Question 1

The average age of a 5-member committee today is the same as it was 3 years ago because one old member was replaced by a new member. What is the difference between the ages of the old and the new member?

Accepted Answer

Introduction: If a group’s average remains unchanged after a certain number of years despite everyone aging, then the replacement must have offset exactly the total aging of the group. This insight allows a quick computation of the difference between the outgoing and incoming member’s ages. Given Data / Assumptions: Committee size remains 5 Average now equals average 3 years ago One old member replaced by one new member Concept / Approach: If no one were replaced, the total age would increase by 5 * 3 = 15 years over 3 years. Since the average (and thus total) is unchanged, the incoming member must be younger than the outgoing member by exactly 15 years to cancel out this increase. Step-by-Step Solution: Total aging over 3 years for 5 members = 15 years Replacement effect must subtract 15 years from the total Age(old) − Age(new) = 15 years Verification / Alternative check: Model totals numerically: if old total was S, after 3 years without replacement it would be S + 15. With replacement, it returns to S, implying a 15-year reduction due to replacement, i.e., the age gap is 15. Why Other Options Are Wrong: 2, 4, 8, and 10 years would not offset the full 15-year increase across five members over three years. Common Pitfalls: Multiplying by 3 only once or forgetting to apply it to all 5 members. The average constraint fixes the exact difference. Final Answer: 15 years

Question 2

In a cricket team, the average age of 11 players decreases by 2 months when two players aged 17 years and 20 years are replaced by two reserves. What is the average age of the two reserves?

Accepted Answer

Introduction: This average-age replacement problem asks you to translate a change in average (2 months) into a change in total age for the whole team, and then use that to determine the combined and average ages of the two incoming reserve players. Given Data / Assumptions: Team size = 11 players. Two outgoing ages: 17 years and 20 years. Average decreases by 2 months = 2/12 years = 1/6 year. Two incoming reserves replace the outgoing players. Concept / Approach: A change in average by Δ across n people changes the total by n * Δ. Here Δ = −1/6 year. So total decreases by 11 * (1/6) = 11/6 years. That total decrease must equal (sum of reserves) − (sum of outgoing). Solve for the reserves’ total, then divide by 2 to get their average. Step-by-Step Solution: Decrease in total = 11 * (−1/6) = −11/6 years. Let S_res be sum of reserves; outgoing sum = 17 + 20 = 37. S_res − 37 = −11/6 ⇒ S_res = 37 − 11/6 = 211/6 years. Average of reserves = (211/6) / 2 = 211/12 = 17 + 7/12 years. 7/12 year = 7 months → 17 years 7 months. Verification / Alternative check: If each player’s average drops by 2 months, the group total must drop by 11 * 2 months = 22 months = 1 year 10 months, exactly the difference between outgoing and incoming totals computed above. Why Other Options Are Wrong: 17 years 1 month, 17 years 11 months, 18 years 3 months, and 16 years 9 months do not produce the exact 22-month reduction in total age when compared to the outgoing pair (17 and 20). Common Pitfalls: Forgetting to multiply the average change by the team size, or averaging the two outgoing ages directly with the 2-month change instead of working with totals. Final Answer: 17 years 7 months

Question 3

Five years ago, the average age of Ram and Shyam was 20 years. Now, the average age of Ram, Shyam, and Mohan is 30 years. What will be Mohan’s age 10 years hence?

Accepted Answer

Introduction: Family or group age problems are most cleanly handled by converting averages to totals and then tracking how totals change over time as members age or new members are added. Given Data / Assumptions: 5 years ago: average of (Ram, Shyam) = 20 ⇒ past total = 40. Now: average of (Ram, Shyam, Mohan) = 30 ⇒ current total = 90. Aging is linear: each of Ram and Shyam gained 5 years since the earlier snapshot. Concept / Approach: Convert both averages to totals and reconcile. Compute Ram + Shyam now, then subtract from the current 3-person total to isolate Mohan’s present age. Add 10 to project 10 years ahead. Step-by-Step Solution: Ram + Shyam (then) = 40; (now) = 40 + 5 + 5 = 50. Current (Ram + Shyam + Mohan) = 90. Mohan (now) = 90 − 50 = 40 years. Mohan in 10 years = 40 + 10 = 50 years. Verification / Alternative check: Present averages: (Ram + Shyam)/2 is not required; totals are sufficient. All values remain consistent with the provided averages after back-substitution. Why Other Options Are Wrong: 45, 49, and 60 years do not follow from Mohan’s present age of 40. 40 years is Mohan’s current age, not his age 10 years later. Common Pitfalls: Averaging the given averages or forgetting that both Ram and Shyam aged for 5 years each, i.e., +10 total years for the pair. Final Answer: 50 years

Question 4

The average of A and B is 20. If C replaces A, the average becomes 19; if C replaces B, the average becomes 21. What are the ages of A, B, and C?

