Free Topic-Wise General Studies MCQs
Master Averages and Mixtures for UPSC CSAT. Practice high-yield MCQs with detailed step-by-step solutions to build your quantitative aptitude for the Civil Services Prelims.
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Explanation: The sum of the first n odd natural numbers is always n². For n = 15, the sum is 15² = 225. To find the average, divide the total sum by the count of numbers: 225 ÷ 15 = 15. In similar problems, remember that the average of the first n odd numbers is always n, because the sum is n² and there are n terms.
Explanation: When the average of several values is known, the total sum equals the average multiplied by the number of values. Here, the average of 5 innings is 72, so the total runs needed are 72 × 5 = 360. Add the known scores: 87 + 94 + 32 + 76 = 289. Subtract this from the required total to find the missing score: 360 − 289 = 71. For any missing-value average problem, always compute Total = Average × Count first, then subtract the known values.
Explanation: The key technique is to convert every average into a total sum. Total of 25 numbers = 25 × 36 = 900. Total of first 12 = 12 × 30 = 360. Total of last 12 = 12 × 40 = 480. The 13th number is the overlap that is excluded when you add the first and last groups together. Therefore: 13th number = Overall Total − (First Group Total + Last Group Total) = 900 − 360 − 480 = 60. This method works for any 'middle number' problem where two subgroups overlap by exactly one term.
Explanation: First, find the total age of the family 4 years ago: 4 members × 20 years = 80 years. Now, each of those 4 members has aged 4 years, so their current total age is 80 + (4 × 4) = 96 years. With the baby, there are 5 members and the average is 20, so the new total age is 5 × 20 = 100 years. The baby's age is the difference: 100 − 96 = 4 years. In similar problems, always account for the time elapsed by adding the appropriate years to each existing member's age before comparing totals.
Explanation: Convert averages to totals. Total of 11 results = 11 × 50 = 550. Total of first 6 = 6 × 49 = 294. Total of last 6 = 6 × 52 = 312. Notice that the 6th result is counted in both the first 6 and the last 6, so adding the two subgroup totals counts it twice. Therefore, the 6th result = (First 6 Total + Last 6 Total) − Overall Total = 294 + 312 − 550 = 56. This formula works whenever two subgroups overlap by exactly one term: Overlap = Sum of subgroup totals − Overall total.
Explanation: For weighted averages, you cannot simply average the two averages; you must weight them by the number of students in each class. Calculate total marks for each class: Class A = 40 × 65 = 2600; Class B = 60 × 75 = 4500. Combined total marks = 7100. Combined students = 100. Combined average = 7100 ÷ 100 = 71. The general formula is: (n₁ × A₁ + n₂ × A₂) ÷ (n₁ + n₂). Always use the group sizes as weights, not just the averages themselves.
Explanation: First find the total cost of the mixture: (30 kg × Rs 40) + (20 kg × Rs 60) = 1200 + 1200 = Rs 2400. Total quantity = 50 kg. Cost per kg of mixture = 2400 ÷ 50 = Rs 48. To earn 25% profit, multiply the cost price by 1.25 (or add 25% to it): 48 × 1.25 = Rs 55 per kg. In mixture problems, always calculate the weighted average cost first, then apply the profit percentage to that cost, not to the individual component prices.
Explanation: Calculate the total weight of each subgroup: Boys = 10 × 50 = 500 kg; Girls = 10 × 40 = 400 kg. Combined weight = 900 kg. When the teacher joins, add her weight: 900 + 45 = 945 kg. Total people = 21. New average = 945 ÷ 21 = 45 kg. For any 'new member joins' average problem, compute the old total, add the new value, and divide by the new count.
Explanation: Original total marks = 30 × 40 = 1200. The 6 students who leave take with them 6 × 30 = 180 marks. The 6 new students bring 6 × 50 = 300 marks. New total = 1200 − 180 + 300 = 1320. The class size remains 30, so the new average = 1320 ÷ 30 = 44. The shortcut method: the net change in total is (6 × 50) − (6 × 30) = +120, so the average increases by 120 ÷ 30 = 4, giving 40 + 4 = 44. This shortcut works whenever the number of leavers equals the number of joiners.
