Question 1

Each window covering can span 15 identical windows in a row. How many such window coverings are needed to span a total of 50 windows?

Accepted Answer

Introduction: This question involves simple division and the idea of rounding up to the next whole number. We want to cover a certain number of windows using window coverings, each of which can cover only a fixed number of windows. Since the number of coverings must be a whole number, any fraction means one more full covering is required. Given Data / Assumptions: - Each window covering is long enough to cover 15 windows. - There are 50 windows that need to be spanned. - We assume partial coverings are not allowed; each covering is a complete unit. - We must find the minimum number of coverings needed to span all 50 windows. Concept / Approach: The basic concept is to divide the total number of windows by the number of windows each covering can span. If the result is an integer, that is the answer. If the result is not an integer, we must round up to the next whole number, because even a small remainder of uncovered windows requires an additional full covering. Step-by-Step Solution: Step 1: Total number of windows to be covered = 50. Step 2: Capacity of one window covering = 15 windows. Step 3: Compute 50 / 15. Step 4: 15 * 3 = 45 and 15 * 4 = 60, so 50 / 15 is between 3 and 4. Step 5: Numerically, 50 / 15 is approximately 3.33. Step 6: Since we cannot use a fraction of a window covering, we must round 3.33 up to the next whole number. Step 7: Therefore, the number of window coverings required is 4. Verification / Alternative Check: If we use 3 coverings, they will span 3 * 15 = 45 windows, which leaves 5 windows uncovered. That is not acceptable. If we use 4 coverings, they span 4 * 15 = 60 windows, which is more than enough to cover 50 windows. Thus 4 coverings are sufficient and also the minimum number required. Why Other Options Are Wrong: Option 3: Only covers 45 windows, leaving some windows uncovered. Options 5, 6 and 7: All of these provide more coverings than required, so they are not the minimum number of coverings. Common Pitfalls: One common mistake is to take the integer part of 50 / 15 as the answer, which would give 3. However, in real life and in such problems, even a small uncovered part requires one additional full unit. The correct method is to round up, not to round down, whenever there is any remainder. Final Answer: The number of window coverings needed is 4.

Question 2

Sixteen children are divided into two groups A and B of 10 and 6 children respectively. The average percentage of marks obtained by the children of group A is 75, and the average percentage of marks of all the 16 children together is 76. What is the…

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Introduction: This problem is about averages of subgroups within a larger group. We know the overall average of all children and the average of one subgroup. From this information we can find the total marks, subtract the marks of group A and then compute the average for group B. This is a standard type of question in aptitude tests dealing with averages and weighted sums. Given Data / Assumptions: - Total number of children = 16. - Group A has 10 children. - Group B has 6 children. - Average marks of group A = 75 percent. - Average marks of all 16 children together = 76 percent. - We need the average marks of group B. Concept / Approach: Average is defined as total marks divided by number of children. We use this definition twice: once to compute the total marks of all 16 children and once to compute the total marks of group A. The marks of group B are obtained by subtracting the sum for group A from the overall sum. Finally, dividing by the number of children in group B gives the required average in that group. Step-by-Step Solution: Step 1: Total marks of all 16 children = 16 * 76 = 1216. Step 2: Total marks of group A = 10 * 75 = 750. Step 3: Total marks of group B = total marks of all children minus total marks of group A. Step 4: So, marks of group B = 1216 - 750 = 466. Step 5: Number of children in group B = 6. Step 6: Average marks of group B = 466 / 6. Step 7: Simplify 466 / 6 by dividing numerator and denominator by 2: 466 / 6 = 233 / 3. Step 8: As a mixed fraction this is 77 and two thirds, which is approximately 77.67 percent. Verification / Alternative Check: We can check by computing the combined average using this group B average. Group A contributes 750 marks and group B contributes 466 marks, total 1216 marks. Dividing by 16 gives 1216 / 16 = 76, which matches the given overall average. This confirms that the value 233 / 3 for group B is correct. Why Other Options Are Wrong: Option 76: This is the average of all 16 children, not of group B alone. Option 77 and option 80 are simple round numbers but they do not satisfy the condition that the combined average is 76. Option 243/4 corresponds to 60.75 percent, which is far from the required higher average for group B. Common Pitfalls: Many students mistakenly average the two given averages 75 and 76 directly, without taking into account the different group sizes. Another error is to forget that the overall average is influenced more by the larger group. Working with total marks rather than directly with averages helps avoid these mistakes. Final Answer: The average marks of the children in group B is 233/3 percent.

Question 3

In a class the average age of the boys is 16 years and the average age of the girls is 15 years. What is the average age of the whole class?

