Question 1

The average of the sequence 100, 200, 300, ..., 1000 is 550. If each term is divided by 5, what is the new average?

Accepted Answer

Introduction / Context: Transformations applied uniformly to every term in a data set apply the same transformation to the average. This property lets us compute new averages quickly without re-summing all terms. Given Data / Assumptions: Original sequence: 100, 200, 300, ..., 1000 (step 100) Original average = 550 Operation: divide each term by 5 Concept / Approach: If each data point is divided by a constant k, the average is also divided by k. Thus New average = Old average / 5. Step-by-Step Solution: Old average = 550 New average = 550 / 5 = 110 Verification / Alternative check: New sequence becomes 20, 40, 60, ..., 200. The first and last terms’ mean is (20 + 200) / 2 = 110, matching the transformation rule. Why Other Options Are Wrong: 450: Incorrect scaling; seems like subtracting 100 instead of dividing. 45 and 55: Both reflect order-of-magnitude mistakes. None of these: Correct, since the true new average 110 is not listed among the numeric options. Common Pitfalls: Confusing division with subtraction, or mistakenly dividing by the number of terms again after transformation. Final Answer: None of these (the correct new average is 110).

Question 2

Five electronics shops have 20, 30, 60, 80, and 50 TV sets respectively. What is the average number of TV sets per shop?

Accepted Answer

Introduction / Context: Average inventory questions test straightforward mean computation. Keeping track of totals and counts is the central skill. Given Data / Assumptions: TV sets: 20, 30, 60, 80, 50 Number of shops = 5 Concept / Approach: Average = total units / number of shops. Grouping numbers helps reduce mistakes during addition. Step-by-Step Solution: Sum = 20 + 30 + 60 + 80 + 50 Pairing: (20 + 80) = 100; (30 + 50) = 80; remaining 60 Total = 100 + 80 + 60 = 240 Average = 240 / 5 = 48 Verification / Alternative check: The mean 48 is close to the middle values and balances the distribution between low (20, 30) and high (60, 80) counts, confirming reasonableness. Why Other Options Are Wrong: 24: Exactly half the correct value; likely divided by 10 by mistake. 50 and 60: Above the computed mean; not supported by the totals. None of these: A correct option (48) exists. Common Pitfalls: Arithmetic slips when multiple terms are added, or dividing by the wrong count of shops. Final Answer: 48

Question 3

Six families in Indira Nagar have 7, 8, 10, 13, 6, and 10 members. If 1 member from each family moves to a hostel, what is the new average family size?

Accepted Answer

Introduction / Context: When the same increment or decrement is applied to each group in a collection, the effect on the overall average is straightforward. Here, one member leaves each family, reducing each family's size by 1. Given Data / Assumptions: Original sizes: 7, 8, 10, 13, 6, 10 Six families total One member leaves each family Concept / Approach: Subtract 1 from each family, sum the new sizes, and divide by 6. Alternatively, compute the original average and then subtract 1 directly from the average (since all groups change equally). Step-by-Step Solution: Original sum = 7 + 8 + 10 + 13 + 6 + 10 = 54 After one leaves each: new sum = 54 - 6 = 48 New average = 48 / 6 = 8 Verification / Alternative check: Original average = 54 / 6 = 9. Decreasing each family by 1 reduces the average by 1 ⇒ 9 - 1 = 8. Matches the computed value. Why Other Options Are Wrong: 9: That was the original average, not the new one. 10 and 13: Both exceed the feasible mean after uniform reduction. None of these: A correct option exists (8). Common Pitfalls: Subtracting only 1 in total rather than 1 per family (which would undercount the reduction). Final Answer: 8

Question 4

Salman, Ambani, and Dalmiya together have an average of 60 shirts. If each buys 6 more shirts, what is the new average number of shirts per person?

