Question 1

A batsman scores 95 runs in his 13th match and thereby increases his average runs per match by 4. What is his average after the 13th match?

Accepted Answer

Introduction: Batting average equals total runs divided by matches played. Here, scoring 95 in the 13th match raises the average by 4. We set up an equation with the old and new averages using totals before and after the match. Given Data / Assumptions: Matches before latest = 12. Runs in 13th match = 95. Increase in average after 13th = 4. Let old average be A; new average = A + 4. Concept / Approach: Total runs before = 12 * A. After the 13th match, total runs = 12A + 95 over 13 matches. So (12A + 95) / 13 = A + 4. Solve for A, then compute A + 4. Step-by-Step Solution: (12A + 95) / 13 = A + 4. 12A + 95 = 13A + 52. 95 − 52 = 13A − 12A ⇒ A = 43. New average = A + 4 = 43 + 4 = 47. Verification / Alternative check: Total runs before = 12 * 43 = 516. After scoring 95, total = 611. New average = 611 / 13 = 47, confirming the result. Why Other Options Are Wrong: 43: This is the old average, not the new one. 45, 46, 49: Do not match the increase of 4 when combined with a feasible old average given a 95-run innings. Common Pitfalls: Averaging 95 with the old average directly or forgetting to divide the new total by 13. Always write the equation using totals and solve systematically. Final Answer: The new batting average is 47 runs per match.

Question 2

In a class of 39 students, there are 26 girls with an average weight of 42 kg. The average weight of the entire class is 48 kg. What is the average weight (in kg) of the boys in the class?

Accepted Answer

Introduction: This is a weighted-average problem. Knowing the average for the whole class and for a subgroup (girls), we can compute the subgroup complement (boys) average by converting to totals and subtracting. Given Data / Assumptions: Total students = 39. Girls = 26 with average 42 kg. Boys = 39 − 26 = 13. Class average = 48 kg. Concept / Approach: Compute total class weight and total girls' weight. The difference is total boys' weight. Divide by number of boys to get boys' average weight. Step-by-Step Solution: Total class weight = 39 * 48 = 1872 kg. Total girls' weight = 26 * 42 = 1092 kg. Total boys' weight = 1872 − 1092 = 780 kg. Boys' average = 780 / 13 = 60 kg. Verification / Alternative check: Weighted average using 26 at 42 and 13 at 60 yields (26*42 + 13*60) / 39 = (1092 + 780) / 39 = 1872 / 39 = 48, consistent with the class average. Why Other Options Are Wrong: 54, 58, 62, 66: None produce the correct overall average when combined with 26 girls at 42 kg. Common Pitfalls: Averaging 42 and 48 directly or dividing by the wrong headcount. Always work with totals for weighted-average questions. Final Answer: The average weight of the boys is 60 kg.

Question 3

A factory buys 7 machines: 2 of type A, 4 of type B, and 1 of type C. Their prices are Rs 56,000, Rs 45,000, and Rs 30,000 respectively. What is the average cost per machine?

Accepted Answer

Introduction: This is a straightforward weighted-average price question. We compute the total cost for all machines by type and divide by the total number of machines to find the average cost per unit. Given Data / Assumptions: Type A: 2 machines at Rs 56,000 each. Type B: 4 machines at Rs 45,000 each. Type C: 1 machine at Rs 30,000. Total machines = 7. Concept / Approach: Average cost = (Total cost of all machines) / (Total number of machines). Compute by summing each category's cost contribution. Step-by-Step Solution: Cost of A = 2 * 56,000 = Rs 112,000. Cost of B = 4 * 45,000 = Rs 180,000. Cost of C = 1 * 30,000 = Rs 30,000. Total cost = 112,000 + 180,000 + 30,000 = Rs 322,000. Average cost = 322,000 / 7 = Rs 46,000. Verification / Alternative check: Multiply back: Rs 46,000 * 7 = Rs 322,000, which matches the computed total cost across types A, B, and C. Why Other Options Are Wrong: Rs 43,666.66, Rs 41,333.33, Rs 44,500, Rs 45,000: None equal the exact quotient 322,000 / 7. Common Pitfalls: Averaging the three listed prices directly without weighting by counts, or forgetting that there is only one machine of type C. Final Answer: The average cost per machine is Rs 46,000.

