Question 1

A buyer purchases 60 quintals of rice of two sorts for Rs. 4642.50 in total. The better sort costs Rs. 80 per quintal and the inferior sort costs Rs. 75.50 per quintal. How many quintals of each sort did he buy?

Accepted Answer

Introduction / Context: This is a classic two-variable mixture cost problem. Knowing the total weight and total cost lets us determine how much was bought at each price point. Given Data / Assumptions: Total weight = 60 quintals. Price (better) = Rs. 80/q; price (worse) = Rs. 75.50/q. Total cost = Rs. 4642.50. Concept / Approach: Let x be the quantity of the better sort. Then the inferior sort is 60 − x. Form the total cost equation 80x + 75.5(60 − x) = 4642.5 and solve for x. Step-by-Step Solution: 80x + 75.5(60 − x) = 4642.580x + 4530 − 75.5x = 4642.54.5x = 112.5 ⇒ x = 25 quintals (better)Inferior sort = 60 − 25 = 35 quintals Verification / Alternative check: Cost check: 25*80 + 35*75.5 = 2000 + 2642.5 = 4642.5, matches the total. Why Other Options Are Wrong: Other splits do not satisfy the cost equation simultaneously with the total weight of 60 quintals. Common Pitfalls: Arithmetic slip on 4.5x or misreading 75.50 as 75 or 75.05. Final Answer: 25 quintals, 35 quintals

Question 2

A petrol pump owner has a 1-liter mixture containing 10% unleaded petrol (the rest is leaded). How much leaded petrol must be added so that the unleaded percentage becomes 5%?

Accepted Answer

Introduction / Context: This is a dilution problem where an inert (leaded with 0% unleaded) is added to reduce the proportion of the active component (unleaded). The amount of unleaded remains fixed while total volume increases. Given Data / Assumptions: Initial mixture = 1 L (1000 ml), unleaded = 10% = 0.1 L. Add x liters of leaded (0% unleaded). Target unleaded% = 5%. Concept / Approach: Unleaded volume stays at 0.1 L. New total volume = 1 + x. Set 0.1 / (1 + x) = 0.05 and solve for x. Step-by-Step Solution: 0.1 / (1 + x) = 0.051 + x = 0.1 / 0.05 = 2 ⇒ x = 1 L = 1000 ml Verification / Alternative check: After adding 1 L leaded, total is 2 L; unleaded = 0.1 L, so percentage = 0.1 / 2 = 5%. Why Other Options Are Wrong: 900 ml or 1800 ml give unleaded percentages not equal to 5%. Only 1000 ml halves the percentage from 10% to 5%. Common Pitfalls: Reducing 10% to 5% by “subtracting” rather than diluting; remember the active amount is unchanged while total grows. Final Answer: 1000 ml

Question 3

Four equal-volume vessels contain spirit concentrations of 60%, 70%, 75%, and 80%. If all are mixed in equal amounts, what is the ratio of spirit to water in the resulting mixture?

Accepted Answer

Introduction / Context: When equal volumes are mixed, the final concentration is simply the arithmetic mean of the individual concentrations. Convert that to a spirit : water ratio by comparing to 100%. Given Data / Assumptions: Spirit percentages: 60, 70, 75, 80. Equal volumes from each vessel. Concept / Approach: Average spirit% = (60 + 70 + 75 + 80)/4. Water% = 100 − spirit%. Ratio spirit : water = spirit% : water%. Step-by-Step Solution: Mean spirit% = 285/4 = 71.25%Water% = 100 − 71.25 = 28.75%Spirit : Water = 71.25 : 28.75 = (285 : 115) = 57 : 23 Verification / Alternative check: Divide both by 5: 285 : 115 → 57 : 23; consistent. Why Other Options Are Wrong: 57 : 13 and 23 : 57 imply impossible >100% or too little spirit; only 57 : 23 matches 71.25% spirit. Common Pitfalls: Taking a weighted average even though volumes are equal, or confusing spirit : water with water : spirit. Final Answer: 57 : 23

Question 4

How much chicory at Rs. 24 per kg must be added to 15 kg of tea priced at Rs. 60 per kg so that the resulting mixture is worth Rs. 39 per kg?

