Question 1

A car owner buys petrol at Rs 7.50, Rs 8.00 and Rs 8.50 per litre for three successive years, spending Rs 4000 in each year. Approximately what is the average cost per litre of petrol over the three-year period?

Accepted Answer

Introduction / Context:This question requires calculating an average price per litre when the buyer spends the same amount of money at different per-litre prices in successive years. This is a classic application of the harmonic mean concept, because the quantity purchased varies inversely with price when the expenditure is fixed. However, we can solve it directly with basic arithmetic by working with total money spent and total litres bought. Given Data / Assumptions: In year 1, petrol price = Rs 7.50 per litre and expenditure = Rs 4000. In year 2, petrol price = Rs 8.00 per litre and expenditure = Rs 4000. In year 3, petrol price = Rs 8.50 per litre and expenditure = Rs 4000. We must find the overall average cost per litre for the entire three-year period. Concept / Approach:The overall average cost per litre is defined as total money spent divided by total litres obtained over the three years. Since the car owner spends the same amount each year, the litres purchased in a year equals (yearly expenditure) / (price per litre). We compute the litres bought each year, sum them to get total litres, sum up the total expenditure, and then divide. Step-by-Step Solution:Total expenditure for three years = 4000 + 4000 + 4000 = Rs 12000.Year 1 price = Rs 7.50 per litre, so litres bought in year 1 = 4000 / 7.5.Year 2 price = Rs 8.00 per litre, so litres bought in year 2 = 4000 / 8.Year 3 price = Rs 8.50 per litre, so litres bought in year 3 = 4000 / 8.5.Compute these quantities: 4000 / 7.5 ≈ 533.33 litres, 4000 / 8 = 500 litres, 4000 / 8.5 ≈ 470.59 litres.Total litres bought ≈ 533.33 + 500 + 470.59 ≈ 1503.92 litres.Average cost per litre = total expenditure / total litres ≈ 12000 / 1503.92 ≈ Rs 7.98 per litre. Verification / Alternative check:We know that when equal amounts of money are spent at different prices, the average price is closer to the lower prices if more litres are bought at those prices. Here, more litres are bought at Rs 7.50 than at Rs 8.50, so the overall average must lie between 7.5 and 8.0. The computed value of about 7.98 fits naturally between these numbers and very close to 8, which is reasonable. Why Other Options Are Wrong:Option a, 6.23, is too low and would imply an unrealistically large number of litres purchased. Options c and d (8.97 and 9.89) are larger than any of the three yearly prices, which is impossible for an average price per litre when all purchases are at or below 8.50. Option e, 7.50, is just the first year's price and fails to account for higher prices in later years. Only 7.98 matches the computed average. Common Pitfalls:A frequent mistake is to simply average the three prices (7.5, 8, 8.5) arithmetically, which would give 8.0, ignoring the different quantities of petrol purchased at each price. The correct method is to use total expenditure divided by total litres. Careful calculation with division ensures that you arrive at the correct approximate figure. Final Answer:The approximate average cost per litre of petrol over the three years is Rs 7.98.

Question 2

The average weight of 45 passengers in a bus is 52 kg. Five of them, whose average weight is 48 kg, leave the bus and another five passengers, whose average weight is 54 kg, board the bus at the same stop. What is the new average weight of the passe…

Accepted Answer

Introduction / Context:This problem explores how the average changes when some members leave a group and others join, each subgroup having different average weights. It is a standard type of question in aptitude exams that reinforces the idea of working with total sums and then recomputing the average after changes in group composition. Given Data / Assumptions: Initial number of passengers = 45. Initial average weight = 52 kg. 5 passengers with average weight 48 kg leave the bus. 5 new passengers with average weight 54 kg board the bus. We need to find the new average weight of the passengers. Concept / Approach:The method is to compute the initial total weight using the initial average, adjust the total weight when some passengers leave, add the weight of the new passengers, and finally divide by the new number of passengers to get the new average. The number of passengers ends up the same (45) because 5 leave and 5 join, but the total weight changes. Step-by-Step Solution:Initial number of passengers = 45.Initial average weight = 52 kg.Initial total weight = 45 * 52 = 2340 kg.Five passengers with average weight 48 kg leave, so their total weight = 5 * 48 = 240 kg.Total weight after they leave = 2340 - 240 = 2100 kg.Five new passengers with average weight 54 kg board the bus, total weight of new passengers = 5 * 54 = 270 kg.New total weight = 2100 + 270 = 2370 kg.Number of passengers remains 45 (45 - 5 + 5).New average weight = 2370 / 45 = 52.666..., which is approximately 52.66 kg. Verification / Alternative check:The net change in total weight is the difference between the weight of the leaving group and the joining group. Leaving group weight = 240 kg, joining group weight = 270 kg, so net gain = 30 kg. Spread over 45 passengers, the average increases by 30 / 45 = 0.666..., so new average = 52 + 0.666... = 52.666..., which confirms the detailed calculation. Why Other Options Are Wrong:Option a (54.21 kg) and option d (53.45 kg) would require a much larger net gain in total weight than is actually happening. Option b (51.07 kg) suggests a decrease in the average even though heavier passengers replaced lighter ones. Option e (52.00 kg) assumes no change at all in total weight. Only 52.66 kg matches the exact arithmetic. Common Pitfalls:Some students incorrectly change the number of passengers (for example, to 40 or 50) or mis-compute the total weights of the leaving and joining groups. Others forget that the difference in subgroup averages directly affects the average of the whole group. Carefully tracking total weight and passenger count avoids these mistakes. Final Answer:The new average weight of the passengers in the bus is approximately 52.66 kg.