Question 4

Five numbers have an average of 72. The first number equals one-eighth of the sum of the other four numbers. What is the value of the first number?

Accepted Answer

Introduction: Averages convert to totals cleanly. If one number is a stated fraction of the sum of the others, set up an equation in terms of the sum of the remaining numbers and solve for the first number directly. Given Data / Assumptions: Average of 5 numbers = 72 ⇒ total T = 5 * 72 = 360. Let S = sum of the other four numbers. First number = (1/8) * S. Concept / Approach: Total T = first + S = (1/8)S + S = (9/8)S. Solve for S from T, then compute first = S / 8. Step-by-Step Solution: (9/8) * S = 360 ⇒ S = 360 * (8/9) = 320. First number = S / 8 = 320 / 8 = 40. Verification / Alternative check: Check totals: First = 40, others sum to 320, total = 360, average = 360 / 5 = 72, consistent with the given average. Why Other Options Are Wrong: 60, 26, 80, 32: None satisfy the relation first = (1/8) of the sum of the other four with total 360. Common Pitfalls: Adding 1/8 directly to the average or misinterpreting “one-eighth of the sum of other four” as one-eighth of the total. Always define variables clearly and use totals. Final Answer: The first number is 40.

Question 5

The average weight of Shivshankar, Gopesh, and Reena is 97 kg. The average of Shivshankar and Gopesh is 93 kg, and the average of Gopesh and Reena is 82 kg. What is the weight of Gopesh (in kg)?

Accepted Answer

Introduction: When multiple averages over overlapping pairs and a triple are given, add the pair sums and compare with the triple sum to isolate the shared person's value. This uses linear relationships among sums efficiently. Given Data / Assumptions: Avg(S, G, R) = 97 ⇒ S + G + R = 291. Avg(S, G) = 93 ⇒ S + G = 186. Avg(G, R) = 82 ⇒ G + R = 164. Concept / Approach: Add (S + G) and (G + R) to get S + 2G + R, which equals (S + G + R) + G. Compare with the triple sum to solve for G. Step-by-Step Solution: (S + G) + (G + R) = 186 + 164 = 350. But S + 2G + R = (S + G + R) + G = 291 + G. Therefore, 350 = 291 + G ⇒ G = 59. Verification / Alternative check: Compute S = 186 − 59 = 127 and R = 164 − 59 = 105. Check: 127 + 59 + 105 = 291 and averages match the given values. Why Other Options Are Wrong: 72, 56, 63, 61: Substituting any of these for G breaks one or more of the provided average relationships. Common Pitfalls: Trying to average the three given averages directly or solving for the wrong variable first. Focus on sums: pair sums vs triple sum lets you isolate G quickly. Final Answer: The weight of Gopesh is 59 kg.

Question 6

The average revenues of a company over seven consecutive years is Rs 79 lakhs. The average of the first four years is Rs 74 lakhs and the average of the last four years is Rs 86 lakhs. What is the revenue in the 4th year?

Accepted Answer

Introduction: This classic sliding-window average problem uses the fact that the 4th year is included in both the first four-year block and the last four-year block. Summing those blocks and subtracting the seven-year total isolates the shared year's revenue. Given Data / Assumptions: Avg of 7 years = 79 ⇒ total of 7 years = 7 * 79 = 553 lakhs. Avg of first 4 years = 74 ⇒ sum(first four) = 296 lakhs. Avg of last 4 years = 86 ⇒ sum(last four) = 344 lakhs. Concept / Approach: The 4th year appears in both four-year sums. Therefore, revenue in year 4 = sum(first four) + sum(last four) − sum(all seven). Step-by-Step Solution: Revenue in 4th year = 296 + 344 − 553 = 87 lakhs. Verification / Alternative check: Constructing a symbolic sequence Y1..Y7 with the two window sums confirms that only Y4 remains after subtraction, validating the formula. Why Other Options Are Wrong: 89, 85, 83, 81 lakhs: None satisfy the overlapping-window identity with the given averages. Common Pitfalls: Averaging the two window averages (74 and 86) to get 80, which ignores the overlap and is incorrect. Use totals and the overlap logic instead. Final Answer: Revenue in the 4th year is Rs 87 lakhs.

