Question 1

A class has average age 35 years. Six new students with average age 33 years join, decreasing the overall average by 0.5 year. What was the original class strength?

Accepted Answer

Introduction / Context: When adding a subgroup changes the average by a small amount, set up an equation using totals before and after the addition. Unknown original strength can be solved from the equality of the new average. Given Data / Assumptions: Original mean = 35; original size = N (unknown). New group: 6 students with mean = 33. New mean after joining = 35 − 0.5 = 34.5. Concept / Approach: Equation: (35N + 6 * 33) / (N + 6) = 34.5. Solve for N with basic algebra. Step-by-Step Solution: 35N + 198 = 34.5N + 207 35N − 34.5N = 207 − 198 0.5N = 9 ⇒ N = 18 Verification / Alternative check: Totals: original 35 * 18 = 630; new group total 6 * 33 = 198; combined = 828 over 24 people gives 828 / 24 = 34.5, confirming the drop by 0.5 year. Why Other Options Are Wrong: 14, 16, 20: Do not satisfy the new-average equation. Common Pitfalls: Using unweighted averaging of 35 and 33 or misplacing the 0.5 decrease on only one student instead of across the whole class. Final Answer: 18

Question 2

The average age of 5 family members is 25 years. Including the servant increases the average age by 40%. What is the servant's age?

Accepted Answer

Introduction / Context: Average increase by a percentage upon adding one member suggests the new average equals old average times (1 + percentage). Use totals to get the added person’s age precisely. Given Data / Assumptions: 5 members, mean 25 ⇒ total = 125. Including servant: 6 persons; mean increases by 40% ⇒ new mean = 25 * 1.4 = 35. Concept / Approach: New total = new mean * new headcount = 35 * 6. Servant's age = new total − old total. Step-by-Step Solution: New total = 35 * 6 = 210 Old total = 125 Servant's age = 210 − 125 = 85 years Verification / Alternative check: With servant 85, total becomes 210; average 210 / 6 = 35, which is exactly 40% above 25. Why Other Options Are Wrong: 90, 73, 55: Do not produce a new average of 35 when recomputed. Common Pitfalls: Adding 40% of the total instead of 40% of the average, or applying the 40% to only one person's age erroneously. Final Answer: 85 yr

Question 3

Replacing a 25-year-old woman with Priyanka decreases the average age of 30 women by 3 months (0.25 year). What is Priyanka's age?

Accepted Answer

Introduction / Context: Average change due to replacement equals (new member's value − old member's value) divided by group size. A decrease in the average indicates the newcomer is younger than the outgoing member by (drop * group size). Given Data / Assumptions: Group size constant = 30. Outgoing age = 25 years. Average decreases by 0.25 year. Concept / Approach: Total decrease = drop in average * group size = 0.25 * 30. Newcomer's age = outgoing age − total decrease, since the new total is lower by that amount across the group. Step-by-Step Solution: Total decrease = 0.25 * 30 = 7.5 years Priyanka's age = 25 − 7.5 = 17.5 years Verification / Alternative check: If the new average is lower by 0.25, across 30 women the total age must be 7.5 years less; replacing 25 by 17.5 decreases the total by 7.5, exactly as required. Why Other Options Are Wrong: 20, 30, 22: None yields a 7.5-year reduction in total; they would cause different average shifts. Common Pitfalls: Subtracting 0.25 directly from 25 (getting 24.75), forgetting that the 0.25 drop applies across all 30 members. Final Answer: 17.5 yr

Question 4

The mean temperature of Monday–Wednesday is 37°C and of Tuesday–Thursday is 34°C. If Thursday's temperature is 4/5 of Monday's, what is Thursday's temperature?

