Q: Find the average of all the numbers between 6 and 34 which are divisible by 5.

Concept & Logic: When a sequence of numbers forms an Arithmetic Progression (AP) with a constant difference, the average (mean) is simply the mathematical midpoint of the series. Average = (First\ Term + Last\ Term/2) Step-by-Step Solution: • Given: We need numbers between 6 and 34 that are perfectly divisible by 5. • Deduction: The first multiple of 5 after 6 is 10. The last multiple of 5 before 34 is 30. • The complete valid sequence is: 10, 15, 20, 25, 30. • Since the common difference is a constant 5, this is an AP. • Calculation: Apply the average formula using the first valid term (10) and last valid term (30). • Average = (10 + 30/2) = (40/2) = 20. Exam Strategy & Shortcut: For any evenly spaced set of numbers, never calculate the total sum. Just identify the first term and the last term, add them together, and divide by 2. It saves massive amounts of time on large sets. Common Pitfall: A standard error is manually adding all terms together (10 + 15 + 20 + 25 + 30) and making an arithmetic calculation mistake. Always default to the AP boundary shortcut to avoid unnecessary addition. Final Answer: Therefore, the correct answer is 20.

Q: A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and $x$. He found the mean to be 12. What should be the number in place of $x$?

Concept & Logic: The mean relates the total sum of all elements to the number of elements. By using the given mean and total count, we can find the required total sum and isolate the unknown variable. Total\ Sum = Mean × Total\ Count Step-by-Step Solution: • Given: The numbers are 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14, and x. • The given mean is 12. • Deduction: Count the total number of terms. There are 11 known numbers plus x, making exactly 12 terms in total. • Calculate the required total sum for the mean to be 12. • Total\ Sum = 12 × 12 = 144. • Calculation: Sum the 11 known numbers. • Known\ Sum = 3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 = 137. • Set up the algebraic equation to find x. • 137 + x = 144 • x = 144 - 137 = 7. Exam Strategy & Shortcut: You can use the "Deviation Method". Compare each number to the mean (12). Deviations: -9, -1, -5, -3, +3, +1, -4, +7, +5, +9, +2. Sum of deviations = (-9+9) + (-1+1) + (-5+5) + (-3+3) + (-4+7+2) = 5. Since the net deviation of known numbers is +5, x must be 5 *below* the mean to balance it out to 0. x = 12 - 5 = 7. This avoids adding up large numbers! Common Pitfall: A common error is miscounting the total number of terms as 11 instead of 12, leading to an incorrect total sum of (11 × 12 = 132). Always double-check your n value when a variable is mixed into a list. Final Answer: Therefore, the correct answer is 7.

Q: The average of 2, 7, 6 and $x$ is 5 and the average of 18, 1, 6, $x$ and $y$ is 10. What is the value of $y$?

Concept & Logic: The formula for the average is the total sum of observations divided by the number of observations. By solving the first equation for the unknown variable, we can substitute it into the second equation to find the final target value. Average = (Sum\ of\ Observations/Total\ Number\ of\ Observations) Step-by-Step Solution: • Given: The average of 2, 7, 6, and x is 5. • The average of 18, 1, 6, x, and y is 10. • Calculation: Set up the first equation to find x. • (2 + 7 + 6 + x/4) = 5 • 15 + x = 20 • x = 5 • Now, substitute x = 5 into the second equation to find y. • (18 + 1 + 6 + 5 + y/5) = 10 • (30 + y/5) = 10 • 30 + y = 50 • y = 20 Exam Strategy & Shortcut: Think in terms of "Total Sum" immediately rather than writing out the fraction. For the first group: 4 numbers averaging 5 means the total sum is 20. The knowns sum to 15, so x is 5. For the second group: 5 numbers averaging 10 means the total sum is 50. The knowns (including x=5) sum to 30. Therefore, y = 50 - 30 = 20. This mental math saves crucial seconds. Common Pitfall: A common mistake is forgetting to count y as one of the numbers in the second set, leading to dividing by 4 instead of 5, which fundamentally ruins the algebraic equation. Always verify your total count n. Final Answer: Therefore, the correct answer is 20.