Explanation: Original total marks = 5 × 55 = 275. The net change in total marks is +10 (Mathematics) − 5 (Science) = +5. New total = 275 + 5 = 280. The number of subjects remains 5, so the new average = 280 ÷ 5 = 56. For similar problems, do not recalculate each subject individually; simply find the net change in the total and divide by the number of subjects to get the change in average.
Explanation: When equal distances are covered at two different speeds, the average speed for the entire journey is the harmonic mean of the two speeds, given by the formula: 2ab ÷ (a + b). Here, a = 12 km/hr and b = 20 km/hr. Average speed = (2 × 12 × 20) ÷ (12 + 20) = 480 ÷ 32 = 15 km/hr. Remember: never use the arithmetic mean (a + b) ÷ 2 for equal-distance round trips; always use the harmonic mean formula because more time is spent at the slower speed.
Explanation: Average speed is always Total Distance ÷ Total Time, not the arithmetic mean of the two speeds. Total distance = 150 + 150 = 300 km. Time for first part = 150 ÷ 50 = 3 hours. Time for second part = 150 ÷ 30 = 5 hours. Total time = 8 hours. Average speed = 300 ÷ 8 = 37.5 km/hr. In similar problems, resist the temptation to average the speeds directly; always compute the total distance and total time separately.
Explanation: Let the distance to the station be d km and the correct time to reach be t hours. At 4 km/hr, he takes d/4 hours, which is 10 minutes (1/6 hour) too much: d/4 = t + 1/6. At 6 km/hr, he takes d/6 hours, which is 5 minutes (1/12 hour) too little: d/6 = t − 1/12. Subtract the second equation from the first: d/4 − d/6 = 1/6 + 1/12 = 1/4. Simplify: d × (1/12) = 1/4, so d = 3 km. For time-and-distance problems with two speeds, set up two equations with distance and time as variables, then eliminate time by subtraction.
Explanation: For equal distances covered at different speeds, average speed = Total Distance ÷ Total Time. Let total distance = D. Each one-third is D/3. Time = (D/3)/60 + (D/3)/40 + (D/3)/20 = D/3 × (1/60 + 1/40 + 1/20). Find common denominator (120): = D/3 × (2 + 3 + 6)/120 = D/3 × 11/120 = 11D/360. Average speed = D ÷ (11D/360) = 360/11 ≈ 32.73 km/hr. For three or more equal segments, use the same principle: total distance divided by the sum of individual times.
Explanation: For a round trip with equal distances, average speed = 2ab ÷ (a + b), where a and b are the two speeds. Here, a = 80 km/hr and b = 120 km/hr. Average speed = (2 × 80 × 120) ÷ (80 + 120) = 19200 ÷ 200 = 96 km/hr. This formula is derived from Total Distance ÷ Total Time and is the standard shortcut for equal-distance round-trip problems. Always verify that the distances are equal before applying it.
Explanation: Use the rule of alligation to find the mixing ratio. The mean price is Rs 75. The cheaper variety is Rs 60 (difference from mean = 75 − 60 = 15). The dearer variety is Rs 90 (difference from mean = 90 − 75 = 15). The ratio in which they must be mixed is the inverse of these differences: dearer difference : cheaper difference = 15 : 15 = 1 : 1. The alligation method states: Ratio = (Price of dearer − Mean price) : (Mean price − Price of cheaper). This works for any mixture problem where you need to find the mixing ratio.
Explanation: First, determine the current quantities. In a 60-litre mixture with ratio 2:1, milk = 60 × 2/3 = 40 litres and water = 20 litres. The milk quantity remains unchanged when only water is added. To get a 1:2 ratio, the water must be twice the milk. Since milk is 40 litres, water must be 80 litres. Water already present = 20 litres, so water to add = 80 − 20 = 60 litres. In similar problems, identify the component that does not change, set up the new ratio using that fixed quantity, and calculate the required addition.