Accepted Answer

Introduction / Context: Overlapping averages with a ratio relation let us set linear equations. Here, two 3-day means and a link between Monday and Thursday produce solvable equations for the daily values needed. Given Data / Assumptions: (M + T + W) / 3 = 37 ⇒ M + T + W = 111 (T + W + Th) / 3 = 34 ⇒ T + W + Th = 102 Th = (4/5) * M Concept / Approach: Subtract the second sum from the first to relate M and Th, then use the ratio to solve M and hence Th. Step-by-Step Solution: (M + T + W) − (T + W + Th) = 111 − 102 ⇒ M − Th = 9 Given Th = 0.8M ⇒ M − 0.8M = 0.2M = 9 ⇒ M = 45 Th = 0.8 * 45 = 36 Verification / Alternative check: Plug into sums: T + W + 36 = 102 ⇒ T + W = 66; M + T + W = 45 + 66 = 111, consistent with the first equation. Why Other Options Are Wrong: 35.5°C, 34°C, 36.5°C: Do not satisfy both the overlap sums and the 4/5 relation simultaneously. Common Pitfalls: Using 4/5 of Tuesday instead of Monday, or averaging the two means (37 and 34) directly without respecting overlaps. Final Answer: 36°C

Question 5

The average of 11 numbers is 36. The average of 9 of them is 34. If the remaining two numbers are in the ratio 2 : 3, find the smaller number.

Accepted Answer

Introduction / Context: Given averages for the whole and a subset, we can compute the total of the remaining items. With a ratio for those two items, convert the total to parts and extract each number. Given Data / Assumptions: 11-number mean = 36 ⇒ total T = 396. 9-number mean = 34 ⇒ subtotal S = 306. Remaining two are in ratio 2 : 3. Concept / Approach: Sum of the remaining two = T − S. If they are 2k and 3k, then 5k equals that sum; solve k to get both, then choose the smaller (2k). Step-by-Step Solution: Sum of two = 396 − 306 = 90 Let numbers be 2k and 3k ⇒ 5k = 90 ⇒ k = 18 Smaller number = 2k = 36 Verification / Alternative check: Other number = 54; together 36 + 54 = 90. Totals reassemble correctly: 306 + 90 = 396 ⇒ average 36. Why Other Options Are Wrong: 18, 54, 48: Do not match the ratio and total simultaneously. 36 is the unique smaller value. Common Pitfalls: Dividing 90 by 2 (getting 45, 45) and ignoring the 2 : 3 ratio, or picking the larger number by mistake. Final Answer: 36

Question 6

The average of 5 consecutive odd positive integers is 9. What is the least of these integers?

Accepted Answer

Introduction / Context: For any odd-count set of consecutive odd integers, the average equals the middle term. From that center, list the two preceding and two following odds to identify the least value. Given Data / Assumptions: Five consecutive odd positive integers. Average = 9 ⇒ middle term = 9. Concept / Approach: Consecutive odd integers differ by 2. With middle 9, the sequence is 9 − 4, 9 − 2, 9, 9 + 2, 9 + 4. The least is the smallest of that set. Step-by-Step Solution: List: 5, 7, 9, 11, 13 Least value = 5 Verification / Alternative check: Average of 5 and 13 is (5 + 13) / 2 = 9, matching the given mean; the set is symmetric around 9. Why Other Options Are Wrong: 3 and 1 are too small; 7 is not the least in the sequence; “None of these” is unnecessary since 5 is correct. Common Pitfalls: Using even spacing of 1 (integers) rather than spacing of 2 (odd integers), which would misplace endpoints. Final Answer: 5

Question 7

Two classes have average test scores 70 (m students) and 91 (n students). Combined, the average is 80. What is n/m?

Accepted Answer

Introduction / Context: Weighted averages yield linear equations in group sizes when subgroup means and combined mean are known. Solving for the ratio avoids needing absolute headcounts. Given Data / Assumptions: Group 1: mean 70 over m students. Group 2: mean 91 over n students. Combined mean = 80. Concept / Approach: Equation: (70m + 91n) / (m + n) = 80 ⇒ 70m + 91n = 80m + 80n ⇒ 11n = 10m ⇒ n / m = 10 / 11. Step-by-Step Solution: 70m + 91n = 80(m + n) 70m + 91n = 80m + 80n 11n = 10m ⇒ n/m = 10/11 Verification / Alternative check: Choose m = 11, n = 10: combined mean = (70*11 + 91*10) / 21 = (770 + 910) / 21 = 1680 / 21 = 80, confirming the ratio. Why Other Options Are Wrong: 11/10, 13/10, 10/13: Do not satisfy the linear relation derived from the weighted mean 80. Common Pitfalls: Taking an unweighted mean of 70 and 91 or inverting the ratio by mistake. Final Answer: 10/11

Question 8

If 47a + 47b = 5452, what is the average of a and b?