Question 7

The average of a, b, and c is 20, and the average of b, c, and d is 25. If d = 30, what is the value of a?

Accepted Answer

Introduction: Two overlapping triples with known averages allow quick recovery of shared pair sums and then the remaining unknown. Translate averages to totals, subtract, and isolate a. Given Data / Assumptions: (a + b + c) / 3 = 20 ⇒ a + b + c = 60. (b + c + d) / 3 = 25 ⇒ b + c + d = 75. d = 30. Concept / Approach: First compute b + c from the second total using d, then substitute into the first total to solve for a. Step-by-Step Solution: b + c = 75 − d = 75 − 30 = 45. a = (a + b + c) − (b + c) = 60 − 45 = 15. Verification / Alternative check: Plugging back: a + b + c = 15 + 45 = 60 (OK). b + c + d = 45 + 30 = 75 (OK). Why Other Options Are Wrong: 25, 30, 45, 20: Do not satisfy both given average relationships when d = 30. Common Pitfalls: Averaging the two averages directly, or treating a as equal to d. Always compute pair sums first from totals. Final Answer: The value of a is 15.

Question 8

The average monthly salary of 19 members of a group is Rs 16,000. One more member with a monthly salary of Rs 20,000 joins the group. What is the new average monthly salary of the group?

Accepted Answer

Introduction: Adding one new data point shifts the average depending on how the new value compares with the old average. Compute the old total, add the newcomer's salary, then divide by the new headcount. Given Data / Assumptions: Old average (19 members) = Rs 16,000. New member's salary = Rs 20,000. New headcount = 20. Concept / Approach: New average = (Old total + newcomer) / New headcount. Old total = 19 * 16,000. Step-by-Step Solution: Old total = 19 * 16,000 = Rs 304,000. New total = 304,000 + 20,000 = Rs 324,000. New average = 324,000 / 20 = Rs 16,200. Verification / Alternative check: Since 20,000 is 4,000 above the old average, the increase in average should be 4,000 / 20 = 200. Old average 16,000 + 200 = 16,200, matching the calculation. Why Other Options Are Wrong: 18,250 and 18,000: Far too high for a single addition of 20,000 to a base near 16,000. 16,250: Does not match the exact computation. 16,000: The average must rise because 20,000 exceeds 16,000. Common Pitfalls: Averaging 16,000 and 20,000 without weighting by the 19 existing members. Always use totals and the new count. Final Answer: The new average salary is Rs 16,200 per month.

Question 9

In a 50-over innings, the run rate in the first 45 overs is 5.8 runs per over. What run rate is required in the remaining 5 overs to reach a target of 295 runs?

Accepted Answer

Introduction: Cricket run-rate problems combine totals and rates. Compute runs already scored, subtract from the target, and divide by remaining overs to get the required rate. Given Data / Assumptions: Target runs = 295. First 45 overs at 5.8 runs per over. Remaining overs = 5. Concept / Approach: Runs scored so far = run rate * overs = 5.8 * 45. Required runs = target − scored. Required run rate = required runs / remaining overs. Step-by-Step Solution: Runs scored = 5.8 * 45 = 261. Runs needed = 295 − 261 = 34. Required rate = 34 / 5 = 6.8 runs per over. Verification / Alternative check: At 6.8 for 5 overs you get 34 runs; 261 + 34 = 295, hitting the target exactly. Why Other Options Are Wrong: 6.6, 6.7: Fall short of the needed 34 runs in 5 overs. 6.9, 7.2: Overshoot the target; higher than necessary. Common Pitfalls: Rounding 5.8 * 45 incorrectly or forgetting to use the remaining overs (5, not 50 − 45 wrongly reapplied). Final Answer: The required run rate is 6.8 runs per over.