Explanation: In a 20-litre mixture with ratio 3:2, milk = 20 × 3/5 = 12 litres and water = 8 litres. When 5 litres of the mixture is removed, the removed portion also has the same 3:2 ratio. Milk removed = 5 × 3/5 = 3 litres; water removed = 5 × 2/5 = 2 litres. After removal: milk = 12 − 3 = 9 litres; water = 8 − 2 = 6 litres. Then 5 litres of pure water is added: water becomes 6 + 5 = 11 litres. New ratio = 9 : 11. The key principle is that any removed portion maintains the original ratio of the mixture.
Explanation: When a quantity is drawn from a pure substance and replaced with a different substance, the remaining pure substance follows the formula: Final Amount = Initial Amount × (1 − Replacement Fraction)^n, where n is the number of repetitions. Here, initial milk = 50 litres, replacement = 10 litres, so the fraction remaining each time is (50 − 10)/50 = 4/5. After 2 replacements: milk = 50 × (4/5)² = 50 × 16/25 = 32 litres. This formula works for any repeated replacement problem where the same quantity is replaced each time.
Explanation: Let the original quantity be 5x litres (using the ratio 4:1). Then spirit = 4x and water = x. When 5 litres is drawn out, the removed spirit = 5 × 4/5 = 4 litres, and removed water = 1 litre. After replacement with 5 litres of water: new spirit = 4x − 4, new water = x − 1 + 5 = x + 4. The new ratio is given as 2:3, so (4x − 4)/(x + 4) = 2/3. Cross-multiply: 12x − 12 = 2x + 8, giving 10x = 20, so x = 2. Original quantity = 5x = 10 litres. For replacement problems with unknown initial quantity, introduce a variable based on the ratio and use the final ratio condition to form an equation.
Explanation: After each replacement, the remaining wine is multiplied by the fraction (Total − Drawn)/Total. Here, total = 60 litres, drawn = 6 litres, so the multiplier is 54/60 = 0.9. After two replacements: wine remaining = 60 × (0.9)² = 60 × 0.81 = 48.6 litres. The general formula for n replacements is: Initial × [(V − x)/V]^n, where V is the total volume and x is the amount replaced each time. This is the most efficient method for repeated replacement problems.
Explanation: When mixing two vessels in a given ratio, calculate the contribution of each component from each vessel separately, then combine them. Vessel A (ratio 5:3) has milk = 5/8 and water = 3/8 per unit. Vessel B (ratio 2:1) has milk = 2/3 and water = 1/3 per unit. Mixed in ratio 2:1: Total milk = 2×(5/8) + 1×(2/3) = 5/4 + 2/3 = (15 + 8)/12 = 23/12. Total water = 2×(3/8) + 1×(1/3) = 3/4 + 1/3 = (9 + 4)/12 = 13/12. Resulting ratio = 23/12 : 13/12 = 23 : 13. For vessel-mixing problems, always work with the fractional parts of each component in each vessel.
Explanation: Current acid = 30 × 7/10 = 21 litres; water = 9 litres. Only water is added, so acid remains 21 litres. For a 3:7 ratio, water must be 7/3 times the acid: 21 × 7/3 = 49 litres. Water to add = 49 − 9 = 40 litres. The strategy is to identify the unchanged component, use the desired ratio to find the required quantity of the other component, and subtract what is already present.
Explanation: Let the mixing ratio be x:y. Milk from A = x × 5/6; milk from B = y × 1/4. Water from A = x × 1/6; water from B = y × 3/4. For the final ratio to be 1:1, milk must equal water: 5x/6 + y/4 = x/6 + 3y/4. Rearranging: 5x/6 − x/6 = 3y/4 − y/4 → 4x/6 = 2y/4 → 2x/3 = y/2 → 4x = 3y → x:y = 3:4. In algebraic mixture problems, express each component in terms of the unknown ratio, set up an equation based on the final condition, and solve for the ratio.
Explanation: Let the vessel capacity be V litres. After two replacements of 10 litres each, the remaining milk = V × ((V − 10)/V)². The final ratio is 16:9, so milk constitutes 16/25 of the total. Therefore: V × (V − 10)²/V² = 16V/25. Simplify: (V − 10)²/V = 16V/25 → 25(V − 10)² = 16V² → 5(V − 10) = 4V → 5V − 50 = 4V → V = 50 litres. For problems where the final ratio is known but the capacity is unknown, set up the replacement formula with V as the variable and solve the resulting equation.