Accepted Answer

Introduction / Context: A linear equation of the form k(a + b) = value immediately reveals a + b when divided by k. The average is then (a + b) / 2. No individual values are required. Given Data / Assumptions: 47a + 47b = 5452 ⇒ 47(a + b) = 5452. Concept / Approach: Compute a + b, then divide by 2 for the mean. This is an application of factoring out common coefficients in linear expressions. Step-by-Step Solution: a + b = 5452 / 47 Since 47 * 100 = 4700 and 47 * 16 = 752, total is 4700 + 752 = 5452 ⇒ a + b = 116 Average = (a + b) / 2 = 116 / 2 = 58 Verification / Alternative check: Plugging back: 47 * 116 = 5452; the arithmetic checks out. Why Other Options Are Wrong: 116 is the sum, not the average; 23.5, 96 are unrelated to the computed values. Common Pitfalls: Reporting a + b instead of the average, or dividing 5452 by 2 first without removing the factor 47. Final Answer: 58

Question 9

The average salary per head of all workers in an institution is ₹60. The average salary per head of 12 officers is ₹400, and the average salary per head of the rest is ₹56. What is the total number of workers?

Accepted Answer

Introduction / Context: This is a weighted-average headcount problem. Knowing subgroup means and the overall mean allows solving for the total size using a single linear equation in N, the total workers. Given Data / Assumptions: Total workers = N (unknown). Officers: 12 with mean ₹400. Others: N − 12 with mean ₹56. Overall mean = ₹60. Concept / Approach: Set (12 * 400 + (N − 12) * 56) / N = 60 and solve for N. Multiply both sides by N to remove the denominator and simplify. Step-by-Step Solution: Total = 4800 + 56(N − 12) = 4800 + 56N − 672 = 56N + 4128 Average equation: (56N + 4128) / N = 60 ⇒ 56N + 4128 = 60N 4128 = 4N ⇒ N = 1032 Verification / Alternative check: Check totals: Officers 12 * 400 = 4800; Others (1032 − 12) * 56 = 1020 * 56 = 57120; Grand total = 61920; Average = 61920 / 1032 = 60. Verified. Why Other Options Are Wrong: 1030, 1060, 1062: Do not satisfy the linear relation required by the overall mean of ₹60. Common Pitfalls: Using an unweighted mean of 400 and 56 or forgetting to include the officers' count separately from the others. Final Answer: 1032

Question 10

Average price across mixed purchases: Roshan buys 5 pants at ₹25 each, 10 shirts at ₹50 each, and 15 ties at ₹35 each. Compute the overall average price per article for all 30 items purchased.

Accepted Answer

Introduction / Context: This problem checks weighted average with unequal counts at different prices. We combine total cost over all items and divide by the total number of articles to obtain a single per-article average. Given Data / Assumptions: Pants: 5 pieces at ₹25 each Shirts: 10 pieces at ₹50 each Ties: 15 pieces at ₹35 each Total number of articles = 5 + 10 + 15 = 30 Concept / Approach: Average price per article = total expenditure / total number of articles. Since quantities differ by item type, compute each subtotal and add them before dividing by 30. Step-by-Step Solution: Cost of pants = 5 * 25 = ₹125Cost of shirts = 10 * 50 = ₹500Cost of ties = 15 * 35 = ₹525Total cost = 125 + 500 + 525 = ₹1150Average price = 1150 / 30 = ₹38.333… ≈ ₹38.33 Verification / Alternative check: Use a calculator to confirm 1150 divided by 30. Rounding to two decimals yields ₹38.33, which is consistent with currency presentation. Why Other Options Are Wrong: ₹45.33, ₹42.00, ₹40.00, ₹45.00: do not equal the precise ratio of total cost to item count (1150/30). Common Pitfalls: Adding the three prices and dividing by 3 ignores the different quantities. Always weight by quantity purchased. Final Answer: ₹ 38.33

Question 11

Weighted class average: A class has 15 boys and 10 girls with an overall average weight of 38.4 kg. If the boys average 40 kg, find the average weight of the girls.