Question 10

The average of 15 numbers is 7. The average of the first 8 numbers is 6.5, and the average of the last 8 numbers is 8.5. What is the middle (8th) number?

Accepted Answer

Introduction: When two overlapping blocks of numbers share only the middle element, the sums of the blocks and the total allow you to isolate the shared middle value. This is a standard technique for consecutive-list average problems. Given Data / Assumptions: Avg of all 15 = 7 ⇒ total T = 15 * 7 = 105. Avg of first 8 ⇒ sum F = 8 * 6.5 = 52. Avg of last 8 ⇒ sum L = 8 * 8.5 = 68. The middle number is counted once in T, but in both F and L. Concept / Approach: Because the 8th number is included in both block sums, F + L double-counts it relative to T. Therefore, middle = F + L − T. Step-by-Step Solution: Middle = 52 + 68 − 105 = 15. Verification / Alternative check: Let the middle be m. Then T = (sum of first 7) + m + (sum of last 7). F = (sum of first 7) + m; L = (sum of last 7) + m. Therefore F + L − T = m, which matches the computation. Why Other Options Are Wrong: 10, 13, 9, 23: None satisfy the identity F + L − T with the given averages. Common Pitfalls: Averaging 6.5 and 8.5 directly or assuming symmetry without overlap arithmetic. Use sums with the overlap formula. Final Answer: The middle number is 15.

Question 11

In a 50-over innings, the run rate in the first 40 overs is 4.8 runs per over. What run rate is required in the final 10 overs to reach a target of 241 runs?

Accepted Answer

Introduction: Compute runs already scored using the initial run rate and overs, find the remaining runs needed, then divide by the overs left to obtain the required run rate for the end phase of the innings. Given Data / Assumptions: Target = 241 runs. First 40 overs at 4.8 runs per over. Remaining overs = 10. Concept / Approach: Runs scored = 4.8 * 40. Runs required = target − scored. Required run rate = required runs / 10. Step-by-Step Solution: Runs scored = 4.8 * 40 = 192. Runs needed = 241 − 192 = 49. Required rate = 49 / 10 = 4.9 runs per over. Verification / Alternative check: At 4.9 for 10 overs you get 49 runs; 192 + 49 = 241, exactly the target. Why Other Options Are Wrong: 6.7, 6.5, 6.3: Much higher than needed for the remaining overs. 5.4: Overshoots the target; higher than the computed requirement. Common Pitfalls: Forgetting that only 10 overs remain, or miscomputing 4.8 * 40. Keep arithmetic precise. Final Answer: The required run rate is 4.9 runs per over.

Question 12

What is the average of all numbers between 11 and 80 (inclusive bounds not required) that are divisible by 6?

Accepted Answer

Introduction: Finding the average of equally spaced numbers (an arithmetic progression) uses a simple midpoint property. For any AP, the average of all terms equals the average of the first and last term. Given Data / Assumptions: Multiples of 6 in the open interval (11, 80): 12, 18, ..., 78. First term a = 12, last term l = 78. Concept / Approach: For an AP, Average = (first + last) / 2. No need to count terms explicitly if first and last are known. Step-by-Step Solution: Average = (12 + 78) / 2 = 90 / 2 = 45. Verification / Alternative check: The list is symmetric around 45: pairs (12,78), (18,72), ..., each summing to 90, confirming the average 45. Why Other Options Are Wrong: 46, 47, 44, 48: Do not equal the AP mean (12 + 78)/2. Common Pitfalls: Enumerating all terms unnecessarily or mistakenly including 6 or 84. Respect the interval and the divisibility rule. Final Answer: The required average is 45.