Accepted Answer

Introduction / Context: We blend a cheaper component (chicory) with a dearer one (tea) to achieve a target price. Set up the value equation: total cost divided by total weight equals the target price. Given Data / Assumptions: Tea: 15 kg at Rs. 60/kg. Chicory: x kg at Rs. 24/kg. Mixture price: Rs. 39/kg. Concept / Approach: Use (total cost)/(total weight) = target price. Solve for x. Step-by-Step Solution: Total cost = 15*60 + 24x = 900 + 24xTotal weight = 15 + x(900 + 24x)/(15 + x) = 39900 + 24x = 585 + 39x ⇒ 315 = 15x ⇒ x = 21 kg Verification / Alternative check: Alligation: 60 to 39 (diff 21) and 39 to 24 (diff 15) gives ratio tea : chicory = 15 : 21 = 5 : 7. For tea 15 kg, chicory = 15*(7/5) = 21 kg. Why Other Options Are Wrong: Other values of x do not satisfy the price equation exactly. Common Pitfalls: Mixing up which side of alligation corresponds to which quantity; use inverse differences. Final Answer: 21 kg

Question 5

A mixture of 150 liters contains 20% Diesel (the rest is Petrol). How many more liters of Diesel must be added so that Diesel becomes 25% of the new mixture?

Accepted Answer

Introduction / Context: We increase the Diesel component by adding pure Diesel. The Diesel amount changes while Petrol stays the same; total volume also increases by the added Diesel. Given Data / Assumptions: Initial total = 150 L; Diesel = 20% = 30 L; Petrol = 120 L. Add y liters Diesel. Target Diesel% = 25%. Concept / Approach: Set (Diesel after addition)/(Total after addition) = 25% and solve for y. Step-by-Step Solution: (30 + y) / (150 + y) = 0.2530 + y = 37.5 + 0.25y ⇒ 0.75y = 7.5 ⇒ y = 10 L Verification / Alternative check: New total = 160 L; Diesel = 40 L; 40/160 = 0.25 = 25%. Why Other Options Are Wrong: 5, 15, 20, 35 produce Diesel percentages different from 25%. Common Pitfalls: Using 25% of the original total or forgetting that the denominator changes after adding Diesel. Final Answer: 10 liter

Question 6

Two vessels contain milk : water as follows — A: 7 : 5 and B: 17 : 7. In what ratio should mixtures from A and B be combined to obtain milk : water = 5 : 3 in the final mixture?

Accepted Answer

Introduction / Context: We target a milk fraction of 5/8 in the final mixture. Use alligation on milk fractions of the two sources to determine the required combining ratio. Given Data / Assumptions: A milk fraction = 7/12 ≈ 0.5833. B milk fraction = 17/24 ≈ 0.7083. Target milk fraction = 5/8 = 0.625. Concept / Approach: For fractions fA and fB, A : B = (fB − m) : (m − fA). Step-by-Step Solution: A : B = (17/24 − 5/8) : (5/8 − 7/12)= (17/24 − 15/24) : (15/24 − 14/24) = (2/24) : (1/24) = 2 : 1 Verification / Alternative check: Take 2 L from A and 1 L from B: milk = 2*(7/12) + 1*(17/24) = 14/12 + 17/24 = 28/24 + 17/24 = 45/24 = 15/8; total 3 L; milk fraction = (15/8)/3 = 5/8 → milk : water = 5 : 3. Why Other Options Are Wrong: Other ratios produce milk fractions above or below 5/8. Only 2 : 1 meets the target exactly. Common Pitfalls: Reversing the differences in alligation and getting 1 : 2. Final Answer: 2 : 1

Question 7

A person covers 100 km in 10 hours, partly walking at 7 km/h and the rest running at 12 km/h. Find the distance covered walking and the distance covered running.