Question 3

Nineteen friends went to a restaurant for a combined weekend dinner party. Thirteen of them spent Rs 79 each on their dinner and the remaining six each spent Rs 4 more than the average expenditure of all 19 friends. What was the total amount of mone…

Accepted Answer

Introduction / Context:This question is more challenging because the unknown average appears both in the definition of the total and inside the description of how much some people spent. It is a typical algebraic average problem where some members spend a fixed amount and the remaining ones spend an amount defined in terms of the overall average. You must set up an equation for the total expenditure in terms of the unknown average and then solve it. Given Data / Assumptions: Total number of friends = 19. 13 friends each spent Rs 79. The remaining 6 friends each spent Rs 4 more than the overall average expenditure of all 19 friends. We are required to find the total amount spent by all 19 friends together. Concept / Approach:Let the average expenditure of all 19 friends be A rupees. Then the total expenditure of the group is 19A. We can also express the total expenditure as the sum of what the first 13 friends spent plus what the remaining 6 friends spent. Since each of the remaining 6 spent A + 4, their total is 6(A + 4). Equating these two expressions for the total leads to a linear equation in A that we can solve, and then we can compute 19A. Step-by-Step Solution:Let the overall average expenditure be A rupees.Then total expenditure of all 19 friends = 19A.Thirteen friends spent Rs 79 each, so their total expenditure = 13 * 79.Compute 13 * 79 = 1027.The remaining 6 friends each spent A + 4 rupees, so their total expenditure = 6(A + 4).Therefore, total expenditure can also be written as 1027 + 6(A + 4).Equate both expressions for total expenditure: 19A = 1027 + 6(A + 4).Expand the right side: 19A = 1027 + 6A + 24 = 1051 + 6A.Rearrange: 19A - 6A = 1051, so 13A = 1051.Hence A = 1051 / 13 ≈ 80.846 rupees.Total expenditure = 19A = 19 * 1051 / 13 = 1051 * (19 / 13) ≈ Rs 1536.07. Verification / Alternative check:We can verify with approximate values. If the average A ≈ 80.85, then the six friends each spent about 84.85. Total for 13 friends = 13 * 79 = 1027. Total for 6 friends ≈ 6 * 84.85 ≈ 509.1. Combined total ≈ 1536.1, which divided by 19 gives about 80.85 again. This matches A, so the calculation is self-consistent and the total of about Rs 1536.07 is correct. Why Other Options Are Wrong:Each other option corresponds to a different assumed average. If you work backwards from those totals and try to recompute the amounts spent by the last 6 friends as A + 4, you will find inconsistencies. The precise algebraic solution shows that only around Rs 1536.07 matches all the constraints of the problem. Common Pitfalls:Many students get confused because the average appears inside the expression A + 4 as well as in 19A. Some forget that the 6 friends depend on the final average, not the 13 friends' spending. Others attempt to guess the average instead of writing the equation. The safest approach is always to define A, express the total in two ways, and solve the resulting linear equation carefully. Final Answer:The total money spent by the 19 friends is approximately Rs 1536.07.

Question 4

If the average height of boys in a hall is 180 cm and the average height of girls in the hall is 150 cm, and the combined average height of everyone present is 165 cm, then how many girls are present if there are 78 boys?

Accepted Answer

Introduction / Context:This question tests the concept of a weighted average in the context of heights of boys and girls. The overall average is exactly midway between 180 cm and 150 cm, which hints at equal contributions from both groups. You are given the number of boys and asked to find the number of girls that makes the combined average equal to 165 cm. Given Data / Assumptions: Average height of boys = 180 cm. Average height of girls = 150 cm. Combined average height of all people present = 165 cm. Number of boys = 78. We must find the number of girls. Concept / Approach:If we let the number of girls be G, then total height of boys is 180 * 78 and total height of girls is 150 * G. The overall average is 165, so the combined total height (180 * 78 + 150 * G) divided by (78 + G) must equal 165. Solving this equation gives G. Because the overall average lies exactly halfway between 180 and 150, we can intuitively expect the numbers of boys and girls to be equal, but we confirm this algebraically. Step-by-Step Solution:Let the number of girls be G.Total height of boys = 180 * 78.Total height of girls = 150 * G.Combined average height = 165, so (180 * 78 + 150 * G) / (78 + G) = 165.Multiply both sides by (78 + G): 180 * 78 + 150G = 165(78 + G).Compute left side constant part: 180 * 78 = 14040.Right side: 165 * 78 + 165G = 12870 + 165G.So the equation is 14040 + 150G = 12870 + 165G.Rearrange: 14040 - 12870 = 165G - 150G.1170 = 15G, so G = 1170 / 15 = 78. Verification / Alternative check:If there are 78 boys and 78 girls, total people = 156. Total boys' height = 180 * 78 = 14040. Total girls' height = 150 * 78 = 11700. Combined total height = 14040 + 11700 = 25740. Average height = 25740 / 156 = 165 cm, which matches the given combined average. This confirms that the number of girls is indeed 78. Why Other Options Are Wrong:Any number of girls different from 78 would shift the weighted average away from 165. More girls (who are shorter on average) would pull the average below 165, while fewer girls would push the average closer to 180. Options such as 56, 64 or 87 do not maintain the overall average at 165 when checked through the weighted average formula. Common Pitfalls:Some students mis-handle the algebra and get G wrong, or they attempt to guess the number of girls without checking the average. Others mistakenly try to average 180 and 150 without weighting by the numbers of boys and girls. Always write the equation for total height and solve it carefully to avoid these mistakes. Final Answer:The number of girls present in the hall is 78.

Question 5

If the arithmetic mean of the three numbers a, b and c is M and it is given that ab + bc + ca = 0, then what is the arithmetic mean of the three numbers a^2, b^2 and c^2?