Accepted Answer

Introduction / Context: The task is a direct application of weighted averages. Knowing the overall mean and one subgroup mean, we can determine the other subgroup mean by working with totals (mean * count). Given Data / Assumptions: Boys = 15, girls = 10 Overall average = 38.4 kg Boys average = 40 kg Concept / Approach: Total weight of all = overall average * total students. Total of boys = boys average * number of boys. Subtract to get girls’ total, then divide by number of girls for girls’ mean. Step-by-Step Solution: Total students = 15 + 10 = 25Overall total = 25 * 38.4 = 960 kgBoys total = 15 * 40 = 600 kgGirls total = 960 - 600 = 360 kgGirls average = 360 / 10 = 36 kg Verification / Alternative check: Check weighted recombination: (600 + 360) / 25 = 960 / 25 = 38.4 kg, matching the given overall average. Why Other Options Are Wrong: 36.5, 35.6, 35, 37 kg: do not satisfy the weighted total when recomputed. Common Pitfalls: Taking an unweighted mean of 40 and an assumed girls mean without weighting by group sizes. Final Answer: 36 kg

Question 12

Pass–fail split via averages: The average of 120 students is 35. If those who passed averaged 39 and those who failed averaged 15, how many students passed?

Accepted Answer

Introduction / Context: This classic mixture problem uses subgroup averages and the overall average to find subgroup sizes. Convert averages to totals and form a simple linear equation in the number of passed students. Given Data / Assumptions: Total students = 120 Overall average = 35 Average (passed) = 39 Average (failed) = 15 Concept / Approach: Let the number of students who passed be x. Then total marks = 39x + 15(120 - x). Equate this to the overall total 120 * 35 and solve for x. Step-by-Step Solution: Overall total = 120 * 35 = 4200Equation: 39x + 15(120 - x) = 420039x + 1800 - 15x = 4200 ⇒ 24x = 2400x = 100 Verification / Alternative check: Compute totals: passed = 100 * 39 = 3900; failed = 20 * 15 = 300; sum = 4200, which matches the overall requirement. Why Other Options Are Wrong: 80, 92, 88, 96: do not satisfy the balance of subgroup totals to 4200. Common Pitfalls: Forgetting that the failed group size is 120 - x or mis-multiplying the subgroup totals. Final Answer: 100

Question 13

Average speed for equal distances: A car goes from A to B at 42 km/h and returns from B to A at 48 km/h. Find the overall average speed for the round trip (equal distances each way).

Accepted Answer

Introduction / Context: For a round trip over equal distances with different speeds, the overall average speed is the harmonic mean of the two speeds, not the arithmetic mean. This accounts for the greater time spent at the slower speed. Given Data / Assumptions: Outbound speed = 42 km/h Return speed = 48 km/h Equal distances each way Concept / Approach: Average speed for equal distances = 2ab / (a + b). This is derived from total distance divided by total time: time = d/a + d/b; distance = 2d. Step-by-Step Solution: a = 42, b = 48Average = 2 * 42 * 48 / (42 + 48)Average = 4032 / 90 = 44.8 km/h Verification / Alternative check: Assume 90 km each way: time = 90/42 + 90/48 = 2.142857… + 1.875 = 4.017857… h. Total distance = 180 km → 180 / 4.017857… ≈ 44.8 km/h. Why Other Options Are Wrong: 45.0, 44.0, 46.0, 48.0: approximations or arithmetic means that ignore correct time weighting. Common Pitfalls: Using the arithmetic mean (42 + 48)/2 = 45, which is incorrect for equal distances at different speeds. Final Answer: 44.8 km/h

Question 14

Four positive integers have average 73.5. The highest is 108 and the lowest is 29. The remaining two differ by 15. Find the smaller of those two remaining integers.

