Question 1

Ages with time shift: Five years ago, the average age of A, B, C, and D was 45 years. Including P now, the present average age of all five is 49 years. Find the present age of P.

Accepted Answer

Introduction / Context: When an average is given for a past time, we can update totals by adding years for each person to obtain present totals. Given Data / Assumptions: Five years ago: average of A, B, C, D = 45 years Now: average of A, B, C, D, P = 49 years Concept / Approach: Compute past total of four, add 5 years per person to get present total of the four, then find P from the five person total. Step-by-Step Solution: Past total (four) = 4 * 45 = 180Present total (same four) = 180 + 4 * 5 = 200Present total (five) = 5 * 49 = 245P present age = 245 - 200 = 45 years Verification / Alternative check: Average check holds with computed values. Why Other Options Are Wrong: 64, 48, 40 years: Do not equal the difference between the five person present total and the updated four person present total. Common Pitfalls: Forgetting to add 5 years for each of the four when moving from past to present totals. Final Answer: 45 years

Question 2

Three linked numbers: The first number is twice the second and half of the third. If the average of the three numbers is 56, find the numbers in order (first, second, third).

Accepted Answer

Introduction / Context: Ratio based constraints allow representation of all numbers in a single variable and then direct solution using the average. Given Data / Assumptions: First = twice the second First = half of the third Average of three = 56 Concept / Approach: Let the second be x. Then first = 2x and third = 4x. Use the average to solve x. Step-by-Step Solution: (2x + x + 4x)/3 = 567x/3 = 56x = 56 * 3 / 7 = 24Numbers: 2x = 48, x = 24, 4x = 96 Verification / Alternative check: Average = (48 + 24 + 96)/3 = 168/3 = 56. Correct. Why Other Options Are Wrong: Other orders fail the given relationships and the computed average simultaneously. Common Pitfalls: Assigning the variable to the wrong number or reversing the given relationships. Final Answer: 48, 24, 96

Question 3

Seven number set: The sum of seven numbers is 235. The average of the first three is 23 and that of the last three is 42. Find the fourth number.

Accepted Answer

Introduction / Context: This is a partition of totals problem where we remove the known blocks to get the central term. Given Data / Assumptions: Total of seven = 235 First three average = 23 Last three average = 42 Concept / Approach: Compute sums of the first and last triplets, subtract from the grand total to obtain the middle (fourth) element. Step-by-Step Solution: Sum first three = 3 * 23 = 69Sum last three = 3 * 42 = 126Fourth number = 235 - (69 + 126) = 40 Verification / Alternative check: Recompose totals to confirm the grand total remains 235. Why Other Options Are Wrong: 126 and 195 are component sums, not the standalone fourth value. 69 is the first triplet sum, not the fourth element. Common Pitfalls: Mixing sums with single values or forgetting there are seven numbers in total. Final Answer: 40

Question 4

Compute the average of these nine scores: 253, 124, 255, 534, 836, 375, 101, 443, 760.

Accepted Answer

Introduction / Context: Finding the mean of a finite set is a direct total divided by count operation. Careful addition avoids arithmetic slips. Given Data / Assumptions: Scores: 253, 124, 255, 534, 836, 375, 101, 443, 760 Concept / Approach: Sum all values accurately, then divide by 9. Step-by-Step Solution: Partial sums: 253 + 124 = 377377 + 255 = 632632 + 534 = 11661166 + 836 = 20022002 + 375 = 23772377 + 101 = 24782478 + 443 = 29212921 + 760 = 3681Average = 3681 / 9 = 409 Verification / Alternative check: Use grouping to cross check: (253 + 443) + (124 + 836) + (255 + 375) + 534 + 101 + 760 = 696 + 960 + 630 + 534 + 101 + 760 = 3681; 3681/9 = 409. Why Other Options Are Wrong: 413, 427, 490: Do not match the exact computed average. Common Pitfalls: Arithmetic errors when summing many numbers; double count or omission mistakes. Final Answer: 409

Question 5

Compute the average of these eight scores: 124, 856, 331, 227, 963, 338, 259, 662.