Accepted Answer

Introduction / Context: When each member of a group adds the same quantity, the group average increases by that quantity as well. This property saves time over recomputing totals from scratch. Given Data / Assumptions: Initial average shirts per person = 60 Three people Each buys 6 more shirts Concept / Approach: If everyone adds the same number x, the average increases by x. No need to know individual counts to find the new mean. Step-by-Step Solution: Initial average = 60 Increase per person = 6 New average = 60 + 6 = 66 Verification / Alternative check: Total initially = 60 * 3 = 180. Added total = 6 * 3 = 18. New total = 198. New average = 198 / 3 = 66. Confirms the shortcut. Why Other Options Are Wrong: 63 and 62: Do not reflect the full +6 change per person. can't be determined: The rule permits determination without individual breakdowns. None of these: A correct exact value exists (66). Common Pitfalls: Adding 6 only once to the total instead of per person, which would understate the increase in the mean. Final Answer: 66

Question 5

A bouquet initially contains 6 flowers on average. If the seller doubles the flowers in each bouquet, what is the new average number of flowers per bouquet?

Accepted Answer

Introduction / Context: Uniform scaling of every data point scales the average by the same factor. Doubling every bouquet's count doubles the mean as well. Given Data / Assumptions: Original average flowers per bouquet = 6 Each bouquet is doubled Concept / Approach: If each value is multiplied by k, the average is multiplied by k. Here k = 2. Step-by-Step Solution: Original average = 6 Scaling factor = 2 New average = 6 * 2 = 12 Verification / Alternative check: Consider three bouquets with counts 5, 6, 7 (average 6). Doubling gives 10, 12, 14 (average 12). Pattern generalizes to any data set. Why Other Options Are Wrong: 3: This would be halving, not doubling. 8: Not consistent with a full factor-of-two increase. none of these and 6: Both ignore the uniform scaling rule. Common Pitfalls: Confusing doubling with adding a fixed number; multiplying is required here. Final Answer: 12

Question 6

The average salary of A and B is Rs. 6000, while the average salary of C, D, and E is Rs. 8000. What is the average salary of all five people together?

Accepted Answer

Introduction / Context: When combining groups with known averages, compute each group’s total, add them, then divide by the total headcount to get the combined average. Given Data / Assumptions: Average of A, B = Rs. 6000 ⇒ 2 people Average of C, D, E = Rs. 8000 ⇒ 3 people Concept / Approach: Group total = average * count. Combined average = (total of group 1 + total of group 2) / 5. Step-by-Step Solution: Total (A, B) = 6000 * 2 = 12000 Total (C, D, E) = 8000 * 3 = 24000 Grand total = 12000 + 24000 = 36000 Combined average = 36000 / 5 = Rs. 7200 Verification / Alternative check: The combined average must lie between 6000 and 8000 and be closer to 8000 because the larger group has the higher average. Rs. 7200 satisfies both conditions. Why Other Options Are Wrong: Rs. 7000 and Rs. 7500: Do not match the exact weighted computation. can't be determined: All required data are present. None of these: A correct option is provided (Rs.7200). Common Pitfalls: Taking the mean of 6000 and 8000 directly (which would be 7000) instead of doing a weighted average based on group sizes. Final Answer: Rs.7200

Question 7

Rajesh, Bahadur, and Amir earn an average income of Rs. 8000 per month. Their average monthly expenditure is Rs. 5000. What is their average monthly savings?

Accepted Answer

Introduction / Context: Average savings can be computed by subtracting average expenditure from average income, or by using totals. Both routes yield the same result because subtraction distributes over averages for equally sized groups. Given Data / Assumptions: Average income per person = Rs. 8000 Average expenditure per person = Rs. 5000 Three people in total Concept / Approach: Average savings = average income - average expenditure. Alternatively, compute total income and total expenditure, then subtract and divide by the count. Step-by-Step Solution: Method 1 (Averages): 8000 - 5000 = Rs. 3000 Method 2 (Totals): Total income = 8000 * 3 = 24000 Total expenditure = 5000 * 3 = 15000 Total savings = 24000 - 15000 = 9000; Average savings = 9000 / 3 = Rs. 3000 Verification / Alternative check: Both methods agree exactly, confirming arithmetic consistency. Why Other Options Are Wrong: Rs. 5000 and Rs. 2500: Do not match the computed difference. Rs. 9000: This is the total savings, not per-person average. None of these: A correct option exists (Rs. 3000). Common Pitfalls: Confusing total and average amounts or mixing per-person figures with totals across the group. Final Answer: Rs. 3000

Question 8

The average salary of A, B, and C is Rs. 10,000. Only A's average monthly expenditure is known to be Rs. 6000. What is the average savings of B and C together?