Accepted Answer

Introduction / Context:This is an algebra-based average question that tests your understanding of how expressions involving a, b and c relate to their mean. You are told that the mean of a, b and c is M and that the sum ab + bc + ca is zero. From this information, you must determine the mean of the squares a^2, b^2 and c^2. The problem requires careful use of algebraic identities. Given Data / Assumptions: The arithmetic mean of a, b, c is M, so (a + b + c) / 3 = M. This implies a + b + c = 3M. We are also given ab + bc + ca = 0. We need the arithmetic mean of a^2, b^2, c^2, which is (a^2 + b^2 + c^2) / 3. Concept / Approach:The main algebraic identity used is (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). Since we know both a + b + c and ab + bc + ca, we can plug them into this identity to find a^2 + b^2 + c^2. Dividing that result by 3 will give the required mean of the squares. Step-by-Step Solution:From the given mean, we have (a + b + c) / 3 = M, so a + b + c = 3M.We are given that ab + bc + ca = 0.Use the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).Substitute a + b + c = 3M and ab + bc + ca = 0 into the identity.Left-hand side: (a + b + c)^2 = (3M)^2 = 9M^2.Right-hand side: a^2 + b^2 + c^2 + 2 * 0 = a^2 + b^2 + c^2.Therefore, a^2 + b^2 + c^2 = 9M^2.The mean of a^2, b^2, c^2 is (a^2 + b^2 + c^2) / 3 = 9M^2 / 3 = 3M^2. Verification / Alternative check:To verify, we can test with simple numbers that satisfy ab + bc + ca = 0 and a + b + c = 3M. For example, choose a = t, b = -t, c = 3M so that ab + bc + ca becomes -t^2 + t(3M) - t(3M) = -t^2. To make this zero, choose t = 0, giving a = 0, b = 0, c = 3M. Then the mean of a, b, c is M and ab + bc + ca = 0. The squares are 0, 0, 9M^2, whose mean is 3M^2, matching our formula. Why Other Options Are Wrong:Option a, M, ignores the effect of squaring the numbers and the zero condition on ab + bc + ca. Option c, 2(M/3), and option d, M^2, do not arise from the correct algebraic identity. Option e, 3M, treats the sum of the squares as if it scaled linearly with M, which is not correct since we are dealing with squares. Only 3M^2 is consistent with the identity and the given condition. Common Pitfalls:Students sometimes forget the correct form of the expansion (a + b + c)^2 or mis-handle the term 2(ab + bc + ca). Others try to guess the answer from dimensional thinking without using the given ab + bc + ca = 0 condition. Always recall the full identity and substitute systematically to avoid errors. Final Answer:The arithmetic mean of a^2, b^2 and c^2 is 3M^2.

Question 6

Twelve boys agreed to contribute Rs 750 each to an orphanage. Before paying, some boys dropped out and so the remaining boys each had to pay Rs 150 more than originally planned. How many boys dropped out of the contribution plan?

Accepted Answer

Introduction / Context:This contribution problem is a classic example of how a fixed total amount, when divided among fewer people, increases the per-head share. The original plan is that 12 boys each pay Rs 750. Then some boys drop out, and the remaining boys have to pay Rs 150 more each to keep the total contribution the same. We must determine how many boys dropped out. Given Data / Assumptions: Originally, there were 12 boys. Each boy was supposed to pay Rs 750. Some number of boys dropped out before paying. Each of the remaining boys then paid Rs 150 more, that is Rs 900 each. The total contribution amount remained the same as originally planned. We must find how many boys dropped out. Concept / Approach:First, compute the original total planned contribution. Then let the number of boys who actually paid be N. These N boys each pay Rs 900 (750 + 150), and the total amount must equal the original total. Equating these totals gives a simple equation in N. Finally, the number of boys who dropped out is 12 minus N. Step-by-Step Solution:Original number of boys = 12.Original planned contribution per boy = Rs 750.So, original total contribution = 12 * 750 = Rs 9000.After some boys dropped out, each remaining boy paid Rs 150 more, that is 750 + 150 = Rs 900.Let the number of boys who actually paid be N.Total contribution with N boys is N * 900.This total must equal the original total, so N * 900 = 9000.Therefore, N = 9000 / 900 = 10.Number of boys who dropped out = 12 - 10 = 2. Verification / Alternative check:We can check by plugging the numbers back. If 10 boys each pay Rs 900, total contribution = 10 * 900 = Rs 9000, which is exactly the same as the original planned total. The drop from 12 to 10 boys is 2 boys, confirming that 2 boys dropped out. The change in per-head contribution is indeed Rs 150, matching the question. Why Other Options Are Wrong:If 3 or more boys had dropped out, the remaining number of boys would be less than 10 and they would have to pay more than Rs 900 each to maintain the same total. Similarly, if only 1 boy had dropped out, 11 boys paying Rs 900 each would contribute Rs 9900, which is more than the original total. Options 4, 5 or 6 would yield even larger discrepancies. Only 2 boys dropped out fits the numbers correctly. Common Pitfalls:Some test takers forget that the total contribution does not change and treat Rs 900 as the new target total per boy without connecting it to the original Rs 9000. Others miscompute the original total or divide by the wrong amount when solving for N. Always compute the original total first, then use the new per-head amount to solve for the new number of contributors. Final Answer:The number of boys who dropped out is 2.

Question 7

The average of six numbers is 3.95. The average of two of these numbers is 3.4, and the average of another two of these numbers is 3.85. What is the average of the remaining two numbers?

Accepted Answer

Introduction / Context: This problem is a direct application of the concept of averages and how they relate to total sums. It checks whether you can move comfortably between average and total, and then separate the contribution of different groups of numbers from the whole set. Given Data / Assumptions: • There are six numbers. • Average of all six numbers = 3.95. • Average of two of the numbers = 3.4. • Average of another two of the numbers = 3.85. • We assume all averages are arithmetic means and all numbers are real values. Concept / Approach: The key idea is that average = total sum / number of items. From this, total sum = average * number of items. Once we know the total sum of all six numbers, and the sums of two separate pairs of numbers, we can subtract those known sums from the overall total to get the combined sum of the remaining two numbers. Finally, we divide that remaining sum by 2 to get their average. Step-by-Step Solution: Total sum of all six numbers = 3.95 * 6 = 23.70. Sum of the first pair with average 3.4 = 3.4 * 2 = 6.80. Sum of the second pair with average 3.85 = 3.85 * 2 = 7.70. Combined sum of these four numbers = 6.80 + 7.70 = 14.50. Sum of the remaining two numbers = 23.70 − 14.50 = 9.20. Average of the remaining two numbers = 9.20 / 2 = 4.60. Verification / Alternative check: One good check is to reconstruct the overall average. If two numbers average 3.4, two numbers average 3.85 and two numbers average 4.6, then total sum = 2 * 3.4 + 2 * 3.85 + 2 * 4.6 = 6.8 + 7.7 + 9.2 = 23.7. Dividing this by 6 gives 23.7 / 6 = 3.95, which matches the original average. This confirms that the derived average of 4.6 for the remaining pair is consistent with the given information. Why Other Options Are Wrong: An average of 3.6 or 6.5 or 7.3 would change the total sum of all six numbers and would no longer give an overall average of 3.95. If you plug any of those values back and recompute the total, you will not get 23.7 as the total sum. Therefore, those options do not satisfy the relationships between the group averages and the global average. Common Pitfalls: A common mistake is to try to average the averages directly without taking into account how many numbers each average represents. Another error is to forget that the remaining two numbers must account for the difference between the known total and the sums of the other four numbers. Always move through sums first, then come back to averages to avoid algebraic confusion. Final Answer: The average of the remaining two numbers is 4.6.

