Question 1

In a class of 55 students, 34 are girls with an average weight of 51 kg and the average weight of the whole class is 55.2 kg. What is the average weight of the boys in the class (in kg)?

Accepted Answer

Introduction / Context:This problem is a classic example of a weighted average question in aptitude tests. We know the overall average weight of the entire class, as well as the number and average weight of one subgroup, namely the girls. The task is to use this information to compute the average weight of the remaining subgroup, the boys. Such questions test understanding of how averages relate to totals and how to separate contributions of different groups. Given Data / Assumptions:- Total number of students in the class = 55- Number of girls = 34- Average weight of girls = 51 kg- Average weight of the whole class = 55.2 kg- Therefore number of boys = 55 minus 34 = 21 boys- We assume all weights are measured accurately and in kilograms Concept / Approach:The key idea is that total weight equals average weight multiplied by the number of students. First we compute the total weight of the entire class using the overall average. Then we compute the total weight of all girls using their average. Subtracting the girls total from the class total gives the total weight of the boys. Finally we divide that boys total by the number of boys to obtain the required average weight. Step-by-Step Solution:Step 1: Compute total weight of the class: total class weight = 55.2 × 55 = 3036 kg.Step 2: Compute total weight of the girls: girls total weight = 34 × 51 = 1734 kg.Step 3: Compute total weight of the boys: boys total weight = 3036 − 1734 = 1302 kg.Step 4: Number of boys = 21, so boys average weight = 1302 ÷ 21 = 62 kg.Step 5: Therefore the required average weight of the boys is 62 kg. Verification / Alternative check:If boys average weight is 62 kg, total boys weight is 62 × 21 = 1302 kg. Adding girls total 1734 kg gives 1302 + 1734 = 3036 kg, which matches the earlier computed total class weight. Dividing 3036 by 55 again gives 55.2 kg, confirming that our result is consistent with the original overall average, so the calculation is correct. Why Other Options Are Wrong:Values like 59.4 kg or 56.8 kg would lead to a boys total weight that does not sum to 3036 kg when combined with the girls total. Similarly, 60 kg or 64 kg give combined totals that do not reproduce the class average of 55.2 kg, so these alternatives are mathematically inconsistent with the given data. Common Pitfalls:A common mistake is to average 51 kg and 55.2 kg directly, which is invalid because the two groups have different sizes. Another frequent error is to divide the difference in total weight by the total number of students rather than by the number of boys. Forgetting to compute the correct number of boys from the difference between total students and girls also leads to wrong answers. Final Answer:The average weight of the boys in the class is 62 kilograms.

Question 2

The average of 19 consecutive even integers is 50. What is the value of the highest integer in this sequence?

Accepted Answer

Introduction / Context:This question tests understanding of averages and properties of consecutive even integers. When numbers are consecutive and equally spaced, the average is equal to the middle term of the sequence. Once we identify the middle term, we can easily move up or down by the common difference to reach the highest or lowest term in the list. Given Data / Assumptions:- We have 19 consecutive even integers.- The average of these 19 integers is 50.- The integers are equally spaced with a common difference of 2 because they are even and consecutive.- We are asked to find the highest integer in this sequence. Concept / Approach:For any arithmetic progression, including consecutive even integers, the average of all terms is equal to the middle term when the number of terms is odd. Since we have 19 terms, the 10th term in the ordered list will be the middle term and equal to the average. Then the highest term will be 9 steps above the middle term, each step increasing by 2, because there are 9 integers after the middle term in a list of 19 terms. Step-by-Step Solution:Step 1: Note that the number of terms is 19, which is odd, so the middle term is the 10th term.Step 2: For an arithmetic progression with an odd number of terms, the average equals the middle term.Step 3: Therefore the 10th term is equal to the average, which is 50.Step 4: There are 9 numbers after the 10th term, and each successive term increases by 2.Step 5: Highest term = 10th term + 9 × 2 = 50 + 18 = 68. Verification / Alternative check:You can list the sequence around the middle term to verify: 10th term is 50, so the sequence looks like 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68. The highest term is clearly 68. If you compute the average of these 19 terms, you again obtain 50, which confirms the method. Why Other Options Are Wrong:Values such as 88, 100, or 126 are far larger than the correct upper term of a sequence centered at 50 with only 19 even numbers and would require many more than 9 steps of size 2 from the middle term. The value 72 is closer but would correspond to 11 steps above 50, implying more than 19 terms, so it does not match the given length of the sequence. Common Pitfalls:Some learners try to build the sequence from an assumed lowest term without using the property that the average equals the middle term. Others mistakenly add 19 × 2 to the average, which is incorrect because that would count too many steps. Remember that for an odd number of equally spaced values, the average is always the central term, which greatly simplifies problems like this. Final Answer:The highest integer in the sequence is 68.