Explanation: Current total age = 5 × 24 = 120 years. After 5 years, the 4 remaining members each age by 5 years, adding 20 years to their combined age. The replaced member is exchanged for a 10-year-old child. New total age = 120 + 20 + 10 − (present age of replaced member). The new average is 22, so new total = 5 × 22 = 110. Therefore: 150 − x = 110, giving x = 40 years. In age-replacement problems, account for the aging of the continuing members and equate the new total to the new average multiplied by the number of members.
Explanation: When one person is replaced and the average increases, the new person's age equals the replaced person's age plus (increase in average × number of people). Here, the average increases by 2 years for 8 people, so the total age increases by 8 × 2 = 16 years. The replaced person was 24, so the new person must be 24 + 16 = 40 years old. The general formula is: New person's age = Old person's age + (Change in average × Total number of people). This applies to any replacement problem where the average changes.
Explanation: Original total age = 6 × 16 = 96 years. New total age with 10 boys = 10 × 18 = 180 years. The total age added by the 4 new boys = 180 − 96 = 84 years. Average age of the new boys = 84 ÷ 4 = 21 years. For any group-expansion average problem, compute the old total and the new total from the averages; their difference gives the total contribution of the new members, which you then divide by their count.
Explanation: Three years ago, total age of A, B, C = 3 × 27 = 81 years. Their present total age = 81 + (3 × 3) = 90 years. Five years ago, total age of B and C = 2 × 20 = 40 years. Their present total age = 40 + (2 × 5) = 50 years. A's present age = Present total of (A+B+C) − Present total of (B+C) = 90 − 50 = 40 years. In multi-timeframe age problems, bring all totals to the same time point (present) before subtracting to isolate one person's age.
Explanation: In an arithmetic progression with 4 terms, let the ages be 6, 6+d, 6+2d, 6+3d. Their sum = 24 + 6d. The average is 12, so the total sum = 4 × 12 = 48. Equating: 24 + 6d = 48 → 6d = 24 → d = 4. The oldest brother's age = 6 + 3d = 6 + 12 = 18 years. For AP average problems, remember that the average of an AP equals its middle term (or average of first and last terms). With 4 terms, use the average to find the total sum, then solve for the common difference.
Explanation: Current total runs = 20 × 45 = 900. Desired total after 21 innings at average 48 = 21 × 48 = 1008. Runs needed in the 21st innings = 1008 − 900 = 108. The general approach for 'raising average' problems is: Required score = (New Average × New Count) − (Old Average × Old Count). This gives the exact score needed to achieve the target average.
Explanation: Total after 11 innings = 11 × 40 = 440. After scoring 88 in the 12th innings, total = 440 + 88 = 528. New average after 12 innings = 528 ÷ 12 = 44 (an increase of 4). To raise the average by another 4, the new target average after 13 innings must be 48. Required total after 13 innings = 13 × 48 = 624. Runs needed in the 13th innings = 624 − 528 = 96. For successive average increases, calculate the new total required at each step and subtract the accumulated total so far.
Explanation: Let the old average after 16 innings be x. Total runs after 16 innings = 16x. After scoring 85 in the 17th innings, total = 16x + 85. This equals 17 times the new average, which is (x + 3). So: 16x + 85 = 17(x + 3) = 17x + 51. Solving: 85 − 51 = 17x − 16x → x = 34. New average = x + 3 = 37. For problems where the average increase is given but the old average is unknown, set up an equation with the old average as the variable and solve.
Explanation: Old total = 8 × 65 = 520. New average = 65 − 5 = 60. New total after 9 tests = 9 × 60 = 540. Score in the 9th test = 540 − 520 = 20. The general method is: New Score = (New Average × New Count) − (Old Average × Old Count). This works for any problem where a new value causes the average to change.