Q: The average of odd numbers up to 100 is

Concept & Formula: The sum of the first n odd natural numbers is exactly n^2. Therefore, the average of the first n odd numbers is (n^2/n), which simplifies perfectly to just n. Average\ of\ first\ n\ odd\ numbers = n Step-by-Step Solution: • Given: We need the average of all odd numbers up to 100. • Deduction: The series is 1, 3, 5, ..., 99. • First, determine how many odd numbers exist up to 100. Exactly half the numbers from 1 to 100 are odd, so n = 50. • Calculation: Since we are finding the average of the first 50 odd numbers, the average is exactly equal to n. • Average = 50. Exam Strategy & Shortcut: You can also use the arithmetic progression (AP) midpoint shortcut. The first odd number is 1, and the last odd number before 100 is 99. The average is the sum of the first and last terms divided by 2. (1 + 99/2) = (100/2) = 50. Common Pitfall: A frequent error is assuming n = 100 because the question says "up to 100," leading to an incorrect answer of 100. You must count *how many* odd numbers are actually in that range. Final Answer: Therefore, the correct answer is 50.

Q: In a family, the average age of a father and a mother is 35 years. The average age of the father, mother and their only son is 27 years. What is the age of the son?

Concept & Logic: When dealing with average ages of a group, calculate the total sum of the ages. By comparing the total age of the group with a member to the total age of the group without that member, you can easily isolate the individual's age. Total\ Age = Average\ Age × Number\ of\ People Step-by-Step Solution: • Given: The average age of the father and mother is 35 years. • The average age of the father, mother, and son is 27 years. • Calculation: Calculate the total age of the parents (2 people). • Total\ Age\ (Father + Mother) = 35 × 2 = 70\ years. • Calculate the total age of the family including the son (3 people). • Total\ Age\ (Father + Mother + Son) = 27 × 3 = 81\ years. • Subtract the total age of the parents from the total age of the whole family to find the son's age. • Son's\ Age = 81 - 70 = 11\ years. Exam Strategy & Shortcut: Use the "Deviation Method." The son joins a group of 2 people whose average is 35, and drags the average down to 27. The drop in average is 35 - 27 = 8 years. This 8-year drop applies to the 2 original members, meaning a total deficit of 8 × 2 = 16 years. The son must be 16 years younger than the new average to cause this drop. So, 27 - 16 = 11 years. Common Pitfall: Miscalculating the total number of people in the first group. Some students read "father and a mother" and forget it represents exactly 2 distinct individuals, incorrectly multiplying 35 by 3 or failing to multiply at all. Final Answer: Therefore, the correct answer is 11 years.

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 kilogramsConcept / 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 studentsConcept / 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.

Question 6

Find the average of the following sets of scores: $385$, $441$, $876$, $221$, $536$, $46$, $291$, $428$

Accepted Answer

Concept & Formula: The average of a discrete dataset is found by taking the total sum of all the data points and dividing it by the total count of the data points. Average = \frac{Sum of scores}{Total number of scores} Step-by-Step Solution: • Given: The scores are 385, 441, 876, 221, 536, 46, 291, and 428. By counting them, we determine there are 8 scores in total. • Calculation: Add all the individual scores together to find the sum: Sum = 385 + 441 + 876 + 221 + 536 + 46 + 291 + 428 Sum = 3224 Now, divide this total sum by the number of scores (8): Average = (3224/8) Average = 403 Exam Strategy & Shortcut: Unit Digit Verification: When adding the numbers, check the unit digits: 5+1+6+1+6+6+1+8 = 34. The sum ends in 4. You are calculating \frac{Sum}{8}. You need a number that, when multiplied by 8, gives a unit digit of 4. Looking at the options, 3 × 8 = 24 (ends in 4). Thus, option (b) 403 is an extremely strong candidate, drastically reducing the need to double-check the entire arithmetic sum manually. Common Pitfall: Skipping a number or misaligning digits during manual addition is the single biggest cause of errors here. Always write the numbers neatly in a vertical column if you choose to sum them manually, or use the unit/tens digit approximation trick to verify. Final Answer: Therefore, the correct answer is 403.