Question 13

The average temperature of Monday–Thursday is 60°. The average of Tuesday–Friday is 63°. If the ratio of Monday to Friday temperatures is 21 : 25, what is Friday’s temperature?

Accepted Answer

Introduction: Overlapping four-day averages allow elimination of common days to get a relation between Monday and Friday. The given ratio then provides absolute values. This is a standard overlapping-averages technique. Given Data / Assumptions: (Mon + Tue + Wed + Thu) / 4 = 60 ⇒ M + T + W + Th = 240. (Tue + Wed + Thu + Fri) / 4 = 63 ⇒ T + W + Th + F = 252. M : F = 21 : 25. Concept / Approach: Subtract the two four-day sums to get F − M. Combine with the ratio M : F to solve for M and F numerically. Step-by-Step Solution: (T + W + Th + F) − (M + T + W + Th) = 252 − 240 ⇒ F − M = 12. Let M = 21k and F = 25k (from the ratio). F − M = 25k − 21k = 4k = 12 ⇒ k = 3. Therefore, F = 25k = 25 * 3 = 75°. Verification / Alternative check: M = 21 * 3 = 63°. Check sums: M + T + W + Th and T + W + Th + F differ by 12, which matches F − M = 75 − 63 = 12. Why Other Options Are Wrong: 70°, 73°, 78°, 69°: Do not satisfy both the difference F − M = 12 and the ratio M : F = 21 : 25. Common Pitfalls: Averaging 60 and 63 directly or using 21 : 25 incorrectly (e.g., reversing M and F). Use the subtraction step first, then apply the ratio. Final Answer: Friday’s temperature is 75°.

Question 14

In a class of 50, the average height of 30 boys is 160 cm. The remaining students have an average height of 165 cm. What is the average height of the whole class (in cm)?

Accepted Answer

Introduction: This is a weighted-average calculation across two subgroups. Convert each subgroup's average to a total, add, and divide by the total headcount to find the overall average. Given Data / Assumptions: Total students = 50. Group 1: 30 at 160 cm. Group 2: 20 at 165 cm (the remaining students). Concept / Approach: Overall average = (Sum of subgroup totals) / 50. Subgroup totals are count * average for each group. Step-by-Step Solution: Total height group 1 = 30 * 160 = 4800 cm. Total height group 2 = 20 * 165 = 3300 cm. Total height overall = 4800 + 3300 = 8100 cm. Overall average = 8100 / 50 = 162 cm. Verification / Alternative check: Weighted average view: (30/50)*160 + (20/50)*165 = 96 + 66 = 162, which matches. Why Other Options Are Wrong: 161, 163, 164, 160: Do not equal the computed weighted average. Common Pitfalls: Averaging 160 and 165 directly to get 162.5 without considering group sizes; correct method uses weights 30 and 20. Final Answer: The class average height is 162 cm.

Question 15

The average marks of a class of 60 students is 65. The average marks of half of the students (30 students) is 85. What is the average marks of the remaining 30 students?

Accepted Answer

Introduction: Overall average with subgroup average known lets you find the complement subgroup average by using totals. Compute totals from averages and subtract to find the remaining group's sum and average. Given Data / Assumptions: Total students = 60, class average = 65 ⇒ total marks = 60 * 65 = 3900. Top half (30 students) average = 85 ⇒ their total = 30 * 85 = 2550. Remaining half = 30 students. Concept / Approach: Remaining group total = overall total − known group total. Divide by 30 to get the remaining group's average. Step-by-Step Solution: Remaining total = 3900 − 2550 = 1350. Remaining average = 1350 / 30 = 45. Verification / Alternative check: Weighted average check: (30 * 85 + 30 * 45) / 60 = (2550 + 1350)/60 = 3900/60 = 65, consistent with the class average. Why Other Options Are Wrong: 35, 55, 65, 40: None match the exact complementary average computed from totals. Common Pitfalls: Averaging 85 and 65 directly or making assumptions about distribution. The complement must satisfy the total-marks constraint. Final Answer: The average of the remaining students is 45.