Explanation: Total after 10 matches = 10 × 40 = 400. After 11th match, total = 400 + 95 = 495. New average = 495 ÷ 11 = 45. Increase in average = 45 − 40 = 5 runs. For 'how much does average increase' problems, simply compute the new average and subtract the old average. No complex formula is needed if you track the totals carefully.
Explanation: Total marks = 30 × 52 = 1560. Top 5 total = 5 × 80 = 400. Bottom 5 total = 5 × 30 = 150. Remaining 20 students' total = 1560 − 400 − 150 = 1010. Their average = 1010 ÷ 20 = 50.5. For subgroup average problems, convert all averages to totals, subtract the known subgroup totals from the overall total, and divide by the remaining count.
Explanation: After a 5-mark bonus, each of the 20 students has their average increased by 5, so their new total = 20 × (50 + 5) = 20 × 55 = 1100. The 5 new students bring 5 × 60 = 300 marks. Combined total = 1400. Total students = 25. New average = 1400 ÷ 25 = 56. When bonuses are applied uniformly, the average increases by exactly the bonus amount before considering new members. Then add the new members' contribution and recalculate.
Explanation: Total weight of A, B, C = 3 × 45 = 135 kg. Total weight of A and B = 2 × 40 = 80 kg, so C = 135 − 80 = 55 kg. Total weight of B and C = 2 × 43 = 86 kg, so B = 86 − C = 86 − 55 = 31 kg. Alternatively, use the formula: B = (A+B) + (B+C) − (A+B+C) = 80 + 86 − 135 = 31 kg. This formula is useful when you have averages of overlapping pairs and the full group.
Explanation: Total of 12 numbers = 12 × 25 = 300. After discarding 15 and 30, the remaining sum = 300 − 15 − 30 = 255. Remaining count = 10. New average = 255 ÷ 10 = 25.5. For 'discard and recalculate' problems, subtract the discarded values from the original total and divide by the new count. Always verify that you subtract the correct number of values from the count.
Explanation: For any odd number of consecutive odd numbers in arithmetic progression, the average equals the middle term. With 5 numbers and average 31, the middle (3rd) number is 31. The consecutive odd numbers are 27, 29, 31, 33, 35. The largest is 35. In similar problems, the average of an odd count of consecutive odd (or even, or natural) numbers is always the middle term. From the middle term, count forward by 2 for each position to reach the largest.
Explanation: Assume the cost price of each article is Rs 100 for simplicity. Total CP = 10 × 100 = Rs 1000. SP of 4 articles at 20% profit = 4 × 120 = Rs 480. SP of 6 articles at 10% loss = 6 × 90 = Rs 540. Total SP = 480 + 540 = Rs 1020. Profit = 1020 − 1000 = Rs 20. Profit percentage = (20/1000) × 100 = 2%. The standard approach is to assume a convenient CP per unit, calculate total CP and total SP, find the difference, and express it as a percentage of total CP.
Explanation: Total CP = 20 × 15 = Rs 300. SP of 12 pens at 20% profit = 12 × 15 × 1.20 = Rs 216. SP of 8 pens at 10% loss = 8 × 15 × 0.90 = Rs 108. Total SP = 216 + 108 = Rs 324. Profit = 324 − 300 = Rs 24. Profit percentage = (24/300) × 100 = 8%. For mixed profit/loss problems, always calculate total CP and total SP first, then determine the overall percentage from the net difference.
Explanation: Total CP = (3 × 2000) + (2 × 5000) = 6000 + 10000 = Rs 16000. Total SP = 5 × 4000 = Rs 20000. Profit = 20000 − 16000 = Rs 4000. Profit percentage = (4000/16000) × 100 = 25%. In average price problems, never average the averages directly. Calculate the actual total cost price and total selling price, then find the percentage profit or loss on the total investment.
Explanation: Let the cost of one apple be a and one orange be o. From the first condition: 15a + 10o = 25 × 35 = 875. From the second condition: 10a + 15o = 25 × 30 = 750. Add both equations: 25a + 25o = 1625. This is the total cost of 25 apples and 25 oranges. The average cost per fruit = 1625 ÷ 50 = Rs 32.50. For systems with two averages of overlapping groups, adding the equations often reveals the combined total directly.