Accepted Answer

Introduction / Context: The traveler uses two constant speeds over a fixed total time and distance. Let walking time be t hours, so running time is 10 − t. Form a single linear equation in t using distance = speed * time. Given Data / Assumptions: Total time = 10 hours. Walking speed = 7 km/h. Running speed = 12 km/h. Total distance = 100 km. Concept / Approach: Total distance = 7t + 12(10 − t). Solve for t, then compute each distance separately: 7t for walking and 12(10 − t) for running. Step-by-Step Solution: 7t + 12(10 − t) = 1007t + 120 − 12t = 100 ⇒ −5t = −20 ⇒ t = 4 hoursWalking distance = 7*4 = 28 kmRunning distance = 12*(10 − 4) = 72 km Verification / Alternative check: Average speed check: total distance/time = 100/10 = 10 km/h. Weighted speeds (7 for 4 h and 12 for 6 h) produce overall 10 km/h, consistent. Why Other Options Are Wrong: They do not satisfy both the time and distance equations simultaneously. Common Pitfalls: Setting up two unknown distances without relating them through time, or assuming equal times at each speed. Final Answer: Walking = 28 km, Running = 72 km

Question 8

A cask contains 25 liters of wine. First 5 liters are withdrawn and replaced by water; this is repeated twice more (three operations in total). Find the quantity of wine left and the ratio of wine to water in the final mixture.

Accepted Answer

Introduction / Context: Repeated replacement problems use the exponential decay formula for the retained fraction after each withdrawal. Each time 5 liters are removed from 25 liters, the same fraction of wine is taken out and replaced by water. Given Data / Assumptions: Initial wine = 25 L. Each operation: remove 5 L, then add 5 L water. Number of operations = 3. Concept / Approach: After each operation, the fraction of wine remaining equals (1 − 5/25) = 4/5 of the previous amount. After n operations: wine_left = 25*(4/5)^n. Step-by-Step Solution: wine_left = 25*(4/5)^3 = 25*(64/125) = 12.8 LWater = total − wine = 25 − 12.8 = 12.2 LWine : Water = 12.8 : 12.2 = 128 : 122 = 64 : 61 Verification / Alternative check: Successive amounts: after 1st = 20 L, 2nd = 16 L, 3rd = 12.8 L; consistent with (4/5)^n scaling. Why Other Options Are Wrong: 46 : 16 reverses values; 61 : 64 is water : wine, not wine : water; 1 : 1 is inaccurate. Common Pitfalls: Subtracting 5 liters of wine each time (linear) instead of multiplying by 4/5 (geometric). Final Answer: 64 : 61

Question 9

A vessel initially contains liquids A and B in the ratio 3 : 2 (by volume). Twenty liters of the mixture are removed and replaced with 20 liters of B, after which the ratio becomes 1 : 4 (A : B). How many liters of A were present initially?

Accepted Answer

Introduction / Context: This is a remove-and-replace problem. Removing a fixed volume of mixture takes out each component in proportion to its current fraction. Then pure B is added, altering the ratio further. Given Data / Assumptions: Initial A : B = 3 : 2; let total = T L → A = 3T/5, B = 2T/5. Remove 20 L mixture; then add 20 L of B. Final A : B = 1 : 4. Concept / Approach: Removal takes out components proportionally: A removed = (3/5)*20 = 12; B removed = (2/5)*20 = 8. Update quantities and then add 20 L B. Use the final ratio to solve for T. Step-by-Step Solution: A after removal = 3T/5 − 12B after removal and addition = 2T/5 − 8 + 20 = 2T/5 + 12(3T/5 − 12) : (2T/5 + 12) = 1 : 44*(3T/5 − 12) = 2T/5 + 12 ⇒ (12T/5 − 48) = 2T/5 + 1210T/5 = 60 ⇒ 2T = 60 ⇒ T = 30 LInitial A = 3T/5 = 18 L Verification / Alternative check: Check final: A = 18 − 12 = 6; B = 12 + (2*30/5) = 12 + 12 = 24; ratio 6 : 24 = 1 : 4. Why Other Options Are Wrong: 12, 24, 22, 20 do not satisfy the final ratio after the remove-and-add operations. Common Pitfalls: Subtracting 20 directly from A or B instead of removing proportionally to the mixture composition. Final Answer: 18 litres

Question 10

An alloy piece of two metals A and B weighs 15 g and is worth Rs. 150. If the weights of A and B are interchanged, the new alloy would be worth Rs. 120. Given price of A is Rs. 6 per g, find the weight of the other metal (B) in the original alloy.