Accepted Answer

Introduction / Context: Sometimes average problems do not supply enough information to isolate unknowns for a subgroup. Recognizing insufficiency is as important as computing a value when possible. Given Data / Assumptions: Average salary of A, B, C = Rs. 10,000 ⇒ total salary = Rs. 30,000 Only A's average expenditure is known: Rs. 6000 Expenditures or savings of B and C are not provided. Concept / Approach: To compute the average savings of B and C, we need their incomes (known only in total with A) and their expenditures (unknown). Without at least the combined expenditure of B and C, infinitely many distributions satisfy the constraints, leading to different savings outcomes. Step-by-Step Solution: Total salary = Rs. 30,000 (for A, B, C) Let expenditures be: A = 6000, B = x, C = y (unknown) Savings(B + C) = (salary(B) + salary(C)) - (x + y) However, salary(B) + salary(C) is not known individually, and x + y is also unknown. Verification / Alternative check: Different choices of B and C expenditures produce different savings while respecting the given information. Hence no unique value exists. Why Other Options Are Wrong: Rs. 5500, Rs. 4500, Rs. 4000: Each assumes extra information not provided. None of these: Insufficiency is explicitly captured by “can't be determined.” Common Pitfalls: Assuming B and C have the same expenditure as A or that the group average for expenditure is known; neither is given. Final Answer: can't be determined

Question 9

An office has 60 employees with an overall average salary of Rs. 12,000 per month. If the number of executives is twice the number of non-executives, what is the average salary of the non-executive employees?

Accepted Answer

Introduction / Context: Weighted-average questions require either the subgroup averages or enough relations between them to solve for unknowns. Knowing only the overall mean and headcount ratio often isn’t sufficient. Given Data / Assumptions: Total employees = 60 Executives : Non-executives = 2 : 1 ⇒ Executives = 40, Non-executives = 20 Overall average salary = Rs. 12,000 No subgroup averages provided. Concept / Approach: Let average salaries be E for executives and N for non-executives. Then overall average gives a single equation: (40*E + 20*N) / 60 = 12000. With two unknowns (E and N) and only one equation, infinitely many solutions exist unless another constraint is given. Step-by-Step Solution: 40*E + 20*N = 60 * 12000 = 720000 This single linear relation cannot determine both E and N uniquely. Therefore the average salary of non-executives (N) is indeterminate. Verification / Alternative check: Pick E = 13000 ⇒ 40*13000 = 520000, then 20*N = 200000 ⇒ N = 10000. Pick E = 11000 ⇒ 40*11000 = 440000, then 20*N = 280000 ⇒ N = 14000. Multiple feasible values show non-uniqueness. Why Other Options Are Wrong: Rs. 9000, Rs. 8000, Rs. 6000: Each is a specific value without justification; many values are possible. None of these: Insufficient data is properly captured by “can't be determined.” Common Pitfalls: Assuming executives are always paid more and plugging a guess; the problem does not state subgroup averages or ordering constraints. Final Answer: can't be determined

Question 10

Average speed for a round trip: Einstein travels from Ahmedabad to Lucknow at 40 km/h and returns at 60 km/h. What is the overall average speed for the entire journey (same distance each way)?