Question 16

The average weight of 10 parcels is 1.7 kg. After adding one more parcel, the average drops by 0.06 kg. What is the weight of the new parcel (in kg)?

Accepted Answer

Introduction: Average change problems with one new item are solved by comparing old and new totals. The difference between the new total and the old total equals the weight of the added parcel. Given Data / Assumptions: Old average (10 parcels) = 1.7 kg ⇒ old total = 10 * 1.7 = 17.0 kg. New count = 11 parcels. New average = 1.7 − 0.06 = 1.64 kg. Concept / Approach: New total = new average * 11. The difference between the new total and old total equals the single added parcel's weight. Step-by-Step Solution: New total = 11 * 1.64 = 18.04 kg. Added parcel weight = 18.04 − 17.00 = 1.04 kg. Verification / Alternative check: If the extra parcel weighs 1.04 kg, the total becomes 18.04 kg and the average 18.04 / 11 = 1.64 kg, confirming the 0.06 kg drop from 1.70 kg. Why Other Options Are Wrong: 1.08, 1.4, 1.8, 0.98: None yield the exact new total 18.04 kg and average 1.64 kg. Common Pitfalls: Subtracting 0.06 directly from 1.7 and assuming that result is the new parcel weight, or forgetting to multiply averages by counts to get totals. Final Answer: The new parcel weighs 1.04 kg.

Question 17

In a class of 50 students, there are 27 boys whose average weight is 72 kg, and the average weight of the whole class is 55.44 kg. What is the average weight in kilograms of the girls in the class?

Accepted Answer

Introduction / Context: This quantitative aptitude problem tests your understanding of weighted averages and how to separate the average of a subgroup (girls) from information about the whole group (entire class) and another subgroup (boys). Such questions are very common in competitive exams because they combine basic arithmetic with logical reasoning about groups and totals. Given Data / Assumptions: Total number of students in the class = 50. Number of boys = 27 and therefore number of girls = 50 - 27 = 23. Average weight of the boys = 72 kg. Average weight of the full class = 55.44 kg. Weights are measured in kilograms and we assume all given averages are correct and precise. Concept / Approach: The key idea is that average = total weight / number of students. From the overall average and the class size, we can find the total weight of all students. From the boys' average and the number of boys, we can find the total weight of the boys. Subtracting the boys' total from the class total gives the total weight of the girls. Dividing that by the number of girls gives the girls' average weight. Step-by-Step Solution: Step 1: Compute total weight of all 50 students. Total class weight = 55.44 * 50 = 2772 kg. Step 2: Compute total weight of the 27 boys. Total boys' weight = 72 * 27 = 1944 kg. Step 3: Find total weight of the 23 girls. Total girls' weight = 2772 - 1944 = 828 kg. Step 4: Compute average weight of the girls. Average girls' weight = 828 / 23 = 36 kg. Verification / Alternative Check: We can verify by recomputing the overall average. Boys contribute 1944 kg and girls contribute 828 kg, giving a total of 2772 kg. Dividing by 50 students gives 2772 / 50 = 55.44 kg, which matches the given class average. This confirms that the girls' average weight of 36 kg is consistent with all the data. Why Other Options Are Wrong: 42 kg and 48 kg are too high; if girls were that heavy on average, the class average would be much closer to the boys' average of 72 kg. That would contradict the relatively low class average of 55.44 kg. 30 kg is too low; it would pull the class average significantly below 55.44 kg once combined with the heavier boys, which is not correct. Common Pitfalls: A common mistake is to simply average 72 kg and 55.44 kg or to guess a value between them, which ignores the different group sizes. Another frequent error is to forget to compute totals and instead directly try to manipulate averages. Always remember that when subgroups are involved, you must work through total sums before returning to averages. Final Answer: The average weight of the girls in the class is 36 kg.

Question 18

What is the average of all the numbers between 9 and 90 (inclusive) that are exactly divisible by 8?