Explanation: Total of 5 numbers = 5 × 27 = 135. Total of 4 numbers = 4 × 25 = 100. The excluded number = 135 − 100 = 35. The general method is: Excluded Value = (Original Average × Original Count) − (New Average × New Count). This applies to any problem where removing one value changes the average of the remaining values.
Explanation: Total cost = (3 × 60) + (4 × 80) + (3 × 100) = 180 + 320 + 300 = Rs 800. Total quantity = 3 + 4 + 3 = 10 kg. Cost per kg of mixture = 800 ÷ 10 = Rs 80. For three-component mixture problems, the method is the same as for two components: calculate the weighted total cost and divide by the total quantity. The weights are the ratio numbers themselves.
Explanation: Take 1 unit of each mixture. Spirit from A = 3/5; spirit from B = 4/5. Total spirit = 3/5 + 4/5 = 7/5. Water from A = 2/5; water from B = 1/5. Total water = 2/5 + 1/5 = 3/5. Resulting ratio = 7/5 : 3/5 = 7 : 3. When mixing equal quantities, simply add the fractional parts of each component from each mixture, then simplify the ratio. The key is to express each mixture's components as fractions of the whole.
Explanation: Original: milk = 5L, water = 5L. Step 1: Remove 5L of 1:1 mixture (2.5L milk + 2.5L water) and add 5L water. New quantities: milk = 2.5L, water = 7.5L. Step 2: The mixture is now in ratio 2.5:7.5 = 1:3. Remove 5L of this mixture: milk removed = 5 × 1/4 = 1.25L; water removed = 5 × 3/4 = 3.75L. Add 5L milk: milk = 2.5 − 1.25 + 5 = 6.25L; water = 7.5 − 3.75 = 3.75L. Ratio = 6.25 : 3.75 = 25 : 15 = 5 : 3. For multi-step replacement problems, track the quantity of each component after each step, using the current ratio to determine what is removed.
Explanation: Current milk = 60 × 5/6 = 50L; water = 10L. Let x litres of the mixture be replaced. Milk removed = x × 5/6; water removed = x × 1/6. After replacement with x litres of pure water: milk = 50 − 5x/6; water = 10 − x/6 + x = 10 + 5x/6. For 1:1 ratio: 50 − 5x/6 = 10 + 5x/6 → 40 = 10x/6 → x = 24 litres. For replacement problems where the amount to replace is unknown, set up equations for the final quantities of each component and solve for x using the desired ratio condition.
Explanation: Vessel 1 (2:1): milk = 2/3, water = 1/3. Vessel 2 (3:1): milk = 3/4, water = 1/4. Vessel 3 (1:1): milk = 1/2, water = 1/2. Mixed in ratio 1:2:3. Total milk = 1×(2/3) + 2×(3/4) + 3×(1/2) = 2/3 + 3/2 + 3/2 = 2/3 + 3 = 11/3. Total water = 1×(1/3) + 2×(1/4) + 3×(1/2) = 1/3 + 1/2 + 3/2 = 1/3 + 2 = 7/3. Ratio = 11/3 : 7/3 = 11 : 7. For multi-vessel mixing, calculate the contribution of each component from each vessel, weighted by the mixing ratio, then combine and simplify.
Explanation: Total of 10 numbers = 10 × 15 = 150. First 4 total = 4 × 12 = 48. Last 5 total = 5 × 18 = 90. The 5th and 6th numbers together = 150 − 48 − 90 = 12. Given that the 5th number is 5 times the 6th, let the 6th be x. Then 5x + x = 12 → 6x = 12 → x = 2. For problems with an additional condition relating two middle numbers, first isolate their combined total using the overall and subgroup totals, then apply the given relationship to solve.
Explanation: Total CP = (3 × 5000) + (2 × 2500) = 15000 + 5000 = Rs 20000. SP of 1 table at 20% loss = 5000 × 0.80 = Rs 4000. SP of 2 tables at 10% profit each = 2 × 5000 × 1.10 = Rs 11000. Total tables SP = 4000 + 11000 = Rs 15000. SP of 2 chairs at 20% profit each = 2 × 2500 × 1.20 = Rs 6000. Total SP = 15000 + 6000 = Rs 21000. Profit = 21000 − 20000 = Rs 1000. Profit % = (1000/20000) × 100 = 5%. For mixed-item profit/loss problems, calculate total CP and total SP across all items, then find the overall percentage.