Accepted Answer

Introduction / Context: Average speed over a round trip with unequal speeds each way is a classic aptitude topic. Because the distances are equal, the correct average is not the simple arithmetic mean. Instead, we must use the harmonic mean of the two speeds to capture time properly. Given Data / Assumptions: Outbound speed = 40 km/h Return speed = 60 km/h Distances in both directions are equal Concept / Approach: The average speed for equal distances is 2ab/(a + b). This comes from total distance divided by total time, where time for each leg is distance/speed. Using a and b for the two speeds ensures the formula accounts for slower segments contributing more time. Step-by-Step Solution: Let a = 40 and b = 60Average speed = 2 * a * b / (a + b)Average speed = 2 * 40 * 60 / (40 + 60)Average speed = 4800 / 100 = 48 km/h Verification / Alternative check: Assume a 60 km one way trip. Time out = 60/40 = 1.5 h. Time back = 60/60 = 1 h. Total distance = 120 km. Total time = 2.5 h. 120/2.5 = 48 km/h. Why Other Options Are Wrong: 50, 52, 56 km/h: arithmetic style guesses that ignore time weighting. 45 km/h: underestimates by overweighting the slower leg incorrectly. Common Pitfalls: Using (40 + 60)/2 = 50 is a common error. The arithmetic mean applies when times are equal, not distances. Final Answer: 48 km/h

Question 11

Average age application: The average of A, B, C, D, and E is 40 years. The average of A and B is 35 years, and the average of C and D is 42 years. Find the exact age of E.

Accepted Answer

Introduction / Context: This problem tests the ability to manipulate averages by converting them into total sums. When multiple subgroup averages are given, convert each to totals and combine to isolate the missing value. Given Data / Assumptions: Average of A, B, C, D, E = 40 years Average of A and B = 35 years Average of C and D = 42 years Concept / Approach: Average * count = total. Sum all five, subtract the totals of the two given pairs, and what remains is the age of E. Step-by-Step Solution: Total of A, B, C, D, E = 5 * 40 = 200Total of A + B = 2 * 35 = 70Total of C + D = 2 * 42 = 84Age of E = 200 - (70 + 84) = 200 - 154 = 46 Verification / Alternative check: Distribute values: If A + B = 70 and C + D = 84, then these four together are 154. The overall of five is 200. The leftover 46 must be E. Why Other Options Are Wrong: 48, 44, 40: incorrect subtractions from the total. 32: likely from mixing up averages with totals. Common Pitfalls: Forgetting to multiply averages by the correct count or mixing up addition and subtraction when isolating the unknown. Final Answer: 46 years

Question 12

Compute the arithmetic mean: Find the average of the five numbers 308, 142, 160, 245, and 25.

Accepted Answer

Introduction / Context: Simple average computation requires adding all values and dividing by the count. This tests care with multi-digit addition and division by a small integer. Given Data / Assumptions: Numbers: 308, 142, 160, 245, 25 Count = 5 Concept / Approach: Average = sum of values / number of values. Keep arithmetic neat and double-check totals. Step-by-Step Solution: Sum = 308 + 142 + 160 + 245 + 25Sum = 880Average = 880 / 5 = 176 Verification / Alternative check: Group numbers for easy addition: (308 + 142) = 450, (160 + 245) = 405, plus 25 gives 450 + 405 + 25 = 880. Divide by 5 gives 176. Why Other Options Are Wrong: 178, 175, 174, 172: all differ from the exact division of 880 by 5. Common Pitfalls: Mistakes typically occur in addition. Verifying with regrouping helps avoid carry errors. Final Answer: 176

Question 13

Average of a group changes when one member leaves: A group of 10 friends has an average age of 15 years. One friend leaves and the new average of the remaining 9 becomes 16 years. Find the exact age of the friend who left.

Accepted Answer

Introduction / Context: This is a reverse-average problem: we know the overall mean before and after removing one value. Converting means to totals allows isolation of the missing age quickly. Given Data / Assumptions: Initial group size = 10, average = 15 years After one leaves: size = 9, average = 16 years Concept / Approach: Total = average * count. The difference between initial total and new total equals the age of the friend who left. Step-by-Step Solution: Initial total = 10 * 15 = 150New total = 9 * 16 = 144Missing age = 150 - 144 = 6 years Verification / Alternative check: If the person who left was 6, the remainder totals 144 with 9 members, average 16, which matches the condition. Why Other Options Are Wrong: 4, 5, 7, 8 years: do not produce the specified post-removal average. Common Pitfalls: Confusing which total to subtract from which. The person who left equals initial total minus new total. Final Answer: 6 years

Question 14

First 79 natural numbers: Using mean properties of a sequence from 1 to 79, determine the average value.