Question 7

The arithmetic mean of $15$ numbers is $41.4$. Then the sum of these numbers is

Accepted Answer

Concept & Formula: The foundational formula for average bridges three variables: Average, Sum, and Count. If you know the average and the total count, you can instantly find the total sum by rearranging the standard formula. Sum = Average × Total Count Step-by-Step Solution: • Given: The arithmetic mean (Average) is 41.4. The total number of values (Count) is 15. • Calculation: Apply the rearranged formula directly: Sum = 41.4 × 15 To calculate easily, break it down: Sum = 41.4 × (10 + 5) Sum = 414 + 207 Sum = 621 Exam Strategy & Shortcut: Unit Digit & Decimal Elimination: You need to multiply 41.4 × 15. You know 4 × 5 = 20, which means the decimal part will result in a whole number (specifically, .4 × 15 = 6). There will be no decimals in the final answer. Furthermore, 41 × 15 is roughly 40 × 15 = 600. The answer must be slightly above 600. Between options (c) 620 and (d) 621, exact unit digit tracking (4 × 5 = 20 \rightarrow adds 0 to unit, but 1 × 5 = 5, plus the 6 from the decimal gives 11, so it ends in 1). The answer must be 621. Common Pitfall: Students sometimes get confused by the decimal and write 41.4 × 15 = 6210, placing the decimal point incorrectly. Breaking the multiplier 15 into (10 + 5) is the safest way to avoid decimal placement errors. Final Answer: Therefore, the correct answer is 621.

Question 8

If $25a + 25b = 115$, what is the average of $a$ and $b$?

Accepted Answer

Concept & Formula: The average of two numbers, a and b, is defined by dividing their sum by 2. We can find the sum (a + b) as a single entity by factoring out the common coefficient from the given algebraic equation. Average = (a + b/2) Step-by-Step Solution: • Given: The equation provided is 25a + 25b = 115. • Calculation: First, factor out the 25 from the left side of the equation: 25(a + b) = 115 Divide both sides by 25 to isolate the sum (a + b): a + b = (115/25) a + b = 4.6 Now, substitute this sum directly into the average formula: Average = (4.6/2) = 2.3 The calculated average is 2.3, which does not match options (a), (b), (c), or (d). Exam Strategy & Shortcut: Instead of calculating the exact decimal immediately, keep the fractions. The sum is (115/25). The average is half of the sum, which means calculating (115/50). Multiplying the numerator and the denominator by 2 gives (230/100), which instantly yields 2.3. This fractional approach is often much faster and less prone to silly division mistakes. Since 2.3 is not present in the options, the answer is apparent. Common Pitfall: A very common mistake is finding the sum (a + b) = 4.6 and hurriedly selecting option (d) without remembering that the question asks for the average of a and b, not their total sum. Always double-check what the question is explicitly asking for before marking the final option. Final Answer: Therefore, the correct answer is None of these.

Question 9

There are five boxes in a cargo hold. The weight of the first box is $200$ kg and the weight of the second box is $20\%$ more than the weight of the third box, whose weight is $25\%$ more than the first box's weight. The fourth box at $350$ kg is $3…