Accepted Answer

Introduction: This question is about understanding when average information is sufficient and when it is not. We are given the average ages of boys and girls separately and asked to find the average age of the entire class. The key issue is whether we know how many boys and how many girls there are. Given Data / Assumptions: - Average age of boys = 16 years. - Average age of girls = 15 years. - The numbers of boys and girls are not given. - We want the average age of all students in the class. Concept / Approach: The average of the whole class is a weighted average of the average age of boys and the average age of girls. The weights are the numbers of boys and girls. Without knowing at least the ratio between the number of boys and the number of girls, we cannot compute a unique value for the overall average. Step-by-Step Reasoning: Step 1: Let the number of boys be b and the number of girls be g. Step 2: Total age of boys = 16b. Step 3: Total age of girls = 15g. Step 4: Total students = b + g, and total age = 16b + 15g. Step 5: Average age of the whole class = (16b + 15g) / (b + g). Step 6: This expression depends on the unknown values b and g. Without a relation between b and g, the result is not a single fixed number. Step 7: For example, if b = g, the average would be (16 + 15) / 2 = 15.5 years, but if there are many more boys than girls, the overall average would be closer to 16 years, and if there are many more girls the overall average would be closer to 15 years. Verification / Alternative Check: Choose specific cases to see the variation. If there are 10 boys and 10 girls, average = (16*10 + 15*10) / 20 = (160 + 150) / 20 = 310 / 20 = 15.5 years. If there are 20 boys and 10 girls, average = (16*20 + 15*10) / 30 = (320 + 150) / 30 = 470 / 30 which is approximately 15.67 years. These examples show that the answer changes with the composition of the class. Why Other Options Are Wrong: Option 15, 15.2, 15.5 and 15.8 each assume a particular ratio of boys to girls. Since that ratio is not given, none of these numerical values can be guaranteed to be correct. Common Pitfalls: A very common error is to directly average 16 and 15 and declare 15.5 as the answer. This only works when the numbers of boys and girls are equal. In general, when you are given the averages of subgroups, you must know the subgroup sizes to find the overall average. Final Answer: The information is insufficient; the correct choice is data inadequate.

Question 4

What is the average of the decimal numbers four tenths and five thousandths?

Accepted Answer

Introduction: This problem checks your understanding of decimal numbers written in words and your ability to compute a simple arithmetic average. You must correctly convert the verbal descriptions four tenths and five thousandths into decimal form, then use the formula for the average of two numbers. Given Data / Assumptions: - One number is four tenths. - The other number is five thousandths. - We are asked to find the average of these two numbers. Concept / Approach: Four tenths means 4 parts out of 10, which is 4/10. Five thousandths means 5 parts out of 1000, which is 5/1000. After converting both to decimal form, we apply the average formula for two numbers: average = (number1 + number2) / 2. Careful handling of decimal places is important to avoid mistakes. Step-by-Step Solution: Step 1: Convert four tenths to decimal form: 4 tenths = 4/10 = 0.4. Step 2: Convert five thousandths to decimal form: 5 thousandths = 5/1000 = 0.005. Step 3: Add the two numbers: 0.4 + 0.005 = 0.405. Step 4: The average of two numbers is their sum divided by 2. Step 5: Compute 0.405 / 2. Step 6: 0.405 divided by 2 equals 0.2025. Step 7: Therefore, the required average = 0.2025. Verification / Alternative Check: We can verify the division 0.405 / 2 by converting 0.405 to a fraction: 0.405 = 405/1000. Dividing by 2 is the same as multiplying by 1/2, so we get 405/1000 * 1/2 = 405/2000. Simplifying 405/2000 by dividing numerator and denominator by 5 gives 81/400. As a decimal, 81/400 = 0.2025. This matches our earlier result. Why Other Options Are Wrong: Option 0.045 comes from incorrectly treating one of the numbers or from dividing by 10 instead of 2. Option 0.205 and 0.225 are close numbers but they do not match the exact result of 0.405 divided by 2. Option 0.4005 wrongly suggests some misplacement of the decimal point when adding or averaging the numbers. Common Pitfalls: Students often confuse tenths with thousandths or misplace the decimal point when converting from fractions to decimals. Another frequent error is to add correctly but then divide by 10 or some other number instead of 2. Writing the fractions explicitly before converting to decimals can help prevent these mistakes. Final Answer: The average of four tenths and five thousandths is 0.2025.

Question 5

The average of any five consecutive odd natural numbers is k. If two more consecutive odd natural numbers, immediately after these five numbers, are added to the set, what will be the new average in terms of k?

Accepted Answer

Introduction: This problem examines the properties of consecutive odd natural numbers and how their average behaves when more numbers from the same sequence are added. Consecutive odd numbers form an arithmetic progression with a common difference of 2, and their average has a simple relationship with the middle term of the sequence. Given Data / Assumptions: - We have five consecutive odd natural numbers. - The average of these five numbers is k. - Two more consecutive odd numbers, immediately following the first five, are added. - We must find the new average in terms of k. Concept / Approach: For any arithmetic progression with an odd number of terms, the average is equal to the middle term. Five consecutive odd numbers can be written symmetrically around their middle value. When two more consecutive odd numbers are added after the first five, the new set has seven consecutive odd numbers. For seven consecutive odd numbers, the average is still the middle term, which can be related back to k. Step-by-Step Solution: Step 1: Let the five consecutive odd numbers be n - 4, n - 2, n, n + 2 and n + 4, where n is the middle odd number. Step 2: For five consecutive odd numbers, the average is the middle term, so the average is n. Step 3: The question states that this average is k, so n = k. Step 4: The next two consecutive odd numbers after n + 4 are n + 6 and n + 8. Step 5: After adding these two, we have seven numbers: n - 4, n - 2, n, n + 2, n + 4, n + 6 and n + 8. Step 6: These seven numbers are still in arithmetic progression with a common difference of 2. Step 7: For seven consecutive odd numbers, the average is again the middle term, which is the fourth number in order. Step 8: The fourth number is n + 2, since the sequence is symmetric around this value. Step 9: Substitute n = k to get the new average: k + 2. Verification / Alternative Check: Take a concrete example. Suppose the five consecutive odd numbers are 3, 5, 7, 9 and 11. Their average is 7, so k = 7. The next two odd numbers are 13 and 15. The seven numbers are 3, 5, 7, 9, 11, 13 and 15. The average of these seven numbers is the middle value 9. Since k = 7, k + 2 = 9, which matches the new average, confirming the formula. Why Other Options Are Wrong: Option k - 2 and k + 1 underestimate the shift in the average when larger numbers are added. Options 2k + 1 and 2k - 3 are not consistent dimensionally; they grow too fast compared to the linear progression of the numbers. Common Pitfalls: Students may incorrectly try to recompute the average using long algebraic expressions for all seven terms instead of using the symmetry of consecutive odd numbers. Remembering that for an odd count of terms in an arithmetic progression the average is equal to the middle term simplifies the problem significantly. Final Answer: The new average after adding the next two odd numbers is k + 2.