Accepted Answer

Introduction / Context: This aptitude question checks your understanding of arithmetic progressions and averages of evenly spaced numbers. When numbers are equally spaced, such as multiples of a fixed number within a range, there is a very simple shortcut for finding their average without adding all of them one by one. Given Data / Assumptions: We consider numbers between 9 and 90 that are divisible by 8. These numbers are positive integers and must be multiples of 8. We take the range as inclusive where applicable, meaning we check if endpoints are multiples of 8. We are required to find the average of all such multiples of 8 within the range. Concept / Approach: Multiples of 8 within a continuous range form an arithmetic progression (AP) with a common difference of 8. For any arithmetic progression, the average (or mean) of all its terms is simply equal to the average of the first and last terms, that is, (first term + last term) / 2. So our task reduces to identifying the smallest and largest multiples of 8 between 9 and 90 and applying this property. Step-by-Step Solution: Step 1: Find the first multiple of 8 greater than 9. Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, ... The first multiple of 8 greater than 9 is 16. Step 2: Find the last multiple of 8 less than or equal to 90. Looking at the list, 88 is the last multiple of 8 below 90. Step 3: Use the AP average shortcut. Average of all these multiples = (first + last) / 2 = (16 + 88) / 2 = 104 / 2 = 52. Verification / Alternative Check: The full list of multiples of 8 between 9 and 90 is 16, 24, 32, 40, 48, 56, 64, 72, 80 and 88. These are 10 terms with constant difference 8. The middle pair is 48 and 56 and their average is (48 + 56) / 2 = 52. In a symmetric arithmetic progression, this must equal the overall average, which confirms our previous result. Why Other Options Are Wrong: 50 and 51 are slightly lower than the correct average and would correspond to a range whose endpoints are closer together than 16 and 88. They do not fit the true arithmetic progression here. 53 is slightly higher than the correct average, again inconsistent with the symmetric placement of numbers around 52. Common Pitfalls: Many students try to add all the numbers and divide, which is time consuming and error prone. Others mistakenly average 9 and 90 or misidentify the first or last multiple of 8 in the interval. Remember to always first list or identify the exact multiples within the range and then use the (first + last) / 2 shortcut for arithmetic progressions. Final Answer: The required average of all numbers between 9 and 90 that are divisible by 8 is 52.

Question 19

An army of 12,000 soldiers consists of Europeans and Indians. The average height of a European soldier is 5 ft 10 inches and the average height of an Indian soldier is 5 ft 9 inches. The average height of the whole army is 5 ft 9 3/4 inches. How man…

Accepted Answer

Introduction / Context: This problem is a classic example of weighted averages, where two groups (Europeans and Indians) with different average heights together form a larger group with a different combined average. The question asks you to work backwards from the combined average to determine the number of soldiers in one subgroup, in this case the number of Indians. Given Data / Assumptions: Total soldiers in the army = 12,000. Average height of a European soldier = 5 ft 10 inches = 70 inches. Average height of an Indian soldier = 5 ft 9 inches = 69 inches. Average height of the whole army = 5 ft 9 3/4 inches = 5 ft 9.75 inches = 69.75 inches. Let the number of European soldiers be E and Indian soldiers be I, with E + I = 12000. Concept / Approach: The combined average height of the entire army is a weighted average of the European and Indian averages, weighted by their respective group sizes. Mathematically, total height of all soldiers divided by the total number of soldiers gives the combined average. We express total height as 70 * E + 69 * I and set it equal to 69.75 * 12000. Solving this system of equations gives the required number of Indians. Step-by-Step Solution: Step 1: Set up the equations. E + I = 12000. Total height equation: 70E + 69I = 69.75 * 12000. Step 2: Compute the right-hand side. 69.75 * 12000 = 837000 inches. So 70E + 69I = 837000. Step 3: Eliminate I using E + I = 12000. I = 12000 - E. Substitute into height equation: 70E + 69(12000 - E) = 837000. This simplifies to 70E + 828000 - 69E = 837000. So E + 828000 = 837000, hence E = 9000. Step 4: Find I. I = 12000 - 9000 = 3000. Verification / Alternative Check: With 9000 Europeans at 70 inches average, total European height = 9000 * 70 = 630000 inches. With 3000 Indians at 69 inches average, total Indian height = 3000 * 69 = 207000 inches. Combined total = 837000 inches. Dividing by 12000 gives 837000 / 12000 = 69.75 inches, which is exactly 5 ft 9 3/4 inches, confirming that the numbers are consistent. Why Other Options Are Wrong: 4000 or 5500 Indians would shift the combined average height noticeably closer to the Indian average of 69 inches, and away from the given 69.75 inches. 2700 Indians would make Europeans even more dominant, pulling the average closer than 69.75 inches to 70 inches, again contradicting the given combined average. Common Pitfalls: Many learners convert feet and inches incorrectly or forget to convert all heights into a single unit (like inches) before forming equations. Others try to directly average the three heights without taking the group sizes into account. Always remember that combined averages are weighted by group sizes, not simply by the number of groups. Final Answer: The number of Indian soldiers in the army is 3000.