Accepted Answer

Concept & Strategy: This problem requires a sequential calculation of weights based on percentage relationships. Once all weights are found, the shortcut to find the difference between the averages of the highest four and lowest four of a five-item set relies on recognizing overlapping values. Step-by-Step Solution: • Given: Box 1 (B_1) = 200 kg B_2 = 1.20 × B_3 B_3 = 1.25 × B_1 B_4 = 350 kg B_4 = (1 - 0.30) × B_5 = 0.70 × B_5 • Calculation: Step 1: Calculate B_3. B_3 = 1.25 × 200 = 250 kg Step 2: Calculate B_2. B_2 = 1.20 × 250 = 300 kg Step 3: Calculate B_5. 350 = 0.70 × B_5 B_5 = (350/0.70) = 500 kg Step 4: List all weights in ascending order. 200, 250, 300, 350, 500 Step 5: Calculate averages. Four heaviest (250, 300, 350, 500): Sum = 1400. Average = (1400/4) = 350 kg. Four lightest (200, 250, 300, 350): Sum = 1100. Average = (1100/4) = 275 kg. Step 6: Find the difference. Difference = 350 - 275 = 75 kg. Exam Strategy & Shortcut: The Overlap Cancellation Trick: When you have 5 ordered items (x_1, x_2, x_3, x_4, x_5), the difference between the average of the top 4 and the bottom 4 is simply the difference of the extreme values divided by 4. The middle three items (x_2, x_3, x_4) overlap and perfectly cancel out when you subtract the two sums! Difference = (x_5 - x_1/4) Difference = (500 - 200/4) = (300/4) = 75 kg. This bypasses the need to sum the heavy and light groups entirely! Common Pitfall: Misinterpreting "30% lighter than the fifth box." Students often calculate 30% of the fourth box (350) and add it back, doing 350 × 1.3 = 455 kg for the fifth box. This is mathematically incorrect. The percentage is based on the *fifth* box's weight, so it must be modeled as 0.70 × B_5 = 350. Final Answer: Therefore, the correct answer is 75 kg.

Question 10

In Arun's opinion, his weight is greater than $65$ kg but less than $72$ kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than $60$ kg but less than $70$ kg. His mother's view is that his weight cannot be greater …

Accepted Answer

Concept & Logic: This is an intersecting sets (inequalities) problem. To satisfy multiple conditions simultaneously, we must find the overlapping range of all given estimations. "Probable weights" implies we are looking for discrete, integer values within this final valid range. Step-by-Step Solution: • Given Conditions: Arun's estimation: 65 65, Brother says > 60. For both to be correct, the weight must be strictly greater than 65. Step 2: Combine the upper bounds (less than or equal to). Arun says < 72, Brother says < 70, Mother says \le 68. For all to be correct, the most restrictive upper limit applies. It must be less than or equal to 68. Step 3: Define the final valid range. 65 < Weight \le 68 Step 4: Identify the probable integer weights. The integers strictly greater than 65 and less than or equal to 68 are 66, 67, and 68. • Calculation: Average of these probable weights: Average = (66 + 67 + 68/3) Since they form an arithmetic progression, the average is the middle number. Average = 67 kg. Exam Strategy & Shortcut: Number Line Visualization: Quickly sketch or mentally picture a number line. Mark the strictest lower bound (65 from Arun) and the strictest upper bound (68 from Mother). The numbers trapped inside are 66, 67, and 68. The middle number of a consecutive sequence is always its average, giving 67 instantly without doing actual addition and division. Common Pitfall: Misinterpreting the phrase "cannot be greater than 68 kg". Students often take this to mean strictly less than 68 (< 68), which would leave only 66 and 67 as probable weights, resulting in an incorrect average of 66.5. "Cannot be greater than" mathematically translates to "less than or equal to" (\le). Final Answer: Therefore, the correct answer is 67 kg.

Question 11

Find the average of all the numbers between 6 and 34 which are divisible by 5.

Accepted Answer

Concept & Logic: When a sequence of numbers forms an Arithmetic Progression (AP) with a constant difference, the average (mean) is simply the mathematical midpoint of the series. Average = (First\ Term + Last\ Term/2) Step-by-Step Solution: • Given: We need numbers between 6 and 34 that are perfectly divisible by 5. • Deduction: The first multiple of 5 after 6 is 10. The last multiple of 5 before 34 is 30. • The complete valid sequence is: 10, 15, 20, 25, 30. • Since the common difference is a constant 5, this is an AP. • Calculation: Apply the average formula using the first valid term (10) and last valid term (30). • Average = (10 + 30/2) = (40/2) = 20. Exam Strategy & Shortcut: For any evenly spaced set of numbers, never calculate the total sum. Just identify the first term and the last term, add them together, and divide by 2. It saves massive amounts of time on large sets. Common Pitfall: A standard error is manually adding all terms together (10 + 15 + 20 + 25 + 30) and making an arithmetic calculation mistake. Always default to the AP boundary shortcut to avoid unnecessary addition. Final Answer: Therefore, the correct answer is 20.

Question 12

A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and $x$. He found the mean to be 12. What should be the number in place of $x$?