Question 6

The average of six numbers is x and the average of three of these numbers is y. If the average of the remaining three numbers is z, which of the following relations between x, y and z is correct?

Accepted Answer

Introduction: This question tests your understanding of how averages of subsets relate to the average of the entire set. We are given the average of all six numbers and the averages of two groups of three numbers each. From this, we can derive a formula connecting x, y and z. Given Data / Assumptions: - There are six numbers in total. - Average of all six numbers = x. - Average of three of these numbers = y. - Average of the remaining three numbers = z. - We must find the correct relation among x, y and z. Concept / Approach: Average is defined as total sum divided by number of items. The total sum of all six numbers can be expressed in two different ways: using x, and using y and z. Equating these expressions allows us to derive a relation between x, y and z. This is a classic weighted average situation where each subgroup has the same size. Step-by-Step Solution: Step 1: Let the six numbers be a1, a2, a3, a4, a5 and a6. Step 2: Average of all six numbers is x, so total sum S = 6x. Step 3: Suppose the first group of three numbers has average y. Then the sum of those three numbers is 3y. Step 4: The remaining three numbers have average z, so their sum is 3z. Step 5: The total sum S can also be written as sum of first three numbers plus sum of last three numbers, that is S = 3y + 3z. Step 6: From Step 2 and Step 5, we have 6x = 3y + 3z. Step 7: Divide both sides by 3: 2x = y + z. Step 8: Therefore, the correct relation is 2x = y + z. Verification / Alternative Check: Consider an example where the three numbers in the first group are 2, 4 and 6 (average y = 4) and the three numbers in the second group are 8, 10 and 12 (average z = 10). The overall average x is (2 + 4 + 6 + 8 + 10 + 12) / 6 = 42 / 6 = 7. Now compute y + z and 2x. We have y + z = 4 + 10 = 14 and 2x = 2 * 7 = 14. Hence 2x = y + z holds for this example. Why Other Options Are Wrong: Option x = y + z is incorrect because it ignores the fact that each of y and z is an average, not a total. Option 3x = 2y - 2z does not follow from any correct manipulation of the sums. Option x = (y + z) / 2 would only be true if the subgroup sizes were equal and x was the average of y and z directly, but here we have already used the sizes in the derivation and obtained 2x = y + z instead. Option 3x = y + z also fails dimensional reasoning, since multiplying x by 3 changes the scale incorrectly. Common Pitfalls: Some learners forget that averages must be multiplied by the count to recover total sums. Others confuse the expression for the total and try to combine x, y and z directly without using the number of items in each group. Always go back to total sums, equate them and then simplify. Final Answer: The correct relationship between x, y and z is 2x = y + z.

Question 7

In a village the average age of n people is 42 years. Later it was found that the age of one person had been recorded as 20 years less than the actual age. After correcting this mistake the average age increased by 1 year. What is the value of n?

Accepted Answer

Introduction: This question explores how an error in one value affects the average of a group and how to reverse the effect to find the number of people in the group. The key idea is that changing one data point changes the total sum, and therefore changes the average in a predictable way. Given Data / Assumptions: - There are n people in a village. - The recorded average age is 42 years. - One person age was recorded as 20 years less than the actual value. - After correcting the mistake, the new average age is 43 years. - We must find the value of n. Concept / Approach: The average age is total age divided by the number of people. Initially, using the incorrect data, the total age is taken as 42n. When the error is corrected, the total age increases by 20 years, because the actual age is 20 years more than what was recorded. The new total, divided by n, gives the corrected average of 43. Equating these expressions allows us to determine n. Step-by-Step Solution: Step 1: Let the initial (incorrect) total age of all n people be S. Step 2: Given that the average age is 42, we have S = 42n. Step 3: One person age was recorded as 20 years less than the true age, so the true total age should be S + 20. Step 4: After correcting the mistake, the average age becomes 43. Step 5: Therefore, the corrected total age is (correct average) * (number of people) = 43n. Step 6: This corrected total must also equal S + 20. Step 7: Substitute S = 42n into S + 20 = 43n, giving 42n + 20 = 43n. Step 8: Rearrange to get 20 = 43n - 42n = n. Step 9: Therefore, n = 20. Verification / Alternative Check: If n = 20, the initial total age is 42 * 20 = 840 years. The corrected total should be larger by 20, so it becomes 860 years. The corrected average is then 860 / 20 = 43 years, which matches the problem statement. This confirms that the value n = 20 fits all the given conditions. Why Other Options Are Wrong: Options 21, 22, 24 and 25 do not satisfy the equation 42n + 20 = 43n. Substituting any of these values for n leads to a mismatch between the corrected total and the required average of 43 years. Common Pitfalls: Students sometimes incorrectly subtract 20 from the total instead of adding it, or they mix up the old and new averages. It is important to recognize that the incorrect recorded age was too low, so the correction increases the total sum. Keeping track of which total corresponds to which average prevents algebraic mistakes. Final Answer: The number of people in the village is 20.