Accepted Answer

Introduction / Context: For consecutive natural numbers from 1 to n, the average is the midpoint between the first and the last term. This is a standard property of arithmetic progressions. Given Data / Assumptions: Sequence: 1, 2, 3, ..., 79 Count = 79 Concept / Approach: Average of an arithmetic progression = (first + last) / 2. This exploits symmetry around the center term for an odd count. Step-by-Step Solution: Average = (1 + 79) / 2Average = 80 / 2 = 40 Verification / Alternative check: Since there are 79 terms, the middle (40th) term is 40. For an odd count the median equals the mean in an arithmetic progression. Why Other Options Are Wrong: 39, 39.5, 40.5, 79: do not match the AP mean formula for 1 to 79. Common Pitfalls: Using n/2 incorrectly or confusing median with mean when the count is even. Here both coincide at 40 due to odd length and symmetry. Final Answer: 40

Question 15

Effect of scaling on average: The average of 8 numbers is 21. If each number is multiplied by 8, what will be the average of the new set?

Accepted Answer

Introduction / Context: This checks understanding that multiplying every value in a set by a constant multiplies the average by the same constant. The count does not change, so the mean scales directly. Given Data / Assumptions: Average of 8 numbers = 21 Each number is multiplied by 8 Concept / Approach: If each x becomes 8x, then new average = 8 * old average. This follows from linearity of sums and the definition of mean. Step-by-Step Solution: Old average = 21Scale factor = 8New average = 21 * 8 = 168 Verification / Alternative check: Let the original total be 21 * 8 = 168. After multiplying each by 8, the total becomes 168 * 8 = 1344. Dividing by the same count 8 gives 168 again. Why Other Options Are Wrong: 8, 21, 29, 84: do not apply the constant scaling rule correctly. Common Pitfalls: Confusing operations on each member with operations on the average. Addition or multiplication across the set affects the mean identically. Final Answer: 168

Question 16

Average adjustment after removing outliers: The average of 50 numbers is 38. If two numbers, 45 and 55, are discarded, what is the new average of the remaining 48 numbers?

Accepted Answer

Introduction / Context: When elements are removed from a dataset, recompute the total and divide by the new count to get the updated average. This problem reinforces careful sum tracking. Given Data / Assumptions: Initial average = 38, count = 50 Removed numbers = 45 and 55 Concept / Approach: Total = average * count. Subtract removed values from the total, then divide by the new count. Step-by-Step Solution: Initial total = 50 * 38 = 1900Remove 45 and 55: new total = 1900 - 100 = 1800New count = 48New average = 1800 / 48 = 37.5 Verification / Alternative check: The two removed values have mean 50 which is above 38. Removing higher values should lower the average, consistent with 37.5. Why Other Options Are Wrong: 36.5, 37.0, 37.52, 38.5: do not match the exact recomputation. Common Pitfalls: Forgetting to adjust the count or mis-adding the removed values, which skews the final mean. Final Answer: 37.5

Question 17

Cricket scores average split: A cricketer averages 43.9 runs over 10 matches. If the average for the first 6 matches is 53, what is the average for the last 4 matches?

Accepted Answer

Introduction / Context: Split-average problems require translating averages into totals for different segments, then finding the residual segment sum and its mean. Given Data / Assumptions: Overall: 10 matches, average 43.9 runs First 6 matches: average 53 runs Concept / Approach: Total = average * count. Subtract the known segment total from the overall total to get the remaining segment total, then divide by its count. Step-by-Step Solution: Overall total = 10 * 43.9 = 439First 6 total = 6 * 53 = 318Last 4 total = 439 - 318 = 121Last 4 average = 121 / 4 = 30.25 Verification / Alternative check: 6 matches at 53 contribute strongly to the overall. The last 4 must average below 43.9 to pull the total down to 439, consistent with 30.25. Why Other Options Are Wrong: 17.15, 29.75, 31, 28.5: do not equal 121 divided by 4. Common Pitfalls: Arithmetic missteps in decimal multiplication or division. Keep one extra decimal place then round at the end if needed. Final Answer: 30.25

Question 18

Weighted average of two subgroups: In a class of 40 girls, the average height of 30 girls is 160 cm and the average height of the remaining 10 girls is 156 cm. Find the overall average height of the class.