Accepted Answer

Concept & Logic: The mean relates the total sum of all elements to the number of elements. By using the given mean and total count, we can find the required total sum and isolate the unknown variable. Total\ Sum = Mean × Total\ Count Step-by-Step Solution: • Given: The numbers are 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14, and x. • The given mean is 12. • Deduction: Count the total number of terms. There are 11 known numbers plus x, making exactly 12 terms in total. • Calculate the required total sum for the mean to be 12. • Total\ Sum = 12 × 12 = 144. • Calculation: Sum the 11 known numbers. • Known\ Sum = 3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 = 137. • Set up the algebraic equation to find x. • 137 + x = 144 • x = 144 - 137 = 7. Exam Strategy & Shortcut: You can use the "Deviation Method". Compare each number to the mean (12). Deviations: -9, -1, -5, -3, +3, +1, -4, +7, +5, +9, +2. Sum of deviations = (-9+9) + (-1+1) + (-5+5) + (-3+3) + (-4+7+2) = 5. Since the net deviation of known numbers is +5, x must be 5 *below* the mean to balance it out to 0. x = 12 - 5 = 7. This avoids adding up large numbers! Common Pitfall: A common error is miscounting the total number of terms as 11 instead of 12, leading to an incorrect total sum of (11 × 12 = 132). Always double-check your n value when a variable is mixed into a list. Final Answer: Therefore, the correct answer is 7.

Question 13

The average of 2, 7, 6 and $x$ is 5 and the average of 18, 1, 6, $x$ and $y$ is 10. What is the value of $y$?

Accepted Answer

Concept & Logic: The formula for the average is the total sum of observations divided by the number of observations. By solving the first equation for the unknown variable, we can substitute it into the second equation to find the final target value. Average = (Sum\ of\ Observations/Total\ Number\ of\ Observations) Step-by-Step Solution: • Given: The average of 2, 7, 6, and x is 5. • The average of 18, 1, 6, x, and y is 10. • Calculation: Set up the first equation to find x. • (2 + 7 + 6 + x/4) = 5 • 15 + x = 20 • x = 5 • Now, substitute x = 5 into the second equation to find y. • (18 + 1 + 6 + 5 + y/5) = 10 • (30 + y/5) = 10 • 30 + y = 50 • y = 20 Exam Strategy & Shortcut: Think in terms of "Total Sum" immediately rather than writing out the fraction. For the first group: 4 numbers averaging 5 means the total sum is 20. The knowns sum to 15, so x is 5. For the second group: 5 numbers averaging 10 means the total sum is 50. The knowns (including x=5) sum to 30. Therefore, y = 50 - 30 = 20. This mental math saves crucial seconds. Common Pitfall: A common mistake is forgetting to count y as one of the numbers in the second set, leading to dividing by 4 instead of 5, which fundamentally ruins the algebraic equation. Always verify your total count n. Final Answer: Therefore, the correct answer is 20.

Question 14

The average of odd numbers up to 100 is

Accepted Answer

Concept & Formula: The sum of the first n odd natural numbers is exactly n^2. Therefore, the average of the first n odd numbers is (n^2/n), which simplifies perfectly to just n. Average\ of\ first\ n\ odd\ numbers = n Step-by-Step Solution: • Given: We need the average of all odd numbers up to 100. • Deduction: The series is 1, 3, 5, ..., 99. • First, determine how many odd numbers exist up to 100. Exactly half the numbers from 1 to 100 are odd, so n = 50. • Calculation: Since we are finding the average of the first 50 odd numbers, the average is exactly equal to n. • Average = 50. Exam Strategy & Shortcut: You can also use the arithmetic progression (AP) midpoint shortcut. The first odd number is 1, and the last odd number before 100 is 99. The average is the sum of the first and last terms divided by 2. (1 + 99/2) = (100/2) = 50. Common Pitfall: A frequent error is assuming n = 100 because the question says "up to 100," leading to an incorrect answer of 100. You must count *how many* odd numbers are actually in that range. Final Answer: Therefore, the correct answer is 50.