Accepted Answer

Introduction / Context: The alloy's value is the sum of values contributed by each metal (price * weight). Interchanging weights creates a second equation. Solve the two linear equations to find the unknown weight of metal B initially. Given Data / Assumptions: Total weight = 15 g. First alloy value = Rs. 150. Second (weights swapped) value = Rs. 120. Price of A = Rs. 6 per g; price of B = unknown p. Concept / Approach: Let initial weights be a (A) and b (B). Then a + b = 15. Equations: 6a + p b = 150 and 6b + p a = 120. Add to eliminate asymmetry and find p. Then solve for a and b. Step-by-Step Solution: (6a + p b) + (6b + p a) = 270 ⇒ (6 + p)(a + b) = 270Since a + b = 15 ⇒ 15(6 + p) = 270 ⇒ 6 + p = 18 ⇒ p = 12Use 6a + 12b = 150 with a + b = 15Multiply the second by 6: 6a + 6b = 90 ⇒ subtract ⇒ 6b = 60 ⇒ b = 10 g Verification / Alternative check: With b = 10, a = 5. Values: first = 6*5 + 12*10 = 30 + 120 = 150; swapped = 6*10 + 12*5 = 60 + 60 = 120. Why Other Options Are Wrong: Other weights fail one or both value equations and do not maintain total 15 g. Common Pitfalls: Assuming the prices are swapped instead of weights; the problem states interchanging weights. Final Answer: 10 gms

Question 11

Mixture in a 90-L container: A vessel initially contains two liquids, A and B, with unknown amounts. From the vessel, 60% of liquid A and 30% of liquid B are withdrawn. After this withdrawal the container is 40% empty (that is, 60% full). Determine …

Accepted Answer

Introduction / Context: This problem combines percentage removal with total volume tracking. The key is to express what remains of each liquid after withdrawing fixed percentages and then use the final fullness (60% of capacity) to solve for the initial split of A and B. Given Data / Assumptions: Total capacity = 90 L. 60% of liquid A withdrawn; 30% of liquid B withdrawn. After withdrawal, the vessel is 60% full ⇒ remaining volume = 54 L. Let initial A = x L, initial B = 90 − x L. Concept / Approach: After withdrawal, remaining A = 40% of x (i.e., 0.4x). Remaining B = 70% of (90 − x) (i.e., 0.7*(90 − x)). Their sum must equal 54 L because the container is 60% full. Step-by-Step Solution: Remaining A = 0.4x.Remaining B = 0.7(90 − x) = 63 − 0.7x.0.4x + (63 − 0.7x) = 54.−0.3x = −9 ⇒ x = 30.Therefore A = 30 L and B = 90 − 30 = 60 L. Verification / Alternative check: Withdrawn A = 60% of 30 = 18 L; remaining A = 12 L. Withdrawn B = 30% of 60 = 18 L; remaining B = 42 L. Total remaining = 12 + 42 = 54 L = 60% of 90 L, consistent. Why Other Options Are Wrong: Other pairs do not satisfy the condition 0.4x + 0.7(90 − x) = 54; they either overfill or underfill relative to the 60% final level. Common Pitfalls: Subtracting percentages from the total instead of per-component; forgetting that 40% empty means 60% full (54 L). Final Answer: A = 30 L, B = 60 L

Question 12

Tea alligation with three varieties: Tea priced at ₹ 126/kg and ₹ 135/kg are mixed with a third variety in the ratio 1 : 1 : 2. If the final mixture is worth ₹ 153/kg, find the price per kg of the third variety.