Accepted Answer

Introduction / Context: Average equals total divided by count. With eight data points, careful addition is essential. Given Data / Assumptions: Scores: 124, 856, 331, 227, 963, 338, 259, 662 Concept / Approach: Sum all values first, then divide by 8. Step-by-Step Solution: 124 + 856 = 980980 + 331 = 13111311 + 227 = 15381538 + 963 = 25012501 + 338 = 28392839 + 259 = 30983098 + 662 = 3760Average = 3760 / 8 = 470 Verification / Alternative check: Group sums for confirmation, for example (124 + 856) = 980 and (331 + 227) = 558 etc., leading to 3760 total. Why Other Options Are Wrong: 350, 570, 660: Do not equal the exact computed average. Common Pitfalls: Misadding or skipping a term which distorts the final mean. Final Answer: 470

Question 6

Average of prime numbers: Find the average of all prime numbers between 60 and 90.

Accepted Answer

Introduction / Context: The task is to list primes in the interval and compute their arithmetic mean. This checks prime identification and mean calculation skills. Given Data / Assumptions: Interval: 60 to 90 inclusive of endpoints only if prime Concept / Approach: Enumerate primes, sum them, divide by count. Step-by-Step Solution: Primes in (60, 90): 61, 67, 71, 73, 79, 83, 89Sum = 61 + 67 + 71 + 73 + 79 + 83 + 89 = 523Count = 7Average = 523 / 7 = 74.7142857, which rounds to 74.7 Verification / Alternative check: Cross check primality quickly: none of the numbers are divisible by 2, 3, 5, 7 except as expected for composites; list is correct. Why Other Options Are Wrong: 72, 74, 73.6: Do not equal the computed mean. 74.7 is the correct rounded value. Common Pitfalls: Missing a prime like 83 or including a composite such as 77 or 87 by mistake. Final Answer: 74.7

Question 7

Recovering a missing value from partial averages: The average of nine numbers is 50. The average of the first five is 54 and the average of the last three is 52. Find the sixth number.

Accepted Answer

Introduction / Context: With the grand average and two block averages known, the single missing term equals the remainder after subtracting the two block totals from the grand total. Given Data / Assumptions: Average of 9 = 50 Average of first 5 = 54 Average of last 3 = 52 Concept / Approach: Translate averages to totals, then compute the sixth term: total9 - total first5 - total last3. Step-by-Step Solution: Total of 9 = 9 * 50 = 450Total of first 5 = 5 * 54 = 270Total of last 3 = 3 * 52 = 156Sixth number = 450 - (270 + 156) = 24 Verification / Alternative check: Recompose totals to confirm the grand total is maintained. Why Other Options Are Wrong: 34, 44, 30: Do not balance the totals to give a grand average of 50. Common Pitfalls: Confusing positions or mistakenly subtracting from the wrong block sum. Final Answer: 24

Question 8

Average decreases after adding an observation: The average of six observations is 12. When a seventh observation is added, the average decreases by 1. What is the seventh observation?

Accepted Answer

Introduction / Context: This mirrors a standard totals method: a new value below the current average pulls the mean down. Compute it via totals. Given Data / Assumptions: Average of 6 = 12 Average of 7 after adding = 11 Concept / Approach: Totals from averages yield the exact seventh value. Step-by-Step Solution: Old total = 6 * 12 = 72New total = 7 * 11 = 77Seventh observation = 77 - 72 = 5 Verification / Alternative check: Rechecking averages confirms correctness. Why Other Options Are Wrong: 1, 3, 6: Do not produce the required new total of 77. Common Pitfalls: Assuming the seventh equals the new average without computing totals. Final Answer: 5

Question 9

Effect of scaling on averages: The average of ten numbers is 7. If each number is multiplied by 12, what is the average of the new numbers?

Accepted Answer

Introduction / Context: Multiplying every term in a dataset by a constant scales the average by the same constant. Given Data / Assumptions: Initial average = 7 Scaling factor = 12 Concept / Approach: If each value becomes 12 * value, then average becomes 12 * average. Step-by-Step Solution: New average = 12 * 7 = 84 Verification / Alternative check: Let the total initially be T = 10 * 7 = 70. After scaling, total = 12 * 70 = 840. New average = 840 / 10 = 84. Matches. Why Other Options Are Wrong: 7 and 9: Ignore scaling property. 82: Arbitrary and unsupported by the constant factor rule. Common Pitfalls: Trying to add instead of multiply the average or confusing the constant's effect. Final Answer: 84