Question 8

In a hotel the average expenditure per guest is Rs 60 when there are 10 guests and the average expenditure per guest is Rs 40 when there are 20 guests. Assume that there are some fixed expenses which do not depend on the number of guests and the rem…

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Introduction: This question studies the effect of fixed and variable expenses on the average cost per guest in a hotel. Some costs remain the same regardless of the number of guests, while others change in direct proportion to the number of guests. By using two given scenarios we can find the fixed and variable parts and then predict the average for a different number of guests. Given Data / Assumptions: - When there are 10 guests, average expenditure per guest = Rs 60. - When there are 20 guests, average expenditure per guest = Rs 40. - There is a fixed expense F that does not depend on the number of guests. - There is a variable expense per guest v, so variable total expense is v times number of guests. - We must find the average expenditure per guest when there are 40 guests. Concept / Approach: Total expenditure for a given number of guests = fixed expense + variable expense per guest times number of guests. Average expenditure per guest = total expenditure divided by number of guests. Using the two given average values for 10 and 20 guests we set up two equations in F and v. Solving these equations gives F and v, which can then be used to compute the average for 40 guests. Step-by-Step Solution: Step 1: Let F be the fixed expense and v be the variable expense per guest. Step 2: For 10 guests, total expenditure = F + 10v. Step 3: Average for 10 guests is 60, so (F + 10v) / 10 = 60, giving F + 10v = 600. Step 4: For 20 guests, total expenditure = F + 20v. Step 5: Average for 20 guests is 40, so (F + 20v) / 20 = 40, giving F + 20v = 800. Step 6: Subtract the equation in Step 3 from the equation in Step 5 to eliminate F. Step 7: (F + 20v) - (F + 10v) = 800 - 600 gives 10v = 200. Step 8: Solve for v: v = 200 / 10 = 20. Step 9: Substitute v = 20 into F + 10v = 600 to find F. Step 10: F + 10 * 20 = 600 gives F + 200 = 600, so F = 400. Step 11: For 40 guests, total expenditure = F + 40v = 400 + 40 * 20 = 400 + 800 = 1200. Step 12: Average expenditure per guest for 40 guests = 1200 / 40 = 30. Verification / Alternative Check: We can check consistency. With F = 400 and v = 20, for 10 guests total = 400 + 10 * 20 = 600; average = 600 / 10 = 60, which matches the given data. For 20 guests total = 400 + 20 * 20 = 800; average = 800 / 20 = 40, again matching. Thus the model is correct and the predicted average for 40 guests is reliable. Why Other Options Are Wrong: Options Rs 25 and Rs 20 would correspond to lower variable expenses or different fixed expenses that do not fit the given data. Option Rs 35 and Rs 40 are too high and ignore the trend that average cost per guest decreases as the number of guests increases due to fixed costs being spread over more people. Common Pitfalls: Some students attempt to average 60 and 40 directly or assume a linear change in average without considering the role of fixed expenses. The correct method is to express the total expenditure algebraically, separate the fixed and variable parts and solve systematically using the given averages. Final Answer: The average expenditure per guest when there are 40 guests is Rs 30.

Question 9

There are 44 students in a hostel. Later 15 new students join the hostel and the total mess expenses increase by Rs 33 per day, while the average expenditure per student per day decreases by Rs 3. What was the original total daily expenditure of the…

Accepted Answer

Introduction: This question deals with average expenditure per person when the number of people changes. The mess expenses increase slightly when more students join, but the average cost per student decreases. We must translate this into equations and find the original total daily expenditure. Given Data / Assumptions: - Original number of students in the hostel = 44. - New students joining the hostel = 15, so new total students = 59. - Total daily mess expenses increase by Rs 33 when the new students join. - Average expenditure per student per day decreases by Rs 3. - We must find the original total daily expenditure of the mess. Concept / Approach: Let the original total daily mess expenditure be E. Then the original average per student is E / 44. After 15 students join, the total expenditure becomes E + 33, and the number of students becomes 59, so the new average is (E + 33) / 59. This new average is Rs 3 less than the original average. Setting up and solving this equation will give us E. Step-by-Step Solution: Step 1: Let E be the original total daily mess expenditure. Step 2: Original average per student = E / 44. Step 3: After 15 students join, total students = 44 + 15 = 59. Step 4: New total daily expenditure = E + 33. Step 5: New average per student = (E + 33) / 59. Step 6: Given that the new average is Rs 3 less than the old average: (E + 33) / 59 = E / 44 - 3. Step 7: Rewrite E / 44 - 3 as (E - 132) / 44. Step 8: Multiply both sides of the equation by 59 * 44 to clear denominators. Step 9: Left side becomes 44(E + 33), right side becomes 59(E - 132). Step 10: So 44E + 44 * 33 = 59E - 59 * 132. Step 11: Compute 44 * 33 = 1452 and 59 * 132 = 7788. Step 12: The equation is 44E + 1452 = 59E - 7788. Step 13: Bring like terms together: 1452 + 7788 = 59E - 44E which gives 9240 = 15E. Step 14: Solve for E: E = 9240 / 15 = 616. Verification / Alternative Check: Original total expenditure E = 616, so original average = 616 / 44 = 14. After 15 students join, total students = 59, new total expenditure = 616 + 33 = 649. New average = 649 / 59 = 11. The difference between the original and new averages is 14 - 11 = 3, which matches the given condition. Hence the value E = 616 is correct. Why Other Options Are Wrong: Options Rs 404, Rs 514, Rs 340 and Rs 480 do not satisfy the equation (E + 33) / 59 = E / 44 - 3. Substituting any of these values for E leads to inconsistent averages. Common Pitfalls: Learners sometimes misinterpret the statement and add 3 instead of subtracting, or they mistakenly apply the change to the total instead of the average. It is very important to write down the expressions for the average both before and after the change and then equate them correctly. Final Answer: The original total daily mess expenditure was Rs 616.