Accepted Answer

Introduction / Context: When three price points are mixed in a fixed quantity ratio, the overall price is the weighted average. Here, the third variety has an unknown price that must make the average equal to ₹ 153/kg across four equal parts (1 + 1 + 2). Given Data / Assumptions: Prices: ₹ 126, ₹ 135, and ₹ x (unknown). Mix ratio: 1 : 1 : 2 (total parts = 4). Mixture price = ₹ 153/kg. Concept / Approach: Use weighted average: (126*1 + 135*1 + x*2) / 4 = 153. Solve for x to get the third variety’s price per kilogram. Step-by-Step Solution: (126 + 135 + 2x) / 4 = 153261 + 2x = 6122x = 351 ⇒ x = 175.50 Verification / Alternative check: Substitute x = 175.50: (126 + 135 + 351) / 4 = 612 / 4 = 153, exactly as required. Why Other Options Are Wrong: Values like ₹ 170 or ₹ 180 shift the weighted average below or above ₹ 153; only ₹ 175.50 balances the 1 : 1 : 2 mix. Common Pitfalls: Taking a simple average of the three prices instead of a weighted average according to 1 : 1 : 2. Final Answer: ₹ 175.50

Question 13

Raising water percentage by adding water: A 20-kg mixture of spirit and water contains 10% water. How much water must be added so that water becomes 25% of the new mixture?

Accepted Answer

Introduction / Context: Increasing the percentage of a component by adding more of that component is a classic proportion problem. Track absolute water mass before and after the addition and form a single equation in the unknown added water. Given Data / Assumptions: Total initial mixture = 20 kg with 10% water ⇒ 2 kg water, 18 kg spirit. Add w kg of water to reach 25% water in the new total. Concept / Approach: New water mass = 2 + w. New total mass = 20 + w. Target fraction = (2 + w)/(20 + w) = 0.25. Solve for w. Step-by-Step Solution: (2 + w)/(20 + w) = 1/44(2 + w) = 20 + w ⇒ 8 + 4w = 20 + w3w = 12 ⇒ w = 4 kg Verification / Alternative check: New mixture = 24 kg with water = 6 kg ⇒ 6/24 = 25%, as required. Why Other Options Are Wrong: 5 kg or 8 kg change the final water percentage to values other than 25%; 30 kg is far too large; 2 kg only raises water to 4/22 ≈ 18.18%. Common Pitfalls: Treating percentages additively rather than using the ratio of new water to new total mass. Final Answer: 4 kg

Question 14

Replacement lowers concentration: A jar of whisky is 40% alcohol. A portion is removed and replaced with another whisky that is 19% alcohol, resulting in a final alcohol percentage of 26%. What fraction of the original whisky was replaced?

Accepted Answer

Introduction / Context: In replacement problems, concentrations combine linearly. If a fraction r of volume is replaced, the final concentration equals the remaining part of the old concentration plus the added part of the new concentration. Volumes cancel, leaving an equation in r only. Given Data / Assumptions: Initial alcohol fraction = 0.40. Replacing fraction = r (of the total volume). Incoming alcohol fraction = 0.19. Final fraction = 0.26. Concept / Approach: Final fraction = (1 − r)*0.40 + r*0.19. Solve for r to find the portion replaced. Step-by-Step Solution: (1 − r)*0.40 + r*0.19 = 0.260.40 − 0.40r + 0.19r = 0.260.40 − 0.21r = 0.26 ⇒ 0.21r = 0.14 ⇒ r = 2/3 Verification / Alternative check: Substitute r = 2/3: final = (1/3)*0.40 + (2/3)*0.19 = 0.1333… + 0.1266… ≈ 0.26, matching the target. Why Other Options Are Wrong: Other fractions produce final concentrations different from 26%; only 2/3 balances the convex combination between 40% and 19%. Common Pitfalls: Averaging 40% and 19% without weighting by r and (1 − r); ignoring that only a fraction of the original remains. Final Answer: 2/3

Question 15

Simple interest split by alligation: ₹ 1000 is lent in two parts at 6% and 8% simple interest per annum. If the total annual interest is ₹ 75, how much was lent at 8%?