Accepted Answer

Introduction: This system uses pairwise averages that translate directly into pairwise sums. Solving the resulting simultaneous linear equations yields unique values for A, B, and C. Given Data / Assumptions: (A + B)/2 = 20 ⇒ A + B = 40. (C + B)/2 = 19 ⇒ B + C = 38. (A + C)/2 = 21 ⇒ A + C = 42. Concept / Approach: Use elimination by subtracting equations to find differences, then combine with a known sum. Specifically, subtract (B + C) from (A + C) to get A − B, then combine with A + B. Step-by-Step Solution: (A + C) − (B + C) = 42 − 38 ⇒ A − B = 4. With A + B = 40 ⇒ 2A = 44 ⇒ A = 22. Then B = 40 − 22 = 18; C from A + C = 42 ⇒ C = 20. Verification / Alternative check: Check: (22 + 18)/2 = 20; (20 + 18)/2 = 19; (22 + 20)/2 = 21. All conditions satisfied. Why Other Options Are Wrong: The other permutations break at least one of the three pairwise average constraints on substitution. Common Pitfalls: Mixing which person is replaced in each scenario or averaging the given averages instead of forming equations on sums. Final Answer: 22, 18, 20

Question 5

What is the average of all even numbers between 11 and 63 (inclusive of the endpoints if even)?

Accepted Answer

Introduction: The average of an arithmetic sequence equals the average of its first and last terms. For even numbers in a range, identify the first even and last even within the interval and take their mean. Given Data / Assumptions: Even numbers from 12 to 62 are within 11 to 63. Sequence: 12, 14, 16, ..., 62. Concept / Approach: For any arithmetic progression with constant difference, mean = (first + last) / 2, independent of the number of terms. This avoids summing the entire list. Step-by-Step Solution: First even ≥ 11 is 12; last even ≤ 63 is 62. Average = (12 + 62) / 2 = 74 / 2 = 37. Verification / Alternative check: The sequence is symmetric around 37: pairs (12,62), (14,60), … all average to 37, confirming the mean. Why Other Options Are Wrong: 37.5 or 42 would be the mean if endpoints were different; 47 and 36 do not match the symmetric pairing check. Common Pitfalls: Including 11 or 63 erroneously (both are odd) or trying to count the total terms unnecessarily. Final Answer: 37

Question 6

What is the average of the cubes of the first five natural numbers?

Accepted Answer

Introduction: You are asked to compute the mean value of 1^3, 2^3, 3^3, 4^3, and 5^3. This can be done either by direct calculation and averaging or by recalling the identity for the sum of cubes. We will use explicit values to keep it transparent. Given Data / Assumptions: First five natural numbers: 1, 2, 3, 4, 5. Cubes: 1, 8, 27, 64, 125. Concept / Approach: Average = (sum of values) / (count). Compute the five cubes, add them, and divide by 5. This is straightforward and avoids memorizing formulas. Step-by-Step Solution: Sum = 1 + 8 + 27 + 64 + 125 = 225. Average = 225 / 5 = 45. Verification / Alternative check: Using the identity (1 + 2 + 3 + 4 + 5)^2 = (sum of cubes) for consecutive starting at 1, we have (15)^2 = 225, confirming the sum before division by 5. Why Other Options Are Wrong: 55, 65, 35, and 50 do not match the computed mean from the explicit cube values. Common Pitfalls: Confusing squares and cubes, or taking an unweighted average by mistake. Ensure all five cubes are included. Final Answer: 45

Question 7

The average of 8 numbers is 14. The average of 6 of these numbers is 16. What is the average of the remaining two numbers?