Question 8

In an NGO, the daily average wage of 20 illiterate employees decreases from Rs. 25 to Rs. 10. As a result, the overall average daily wage of all employees (literate and illiterate together) decreases by Rs. 10. How many educated (literate) employees…

Accepted Answer

Introduction / Context: This question uses the idea of how changing the income of a subgroup affects the overall average. It is a classic average and total change problem, useful for understanding how a change in one part of a group influences the combined average of the entire group of people. Given Data / Assumptions: • Number of illiterate employees = 20. • Their daily wage decreases from Rs. 25 to Rs. 10. • So each illiterate employee loses Rs. 15 per day. • The average daily wage of all employees (literate and illiterate together) decreases by Rs. 10. • Wages of educated employees are assumed unchanged. Concept / Approach: When part of a group experiences a change, the overall average changes according to the total change divided by the total number of individuals. Here, we first compute the total daily loss in wages for the illiterate group. Then we relate that total loss to the given drop in the overall average to find how many total employees there are. Subtracting the known 20 illiterate employees will give the number of educated employees in the NGO. Step-by-Step Solution: Daily loss per illiterate employee = 25 − 10 = Rs. 15. Total loss for 20 illiterate employees = 20 * 15 = Rs. 300 per day. Let total number of employees (literate + illiterate) = N. Overall average wage decreases by Rs. 10, so total daily loss = 10 * N. Equate losses: 10 * N = 300. N = 300 / 10 = 30 employees in total. Number of educated employees = total employees − illiterate employees = 30 − 20 = 10. Verification / Alternative check: To verify, suppose there are 10 educated employees whose wages stay the same. The 20 illiterate employees together lose Rs. 300 per day. Spread over 30 employees, the average drop is 300 / 30 = Rs. 10 per person per day, which matches the given reduction in the overall average. This confirms that the count of educated employees must be 10 for the numbers to be consistent. Why Other Options Are Wrong: If there were 15 or 20 educated employees, the total number of employees would change and the average drop would no longer be exactly Rs. 10 when we divide the total loss of Rs. 300 by that number. The option "Data Insufficient" is incorrect because we have enough information: the number of illiterate employees, the wage change and the change in overall average fully determine the total number of employees. Common Pitfalls: One common error is to work only with averages and forget to translate them into total amounts. Another mistake is to think that the number of educated employees is directly related to the original average, when in fact the crucial link is the total loss and the new average. Always remember that average change = total change divided by total count, which is the key relationship in such problems. Final Answer: The number of educated employees working in the NGO is 10.

Question 9

A man travels from Meerut to Delhi by car at an average speed of 40 km/h. He returns from Delhi to Meerut by scooter at an average speed of 24 km/h over the same route. What is his average speed for the entire round trip?

Accepted Answer

Introduction / Context: This is a classic average speed problem where a person travels the same distance at two different speeds. The average speed for a round trip is not the simple arithmetic mean of the two speeds; instead, it is a harmonic mean because time taken at each speed is different. Many exam questions test this subtle but important concept. Given Data / Assumptions: • Speed from Meerut to Delhi = 40 km/h. • Speed from Delhi back to Meerut = 24 km/h. • Distance between Meerut and Delhi is the same in both directions. • There are no stoppages or additional delays, and speeds are constant over each leg. Concept / Approach: Let the one way distance be d kilometres. Then the total distance for the round trip is 2d. Average speed = total distance / total time. Time taken to go from Meerut to Delhi is d / 40 hours, and the time taken to return is d / 24 hours. The overall average speed is obtained by dividing 2d by the sum of these two times. This effectively becomes the harmonic mean of the two speeds when the distances are equal. Step-by-Step Solution: Let one way distance = d km. Time for onward journey = d / 40 hours. Time for return journey = d / 24 hours. Total distance for round trip = 2d km. Total time = d / 40 + d / 24. Compute the sum: d / 40 + d / 24 = d * (1 / 40 + 1 / 24). 1 / 40 + 1 / 24 = (3 / 120) + (5 / 120) = 8 / 120 = 1 / 15. Total time = d * (1 / 15) = d / 15 hours. Average speed = total distance / total time = 2d / (d / 15) = 2d * 15 / d = 30 km/h. Verification / Alternative check: Recall the formula for average speed when travelling the same distance at speeds v1 and v2: average speed = 2 * v1 * v2 / (v1 + v2). Here v1 = 40 and v2 = 24. So average speed = 2 * 40 * 24 / (40 + 24) = 1920 / 64 = 30 km/h. This matches the value calculated using total distance and total time, confirming the result. Why Other Options Are Wrong: The simple arithmetic mean of 40 and 24 is (40 + 24) / 2 = 32 km/h, which is incorrect because it ignores the differing times at each speed. Values like 27 km/h or 31 km/h do not correspond to the correct harmonic mean or the detailed time based calculation. Only 30 km/h satisfies the correct average speed formula for equal distances at different speeds. Common Pitfalls: A very common mistake is to simply average the two speeds arithmetically. Another error is to forget that average speed is defined using total distance and total time, not by averaging the speeds directly. To avoid such mistakes, always write the expressions for total distance and total time first, and only then compute the average speed. Final Answer: His average speed for the entire round trip is 30 km/h.