Accepted Answer

Introduction / Context: This is a weighted average scenario: two groups have different averages and sizes. The overall average depends on both group means and their counts. Given Data / Assumptions: Total students = 40 Group 1: 30 girls, average 160 cm Group 2: 10 girls, average 156 cm Concept / Approach: Overall average = (sum of group totals) / total count. Compute each group total by mean * size and add them. Step-by-Step Solution: Group 1 total height = 30 * 160 = 4800 cmGroup 2 total height = 10 * 156 = 1560 cmCombined total = 4800 + 1560 = 6360 cmOverall average = 6360 / 40 = 159 cm Verification / Alternative check: Since the larger group average (160) is above the smaller group average (156), the final average should be between 156 and 160 and closer to 160, that is 159. Why Other Options Are Wrong: 158.5, 157: too low relative to the 30 at 160. 160, 161: do not account for the lower subgroup mean at 156. Common Pitfalls: Using plain mean of 160 and 156 without weighting by group sizes. Weighted totals are essential. Final Answer: 159 cm

Question 19

Five consecutive odd numbers: If the numbers are a, a+2, a+4, a+6, and a+8, what is their average in terms of a?

Accepted Answer

Introduction / Context: For equally spaced numbers (an arithmetic progression), the average equals the middle term. Five consecutive odd numbers centered around the third term follow this rule directly. Given Data / Assumptions: Numbers: a, a+2, a+4, a+6, a+8 Common difference = 2 Concept / Approach: The average of an arithmetic progression equals (first + last) / 2. With an odd count, it also equals the middle element. Step-by-Step Solution: Average = (a + (a + 8)) / 2 = (2a + 8) / 2 = a + 4Middle term is a + 4, which matches the computed mean Verification / Alternative check: Pick a = 1. The set is 1, 3, 5, 7, 9. Average = 25 / 5 = 5. Middle term = 5 = 1 + 4. Why Other Options Are Wrong: (a + 8) / 2: ignores symmetry and double counts incorrectly. (a + (a+8)) / 2 is algebraically equal to a + 4, but many candidates misuse parentheses; we use the simpler a + 4 as the canonical form. (a + 2 + a + 4 + a + 6) / 5 omits two terms and is incomplete. a + 5 is off by one from the correct midpoint. Common Pitfalls: Forgetting to average first and last or not recognizing the middle term identity for an odd count. Final Answer: a + 4

Question 20

Three-number relation and average: The first number is twice the second and half of the third. If the average of the three numbers is 56, find the ordered triple (first, second, third).

Accepted Answer

Introduction / Context: This problem blends proportional relationships with an average constraint. Setting variables based on the relationships simplifies solving the system. Given Data / Assumptions: First = twice the second First = half of the third Average of three numbers = 56 Concept / Approach: Let the second be x. Then first = 2x and third = 4x. Use the average to find x and build the ordered triple. Step-by-Step Solution: Numbers: 2x, x, 4xAverage = (2x + x + 4x) / 3 = 7x / 3Set 7x / 3 = 56, so 7x = 168, x = 24First = 48, Second = 24, Third = 96 Verification / Alternative check: Check relations: 48 is twice 24 and half of 96. Average = (48 + 24 + 96) / 3 = 168 / 3 = 56, correct. Why Other Options Are Wrong: All other permutations break at least one relation or do not average to 56. Common Pitfalls: Assigning inconsistent variables (e.g., mixing up which is half or double) or forgetting to apply average to three numbers. Final Answer: 48, 24, 96

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