Accepted Answer

Introduction: This is a sum-partition average problem. Converting averages to totals lets you subtract the known subset and then average what remains. It is a standard technique for decomposing composite averages. Given Data / Assumptions: Overall: 8 numbers, average 14 ⇒ total = 8 * 14 = 112. Subset: 6 numbers, average 16 ⇒ subset total = 6 * 16 = 96. Concept / Approach: The remaining two have total = overall total − subset total. Divide by 2 to get their average. This avoids hunting individual values and relies only on aggregate information. Step-by-Step Solution: Remaining total = 112 − 96 = 16. Average of remaining two = 16 / 2 = 8. Verification / Alternative check: If you construct a concrete set matching these sums, the two leftover values will always average to 8, regardless of their individual spread, since the aggregate constraints are fixed. Why Other Options Are Wrong: 12, 6, 10, and 14 conflict with the fixed totals (112 overall and 96 for the six). Only 8 divides the remainder 16 evenly for two numbers. Common Pitfalls: Averaging 14 and 16 directly (which would be meaningless here) or dividing the difference of averages by 2 without using totals. Final Answer: 8

Question 8

The average age of 5 boys is 16 years, while the average age of 4 of those boys is 16 years 3 months. What is the age of the 5th boy?

Accepted Answer

Introduction: Mixed averages can be separated by converting them into totals. Once you know the total age of all 5 and the total age of 4, the 5th boy’s age is simply the difference of those totals. Given Data / Assumptions: Average (5 boys) = 16 years ⇒ total_5 = 5 * 16 = 80. Average (4 boys) = 16 years 3 months = 16.25 years ⇒ total_4 = 4 * 16.25 = 65. Concept / Approach: Age of 5th = total_5 − total_4. The quarter-year conversion (3 months = 0.25 years) ensures accurate computation without mixing units. Step-by-Step Solution: total_5 = 80; total_4 = 65. 5th boy’s age = 80 − 65 = 15 years. Verification / Alternative check: Recombine: 65 + 15 = 80; 80 / 5 = 16 years, matching the stated average. All units are consistent after converting months to fractional years. Why Other Options Are Wrong: 15 yr 6 months, 15 yr 4 months, and 15 yr 2 months would change the 5-boy total away from 80. 16 yr contradicts the computed difference of totals. Common Pitfalls: Forgetting to convert 3 months to 0.25 year, or averaging 16 and 16 years 3 months directly rather than using totals. Final Answer: 15 yr

Question 9

Eight consecutive integers are given. If the average of the two middle numbers is 6, what is the sum of all eight numbers?

Accepted Answer

Introduction: For any set of evenly spaced numbers, the overall average equals the average of the two middle terms (when the count is even). Once you know the average of the two central numbers, you immediately know the mean of the entire set and hence the sum. Given Data / Assumptions: Eight consecutive integers. Average of the two middle integers = 6. Concept / Approach: For 8 consecutive integers, the two middle positions are the 4th and 5th. Their average equals the overall average. Therefore, the sum of all eight numbers equals (average) * (count) = 6 * 8. Step-by-Step Solution: Overall average = average of middle two = 6. Sum of 8 numbers = 6 * 8 = 48. Verification / Alternative check: Example: integers 3,4,5,6,7,8,9,10 have middle pair (6,7) with average 6.5; adjust down by 0.5 to center at 6 gives sum 48, confirming the method conceptually by symmetry. Why Other Options Are Wrong: 36, 54, and 64 correspond to averages 4.5, 6.75, and 8, not 6; 40 implies average 5, contradicting the given average of the middle pair. Common Pitfalls: Attempting to list all eight numbers explicitly or misidentifying the middle positions; recognizing symmetry saves time and avoids errors. Final Answer: 48

Question 10

Four consecutive even integers have an average of 27. What is the greatest of these four even numbers?

Accepted Answer

Introduction / Context: This problem checks understanding of averages and evenly spaced numbers. For consecutive even integers, values are equally spaced by 2, so the mean equals the midpoint of the set, and endpoints can be rebuilt from that midpoint systematically. Given Data / Assumptions: Four consecutive even integers. Average (arithmetic mean) = 27. We must find the greatest of the four. Concept / Approach: If the smallest even integer is x, the set is x, x + 2, x + 4, x + 6. The average of equally spaced terms equals the average of the first and last term, or equivalently x + 3. Setting x + 3 = 27 gives x directly; the greatest term is x + 6. Step-by-Step Solution: Let smallest = x ⇒ numbers: x, x + 2, x + 4, x + 6 Average = (x + (x + 2) + (x + 4) + (x + 6)) / 4 = (4x + 12) / 4 = x + 3 Set x + 3 = 27 ⇒ x = 24 Greatest number = x + 6 = 24 + 6 = 30 Verification / Alternative check: Numbers are 24, 26, 28, 30. Their mean is (24 + 30)/2 = 27, consistent with the given average. Why Other Options Are Wrong: 28, 26, 32: Do not match the computed greatest value. None of these: A correct option (30) exists. Common Pitfalls: Using odd increments or treating 27 as a member of the set (it cannot be, since all terms are even). Final Answer: 30

Question 11

A businessman bought 30 TV sets for ₹300000 in total. The average cost of 15 TV sets is ₹9000. What is the average cost of the remaining 15 TV sets?