Question 10

Average weight correction in a class of 60 students: The average weight of 60 students was computed as 40 kg. Later, it was discovered that one student's weight had been entered as 36 kg instead of his actual 33 kg (i.e., recorded 3 kg too high). Wh…

Accepted Answer

Introduction / Context: The question tests weighted averages and error correction when one item in the dataset was misrecorded. We adjust the total sum by the discovered error and then recompute the average. Given Data / Assumptions: Number of students = 60 Originally computed average = 40 kg One entry recorded 36 kg but actual is 33 kg (3 kg over-reported) Concept / Approach: An average is total sum / count. If one item was over-reported by 3, the true total is original total minus 3. Recalculate the average using the corrected total and the same count (60). Step-by-Step Solution: Original total = 60 * 40 = 2400 kgError amount = 36 - 33 = 3 kgCorrected total = 2400 - 3 = 2397 kgCorrected average = 2397 / 60 = 39.95 kg Verification / Alternative check: Replacing 36 with 33 lowers the average by 3/60 = 0.05. So 40 - 0.05 = 39.95 kg, confirming the result. Why Other Options Are Wrong: 39.90 kg and 39.80 kg subtract more than the 3 kg error allows; 40.00 kg ignores the correction; 39.65 kg overcorrects the total. Common Pitfalls: Forgetting that the error affects only one entry, or dividing the error by the wrong count. Always adjust total first, then divide by n. Final Answer: 39.95 kg

Question 11

Cricket batting average adjustment: A batsman has a certain average for 14 innings. In the 15th innings he scores 72 runs, and his average increases by 2 runs. What is his new average after 15 innings?

Accepted Answer

Introduction / Context: This problem tests average (arithmetic mean) update when one additional data point is appended and the average changes by a known amount. Given Data / Assumptions: Average after 14 innings = x 15th score = 72 New average increases by 2, so new average = x + 2 Concept / Approach: For n items with average x, total is n * x. After adding a new score s, new average is (n * x + s) / (n + 1). Set this equal to x + 2 and solve for x, then compute the new average. Step-by-Step Solution: (14 * x + 72) / 15 = x + 214x + 72 = 15x + 3072 - 30 = 15x - 14xx = 42New average = x + 2 = 44 Verification / Alternative check: Total after 14 innings = 14 * 42 = 588. Add 72 gives 660. 660 / 15 = 44. Correct. Why Other Options Are Wrong: 46, 42, 48, 40 are inconsistent with the equation and totals computed above. Common Pitfalls: Confusing the previous average with the new one or dividing by 14 instead of 15 after the last innings. Final Answer: 44

Question 12

Average update after adding observations: The average of 5 observations is 15. Three new observations 16.5, 18, and 14.5 are added. What is the new average of all 8 observations?

Accepted Answer

Introduction / Context: This tests recomputing an average after adding additional values. Given Data / Assumptions: Initial count = 5, average = 15 New observations: 16.5, 18, 14.5 Concept / Approach: Compute the original total from average * count. Add the sum of the new three values. Divide by the new count (8) to obtain the updated average. Step-by-Step Solution: Original total = 5 * 15 = 75Sum of new = 16.5 + 18 + 14.5 = 49New total = 75 + 49 = 124New average = 124 / 8 = 15.5 Verification / Alternative check: Compute each value's deviation from 15 and average deviations; total deviation is 0.5 per item on average across eight, confirming 15.5. Why Other Options Are Wrong: 14.8, 16.2, 15.0, 15.6 do not match the exact arithmetic above. Common Pitfalls: Using 5 as the divisor after adding three items; forgetting to add all three values correctly. Final Answer: 15.5

Question 13

Solving with averages and linear relations: The average of x1, x2, and x3 is 14 (so x1 + x2 + x3 = 42). Twice the sum of x2 and x3 is 30 (i.e., 2 * (x2 + x3) = 30). What is the value of x1?