Question 10

On the annual day of a school sweets are to be distributed equally among 112 children. On that day 32 children are absent, so each of the remaining children receives 6 extra sweets. How many sweets was each child originally supposed to get?

Accepted Answer

Introduction: This question is a classic example of how absence of some members changes the distribution when a fixed total quantity is shared equally. The key is to express the total number of sweets in terms of the original per child share and then relate it to the increased share when fewer children are present. Given Data / Assumptions: - Total number of children initially planned for distribution = 112. - On the actual day, 32 children are absent. - Therefore, number of children present = 112 - 32 = 80. - Each present child receives 6 sweets more than originally planned. - We must find the original number of sweets per child. Concept / Approach: Let the original number of sweets per child be x. Then the total number of sweets is 112x. When only 80 children are present, these same 112x sweets are redistributed so that each present child gets x + 6 sweets. Equating the two expressions for total sweets leads to an equation in x which we can solve. Step-by-Step Solution: Step 1: Let the original number of sweets that each child was supposed to receive be x. Step 2: Then total number of sweets = 112x. Step 3: On the day of distribution, number of children present = 112 - 32 = 80. Step 4: Each present child now receives x + 6 sweets. Step 5: Total sweets distributed to the present children = 80 * (x + 6). Step 6: The total number of sweets is the same whether all 112 children are present or only 80 are present. Step 7: Therefore, 112x = 80(x + 6). Step 8: Expand the right side: 112x = 80x + 480. Step 9: Subtract 80x from both sides: 112x - 80x = 480 gives 32x = 480. Step 10: Solve for x: x = 480 / 32 = 15. Verification / Alternative Check: If each child was originally supposed to get 15 sweets, the total sweets = 112 * 15 = 1680. On the actual day, with 80 children present, each child receives 15 + 6 = 21 sweets. The total sweets distributed now are 80 * 21 = 1680, which matches the original total. This confirms that the value x = 15 is correct. Why Other Options Are Wrong: Options 12, 14, 16 and 18 do not satisfy the equation 112x = 80(x + 6). Substituting any of these values for x leads to unequal totals on both sides. Common Pitfalls: Students sometimes incorrectly set up the equation or forget that the total number of sweets remains fixed. Another common error is miscomputing 80(x + 6) or making arithmetic mistakes when simplifying 32x = 480. Writing all steps clearly helps avoid these mistakes. Final Answer: Each child was originally supposed to get 15 sweets.

Question 11

The average monthly salary of 24 employees in an organisation is Rs 1500. When the manager salary is included the average monthly salary increases by Rs 400. What is the monthly salary of the manager?

Accepted Answer

Introduction: This question tests your understanding of how adding one more data point affects the average of a group. The average salary of employees is known, and we are told how the average changes after including the manager. Using these values, we can find the manager salary. Given Data / Assumptions: - Number of employees initially considered = 24. - Average salary of these 24 employees = Rs 1500 per month. - After including the manager, number of people considered = 25. - New average salary after including the manager is Rs 1500 + Rs 400 = Rs 1900 per month. - We must find the manager monthly salary. Concept / Approach: Average salary is total salary divided by number of people. First, we compute the total salary of the 24 employees using the initial average. Then we compute the new total salary of 25 people using the new average. The difference between the new total and the old total is the salary of the manager. Step-by-Step Solution: Step 1: Total salary of 24 employees = 24 * 1500. Step 2: 24 * 1500 = 36000, so the sum of salaries of the 24 employees is Rs 36000. Step 3: After including the manager, there are 25 people with an average salary of Rs 1900. Step 4: Total salary of 25 people = 25 * 1900. Step 5: 25 * 1900 = 47500, so the new total salary is Rs 47500. Step 6: The manager salary is the difference between the new total and the old total. Step 7: Manager salary = 47500 - 36000 = 11500. Verification / Alternative Check: Check the average after including the manager. If the manager salary is Rs 11500, the total salary of 25 people is 36000 + 11500 = 47500. Dividing by 25 gives 47500 / 25 = 1900, which matches the new average stated in the problem. This confirms that the manager salary is correctly calculated. Why Other Options Are Wrong: Options Rs 8840, Rs 12800, Rs 9450 and Rs 15000 do not produce the required new average of Rs 1900 when added to Rs 36000 and divided by 25. They either give a higher or lower average than specified. Common Pitfalls: Some candidates may forget to add the increase in average to the original average, or they might incorrectly multiply the new average by 24 instead of 25. Another mistake is to average the original average and the new average instead of working with totals. Always remember that total = average * number of items is the key relation. Final Answer: The monthly salary of the manager is Rs 11500.