Accepted Answer

Introduction / Context: Averages with grouped data often need a total-first approach: compute the grand total, subtract the known subgroup total, and divide the remainder by the remaining count to get the second subgroup’s average. Given Data / Assumptions: Total sets = 30; total cost = ₹300000. Average cost of first 15 sets = ₹9000 ⇒ subtotal known. Find average cost of remaining 15 sets. Concept / Approach: Let S1 be the cost of the first 15 and S2 the cost of the remaining 15. Then S1 = 15 * 9000. Since S1 + S2 = 300000, we get S2, and average for remaining = S2 / 15. Step-by-Step Solution: S1 = 15 * 9000 = ₹135000 S2 = ₹300000 − ₹135000 = ₹165000 Average (remaining 15) = ₹165000 / 15 = ₹11000 Verification / Alternative check: Grand average is ₹300000 / 30 = ₹10000. With one subgroup below at ₹9000, the other must be above ₹10000 to balance to ₹10000 overall; ₹11000 satisfies this symmetry. Why Other Options Are Wrong: ₹20000, ₹12000: Do not align with the computed remainder. ₹10000: That is the overall average, not the remaining subgroup’s mean. None of these: A correct value exists (₹11000). Common Pitfalls: Dividing the total by 15 or mixing up subgroup and overall averages. Final Answer: ₹ 11000

Question 12

The average age of 30 girls is 13 years. The average age of the first 18 girls is 15 years. What is the average age of the remaining 12 girls?

Accepted Answer

Introduction / Context: When averages of a whole group and a subgroup are given, the complementary subgroup’s average can be derived by working with totals (sum = average * count). This is a staple aptitude technique. Given Data / Assumptions: Total girls = 30; average = 13 years. First 18 girls’ average = 15 years. Find average of remaining 12 girls. Concept / Approach: Let T be total age of 30 girls, T1 be total of first 18 girls, and T2 be total of the remaining 12. Then T = 30 * 13, T1 = 18 * 15, and T2 = T − T1. The required average = T2 / 12. Step-by-Step Solution: T = 30 * 13 = 390 T1 = 18 * 15 = 270 T2 = 390 − 270 = 120 Average (remaining 12) = 120 / 12 = 10 years Verification / Alternative check: The overall mean 13 lies between subgroup means 15 and 10, which is consistent with weighted averages (heavier weight on the 18 with mean 15 pulls the total above 11–12). Why Other Options Are Wrong: 12yr, 16yr, 10.5 yr.: Do not match the derived total. None of these: Correct option exists (10yr). Common Pitfalls: Dividing 390 − 270 by 30 instead of 12, or averaging 13 and 15 directly without weights. Final Answer: 10yr

Question 13

The average of four consecutive integers A, B, C, and D is 49.5. Find the product B × D.

Accepted Answer

Introduction / Context: Consecutive integers form an arithmetic progression with common difference 1. The mean equals the midpoint. Once one term is known, the others follow directly, enabling computation of any requested combination such as a product. Given Data / Assumptions: Four consecutive integers A, B, C, D. Average = 49.5. Find B × D. Concept / Approach: Let A be the smallest; then A, A + 1, A + 2, A + 3 are the four numbers. Their mean is A + 1.5 = 49.5, which gives A immediately. Then B = A + 1, D = A + 3, and compute the product B * D conveniently using the identity (n − 1)(n + 1) = n^2 − 1 when centered around n = A + 2 = 50. Step-by-Step Solution: A + 1.5 = 49.5 ⇒ A = 48 Thus B = 49, C = 50, D = 51 Compute B * D = 49 * 51 = (50 − 1)(50 + 1) = 50^2 − 1 = 2500 − 1 = 2499 Verification / Alternative check: Direct multiplication 49 * 51 also yields 2499; both routes agree. Why Other Options Are Wrong: 2352, 2450, 2550: None matches the exact product; each reflects arithmetic slips or different pairs multiplied. Common Pitfalls: Taking 49.5 as a member of the integer set, or pairing B with C instead of D. Final Answer: 2499

Question 14

Team A's average height is 5 ft 11 in and Team B's is 6 ft 2 in. There are 20 players in A and 18 players in B. What is the overall average height (in inches)?