Explanation: For 7 consecutive natural numbers, the average equals the middle (4th) number. So the 4th number is 20. The sequence is 17, 18, 19, 20, 21, 22, 23. Product of first and last = 17 × 23. Use the difference of squares pattern: (20 − 3)(20 + 3) = 20² − 3² = 400 − 9 = 391. For consecutive number average problems, the average always reveals the middle term. From there, count backward and forward symmetrically to find the first and last terms.
Explanation: Total weight = 40 × 50 = 2000 kg. Boys total = 8 × 60 = 480 kg. Girls total = 12 × 45 = 540 kg. Remaining students = 40 − 8 − 12 = 20. Their total weight = 2000 − 480 − 540 = 980 kg. Average = 980 ÷ 20 = 49 kg. For multi-subgroup average problems, subtract each known subgroup total from the overall total to isolate the remaining group, then divide by their count.
Explanation: Average speed = Total Distance ÷ Total Time. Let total distance = D. Each one-third is D/3. Time = D/3 × (1/30 + 1/60 + 1/90) = D/3 × (6 + 3 + 2)/180 = D/3 × 11/180 = 11D/540. Average speed = D ÷ (11D/540) = 540/11 ≈ 49.09 km/hr. For equal-distance segments at different speeds, the harmonic approach using total time is the only correct method. The arithmetic mean of the speeds would give an incorrect answer because time spent at each speed differs.
Explanation: When every number in a set is multiplied by a constant, the average is also multiplied by that constant. When a constant is added to every number, the average increases by that constant. Original average = 50. After multiplying by 2: 50 × 2 = 100. After adding 10 to each: 100 + 10 = 110. This linear transformation property applies to any set of numbers: New Average = (Old Average × Multiplier) + Added Constant.
Explanation: Current milk = 45 × 2/3 = 30L; water = 15L. To get a 1:2 ratio in 45L total, milk should be 15L and water 30L. Since we are only removing milk and adding water (keeping total at 45L), we need to remove 15L of milk and add 15L of water. In replacement problems where the total volume is fixed, calculate the target quantity of each component from the desired ratio, then determine the difference from the current quantities to find what must be removed and added.
Explanation: Let capacity = V. After two replacements of 8 litres, remaining milk = V × ((V − 8)/V)². The ratio 9:7 means milk is 9/16 of the total. So V × (V − 8)²/V² = 9V/16. Simplify: (V − 8)²/V = 9V/16 → 16(V − 8)² = 9V² → 4(V − 8) = 3V → 4V − 32 = 3V → V = 32 litres. For unknown-capacity replacement problems, set up the standard replacement formula with V, equate to the known final fraction, and solve the resulting equation. Taking square roots of both sides simplifies the quadratic significantly.
Explanation: If the average of 20 numbers is zero, their sum must be zero. To maximize the count of positive numbers, we need exactly one negative number whose value equals the sum of all the positive numbers (but with opposite sign). Therefore, at most 19 numbers can be positive, with the 20th number being negative enough to cancel their total. The general principle is: if the average is zero, the sum is zero, so for every positive number there must be compensating negative value. The maximum positive count is (Total Count − 1).
Explanation: Original: milk = 100 × 3/5 = 60L; water = 40L. Step 1: Remove 20L of 3:2 mixture (12L milk + 8L water), add 20L milk. New: milk = 60 − 12 + 20 = 68L; water = 40 − 8 = 32L. Step 2: Remove 25L of 68:32 mixture. Milk removed = 25 × 68/100 = 17L; water removed = 25 × 32/100 = 8L. Add 25L water. Final: milk = 68 − 17 = 51L; water = 32 − 8 + 25 = 49L. Ratio = 51 : 49. For multi-step mixture problems, carefully track each component through every replacement, using the current ratio to calculate what is removed at each step.