Question 12

In an experiment Sriram records 11 observations whose average is 92. The average of the first five observations is 89 and the average of the last five observations is 86. What is the value of the sixth observation?

Accepted Answer

Introduction: This question deals with averages of overlapping groups of observations. We know the average of all 11 observations, the average of the first five and the average of the last five. Using these values, we can find the value of the middle or sixth observation that is not part of either group. Given Data / Assumptions: - Total number of observations = 11. - Average of all 11 observations = 92. - Average of the first five observations = 89. - Average of the last five observations = 86. - We must find the sixth observation. Concept / Approach: The average of a set of numbers gives us their total sum. So, from the given averages we can compute three sums: the sum of all 11 observations, the sum of the first five observations and the sum of the last five observations. The sixth observation is included in the total sum but not in either of the sums of the first or last five observations. Therefore, we can find it by subtracting these partial sums from the total. Step-by-Step Solution: Step 1: Let the observations be O1, O2, O3, O4, O5, O6, O7, O8, O9, O10 and O11. Step 2: Total sum of all 11 observations = 11 * 92. Step 3: Compute 11 * 92 = 1012, so the total sum S = 1012. Step 4: Sum of the first five observations S1 = 5 * 89 = 445. Step 5: Sum of the last five observations S2 = 5 * 86 = 430. Step 6: The total sum S can be written as S1 + O6 + S2. Step 7: Therefore, 1012 = 445 + O6 + 430. Step 8: Combine 445 and 430 to get 875, so 1012 = 875 + O6. Step 9: Solve for O6: O6 = 1012 - 875 = 137. Verification / Alternative Check: Check that the sums are consistent. S1 = 445, S2 = 430 and O6 = 137. Adding them gives 445 + 430 + 137 = 1012, which matches the total sum of all 11 observations. The averages of the first and last five observations computed from these sums also remain 89 and 86 respectively, confirming the result. Why Other Options Are Wrong: Options 134, 139, 141 and 143 would give total sums different from 1012 when combined with 445 and 430, contradicting the given overall average of 92. Common Pitfalls: Some learners mistakenly divide the overall average by 2 or attempt to average 89 and 86 directly. Others may forget that the sixth observation is not part of either group of five, and incorrectly count some observations twice. The correct approach is to work with total sums and subtract the known partial sums from the overall sum. Final Answer: The sixth observation recorded by Sriram is 137.

Question 13

Eight years ago there were 5 members in the family of Arthur and at that time the average age of the family members was 36 years. Since then Arthur has married and has had a child, and even now the average age of the family is the same as before. If…

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Introduction: This problem tests understanding of average age over time and how adding new members affects the average. It also involves a typical age difference relation between a mother and her child. By using the fact that the average age of the family remains unchanged after some years and new additions, we can find the current ages and then the age of the wife at the time of the child birth. Given Data / Assumptions: - Eight years ago there were 5 members in Arthur family. - Eight years ago the average age of the family was 36 years. - Since then, Arthur has married and has had a child, so now there are 7 members. - The present average age of the family is still 36 years. - The present age of the wife of Arthur is 26 years more than the present age of the child. - We need the age of the wife at the time of the child birth. Concept / Approach: Eight years ago, we can find the total age of the original 5 members from the average. After 8 years, these original members are all 8 years older. The wife and child were not counted then, but they are now. The total current age of all 7 members can be found using the current average. The difference between this current total and the total age of the original members now gives the combined present age of the wife and child. Using the difference between their ages, we can solve for both current ages and then subtract the child current age from the wife current age to find her age at birth of the child. Step-by-Step Solution: Step 1: Eight years ago, total age of the 5 members = 5 * 36 = 180 years. Step 2: Eight years later, each of these 5 members is 8 years older, so their total present age = 180 + 5 * 8 = 180 + 40 = 220 years. Step 3: At present, the family has 7 members (the original 5 members, the wife and the child). Step 4: Present average age of the family is 36 years, so total present age of all 7 members = 7 * 36 = 252 years. Step 5: Combined present age of the wife and the child = total present age of 7 members minus present age of original 5 members = 252 - 220 = 32 years. Step 6: Let the present age of the wife be W years and the present age of the child be C years. Step 7: We have two equations: W + C = 32 and W - C = 26. Step 8: Add the equations: (W + C) + (W - C) = 32 + 26 gives 2W = 58. Step 9: Therefore, W = 58 / 2 = 29 years. Step 10: Substitute W = 29 into W + C = 32 to get C = 3 years. Step 11: The age of the wife at the time of the child birth = present age of wife minus present age of child = 29 - 3 = 26 years. Verification / Alternative Check: If the wife is now 29 and the child is 3, their combined age is 32, matching the value found from the total averages. The age difference is 29 - 3 = 26 years, which matches the condition. The original 5 members have a combined present age of 220 years. Adding 32 for the wife and child gives 252 years. Dividing 252 by 7 gives 36, which confirms the present average. Thus all conditions are satisfied. Why Other Options Are Wrong: Options 25, 20 and 27 years do not match the age difference or the combined age constraint of 32 years. Option 29 years is the present age of the wife, not her age at the time of the child birth. Common Pitfalls: Some students may forget that the original members age increases over time and incorrectly keep their total age fixed. Others may misinterpret the age difference condition or confuse present ages with ages at the time of birth. Carefully setting up equations for the sums and differences of ages helps avoid these errors. Final Answer: The age of the wife of Arthur at the time of the child birth was 26 years.