Question 3

The average marks of 45 students was initially shown as 65, but one entry was wrongly recorded as 84 instead of 48. What is the correct average of the 45 students after rectifying this mistake?

Accepted Answer

Introduction / Context:This question is about correcting an average when one of the data values has been misrecorded. We are given the original average based on an incorrect value and then told what the value should have been. The task is to adjust the total marks accordingly and recompute the true average. This type of problem appears frequently in aptitude tests to check clarity on how averages relate to total sums. Given Data / Assumptions:- Number of students = 45- Incorrect average shown = 65 marks- One student's mark was recorded as 84 instead of the correct value 48- All other entries are correct- We need the corrected average of the same 45 students Concept / Approach:The average of a set of numbers is total sum divided by the number of observations. If one item is wrong, the initial total sum is also wrong. We first reconstruct the wrong total by multiplying the reported average by the number of students. Then we remove the wrong entry and add the correct one to obtain the corrected total. Dividing this corrected total by the same number of students gives the true average. Step-by-Step Solution:Step 1: Compute the total marks based on the wrong average: wrong total = 65 × 45 = 2925 marks.Step 2: The wrong entry is 84, but the correct entry should be 48. So we must subtract 84 and then add 48.Step 3: Correct total = 2925 − 84 + 48 = 2925 − 36 = 2889 marks.Step 4: Correct average = 2889 ÷ 45 = 64.2 marks.Step 5: Therefore, the accurate class average after correction is 64.2 marks. Verification / Alternative check:The change in the total marks is simply the difference between the wrong and correct entries: 84 − 48 = 36 marks extra were counted. Dividing this extra 36 marks by 45 students gives 36 ÷ 45 = 0.8. So the correct average must be 65 − 0.8 = 64.2, which matches the value obtained by recomputing the total. This cross check confirms that the corrected average is consistent. Why Other Options Are Wrong:Values like 65.8 or 67.4 would require an increase in the total marks, but the correction actually reduces the total because 84 is larger than 48. The option 62.6 implies a much larger reduction than the 36 marks difference, and 66.0 also does not match the precise adjustment of 0.8 we computed. Only 64.2 aligns with the corrected total of 2889 marks. Common Pitfalls:Learners sometimes subtract the difference 36 directly from the average instead of dividing it by the number of students, which would give 65 − 36 = 29, an impossible result. Another mistake is to use 84 − 48 with the wrong sign, adding 0.8 rather than subtracting it. Always remember that the correction to the average is equal to the net change in total marks divided by the number of observations. Final Answer:The correct average marks of the 45 students is 64.2 marks.

Question 4

A factory buys 10 machines: 1 machine of type A, 4 machines of type B and the remaining machines of type C. Their prices are Rs 90000 for type A, Rs 40000 for each type B and Rs 40000 for each type C. What is the average cost of these 10 machines (i…