Question 10

In a school examination, the average score of the boys is 81 and the average score of the girls is 83. The combined average score of all students in the school is 81.8. What is the ratio of the number of girls to the number of boys who took the exam…

Accepted Answer

Introduction / Context: This question explores the relationship between group averages and the overall average of a combined group. By forming an equation using weighted averages, we can determine the relative number of boys and girls in the examination. This concept is commonly tested in aptitude exams under the topic of alligation and averages. Given Data / Assumptions: • Average score of boys = 81. • Average score of girls = 83. • Combined average score of all students = 81.8. • Let the number of boys be B and the number of girls be G. • All scores are assumed to be on the same scale and measured in marks. Concept / Approach: The overall average is a weighted average of the averages of the boys and the girls. We can express the total marks of boys as 81 * B and the total marks of girls as 83 * G. The combined total marks are then divided by the total number of students B + G to give the overall average 81.8. Setting up this equation and simplifying will allow us to find the ratio G : B directly without needing the exact numbers of students. Step-by-Step Solution: Total marks of boys = 81 * B. Total marks of girls = 83 * G. Combined average = (81B + 83G) / (B + G) = 81.8. Multiply both sides by (B + G): 81B + 83G = 81.8B + 81.8G. Rearrange terms: 81B − 81.8B + 83G − 81.8G = 0. This simplifies to −0.8B + 1.2G = 0. So, 1.2G = 0.8B. Divide both sides by 0.4: 3G = 2B. Therefore, G : B = 2 : 3. Verification / Alternative check: To verify the ratio, assume a convenient multiple, for example B = 3k and G = 2k. Then total marks = 81 * 3k + 83 * 2k = 243k + 166k = 409k. Total students = 5k. Combined average = 409k / 5k = 81.8, which matches the given overall average. Thus the assumed ratio 2 : 3 for girls to boys is correct. Why Other Options Are Wrong: Ratios such as 3 : 2, 4 : 3 or 3 : 4 do not satisfy the weighted average equation. Substituting those ratios into the same calculation would yield average scores that are either higher or lower than 81.8. Only the ratio 2 : 3 gives exactly the required overall average. Common Pitfalls: Sometimes students confuse the ratio of averages with the ratio of the numbers of students, which are different concepts. Another common mistake is to try to solve using guesswork without setting up the weighted average equation. Writing out the total marks expression and carefully simplifying the equation is the safest and most systematic method. Final Answer: The ratio of the number of girls to the number of boys is 2 : 3.

Question 11

A student's average marks in five subjects is 82. The average of the marks in the first two subjects is 86.5, and the average of the marks in the last two subjects is 84. What are the marks obtained in the third subject?

Accepted Answer

Introduction / Context: This question examines how to use averages to recover the contribution of a single unknown subject when other grouped averages are known. It is a straightforward application of converting between averages and total marks and then isolating the missing part from the whole. Given Data / Assumptions: • Average of 5 subjects = 82. • Average of first two subjects = 86.5. • Average of last two subjects (fourth and fifth) = 84. • All marks are on the same scale and are additive. • We need to find the marks in the third subject. Concept / Approach: First, compute the total marks across all five subjects using the overall average. Then compute the combined marks for the first two subjects and the last two subjects using their respective averages. The marks in the third subject are what remain after subtracting the sums of these four known subject marks from the grand total. This uses the basic identity: total of all items = sum of known items + unknown item, which we can rearrange to solve for the unknown item. Step-by-Step Solution: Total marks in 5 subjects = average * number of subjects = 82 * 5 = 410. Sum of first two subjects = 86.5 * 2 = 173. Sum of last two subjects (fourth and fifth) = 84 * 2 = 168. Combined marks of subjects 1, 2, 4 and 5 = 173 + 168 = 341. Marks in the third subject = total marks − marks of the other four subjects. Marks in third subject = 410 − 341 = 69. Verification / Alternative check: We can quickly verify by reconstructing the overall average. Assume the marks are M1, M2, M3, M4 and M5. We now know that M1 + M2 = 173, M4 + M5 = 168 and M3 = 69. Total = 173 + 69 + 168 = 410. Dividing 410 by 5 gives 82 as the average, which matches the given overall average, confirming that 69 is correct for the third subject. Why Other Options Are Wrong: If the third subject were 67, 71 or 73, the total marks would be 408, 412 or 414 respectively, which would lead to overall averages of 81.6, 82.4 or 82.8, not 82. The only value that preserves the original average when recomputed is 69, so the other options are inconsistent with the data. Common Pitfalls: A typical mistake is to try to average the averages incorrectly, or to miscount the number of subjects in each group. Another frequent error is arithmetic slip when multiplying the averages by 2. Always recompute the final average as a check to ensure that the answer maintains the given conditions. Final Answer: The marks obtained in the third subject are 69.

Question 12

A batsman scores 85 runs in his 17th innings, and as a result his batting average increases by 3 runs. What is his batting average after completing the 17th innings?

Accepted Answer

Introduction / Context: Batting average questions are standard applications of average formulas. The important idea is that when a new score is added, the new average depends on both the old average and the new score. This question asks you to determine the new average from the information that a single innings increased the average by a known amount. Given Data / Assumptions: • Before the 17th innings, the batsman has played 16 innings. • His average before the 17th innings is some value A (in runs per innings). • In the 17th innings, he scores 85 runs. • After this innings, his average increases by 3 runs, so new average = A + 3. • We need to find the new average after 17 innings. Concept / Approach: Let A be the old average. Then the total runs scored in the first 16 innings is 16 * A. After scoring 85 in the 17th innings, his new total runs are 16 * A + 85. His new average after 17 innings is given as A + 3, so we set up the equation (16A + 85) / 17 = A + 3 and solve for A. Once A is known, we add 3 to obtain the new average after 17 innings. Step-by-Step Solution: Let old average = A runs per innings. Total runs in 16 innings = 16 * A. Runs in 17th innings = 85. Total runs after 17 innings = 16A + 85. New average after 17 innings = A + 3. So, (16A + 85) / 17 = A + 3. Multiply both sides by 17: 16A + 85 = 17A + 51. Rearrange: 85 − 51 = 17A − 16A, so 34 = A. Old average A = 34, therefore new average = A + 3 = 37. Verification / Alternative check: Using A = 34, total runs before 17th innings = 16 * 34 = 544. Adding the 17th innings score: 544 + 85 = 629 runs. New average = 629 / 17. Since 17 * 37 = 629, the new average is exactly 37 runs per innings. This confirms that the calculations are consistent and the answer is correct. Why Other Options Are Wrong: A new average of 34, 35 or 36 would imply a smaller increase than the stated 3 runs, given a score of 85. Substituting any of those values into the equation would give inconsistent values of the old average or total runs. Only 37 satisfies both the increase of 3 runs and the specific additional score of 85 runs in the 17th innings. Common Pitfalls: A common mistake is to assume the given increase applies directly to the total runs instead of the average, or to misapply the formula total = average * number of innings. Some students also forget to divide total runs by the new number of innings, which is 17, not 16. Setting up the equation carefully and checking each step helps prevent these errors. Final Answer: The batsman's batting average after the 17th innings is 37 runs per innings.