Question 14

Twenty men go to a restaurant. Nineteen of them pay Rs 70 each and the twentieth man pays Rs 90.25 more than the average amount paid by all 20 men. What is the total bill for the group?

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Introduction: This question tests your ability to work with averages when one data point is defined relative to the average itself. Nineteen men pay a fixed amount, and the last man pays some amount more than the overall average that includes his own payment. We must set up and solve an equation involving the unknown average and then find the total bill. Given Data / Assumptions: - Total number of men in the group = 20. - Nineteen men each pay Rs 70. - The twentieth man pays Rs 90.25 more than the average payment of all 20 men. - We must find the total bill for all 20 men together. Concept / Approach: Let the average payment of the 20 men be A rupees. The nineteenth men contribute a fixed amount, and the twentieth man contributes A + 90.25. The total bill is the sum of all individual payments, which is also equal to 20A because A is the average. By equating these two expressions, we can solve for A and then compute the total bill 20A. Step-by-Step Solution: Step 1: Let A be the average amount paid by each of the 20 men. Step 2: Payment by each of the nineteen men = Rs 70, so total from these nineteen men = 19 * 70 = 1330. Step 3: Payment by the twentieth man = A + 90.25. Step 4: Total bill T can be written as T = 1330 + (A + 90.25). Step 5: By definition of average, T is also equal to 20A. Step 6: Therefore, 20A = 1330 + A + 90.25. Step 7: Rearrange to collect A terms: 20A - A = 1330 + 90.25. Step 8: This gives 19A = 1420.25. Step 9: Solve for A: A = 1420.25 / 19 = 74.75. Step 10: Total bill T = 20A = 20 * 74.75 = 1495. Verification / Alternative Check: Check the twentieth man payment. If A = 74.75, he pays A + 90.25 = 74.75 + 90.25 = 165. The nineteen men pay a total of 1330, so the total bill is 1330 + 165 = 1495. The average from this total is 1495 / 20 = 74.75, which matches A, so the calculation is consistent. Why Other Options Are Wrong: Options Rs 548, Rs 850, Rs 1750 and Rs 1400 do not satisfy the condition that the twentieth man pays Rs 90.25 more than the average of all 20 men. Substituting these totals and recomputing the implied average leads to contradictions with the problem statement. Common Pitfalls: A common mistake is to assume the twentieth man pays Rs 90.25 more than 70, rather than more than the overall average. Another error is to forget that the total is equal to 20A and only count the nineteen men payments. Always connect the average back to the total sum via total = average * number of items. Final Answer: The total bill for the group of 20 men is Rs 1495.

Question 15

In a class of 100 students the average weight is 30 kg. If the average weight of the girls is 24 kg and the average weight of the boys is 32 kg, how many girls are there in the class?

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Introduction: This problem is about using weighted averages to find how many members belong to each subgroup. The overall average weight of the class and the separate average weights of boys and girls are given. We must find the number of girls in the class. Given Data / Assumptions: - Total number of students in the class = 100. - Overall average weight of the class = 30 kg. - Average weight of the girls = 24 kg. - Average weight of the boys = 32 kg. - We must find how many girls are in the class. Concept / Approach: Let the number of girls be g and the number of boys be 100 - g. The total weight of the girls is 24g and the total weight of the boys is 32(100 - g). The total weight of all 100 students is 30 * 100. Equating the expression from the subgroup weights to the expression from the overall average gives an equation in g which can be solved. Step-by-Step Solution: Step 1: Let the number of girls be g. Step 2: Then the number of boys is 100 - g. Step 3: Total weight of all the girls = 24g kg. Step 4: Total weight of all the boys = 32(100 - g) kg. Step 5: Total weight of all 100 students based on overall average = 100 * 30 = 3000 kg. Step 6: Set up the equation: 24g + 32(100 - g) = 3000. Step 7: Expand the right part: 32(100 - g) = 3200 - 32g. Step 8: Substitute back: 24g + 3200 - 32g = 3000. Step 9: Combine like terms: (24g - 32g) + 3200 = 3000 gives -8g + 3200 = 3000. Step 10: Subtract 3200 from both sides: -8g = 3000 - 3200 = -200. Step 11: Divide by -8: g = (-200) / (-8) = 25. Verification / Alternative Check: If there are 25 girls, then there are 75 boys. Total weight of girls = 24 * 25 = 600 kg. Total weight of boys = 32 * 75 = 2400 kg. Combined total = 600 + 2400 = 3000 kg. Dividing by 100 students gives an average weight = 3000 / 100 = 30 kg, which matches the given overall average. This verifies that g = 25 is correct. Why Other Options Are Wrong: Options 26, 27, 28 and 30 do not satisfy the equation 24g + 32(100 - g) = 3000. For each of these values, the combined total weight would differ from 3000 kg, giving a different overall average than 30 kg. Common Pitfalls: A typical error is to attempt to average 24 and 32 directly without accounting for the numbers of boys and girls, or to mis-handle the algebra when combining terms. Keeping separate expressions for total weights and using the overall average formula helps to set up the correct equation. Final Answer: The number of girls in the class is 25.