Accepted Answer

Introduction / Context: This question combines an average-to-total conversion with constraints on two unknown integers. Setting variables for the two middle integers converts the text into a simple linear equation. Given Data / Assumptions: Average of 4 integers = 73.5 Highest = 108 Lowest = 29 Difference between the remaining two = 15 Concept / Approach: Let the remaining two numbers be x and x + 15. Use the total from the average and subtract the known extremes to isolate x. Step-by-Step Solution: Total of all four = 4 * 73.5 = 294Known extremes = 108 and 29 → sum = 137Equation: 137 + x + (x + 15) = 2942x + 152 = 294 ⇒ 2x = 142 ⇒ x = 71 Verification / Alternative check: Numbers are 29, 71, 86, 108. Average = (29 + 71 + 86 + 108) / 4 = 294 / 4 = 73.5; difference between middle pair = 15, all conditions met. Why Other Options Are Wrong: 73, 80, 86: do not satisfy the total and the exact difference simultaneously. Cannot be determined: the information is sufficient to solve uniquely. Common Pitfalls: Forgetting to include the +15 when forming the equation for the two unknowns or miscalculating the total from the average. Final Answer: 71

Question 15

Batting average boost: A cricketer has some average after 9 innings. He scores 100 in the 10th innings and his average increases by 8. What is his average after the 10th innings?

Accepted Answer

Introduction / Context: Average increase problems can be solved algebraically by expressing totals before and after the new data point. The single new score is the difference between the new total and the old total. Given Data / Assumptions: Score in 10th innings = 100 Average increases by 8 runs Concept / Approach: Let A be the new average after 10 innings. Then previous average was A - 8. Relate totals across 9 and 10 innings and equate the 10th score to the difference in totals. Step-by-Step Solution: Total after 10 = 10ATotal after 9 = 9(A - 8) = 9A - 7210th score = 10A - (9A - 72) = A + 72Set A + 72 = 100 ⇒ A = 28 Verification / Alternative check: If A = 28, previous average = 20. Totals: 10 * 28 = 280; 9 * 20 = 180; difference = 100 for the 10th, as given. Why Other Options Are Wrong: 20, 24, 26, 32: do not satisfy A + 72 = 100. Common Pitfalls: Setting the increase relative to the old total incorrectly, or assuming the increase applies to just one innings rather than the cumulative average. Final Answer: 28

Question 16

Family average unchanged after addition: Three years ago, the average age of a 5-member family was 27 years. Today, after adding a child, the average is still 27 years. Find the present age of the child.

Accepted Answer

Introduction / Context: Average age problems over time can be handled using totals. Every existing member’s age increases by the elapsed years, and adding a new member increases count and total simultaneously. Given Data / Assumptions: 3 years ago: 5 members, average = 27 Today: 6 members (child added), average = 27 Concept / Approach: Compute the old total from the past average, add the 3-year growth for the original five to get their total today, then compare with the current grand total to get the child’s age now. Step-by-Step Solution: Past total (5 members) = 5 * 27 = 135Growth over 3 years for 5 members = 5 * 3 = 15Total of original five today = 135 + 15 = 150Current grand total = 6 * 27 = 162Child’s present age = 162 - 150 = 12 years Verification / Alternative check: The unchanged average implies the child’s age offsets the total 3-year increase across the five members (15 years) plus keeps the mean at 27 for six members, which is satisfied by 12 years. Why Other Options Are Wrong: 16, 24, 20, 10: do not yield the required overall total of 162. Common Pitfalls: Multiplying the average change by the new count or forgetting to add the 3-year increment for all original five members. Final Answer: 12 yr

Question 17

Proportional ages with known mean: The average age of three boys is 15 years and their ages are in the ratio 3 : 5 : 7. Find the age of the oldest boy.