Question 13

Five boxes of bananas cost Rs. 30 in total. At the same rate, how many boxes of bananas can you buy for Rs. 9?

Accepted Answer

Introduction / Context: This question is a simple unitary method or proportion problem. It tests your ability to work with direct proportion between cost and quantity, which is a basic but important skill in arithmetic and everyday calculations involving prices and rates. Given Data / Assumptions: • Cost of 5 boxes of bananas = Rs. 30. • The cost per box is constant; there are no discounts or bulk offers. • We want to find how many boxes can be bought for Rs. 9 at the same rate. Concept / Approach: Since cost is directly proportional to the number of boxes, we can first find the cost of one box by dividing the total cost by the number of boxes. Once we have the cost per box, we divide the given amount of money (Rs. 9) by the cost per box to determine how many boxes can be bought. This is the standard unitary method: find the value for one unit, then scale up or down as required. Step-by-Step Solution: Total cost of 5 boxes = Rs. 30. Cost per box = 30 / 5 = Rs. 6. Available money = Rs. 9. Number of boxes that can be bought = 9 / 6. 9 / 6 = 1.5 boxes. Verification / Alternative check: To verify, consider how much 1.5 boxes would cost at Rs. 6 per box. Cost = 1.5 * 6 = Rs. 9, which matches the available money exactly. This confirms that the calculation is consistent. Note that mathematically, fractional boxes are allowed in the calculation, even though in real life you might only be able to buy whole boxes. Why Other Options Are Wrong: If you buy 2 boxes, you would need 2 * 6 = Rs. 12, which exceeds Rs. 9. For 1.25 boxes, the cost would be 1.25 * 6 = Rs. 7.5, which is less than Rs. 9 and does not use the full amount. Similarly, 2.5 boxes would cost 2.5 * 6 = Rs. 15, which is too much. Only 1.5 boxes correspond exactly to Rs. 9 at this rate. Common Pitfalls: Students sometimes try to use ratio directly without computing the cost per box, which can lead to mistakes if done hurriedly. Another common error is to reverse the division and compute 6 / 9 instead of 9 / 6. To avoid confusion, always determine the cost for one unit first, then divide the available money by that unit cost. Final Answer: At the given rate, you can buy 1.5 boxes of bananas for Rs. 9.

Question 14

Ian has 14 boxes of paper and wants to divide them equally among 4 coworkers. If each coworker must get only whole boxes, how many whole boxes does each coworker receive?

Accepted Answer

Introduction / Context: This is an equal sharing and division problem involving whole units. It checks your ability to perform integer division and interpret remainders correctly, especially when the context restricts you to whole items such as boxes that cannot be split among coworkers. Given Data / Assumptions: • Total number of boxes of paper = 14. • Number of coworkers to receive the boxes = 4. • Each coworker must receive an equal number of boxes. • Only whole boxes can be given to each coworker; fractional boxes are not allowed. Concept / Approach: We divide the total number of boxes by the number of coworkers to see how many boxes each coworker would receive. Since the problem specifies whole boxes, we are interested in the integer part of the division. Any remainder will represent boxes left over that cannot be evenly distributed without splitting a box, which the question does not allow. Step-by-Step Solution: Total boxes = 14. Number of coworkers = 4. Compute 14 ÷ 4. 14 ÷ 4 = 3 with a remainder of 2, since 4 * 3 = 12 and 14 − 12 = 2. Therefore, each coworker can receive 3 whole boxes, and 2 boxes remain undistributed if we do not split them. Verification / Alternative check: Check by multiplying back: if each of the 4 coworkers gets 3 boxes, then total distributed = 4 * 3 = 12 boxes. This leaves 14 − 12 = 2 boxes, which matches the remainder found in the division. The sharing is equal and respects the requirement to use whole boxes, confirming that 3 is the correct number per coworker. Why Other Options Are Wrong: If each coworker got 2 boxes, only 8 boxes would be used, leaving 6 unused, which is not the maximum equal distribution. Options 2.5 and 3.5 involve fractional boxes, which are not allowed by the wording that each coworker gets whole boxes. Therefore, 3 is the only option that uses whole boxes and distributes them as evenly as possible. Common Pitfalls: A common slip is to give the exact division result 3.5 without considering the whole box condition. Another mistake is to ignore the remainder and select a smaller integer like 2, which does not distribute the boxes as much as possible. Always read the phrase about whole items carefully and choose the integer part of the division when splitting discrete objects. Final Answer: Each coworker receives 3 whole boxes of paper.

Question 15

The average price of 10 books is Rs. 12. The average price of 8 of these books is Rs. 11.75. Of the remaining two books, the price of one book is 60% more than the price of the other. What is the price of each of these two books?

Accepted Answer

Introduction / Context: This question combines average price calculations with percentage increase. It tests whether you can handle average to total conversions and then apply a percentage relationship between two unknown prices in order to determine their individual values from a known combined total. Given Data / Assumptions: • There are 10 books in total. • Average price of all 10 books = Rs. 12. • Average price of 8 books (a subset of these) = Rs. 11.75. • The remaining 2 books have prices related by: price of one book is 60% more than the price of the other. • We need the prices of the two remaining books. Concept / Approach: First, we convert each average into total price. From the overall average we find the total cost of 10 books. From the average of 8 books we find the total cost of those 8 books. The difference between these totals gives the combined cost of the remaining two books. Next, if the cheaper book costs x rupees, then the costlier book costs x + 0.6x = 1.6x. Their combined price is x + 1.6x = 2.6x, which we set equal to the known total for the two books. Solving this gives the individual prices. Step-by-Step Solution: Total price of 10 books = 10 * 12 = Rs. 120. Total price of 8 books = 8 * 11.75 = Rs. 94. Total price of the remaining two books = 120 − 94 = Rs. 26. Let the cheaper book cost = x rupees. Then the costlier book is 60% more, so its price = x + 0.6x = 1.6x. Combined price = x + 1.6x = 2.6x. Set 2.6x = 26, so x = 26 / 2.6 = 10. Therefore, cheaper book price = Rs. 10 and costlier book price = 1.6 * 10 = Rs. 16. Verification / Alternative check: Check by recomputing the totals: 8 books cost Rs. 94, the remaining two cost Rs. 10 and Rs. 16, so total = 94 + 10 + 16 = Rs. 120. Average price = 120 / 10 = Rs. 12, which matches the given overall average. Also, 16 is exactly 60% more than 10 since 10 * 0.6 = 6 and 10 + 6 = 16. Both conditions are satisfied, confirming the prices are correct. Why Other Options Are Wrong: Pairs such as 16 and 12 or 23 and 9 or 8 and 7 do not sum to Rs. 26, or do not maintain a 60% increase relationship between the two prices. For example, 16 and 12 sum to 28 and 16 is not 60% more than 12. Therefore, these options conflict with either the total price or the percentage condition. Only 10 and 16 fit both constraints. Common Pitfalls: A frequent mistake is to miscalculate the total for the 8 books or to forget to work with the combined price of two books. Another error is to misinterpret 60% more as 1.6 plus x instead of 1.6 times x. Always translate percentage increase as new price = original price * (1 + percentage in decimal). Using this correctly will prevent algebraic mistakes. Final Answer: The two remaining books are priced at Rs. 10 and Rs. 16 respectively.