Accepted Answer

Introduction / Context: When combining groups with different averages and sizes, use weighted averages: convert all measures to a common unit, multiply each group’s average by its headcount, add, then divide by the total headcount. Given Data / Assumptions: Team A: 20 players, average 5 ft 11 in = 71 inches. Team B: 18 players, average 6 ft 2 in = 74 inches. Find combined average in inches. Concept / Approach: Overall average = (sum of all individual heights) / (total players). Equivalently, (average_A * count_A + average_B * count_B) / (count_A + count_B). Step-by-Step Solution: Convert to inches: 5 ft 11 in = 60 + 11 = 71; 6 ft 2 in = 72 + 2 = 74 Total inches (A) = 71 * 20 = 1420 Total inches (B) = 74 * 18 = 1332 Grand total = 1420 + 1332 = 2752 Total players = 20 + 18 = 38 Combined average = 2752 / 38 ≈ 72.421... ≈ 72.42 inches Verification / Alternative check: The result must lie between 71 and 74 and lean toward 71–74 weighted by 20 vs 18; 72.42 is consistent and closer to 71 because A has slightly more players. Why Other Options Are Wrong: 72, 70, 70.22 inches: Do not match weighted computation. None of these: A precise correct value is available. Common Pitfalls: Using an unweighted mean of 71 and 74 (which would be 72.5) or mixing feet and inches without converting. Final Answer: 72.42 inches

Question 15

The average of 13 results is 60. The average of the first 7 results is 59, and the average of the last 7 results is 61. What is the seventh result?

Accepted Answer

Introduction / Context: This classic overlap-average problem uses totals with an overlapping term. The 7th result belongs to both the "first 7" and the "last 7", allowing a direct equation to isolate it. Given Data / Assumptions: Average of 13 = 60 ⇒ total T = 13 * 60. Average of first 7 = 59 ⇒ total T1 = 7 * 59. Average of last 7 = 61 ⇒ total T2 = 7 * 61. Only the 7th result overlaps between these two groups. Concept / Approach: Since the 7th result x is counted in both T1 and T2, but only once in T, we have T = T1 + T2 − x. Solve for x to obtain the overlapped value. Step-by-Step Solution: T = 13 * 60 = 780 T1 = 7 * 59 = 413 T2 = 7 * 61 = 427 780 = 413 + 427 − x ⇒ x = 413 + 427 − 780 = 60 Verification / Alternative check: Using the equation T1 − x covers the first 6, T2 − x covers the last 6; adding those plus x and x reconstructs all 13, confirming the overlap logic. Why Other Options Are Wrong: 90, 50, 75: Do not satisfy the totals relationship; they would distort one or both subgroup means. Common Pitfalls: Adding T1 and T2 without subtracting x once, which double-counts the 7th value. Final Answer: 60

Question 16

Average weight of 19 men is 74 kg and average weight of 38 women is 63 kg. Find the average weight (rounded to the nearest integer) of all 57 people together.

Accepted Answer

Introduction / Context: Weighted averages combine groups with different sizes and means. Always multiply each group’s average by its count to get totals, then divide by the grand count and round if asked. Given Data / Assumptions: Men: 19 with mean 74 kg. Women: 38 with mean 63 kg. Round the final mean to the nearest integer. Concept / Approach: Overall average = (total men's weight + total women's weight) / total people. Rounding applies only after exact computation. Step-by-Step Solution: Men's total = 19 * 74 = 1406 Women's total = 38 * 63 = 2394 Grand total = 1406 + 2394 = 3800 Headcount = 19 + 38 = 57 Average = 3800 / 57 ≈ 66.666... ⇒ nearest integer = 67 kg Verification / Alternative check: The average must lie between 63 and 74 and bias closer to 63 since women are double the men; 66–67 is plausible, and exact rounding gives 67. Why Other Options Are Wrong: 59, 65, 69: Do not match the precise weighted result and rounding rule. "None of these" is unnecessary because a correct choice exists. Common Pitfalls: Taking an unweighted mean of 74 and 63 or rounding too early before dividing. Final Answer: 67 kg

Question 17

A cricketer has a mean score of 60 runs in 10 innings. How many runs must be scored in the 11th inning to raise the overall mean to 62?