Accepted Answer

Introduction / Context:This is a straightforward average cost question involving different categories of machines purchased at given prices. By calculating the total cost of all machines and dividing by the number of machines, we obtain the average cost per machine. Problems like this appear in profit and loss or basic arithmetic sections to check whether candidates can systematically handle grouped data. Given Data / Assumptions:- Total number of machines = 10- Number of type A machines = 1, price per machine = Rs 90000- Number of type B machines = 4, price per machine = Rs 40000- Remaining machines are type C, so number of type C machines = 10 − 1 − 4 = 5- Price per type C machine = Rs 40000- We assume all prices are in rupees and there are no extra costs such as tax or transport charges. Concept / Approach:The average cost per machine is given by total cost of all machines divided by total number of machines. We therefore find the cost contribution from each type separately: type A contributes one expensive machine, types B and C contribute several machines at a lower but equal price. Then we sum these contributions and divide by 10 to get the required average. Step-by-Step Solution:Step 1: Cost of type A machines = 1 × 90000 = Rs 90000.Step 2: Cost of type B machines = 4 × 40000 = Rs 160000.Step 3: Cost of type C machines = 5 × 40000 = Rs 200000.Step 4: Total cost of all machines = 90000 + 160000 + 200000 = Rs 450000.Step 5: Average cost per machine = total cost ÷ number of machines = 450000 ÷ 10 = Rs 45000. Verification / Alternative check:We can check by inspection that most machines (9 out of 10) cost Rs 40000, and one machine costs Rs 90000, which is Rs 50000 more than 40000. When this extra Rs 50000 is spread over 10 machines, each machine effectively carries an extra Rs 5000 over 40000. Therefore average cost = 40000 + 5000 = Rs 45000, which agrees with the detailed calculation above. Why Other Options Are Wrong:An average like 56666.6 or 68333.3 is far too high, since only one machine is more expensive than Rs 40000 and the remaining nine machines are at 40000. Similarly, values like 43500 or 40000 underestimate the effect of the single Rs 90000 machine. Only Rs 45000 correctly reflects the influence of one high cost machine among nine lower cost machines. Common Pitfalls:Students sometimes average the three prices 90000 and 40000 and 40000 directly without weighting them by the number of machines, which is incorrect because different types occur in different quantities. Others may forget to count how many machines are of type C and instead assume an incorrect split. Always ensure that each price is multiplied by its correct quantity when calculating total cost. Final Answer:The average cost of the 10 machines is 45000 rupees per machine.

Question 5

The mean marks of 45 students of division A of class X is 56, of 40 students of division B is 52 and of 40 students of division C is 50. What is the combined mean of marks of all the students of the three divisions of class X?

Accepted Answer

Introduction / Context:This question is about finding a combined mean when we know the separate means and sizes of multiple groups. Here, three different divisions of a class have their own average marks and numbers of students. We must merge these to obtain the overall average for all students together. Such problems test the concept of weighted averages, where each group contributes in proportion to its size. Given Data / Assumptions:- Division A: 45 students, mean marks = 56- Division B: 40 students, mean marks = 52- Division C: 40 students, mean marks = 50- Total number of students = 45 + 40 + 40 = 125- We assume all marks are from the same examination and on the same scale. Concept / Approach:The combined mean is not the simple average of 56, 52 and 50, because each division has a different number of students. Instead, we calculate the total marks in each division using total = mean × number of students, sum these totals to get grand total marks, and then divide by the total number of students. This process gives the correct weighted average that reflects contributions of all three divisions accurately. Step-by-Step Solution:Step 1: Total marks of division A = 45 × 56 = 2520.Step 2: Total marks of division B = 40 × 52 = 2080.Step 3: Total marks of division C = 40 × 50 = 2000.Step 4: Grand total marks for all students = 2520 + 2080 + 2000 = 6600.Step 5: Total number of students = 125, so combined mean = 6600 ÷ 125 = 52.8. Verification / Alternative check:We can quickly check reasonableness. Most students are in division A and B with means 56 and 52, while division C has a slightly smaller mean of 50. Therefore the combined mean should be a bit above 52 rather than near 50 or 56. The computed value 52.8 lies between the highest and lowest divisional means and is slightly closer to 56 because division A has the largest number of students, which is consistent with intuition. Why Other Options Are Wrong:An answer like 51.4 or 52.1 is too low given that two of the three divisions have means 52 and 56. On the other hand, 54.2 is too high, as that would ignore the influence of division C with average 50. The option 53.6 also does not match the exact weighted calculation. Only 52.8 coincides with the correctly computed combined mean of all 125 students. Common Pitfalls:A frequent error is to take the simple average of 56, 52 and 50, which would be 52.67, without accounting for different numbers of students in each division. Another mistake is miscounting the total number of students or miscalculating one of the partial totals. Careful multiplication and addition followed by division by the correct total size avoids these issues. Final Answer:The combined mean of marks of all students in the three divisions is 52.8 marks.

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