Accepted Answer

Introduction / Context: With two rates and a known total interest, set up one linear equation in the unknown portion invested at the higher rate. This is a direct application of simple interest with a single time period (1 year). Given Data / Assumptions: Total principal = ₹ 1000. Rates: 6% and 8% per annum. Total annual interest = ₹ 75. Let x be the amount at 8%. Concept / Approach: Interest = 0.08x + 0.06(1000 − x) = 75. Solve for x to get the principal at 8%. Step-by-Step Solution: 0.08x + 60 − 0.06x = 750.02x = 15 ⇒ x = 750 Verification / Alternative check: Check: 8% of 750 = 60; 6% of 250 = 15; total = 75, consistent. Why Other Options Are Wrong: The other values produce interest totals different from ₹ 75 upon substitution into 0.08x + 0.06(1000 − x). Common Pitfalls: Assuming equal split or averaging rates without regard to the interest total constraint. Final Answer: ₹ 750

Question 16

Equal whisky quantities mixed with water to reach a lower average price: Two lots of whisky priced at ₹ 22/L and ₹ 18/L are taken in equal volumes and mixed with water to make 50 L of mixture worth ₹ 16/L. How many liters of water are in the mixture?

Accepted Answer

Introduction / Context: Water contributes no cost, so the total value of the mixture equals the value of just the whisky present. Using equal quantities of the two whisky lots simplifies the value calculation before matching the target average price over 50 L. Given Data / Assumptions: Equal volumes of ₹ 22/L and ₹ 18/L whisky ⇒ each q L. Water volume = w L. Total volume = 2q + w = 50 L; mixture price = ₹ 16/L ⇒ total value = ₹ 800. Concept / Approach: Total value arises only from whisky: 22q + 18q = 40q. Set 40q = 800 to find q, then compute water as the remainder to 50 L. Step-by-Step Solution: 40q = 800 ⇒ q = 20 L (of each whisky).Total whisky = 40 L ⇒ water w = 50 − 40 = 10 L. Verification / Alternative check: Value from whisky = 20*22 + 20*18 = 440 + 360 = 800; average over 50 L = 800/50 = ₹ 16/L, as required. Why Other Options Are Wrong: 5, 15, or 20 L water would not yield the ₹ 16/L average given the whisky prices and equal whisky amounts. Common Pitfalls: Averaging prices without accounting for added water or forgetting water carries zero cost. Final Answer: 10 L

Question 17

Milk with water yielding profit despite lower selling price: Milk costs ₹ 5 per liter. It is mixed with water and sold at ₹ 4 per liter, yet the seller earns a 12.5% profit on outlay. How much water is present per liter of mixture?

Accepted Answer

Introduction / Context: Even though the selling price is lower than the pure milk cost, profit is possible because water is free. Let the milk fraction in 1 L of mixture be m; the cost per liter is 5*m and revenue is ₹ 4. The profit condition sets a relationship between m and the selling price. Given Data / Assumptions: Cost of milk = ₹ 5/L; water is free. Selling price = ₹ 4/L. Profit = 12.5% ⇒ revenue = 1.125 * cost. Let milk fraction = m ⇒ water fraction = 1 − m. Concept / Approach: Equation: 4 = 1.125 * (5*m). Solve for m, then compute water fraction (1 − m), which is the liters of water per 1 L of mixture. Step-by-Step Solution: 4 = 1.125 * 5m = 5.625mm = 4 / 5.625 = 0.711111… = 32/45.Water per liter = 1 − m = 1 − 32/45 = 13/45 L. Verification / Alternative check: Cost per liter = 5*(32/45) = ₹ 3.555…; 12.5% profit on this is ₹ 0.444…; sum = ₹ 4 exactly. Why Other Options Are Wrong: 32/45 L is the milk amount, not water; 32/13 L is nonsensical here; 1/4 L does not satisfy the profit equation. Common Pitfalls: Comparing prices directly without accounting for water’s zero cost; switching milk and water fractions. Final Answer: 13/45 L

Question 18

Combining three mixtures with different milk strengths: Glasses of 2 L (90% milk), 5 L (80% milk), and 9 L (70% milk) are poured into one vessel. Find the milk concentration and the milk:water ratio in the final mixture.