Question 16

The third proportional to 10 and 20 is the number x such that 10 : 20 = 20 : x. What is the value of this third proportional?

Accepted Answer

Introduction / Context: This question is about proportion and specifically the concept of a third proportional. It tests your understanding of ratios and how to convert a statement such as “a is to b as b is to c” into an algebraic equation that you can solve for the unknown value c. Given Data / Assumptions: • We are given two numbers: 10 and 20. • A third proportional x is defined so that 10 : 20 = 20 : x. • All numbers are positive real numbers. • We need to find the value of x that satisfies this proportion. Concept / Approach: If a : b = b : c, then b is called the mean proportional and c is called the third proportional to a and b. The ratio a : b = b : c can be rewritten as a / b = b / c. From this, we can cross multiply to form an equation a * c = b * b. Solving this equation for c gives c = b^2 / a. Here a = 10, b = 20 and c = x, so we apply this formula directly. Step-by-Step Solution: Given 10 : 20 = 20 : x. Write it as a fraction: 10 / 20 = 20 / x. Simplify 10 / 20 to 1 / 2. So, 1 / 2 = 20 / x. Cross multiply: 1 * x = 2 * 20. x = 40. Alternatively, use the formula c = b^2 / a = 20^2 / 10 = 400 / 10 = 40. Verification / Alternative check: Check the proportion: 10 : 20 equals 1 : 2. Now check 20 : 40, which is also 1 : 2. Since both ratios are equal, the proportion 10 : 20 = 20 : 40 holds true. Therefore, 40 correctly acts as the third proportional to 10 and 20 according to the definition. Why Other Options Are Wrong: If x were 30, then 20 : 30 simplifies to 2 : 3, not 1 : 2, so the ratios would not match. For x = 25, 20 : 25 = 4 : 5, and for x = 20, 20 : 20 = 1 : 1. None of these equal 10 : 20, which is 1 : 2. Hence those values do not satisfy the proportional relationship required for a third proportional. Common Pitfalls: A common mistake is to confuse third proportional with mean proportional (geometric mean), where the relationship is different. Another error is to forget to cross multiply and simplify ratios before solving. To avoid confusion, always start by writing the ratio as a fraction and then apply cross multiplication carefully. Final Answer: The third proportional to 10 and 20 is 40.

Question 17

The average of five numbers is 27. If one number is removed from the set, the average of the remaining four numbers becomes 25. What is the value of the number that was excluded?

Accepted Answer

Introduction / Context: This question uses basic average and total relationships to identify a single missing or excluded number. It is a very common pattern in aptitude tests, where the overall average and a new average after removing an item are given, and you need to recover the removed item from these two pieces of information. Given Data / Assumptions: • There are initially 5 numbers. • Average of all 5 numbers = 27. • After removing one number, the average of the remaining 4 numbers = 25. • All numbers are real values and the arithmetic mean is used. Concept / Approach: The average of a set is the total sum divided by the number of elements. From this, we can find the total sum for the 5 numbers and the total sum for the remaining 4 numbers. The excluded number is exactly the difference between these two sums. So the plan is: compute total for 5, compute total for 4, and subtract to find the excluded number. Step-by-Step Solution: Total of 5 numbers = 27 * 5 = 135. Total of remaining 4 numbers = 25 * 4 = 100. Excluded number = total of 5 numbers − total of 4 numbers. Excluded number = 135 − 100 = 35. Verification / Alternative check: Assume the numbers are 4 unknowns plus 35. Their total is 100 + 35 = 135. The average with all 5 numbers is 135 / 5 = 27, which matches the original average. When you remove 35, you are left with the 4 numbers totalling 100, giving an average of 100 / 4 = 25, which matches the given new average. This confirms that 35 is indeed the excluded number. Why Other Options Are Wrong: If the excluded number were 25, the remaining total would be 135 − 25 = 110 and the new average for 4 numbers would be 110 / 4 = 27.5, not 25. Similarly, excluding 45 would leave 90 and give an average of 22.5, and excluding 55 would leave 80 and yield an average of 20. None of these match the required average of 25, so those options are inconsistent with the given data. Common Pitfalls: Some students try to average the averages directly or attempt to distribute the change in average without computing totals first, which can be confusing. Others accidentally divide the total by the wrong number of elements. To avoid these errors, always convert each average into a total, perform subtraction or addition as needed, and then recheck by recomputing the average. Final Answer: The excluded number is 35.

Question 18

Navaneet has three times as much money as Poonam. Poonam has 25% more money than Hebah. If the average amount of money held by Navaneet, Poonam and Hebah is Rs. 350, how much money does Navaneet have?