Question 15

In a family, the average age of a father and a mother is 35 years. The average age of the father, mother and their only son is 27 years. What is the age of the son?

Accepted Answer

Concept & Logic: When dealing with average ages of a group, calculate the total sum of the ages. By comparing the total age of the group with a member to the total age of the group without that member, you can easily isolate the individual's age. Total\ Age = Average\ Age × Number\ of\ People Step-by-Step Solution: • Given: The average age of the father and mother is 35 years. • The average age of the father, mother, and son is 27 years. • Calculation: Calculate the total age of the parents (2 people). • Total\ Age\ (Father + Mother) = 35 × 2 = 70\ years. • Calculate the total age of the family including the son (3 people). • Total\ Age\ (Father + Mother + Son) = 27 × 3 = 81\ years. • Subtract the total age of the parents from the total age of the whole family to find the son's age. • Son's\ Age = 81 - 70 = 11\ years. Exam Strategy & Shortcut: Use the "Deviation Method." The son joins a group of 2 people whose average is 35, and drags the average down to 27. The drop in average is 35 - 27 = 8 years. This 8-year drop applies to the 2 original members, meaning a total deficit of 8 × 2 = 16 years. The son must be 16 years younger than the new average to cause this drop. So, 27 - 16 = 11 years. Common Pitfall: Miscalculating the total number of people in the first group. Some students read "father and a mother" and forget it represents exactly 2 distinct individuals, incorrectly multiplying 35 by 3 or failing to multiply at all. Final Answer: Therefore, the correct answer is 11 years.

Question 16

The total marks obtained by a student in Physics, Chemistry and Mathematics together is 120 more than the marks obtained by him in Chemistry. What is the average marks obtained by him in Physics and Mathematics together?

Accepted Answer

Concept & Logic: This problem requires basic algebraic translation. By converting the English sentence into a mathematical equation, variables will naturally cancel out, leaving exactly the sum needed to compute the average. Average = (Sum\ of\ Observations/Total\ Number\ of\ Observations) Step-by-Step Solution: • Given: The total marks in Physics, Chemistry, and Mathematics is 120 more than the marks in Chemistry alone. • Calculation: Let the marks be P, C, and M. Write the equation based on the given statement. • P + C + M = C + 120 • Subtract C from both sides of the equation. • P + M = 120 • We need the average marks obtained in Physics and Mathematics together. • Average = (P + M/2) • Average = (120/2) = 60 Exam Strategy & Shortcut: Read the problem carefully: if total marks of 3 subjects are 120 more than Chemistry, it means the *other two* subjects (Physics and Math) must perfectly sum to that "extra" 120. If their sum is 120, the average of these two subjects is simply half of that sum, which is 60. Common Pitfall: Students often assume "Cannot be determined" because the individual marks for Physics and Math are unknown. You don't need individual values to find an average if you have their combined sum. Final Answer: Therefore, the correct answer is 60.

Question 17

The average of the first 100 positive integers is

Accepted Answer

Concept & Formula: The set of positive integers starting from 1 forms an Arithmetic Progression (AP) with a common difference of 1. The average of the first n natural numbers (positive integers) can be found using a direct formula. Average = (n + 1/2) Step-by-Step Solution: • Given: We need the average of the first 100 positive integers. • Deduction: The sequence is 1, 2, 3, ..., 100. • Here, the total number of terms n is 100. • Calculation: Plug n = 100 into the standard AP average formula. • Average = (100 + 1/2) • Average = (101/2) • Average = 50.5 Exam Strategy & Shortcut: For any sequence with a constant difference (like consecutive integers), the average is simply the midpoint between the first and last term. First term is 1, last term is 100. Average is (1 + 100) / 2 = 50.5. Memorize this property for instant answers. Common Pitfall: Confusing "positive integers" with "whole numbers." Whole numbers start at 0, which would make the first 100 numbers range from 0 to 99, changing the average to 49.5. Positive integers strictly begin at 1. Final Answer: Therefore, the correct answer is 50.5.