Accepted Answer

Introduction / Context: Averages often give a sum; paired with another relation, we can isolate a specific variable. Given Data / Assumptions: (x1 + x2 + x3) / 3 = 14 → x1 + x2 + x3 = 42 2 * (x2 + x3) = 30 → x2 + x3 = 15 Concept / Approach: Use the two equations to solve for x1 directly by subtracting the sum of x2 and x3 from the total. Step-by-Step Solution: x1 + x2 + x3 = 42x2 + x3 = 15x1 = 42 - 15 = 27 Verification / Alternative check: Check that 2 * (x2 + x3) equals 30 when x1 is 27; since x2 + x3 = 15, this matches the given condition. Why Other Options Are Wrong: 20, 16, 2, 12 do not satisfy both equations simultaneously. Common Pitfalls: Dividing 30 by 2 incorrectly or mixing up which pair is summed; always write both equations first. Final Answer: 27

Question 14

Weighted average of ages after new admissions: The average age of 45 students in a class is 13 years. Then 15 new students with an average age of 18 years join the class. What is the average age of all 60 students now?

Accepted Answer

Introduction / Context: This is a weighted average across two groups with different sizes and averages. Given Data / Assumptions: Group 1: 45 students, average 13 yr Group 2: 15 students, average 18 yr Total after joining = 60 students Concept / Approach: Weighted average = (sum1 + sum2) / total count, where sum is average * count for each group. Step-by-Step Solution: Sum1 = 45 * 13 = 585Sum2 = 15 * 18 = 270Combined sum = 585 + 270 = 855Average = 855 / 60 = 14.25 yr Verification / Alternative check: Note that the final average must lie between 13 and 18 and closer to 13 because the larger group is at 13; 14.25 is reasonable. Why Other Options Are Wrong: 13.15 and 12.28 are too low; 16.42 is too high; 15.00 does not match the weighted calculation. Common Pitfalls: Taking the average of 13 and 18 directly (15.5) without weighting by group sizes. Final Answer: 14.25 yr

Question 15

Staff salary mix problem (officers vs non-officers): The average daily salary of the entire staff is ₹200. Officers average ₹550 per day and non-officers average ₹120 per day. If there are 16 officers, how many non-officers are there?

Accepted Answer

Introduction / Context: This is a classic mixture (weighted average) problem involving two subgroups with different averages contributing to an overall average. Given Data / Assumptions: Total average = ₹200 Officer average = ₹550; Non-officer average = ₹120 Number of officers = 16; number of non-officers = n Concept / Approach: Overall average = (sum of subgroup totals)/(total people). Set up equation with n unknown and solve. Step-by-Step Solution: (16 * 550 + n * 120) / (16 + n) = 2008800 + 120n = 200 * (16 + n)8800 + 120n = 3200 + 200n8800 - 3200 = 200n - 120n5600 = 80n → n = 70 Verification / Alternative check: Plug n = 70 back to confirm overall average = ₹200 exactly. Why Other Options Are Wrong: 60, 65, 68, 72 do not satisfy the balance of totals implied by the given averages. Common Pitfalls: Setting denominator incorrectly or mixing rupee totals with averages; always convert to totals first. Final Answer: 70

Question 16

Average property value with symmetric deviation: The average property value of A, B, and C is ₹11111 crore. The property of C is as much less than the mean of A and C as the property of A is greater than that same mean (i.e., A and C are equally spa…

Accepted Answer

Introduction / Context: The problem features symmetry about the mean for A and C. It asks for B given that the overall average of A, B, C is fixed. Given Data / Assumptions: Average of A, B, C = ₹11111 crore A and C are equidistant around their mean (their deviations cancel) Concept / Approach: Let m = (A + C) / 2. If A is m + d and C is m - d, then A + C = 2m regardless of d. The total sum A + B + C = 3 * 11111, so B is determined solely by the overall average, independent of d. Step-by-Step Solution: Sum of all three = 3 * 11111 = 33333A + C = 2m (unknown, but fixed by their mean)B = 33333 - (A + C)Since A and C are symmetric, no net shift from their mean; the only value ensuring the overall average is B = 11111 Verification / Alternative check: Pick any symmetric pair (m ± d); their sum is 2m. Choosing B = 11111 preserves the overall average of 11111. Why Other Options Are Wrong: Values like 110 crore or 21410 crore would break the overall mean of 11111 unless A + C were contrived accordingly, contradicting the symmetry context described. Common Pitfalls: Overcomplicating the symmetric condition; the key is that the overall average fixes B directly to the same average amount. Final Answer: 11111 crore

Question 17

Replacement effect on class average age: In a class of 25 girls, the average age decreases by 6 months when a new girl replaces a 20-year-old. What is the age of the new girl?