Question 20

The average weight of three men A, B and C is 84 kg. Another man D joins the group and the average weight of the four men becomes 80 kg. Then another man E, whose weight is 3 kg more than that of D, replaces A. The average weight of B, C, D and E no…

Accepted Answer

Introduction / Context: This problem focuses on average weight and how it changes when people join or leave a group. It is a multi-step average problem where you need to track totals and substitutions carefully. Such questions are common in aptitude tests to check your control over algebraic expressions and averages when several conditions are given in sequence. Given Data / Assumptions: Average weight of A, B and C = 84 kg. Average weight of A, B, C and D = 80 kg. Weight of E is 3 kg more than the weight of D. After E replaces A, the average weight of B, C, D and E = 79 kg. All weights are in kilograms; we assume no other changes in membership apart from what is stated. Concept / Approach: The key is to convert each average into an equation involving total weights. For each group, total weight = average * number of people. We first use the averages to find the total weight of A, B, C and the weight of D. Then we use the second condition involving B, C, D and E to find B + C and then isolate A by subtraction. Systematic manipulation of these totals leads directly to the required weight of A. Step-by-Step Solution: Step 1: Compute the total weight of A, B and C. A + B + C = 84 * 3 = 252 kg. Step 2: Use the second average to find D. (A + B + C + D) / 4 = 80, so A + B + C + D = 320. Substitute A + B + C = 252 to get 252 + D = 320, so D = 68 kg. Step 3: Express E in terms of D. E = D + 3 = 68 + 3 = 71 kg. Step 4: Use the third average involving B, C, D and E. (B + C + D + E) / 4 = 79, so B + C + D + E = 316. Substitute D = 68 and E = 71 to get B + C + 68 + 71 = 316. So B + C + 139 = 316, hence B + C = 177. Step 5: Find A. From A + B + C = 252 and B + C = 177, we get A = 252 - 177 = 75 kg. Verification / Alternative Check: With A = 75, B + C = 177, so A + B + C = 252 and average of A, B, C is 252 / 3 = 84 kg as given. Adding D = 68 gives 320, giving average 80 kg for A, B, C, D. Replacing A by E = 71 gives B + C + D + E = 177 + 68 + 71 = 316 and average 79 kg, matching all conditions. This confirms that A = 75 kg is correct. Why Other Options Are Wrong: 70 kg or 72 kg would change the total of A, B and C, making it impossible to satisfy all three average conditions simultaneously. 80 kg would make A too heavy and would push earlier averages above the given values once the other weights are fixed by the equations. Common Pitfalls: Many students attempt to solve by guessing weights rather than forming equations, which is unreliable. Another common mistake is forgetting that replacing a member affects which people are included in the final average. Mixing up totals for three-person and four-person groups can also cause errors. Writing down each equation clearly and solving step by step avoids these issues. Final Answer: The weight of A is 75 kg.

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