Accepted Answer

Introduction / Context: This problem combines average calculations with percentage and ratio relationships between the amounts of money held by three people. It tests your ability to express values in terms of a single variable and then use the given average to determine the actual amounts for each person, especially the one with the largest share. Given Data / Assumptions: • Let Hebah's money be H rupees. • Poonam has 25% more than Hebah, so Poonam's money = H + 25% of H. • Navaneet has three times as much as Poonam. • The average money of the three people is Rs. 350. • We need Navaneet's amount of money. Concept / Approach: First, we express the money of each person in terms of H. Then we add these expressions to find the total amount of money for all three people. Since the average is given, total money = 3 * average. Setting our expression for the total equal to this value lets us solve for H. Once we know H, we can compute Poonam's money and then Navaneet's money, which is three times Poonam's amount. Step-by-Step Solution: Let Hebah's money = H. Poonam has 25% more than Hebah, so Poonam's money = H + 0.25H = 1.25H = 5H / 4. Navaneet has three times Poonam's money, so Navaneet's money = 3 * (5H / 4) = 15H / 4. Total money of three people = H + 5H / 4 + 15H / 4. Combine over a common denominator: H = 4H / 4, so total = (4H + 5H + 15H) / 4 = 24H / 4 = 6H. Average money = total / 3 = (6H) / 3 = 2H. Given average = Rs. 350, so 2H = 350, hence H = 175. Poonam's money = 1.25 * 175 = Rs. 218.75. Navaneet's money = 3 * 218.75 = Rs. 656.25. Verification / Alternative check: Check the total money: Hebah has Rs. 175, Poonam has Rs. 218.75 and Navaneet has Rs. 656.25. Total = 175 + 218.75 + 656.25 = 1050. Dividing by 3 gives 1050 / 3 = 350, which matches the given average. The ratios also hold: Poonam has 25% more than Hebah because 175 * 1.25 = 218.75, and Navaneet has three times Poonam's money because 3 * 218.75 = 656.25. Why Other Options Are Wrong: Any other figure for Navaneet's money would change the total and therefore the average. For example, if Navaneet had only Rs. 175, the sum and resulting average would be far lower than Rs. 350. The other decimal figures in the options do not satisfy the relationships or the total sum requirement. Only Rs. 656.25 is consistent with both the given average and the stated relationships between the three amounts. Common Pitfalls: Mistakes often occur in handling the percentage increment or in converting 25% to a fraction. Some students also forget that the average of three people means the total is three times the average. Carefully organizing the relationships in terms of a single variable and working step by step helps prevent algebraic errors. Final Answer: Navaneet has Rs. 656.25.

Question 19

A student's average marks in 7 subjects is 85. His average in 6 subjects, excluding English, is 83. What marks did he score in English?

Accepted Answer

Introduction / Context: This is another average and total marks problem where we must isolate one subject's marks given two different averages. It reinforces the idea that averages can always be converted back into totals, which then allows us to find the missing value by simple subtraction. Given Data / Assumptions: • The student has 7 subjects in total. • Average marks in all 7 subjects = 85. • Average marks in 6 subjects, excluding English, = 83. • All marks are on the same scale and additive. • We need the marks obtained in English. Concept / Approach: We first compute the grand total of marks for all 7 subjects from the given overall average. Next, we compute the total marks for the 6 subjects other than English using their average. The difference between the grand total and the 6 subject total must be the marks in English. This exploits the relationship total = average * number of items and the fact that the overall total is the sum of all individual subject scores. Step-by-Step Solution: Total marks in 7 subjects = 7 * 85 = 595. Total marks in the 6 subjects excluding English = 6 * 83 = 498. Marks in English = total marks for 7 subjects − total marks for the other 6 subjects. Marks in English = 595 − 498 = 97. Verification / Alternative check: We can reconstruct the averages as a check. Suppose English marks are 97, and the sum of the other 6 subjects is 498. Then total = 498 + 97 = 595. The overall average = 595 / 7 = 85, which matches the original data. For the 6 subjects excluding English, the average is 498 / 6 = 83, also matching the given information. This double confirmation shows that 97 is correct. Why Other Options Are Wrong: If English were 98, then the total marks would be 498 + 98 = 596 and the average over 7 subjects would be 596 / 7, which is not 85. Similarly, English marks of 92 or 93 would lead to totals and averages that do not satisfy both the 7 subject and 6 subject averages. Only 97 maintains both conditions at the same time. Common Pitfalls: Some students mistakenly divide rather than subtract or try to re average the 6 subjects in a complicated way. Others miscalculate the totals due to small arithmetic errors. It is helpful to always write down the two totals clearly and then subtract them, followed by a quick recomputation of the averages to verify the result. Final Answer: The student scored 97 marks in English.

Question 20

Find the average of the first 50 natural numbers (1, 2, 3, ..., 50).

Accepted Answer

Introduction / Context: This question asks for the average of the first 50 natural numbers. It is a standard problem involving arithmetic progressions. The first n natural numbers form a simple arithmetic sequence with a well known sum formula, and their average can be obtained directly using symmetry or the sum formula divided by n. Given Data / Assumptions: • The numbers are 1, 2, 3, ..., 50. • These numbers form an arithmetic progression with first term 1 and last term 50. • We are using the usual definition of natural numbers starting from 1. Concept / Approach: For an arithmetic progression with first term a, last term l and n terms, the average is simply (a + l) / 2. This is because the terms are symmetrically placed around the mid value. Therefore, for the first 50 natural numbers, the average is (1 + 50) / 2. Alternatively, we can use the sum formula n * (a + l) / 2 and then divide by n, which yields the same result for the mean. Step-by-Step Solution: First term a = 1. Last term l = 50. Number of terms n = 50. Average of an arithmetic progression = (first term + last term) / 2. So, average = (1 + 50) / 2. Average = 51 / 2 = 25.5. Verification / Alternative check: We can verify by using the sum formula. Sum of first n natural numbers = n * (n + 1) / 2. For n = 50, sum = 50 * 51 / 2 = 25 * 51 = 1275. Average = sum / n = 1275 / 50 = 25.5. This matches the simple average formula result, confirming the correctness. Why Other Options Are Wrong: An average of 25 would correspond to numbers symmetric around 25, for example from 1 to 49, not 1 to 50. Averages of 26 or 26.5 would require a higher centre than 25.5 and do not match the actual spread of the numbers from 1 to 50. Only 25.5 is exactly mid way between 1 and 50 and is backed by the arithmetic progression formulas. Common Pitfalls: Students may mistakenly add 50 and 1 and divide by something other than 2, or confuse the formula with that for the sum of the first n numbers. Another typical confusion is thinking that the average equals n / 2 (which here would be 25), ignoring the shift due to the sequence starting at 1. Always remember that for 1 to n, the average is (1 + n) / 2, not simply n / 2. Final Answer: The average of the first 50 natural numbers is 25.5.

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