Accepted Answer

Introduction / Context: When ages are proportional and the average is given, scale the ratio so the sum matches the total from the average. The largest part of the ratio yields the oldest age directly after scaling. Given Data / Assumptions: Average of 3 = 15 → total = 45 Age ratio = 3 : 5 : 7 Concept / Approach: Let the common factor be k. Then ages are 3k, 5k, 7k and 3k + 5k + 7k = 15k must equal 45. Solve for k, then multiply to get the oldest age (7k). Step-by-Step Solution: 15k = 45 ⇒ k = 3Oldest age = 7k = 7 * 3 = 21 years Verification / Alternative check: Ages are 9, 15, 21 with average (9 + 15 + 21)/3 = 45/3 = 15, satisfying all conditions. Why Other Options Are Wrong: 7, 14, 20, 24: do not match the scaled 7k value with k = 3. Common Pitfalls: Using the mean directly as one of the ages or adding ratio numbers incorrectly (sum must be 15 parts). Final Answer: 21 yr

Question 18

Self-referential average spending: Nine friends dine together. Eight spend ₹12 each. The ninth spends ₹16 more than the average expenditure of all nine. Find the total money spent.

Accepted Answer

Introduction / Context: This problem defines one person’s spend in terms of the unknown overall average, producing an equation in that average. Solve for the average first, then get the total and check consistency. Given Data / Assumptions: Eight people: ₹12 each Ninth person: average of all nine + ₹16 Concept / Approach: Let the average be x. Then total = 9x. Known spend = 8 * 12 + (x + 16). Set equal to 9x and solve for x, then multiply back to get the total amount. Step-by-Step Solution: Equation: 8*12 + (x + 16) = 9x96 + x + 16 = 9x ⇒ 112 + x = 9x8x = 112 ⇒ x = 14Total money = 9 * 14 = ₹126 Verification / Alternative check: Ninth person’s spend = x + 16 = 14 + 16 = ₹30; eight others = ₹96; total = ₹126. Average 126/9 = 14 matches x. Why Other Options Are Wrong: ₹135, ₹111, ₹141, ₹129: do not equal 9 * 14 under the self-referential condition. Common Pitfalls: Assuming the ninth spent ₹16 more than the others rather than more than the average, which changes the equation structure. Final Answer: ₹ 126

Question 19

Average per paper shift: An examinee averages 64 per paper. If he had scored 18 more in Mathematics and 4 more in English, his average would be 66. How many papers were there?

Accepted Answer

Introduction / Context: Average increases caused by a known total increment distribute over the number of papers. The increment per paper equals (increase in total marks) / (number of papers). Given Data / Assumptions: Original average = 64 New average (if extra marks added) = 66 Extra marks: 18 (Math) + 4 (English) = 22 Concept / Approach: If n is the number of papers, then adding 22 to the total raises the average by 22/n. This must equal 2 (from 64 to 66), allowing a direct solve for n. Step-by-Step Solution: 22 / n = 2n = 22 / 2 = 11 Verification / Alternative check: Original total = 64n. After adding 22, total becomes 64n + 22. New average = (64n + 22)/n = 64 + 22/n = 66 ⇒ n = 11, consistent. Why Other Options Are Wrong: 9, 13, 15, 10: do not satisfy the condition 22/n = 2. Common Pitfalls: Confusing the number of affected papers with total papers; the average shift is spread over all papers, not just the two affected subjects. Final Answer: 11

Question 20

Pass–fail composition: The average of 120 students is 30. Passed students average 40 and failed students average 10. What is 25% of the number of passed students?

Accepted Answer

Introduction / Context: Convert overall and subgroup averages to totals to find the number who passed. Once the passed count is known, take one quarter (25%) as requested. Given Data / Assumptions: Total students = 120 Overall average = 30 → total = 3600 Passed average = 40 Failed average = 10 Concept / Approach: Let x be the number who passed. Then 40x + 10(120 - x) = 3600. Solve for x, then compute 0.25x. Step-by-Step Solution: 40x + 1200 - 10x = 360030x = 2400 ⇒ x = 8025% of passed = 0.25 * 80 = 20 Verification / Alternative check: Totals: passed = 80 * 40 = 3200; failed = 40 * 10 = 400; sum = 3600, consistent with the overall total. Why Other Options Are Wrong: 30, 40, 15, 25: not equal to one quarter of 80. Common Pitfalls: Using an unweighted average or forgetting to subtract x when computing the failed count 120 - x. Final Answer: 20

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