water amount. Step-by-Step Solution: Step 1: After first removal (from pure ethanol).Remove 20 L ethanol, remaining ethanol = 80 - 20 = 60 LAdd 20 L water, so water = 20 LAfter step 1: ethanol = 60 L, water = 20 LStep 2: Find composition before second removal.Total = 80 L, ethanol fraction = 60/80 = 3/4, water fraction = 20/80 = 1/4Step 3: Remove 20 L of mixture in step 2.Ethanol removed = 20*(3/4)=15 LWater removed = 20*(1/4)=5 LRemaining ethanol = 60 - 15 = 45 LRemaining water = 20 - 5 = 15 LStep 4: Replace removed 20 L with water.Water becomes 15 + 20 = 35 LFinal: ethanol = 45 L, water = 35 L Verification / Alternative check: Total volume remains 80 L throughout (because we remove 20 and add 20). Final ethanol + water = 45 + 35 = 80, consistent. Also, water must be less than 40 here because the second removal removed some water too (5 L), which prevents water from reaching 40. This sanity check matches the computed 35 L. Why Other Options Are Wrong: 40 assumes no water was removed in the second step, which is incorrect.44 and 45 are impossible because total water cannot exceed 80 and two-step dilution here does not add that much net water.30 is too low; even after removing some water, a significant amount is added twice. Common Pitfalls: • Assuming the second 20 L removed is pure ethanol.• Forgetting that some water is removed in the second operation.• Not tracking volumes after each step and jumping to a wrong shortcut. Final Answer: The amount of water present in the drum after two replacements is 35 litres.

Question 9

A beaker contains a mixture of acid and water in the ratio 1 : x (acid : water). When 300 ml of this mixture is taken and mixed with an additional 50 ml of water, the new ratio of acid to water becomes 2 : 5. What is the value of x?

Accepted Answer

Introduction: This question combines ratio composition with a change caused by adding pure water. The mixture starts with acid and water in the ratio 1:x. From this mixture, 300 ml is taken (so acid and water in that sample remain in the same ratio 1:x). Then 50 ml of pure water is added, changing the acid-to-water ratio to 2:5. The goal is to find x by expressing acid and water amounts using fractions and setting up the final ratio equation. Given Data / Assumptions: • Initial ratio (acid:water) = 1 : x• Quantity taken from mixture = 300 ml• Additional water added = 50 ml• Final ratio (acid:water) = 2 : 5 Concept / Approach: If acid:water = 1:x, then in any amount of this mixture, acid fraction = 1/(1+x) and water fraction = x/(1+x). Multiply these fractions by 300 ml to get the acid and water in the taken sample. Then add 50 ml water and enforce final ratio 2:5 by equating acid/water = 2/5. Step-by-Step Solution: Step 1: Compute acid and water in 300 ml of the mixture.Acid in 300 ml = 300 * (1/(1+x)) = 300/(1+x)Water in 300 ml = 300 * (x/(1+x)) = 300x/(1+x)Step 2: Add 50 ml water.New water amount = 300x/(1+x) + 50Acid amount remains = 300/(1+x)Step 3: Apply final ratio 2:5.Acid / Water = 2/5[300/(1+x)] / [300x/(1+x) + 50] = 2/5Step 4: Multiply numerator and denominator to simplify.300 / (300x + 50(1+x)) = 2/5300 / (300x + 50 + 50x) = 2/5300 / (350x + 50) = 2/5Step 5: Cross-multiply and solve.300*5 = 2*(350x + 50)1500 = 700x + 1001400 = 700xx = 2 Verification / Alternative check: If x=2, initial ratio is 1:2. In 300 ml, acid = 100 ml and water = 200 ml. Add 50 ml water to get water = 250 ml. Then acid:water = 100:250 = 2:5, matching exactly. This confirms x = 2 is correct. Why Other Options Are Wrong: x = 1 gives 1:1; then 300 ml contains 150 ml water, adding 50 gives 200, ratio becomes 150:200 = 3:4, not 2:5.x = 3 or 4 makes the mixture too water-heavy initially, and after adding more water the ratio becomes even smaller than 2:5.x = 5 is far too water-heavy, producing acid:water less than 2:5. Common Pitfalls: • Treating 300 ml as 300 parts instead of using fractions of the ratio.• Forgetting to add 50 ml only to water.• Setting up the ratio backwards (water/acid instead of acid/water). Final Answer: The value of x is 2.

Question 10

A mixture is made of 11 parts of pure milk and 2 parts of water. If 35 litres of water are added to this mixture, the new mixture contains twice as much milk as water. How many litres of pure milk were present in the original mixture?

Accepted Answer

Introduction: This question tests mixture ratios and how adding a pure component changes the relationship between components. The mixture starts with milk and water in the ratio 11:2, which means milk is dominant. Then 35 litres of water is added, increasing only the water quantity. After indicating the new condition that milk becomes twice water, we solve for the original quantities. The ratio method is efficient because it avoids guessing and keeps the relationship exact. Given Data / Assumptions: • Initial milk : water = 11 : 2• Water added = 35 litres• After adding: milk = 2 * water Concept / Approach: Let initial milk = 11x and initial water = 2x for some x. When 35 litres of water is added, milk remains 11x but water becomes 2x + 35. The condition “milk is twice water” gives the equation 11x = 2(2x + 35). Solve for x and then compute milk = 11x. Step-by-Step Solution: Step 1: Represent original mixture using the ratio.Milk = 11xWater = 2xStep 2: Add 35 litres water.New water = 2x + 35Milk stays = 11xStep 3: Apply the final condition.Milk = 2 * Water11x = 2*(2x + 35)11x = 4x + 70Step 4: Solve for x.7x = 70x = 10Step 5: Find original milk quantity.Milk = 11x = 11*10 = 110 litres Verification / Alternative check: Original mixture: milk 110 L, water 20 L. After adding 35 L water, water becomes 55 L. Milk remains 110 L. Milk is exactly twice water because 110 = 2*55. This confirms the answer without ambiguity. Why Other Options Are Wrong: 55 L would correspond to half the milk value, but then milk would not be twice the new water.220 L is too large and would make milk much more than twice water after adding only 35 L.70 L does not preserve the 11:2 structure with an integer scaling that understands the final condition.140 L fails verification because it does not satisfy 11x = 2(2x + 35) for an integer x. Common Pitfalls: • Adding 35 litres to the total instead of only to water.• Misreading “milk is twice water” as “water is twice milk”.• Forgetting to scale both milk and water by the same x in the initial ratio. Final Answer: The original mixture contained 110 litres of pure milk.

Alligation or Mixture Questions