Accepted Answer

Introduction / Context: To combine different concentrations, compute absolute milk volumes from each container, add them, and compare against the total volume to get both concentration and milk:water ratio. Given Data / Assumptions: 2 L at 90% ⇒ milk = 1.8 L. 5 L at 80% ⇒ milk = 4.0 L. 9 L at 70% ⇒ milk = 6.3 L. Total volume = 16 L. Concept / Approach: Sum milk volumes to get total milk; water is total minus milk. Ratio milk:water yields an integer pair on scaling (here by 10). Step-by-Step Solution: Total milk = 1.8 + 4.0 + 6.3 = 12.1 L.Total water = 16 − 12.1 = 3.9 L.Milk : Water = 12.1 : 3.9 = 121 : 39. Verification / Alternative check: Milk fraction = 12.1/16 = 0.75625 = 75.625%; consistent with ratio 121 : 39 since 121/(121+39) = 121/160 = 75.625%. Why Other Options Are Wrong: 131 : 49 or 49 : 131 correspond to different milk totals; 39 : 121 inverts the ratio. Common Pitfalls: Averaging percentages directly; always convert to absolute amounts before combining. Final Answer: 121 : 39

Question 19

Successive dilution of wine with water: A 64-L cask of wine has 8 L withdrawn and replaced with water. This process is repeated two more times. How many liters of wine remain after the third replacement?

Accepted Answer

Introduction / Context: Repeatedly removing a fixed volume and refilling leads to a multiplicative retention factor each time. For volume V and withdrawal w, the fraction retained after one cycle is (1 − w/V). After n cycles, the retained fraction is (1 − w/V)^n of the original amount. Given Data / Assumptions: Initial wine = 64 L. Each cycle: remove 8 L, refill with water. Cycles n = 3. Concept / Approach: Remaining wine = 64 * (1 − 8/64)^3 = 64 * (7/8)^3. Compute the value to get the exact liters remaining of wine (water makes up the rest). Step-by-Step Solution: (7/8)^3 = 343/512.Remaining wine = 64 * 343/512 = (64/512) * 343 = (1/8) * 343 = 42 7/8 L. Verification / Alternative check: Numeric check: 42.875 L; options present 42 7/8 L which is identical. Why Other Options Are Wrong: Other fractional values correspond to incorrect powers or arithmetic and do not match the retention factor (7/8)^3. Common Pitfalls: Subtracting 8 L of wine each time (linear) instead of applying the retention fraction (multiplicative). Final Answer: 42 7/8 L

Question 20

Replacing mixture with pure component changes ratio: A can initially has liquids A and B in the ratio 7 : 5. After drawing off 9 L of the mixture and refilling with pure B, the ratio becomes 7 : 9. How many liters of A were in the can originally?

Accepted Answer

Introduction / Context: When a fixed volume of a mixture is removed, each component is reduced proportionally. Adding back pure B increases B only. Using initial ratio parts and total volume lets us model the new amounts and match the final ratio to solve for the scale factor and the initial amount of A. Given Data / Assumptions: Initial A : B = 7 : 5 ⇒ A0 = 7k, B0 = 5k, total = 12k. Draw off 9 L; refill with 9 L of pure B. Final A : B = 7 : 9. Concept / Approach: After drawing 9 L, remaining fraction = 1 − 9/(12k). Thus A1 = 7k*(1 − 9/(12k)); B1 = 5k*(1 − 9/(12k)) + 9. Set A1/B1 = 7/9 and solve for k, then compute A0 = 7k. Step-by-Step Solution: A1 = 7k − 63/12 = 7k − 5.25B1 = 5k − 45/12 + 9 = 5k + 5.25(7k − 5.25)/(5k + 5.25) = 7/9 ⇒ 9(7k − 5.25) = 7(5k + 5.25)63k − 47.25 = 35k + 36.75 ⇒ 28k = 84 ⇒ k = 3Hence A initially = 7k = 21 L. Verification / Alternative check: Total 12k = 36 L; removing 9 L leaves 27 L; adding 9 L B restores 36 L. Final A : B computed equals 7 : 9 as required. Why Other Options Are Wrong: They do not satisfy the transformed ratio equation after removal and refill. Common Pitfalls: Subtracting 9 L directly from A and B parts instead of reducing them proportionally to total. Final Answer: 21 L

Alligation or Mixture Questions