Question 18

The average of odd numbers up to 100 is

Accepted Answer

Concept & Formula: The sum of the first n odd natural numbers is exactly n^2. Therefore, the average of the first n odd numbers is (n^2/n), which simplifies perfectly to just n. Average\ of\ first\ n\ odd\ numbers = n Step-by-Step Solution: • Given: We need the average of all odd numbers up to 100. • Deduction: The series is 1, 3, 5, ..., 99. • First, determine how many odd numbers exist up to 100. Exactly half the numbers from 1 to 100 are odd, so n = 50. • Calculation: Since we are finding the average of the first 50 odd numbers, the average is exactly equal to n. • Average = 50. Exam Strategy & Shortcut: You can also use the arithmetic progression (AP) midpoint shortcut. The first odd number is 1, and the last odd number before 100 is 99. The average is the sum of the first and last terms divided by 2. (1 + 99/2) = (100/2) = 50. Common Pitfall: A frequent error is assuming n = 100 because the question says "up to 100," leading to an incorrect answer of 100. You must count *how many* odd numbers are actually in that range. Final Answer: Therefore, the correct answer is 50.

Question 19

In a family, the average age of a father and a mother is 35 years. The average age of the father, mother and their only son is 27 years. What is the age of the son?

Accepted Answer

Concept & Logic: When dealing with average ages of a group, calculate the total sum of the ages. By comparing the total age of the group with a member to the total age of the group without that member, you can easily isolate the individual's age. Total\ Age = Average\ Age × Number\ of\ People Step-by-Step Solution: • Given: The average age of the father and mother is 35 years. • The average age of the father, mother, and son is 27 years. • Calculation: Calculate the total age of the parents (2 people). • Total\ Age\ (Father + Mother) = 35 × 2 = 70\ years. • Calculate the total age of the family including the son (3 people). • Total\ Age\ (Father + Mother + Son) = 27 × 3 = 81\ years. • Subtract the total age of the parents from the total age of the whole family to find the son's age. • Son's\ Age = 81 - 70 = 11\ years. Exam Strategy & Shortcut: Use the "Deviation Method." The son joins a group of 2 people whose average is 35, and drags the average down to 27. The drop in average is 35 - 27 = 8 years. This 8-year drop applies to the 2 original members, meaning a total deficit of 8 × 2 = 16 years. The son must be 16 years younger than the new average to cause this drop. So, 27 - 16 = 11 years. Common Pitfall: Miscalculating the total number of people in the first group. Some students read "father and a mother" and forget it represents exactly 2 distinct individuals, incorrectly multiplying 35 by 3 or failing to multiply at all. Final Answer: Therefore, the correct answer is 11 years.

Question 20

If the mean of 5 observations $x$, $x + 2$, $x + 4$, $x + 6$ and $x + 8$ is 11, then the mean of the last three observations is

Accepted Answer

Concept & Strategy: When a sequence of numbers increases by a constant amount, it forms an Arithmetic Progression (AP). The most powerful property of an AP with an odd number of terms is that its mean (average) is exactly equal to its middle term. Mean\ of\ AP = Middle\ Observation Step-by-Step Solution: • Given: The 5 observations are x, x + 2, x + 4, x + 6, x + 8. • The mean of these 5 observations is 11. • Calculation: Verify the sequence is an AP. The difference between consecutive terms is consistently 2. • The middle term of these 5 observations is the 3rd term: x + 4. • Because it is an AP, the mean equals the middle term. • x + 4 = 11 • We need the mean of the last three observations: x + 4, x + 6, x + 8. • This subset is also an AP. Its middle term is x + 6. • Since we know x + 4 = 11, we can find x + 6 by simply adding 2. • x + 6 = 11 + 2 = 13. Exam Strategy & Shortcut: You do not need to find the value of x. The mean of the 5 terms is x+4. The mean of the last 3 terms is x+6. The difference between these means is exactly 2. So, 11 + 2 = 13. This bypasses algebraic solving entirely. Common Pitfall: The slow approach is calculating the full sum: (5x + 20) / 5 = 11, solving for x = 7, then substituting x into the last three terms (11, 13, 15), summing them (39), and dividing by 3 to get 13. While correct, this wastes massive amounts of time on a competitive exam. Final Answer: Therefore, the correct answer is 13.

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