Accepted Answer

Introduction / Context: This explores how replacing one member in a group changes the group average and what that implies about the new member's value. Given Data / Assumptions: Class size = 25 Average decreases by 6 months = 0.5 years Replaced girl had age = 20 years Concept / Approach: Change in average * group size equals change in total. Here, the new total is lower by 25 * 0.5 = 12.5 years relative to before replacement. That drop equals (new age - old age). Step-by-Step Solution: Change in total = 25 * (-0.5) = -12.5 yearsLet new age = xx - 20 = -12.5 → x = 7.5 years Verification / Alternative check: New average reduction per student is 0.5 years; across 25 students this matches a 12.5-year reduction in total, consistent with replacing 20 by 7.5. Why Other Options Are Wrong: 8.5, 9, 6, 10 years don’t produce the exact 12.5-year reduction in total required. Common Pitfalls: Using 24 instead of 25 for the multiplier or converting months to years incorrectly. Final Answer: 7.5 yr

Question 18

Family average by subgroup weights: The average age of all members of a family is 25 years. The average of males is 30 years and the average of females is 20 years. If there are 15 females, how many males are there?

Accepted Answer

Introduction / Context: Another weighted average problem using two subgroup means to deduce the unknown count. Given Data / Assumptions: Overall average = 25 Male average = 30; Female average = 20 Females = 15; let males = m Concept / Approach: Use (30m + 20 * 15) / (m + 15) = 25 and solve for m. Step-by-Step Solution: (30m + 300) / (m + 15) = 2530m + 300 = 25m + 3755m = 75 → m = 15 Verification / Alternative check: Totals: males 30 * 15 = 450; females 20 * 15 = 300; combined 750 over 30 persons gives 25 average. Why Other Options Are Wrong: Other values do not satisfy the overall average constraint. Common Pitfalls: Placing the wrong subgroup count in the numerator or forgetting to add both groups in the denominator. Final Answer: 15

Question 19

Target overall average across three equal-maximum exams: Out of three annual exams (each out of 600), Lalit scored 55% in the first and 60% in the second. How many marks must he score in the third to make his overall average 70%?

Accepted Answer

Introduction / Context: This involves translating percentages to marks, then working backward from a target overall percentage across equal-maximum papers. Given Data / Assumptions: Each exam max = 600 First = 55% of 600, Second = 60% of 600 Target overall average = 70% of 600 Concept / Approach: Compute current total, compute required total for three papers, subtract to get the third paper requirement. Step-by-Step Solution: First = 0.55 * 600 = 330Second = 0.60 * 600 = 360Required total for overall 70% = 3 * (0.70 * 600) = 3 * 420 = 1260Third needed = 1260 - (330 + 360) = 1260 - 690 = 570 Verification / Alternative check: Totals become 330 + 360 + 570 = 1260. Average per exam = 1260 / 3 = 420, i.e., 70%. Why Other Options Are Wrong: 560, 590, 500, 540 do not yield an overall 70% when summed with the first two scores. Common Pitfalls: Taking 70% of the current average rather than recalculating totals across all three equal-weight components. Final Answer: 570

Question 20

Correcting a misrecorded mark in a class mean: The mean marks of 30 students is 58.5. Later, it was found that a mark of 75 had been wrongly recorded as 57. What is the correct mean?

Accepted Answer

Introduction / Context: We correct the class total by replacing a misrecorded value, then recompute the mean. Given Data / Assumptions: n = 30 Original mean = 58.5 → original total = 58.5 * 30 Recorded 57 should be 75 (increase by 18) Concept / Approach: Corrected total = original total + 18. Correct mean = corrected total / 30. Step-by-Step Solution: Original total = 58.5 * 30 = 1755Corrected total = 1755 + 18 = 1773Correct mean = 1773 / 30 = 59.1 Verification / Alternative check: The net increase per student = 18 / 30 = 0.6. 58.5 + 0.6 = 59.1, consistent. Why Other Options Are Wrong: 57.4, 57.5, 58.1, 58.9 are not the corrected mean after adding 18 to the total. Common Pitfalls: Subtracting 18 instead of adding; forgetting to keep the count at 30. Final Answer: 59.1

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