Accepted Answer

Introduction / Context: Average-raising problems hinge on total runs needed for the new mean over the new count. Compute the new required total, subtract the current total, and the difference is the runs needed in the next inning. Given Data / Assumptions: Current mean = 60 over 10 innings. Target mean = 62 over 11 innings. Concept / Approach: Total = average * number of innings. Required 11th score = (target average * 11) − (current average * 10). Step-by-Step Solution: Current total = 60 * 10 = 600 Target total = 62 * 11 = 682 Required 11th-inning runs = 682 − 600 = 82 Verification / Alternative check: After scoring 82, new total 682 / 11 = 62 exactly; condition satisfied. Why Other Options Are Wrong: 80, 81, 83: Each yields a mean slightly below or above 62 when divided by 11. Common Pitfalls: Using 62 − 60 = 2 and adding 2 runs (incorrect), or dividing by 10 instead of 11 for the target total. Final Answer: 82

Question 18

The mean weight of 34 students is 42 kg. Including the teacher raises the mean by 0.4 kg (400 g). What is the teacher's weight?

Accepted Answer

Introduction / Context: When one more person is added and the average changes by a known amount, compute the new total (using the new average and new headcount) and subtract the old total to get the added person's value. Given Data / Assumptions: Students: 34; mean = 42 kg. Including teacher: 35 people; mean = 42.4 kg. Find teacher's weight. Concept / Approach: Let T_students be total of students; T_all be total including teacher. Teacher weight = T_all − T_students. Step-by-Step Solution: T_students = 34 * 42 = 1428 kg T_all = 35 * 42.4 = 1484.0 kg Teacher weight = 1484.0 − 1428 = 56 kg Verification / Alternative check: Check resulting mean: (1428 + 56) / 35 = 1484 / 35 = 42.4, which matches the stated increase. Why Other Options Are Wrong: 66, 55, 57: Do not match the exact difference between new and old totals. Common Pitfalls: Applying the 0.4 kg increase to only one person directly (which would be wrong) instead of recalculating totals across the new headcount. Final Answer: 56

Question 19

The average weight of 21 boys is 64 kg. When the teacher is included (22 people), the average increases by 1 kg. What is the teacher's weight?

Accepted Answer

Introduction / Context: Adding one person and seeing the mean increase by a fixed amount allows us to compute the added person's value from new total minus old total. This is a standard trick in averages. Given Data / Assumptions: 21 boys; mean = 64 kg. With teacher: 22 people; mean = 65 kg. Find teacher's weight. Concept / Approach: Totals method: teacher's weight = (new mean * new count) − (old mean * old count). Step-by-Step Solution: Old total = 21 * 64 = 1344 New total = 22 * 65 = 1430 Teacher's weight = 1430 − 1344 = 86 kg Verification / Alternative check: Plug back: (1344 + 86) / 22 = 1430 / 22 = 65, confirming the +1 kg increase in the average. Why Other Options Are Wrong: 64kg, 72 kg, 98 kg: Do not satisfy the totals calculation; 64 kg would not increase the mean at all. Common Pitfalls: Adding 1 kg directly to 64 for the teacher (incorrect), rather than computing via totals. Final Answer: 86kg

Question 20

The average age of 14 girls plus their teacher is 15 years. Excluding the teacher reduces the average by 1 year. What is the teacher's age?

Accepted Answer

Introduction / Context: Removing one person and observing the average drop lets us compute that person’s value by comparing totals with and without that person. This is the inverse of the “added person” scenario. Given Data / Assumptions: With teacher: 15 people; mean = 15 years. Without teacher (14 girls): mean = 14 years. Find teacher's age. Concept / Approach: Teacher age = (15 * 15) − (14 * 14). Derivation: total_with − total_without. Step-by-Step Solution: Total with teacher = 15 * 15 = 225 Total of 14 girls = 14 * 14 = 196 Teacher's age = 225 − 196 = 29 years Verification / Alternative check: Average of girls alone is 14; adding a 29-year-old to 14 girls yields total 225 and average 15, matching the prompt. Why Other Options Are Wrong: 35 yr, 32 yr, 30 yr: Do not produce the stated averages when recombined with the 14 girls. Common Pitfalls: Subtracting the averages directly (15 − 14 = 1) and reporting 1 year (nonsense). The drop of 1 year applies across 14 people, not as the teacher's age. Final Answer: 29 yr

Average Questions