Question 16

Consider all two digit natural numbers that remain the same when their digits are interchanged, such as 11, 22 and 33. What is the average of all such two digit numbers?

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Introduction: This question asks for the average of all two digit numbers that remain the same when their digits are interchanged. Such numbers have identical tens and units digits and are examples of simple numerical palindromes in the two digit range. Given Data / Assumptions: - We consider all two digit numbers where the tens digit and the units digit are the same. - Examples include 11, 22, 33 and so on up to 99. - We must find the average of all these numbers. Concept / Approach: The two digit numbers with identical digits form a simple arithmetic progression. Writing them out explicitly, we can observe the first term, the last term and the common difference. For any arithmetic progression, the average of all the terms is equal to the average of the first and last term, that is (first term + last term) / 2. Step-by-Step Solution: Step 1: List the numbers: 11, 22, 33, 44, 55, 66, 77, 88 and 99. Step 2: Confirm that these numbers form an arithmetic progression with first term 11, common difference 11 and last term 99. Step 3: There are 9 such numbers, corresponding to digits from 1 to 9 repeated twice. Step 4: For an arithmetic progression, the average of all terms is (first term + last term) / 2. Step 5: Compute (11 + 99) / 2. Step 6: 11 + 99 = 110. Step 7: 110 / 2 = 55. Step 8: Therefore, the average of all such two digit numbers is 55. Verification / Alternative Check: We can also sum all the numbers and divide by 9. Sum = 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99. Pairing terms symmetrically, (11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55 gives (110) + (110) + (110) + (110) + 55 = 440 + 55 = 495. Dividing by 9 gives 495 / 9 = 55, confirming the result. Why Other Options Are Wrong: Options 53, 54, 56 and 57 do not match the average computed using either the arithmetic progression property or the explicit sum. They arise from incorrect calculations or misunderstandings of which numbers are included. Common Pitfalls: Some students may incorrectly include numbers like 10 or 12 that do not have identical digits, or they might forget one of the valid numbers in the list. Others may misapply the formula for the average of an arithmetic progression. Carefully listing the terms and using the formula (first + last) / 2 avoids these issues. Final Answer: The average of all two digit numbers whose digits are the same is 55.

Question 17

The average of the five numbers 26, 29, n, 35 and 43 lies between 25 and 35. If n is an integer greater than this average, what is the value of n?

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Introduction / Context: This aptitude question tests your understanding of numerical averages with an unknown term and basic inequality reasoning. You must express the average in terms of n, apply the condition about the range of the average, and also use the condition that n itself is greater than this average. Finally, you have to choose the only option that satisfies all the conditions. Given Data / Assumptions: - The five numbers are 26, 29, n, 35 and 43. - Their average lies between 25 and 35. - n is an integer and n is greater than the average of these five numbers. - You must pick n from the options 30, 31, 33 and 34. Concept / Approach: The average of k numbers is the sum of the numbers divided by k. Here, the average is a simple linear expression in n. Once we write the average, we can compare it with the boundary values 25 and 35. Separately, the condition that n is greater than the average eliminates some possible values. Finally, we test the remaining candidate values from the options. Step-by-Step Solution: Step 1: Write the sum of the five numbers: 26 + 29 + n + 35 + 43. Step 2: Compute the constant part: 26 + 29 + 35 + 43 = 133, so the sum is 133 + n. Step 3: The average is therefore (133 + n) / 5. Step 4: The condition that the average lies between 25 and 35 gives 25 < (133 + n) / 5 < 35, which implies a wide acceptable range of n, including all the options. Step 5: Now use the condition n > average. That is n > (133 + n) / 5. Step 6: Multiply both sides by 5: 5n > 133 + n, so 4n > 133, which gives n > 33.25. Step 7: Since n is an integer, n must be at least 34. Step 8: Among the options 30, 31, 33 and 34, only 34 satisfies n > 33.25 and also keeps the average between 25 and 35. Verification / Alternative check: For n = 34, the average is (26 + 29 + 34 + 35 + 43) / 5 = 167 / 5 = 33.4. This value is clearly between 25 and 35, and n = 34 is greater than 33.4, so both conditions are satisfied. No other listed option satisfies n greater than the average. Why Other Options Are Wrong: - Option 30: For n = 30, the average is (163) / 5 = 32.6, which is greater than 30, so n is not greater than the average. - Option 31: For n = 31, the average is 164 / 5 = 32.8, again larger than 31, so the condition n greater than average fails. - Option 33: For n = 33, the average is 166 / 5 = 33.2, which is still greater than 33, so the inequality is again not satisfied. Common Pitfalls: Many learners only apply the condition that the average lies between 25 and 35 and forget to check the extra condition that n must be greater than this average. Another common mistake is to treat “lies between 25 and 35” as giving a unique value of n instead of a range. You must always use every condition in the problem. Final Answer: Therefore, the only integer value of n that satisfies all the given conditions is 34.

Question 18

A batsman is playing his twelfth innings. After scoring 26 runs in this innings, his batting average increases from 14 to 15 runs per innings. How many runs must he score in total in this same innings so that his average becomes 19 runs per innings?

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Introduction / Context: This question checks your understanding of averages in the context of cricket scores. The key ideas are how an additional innings affects the batting average and how to use the formula for average to move from the old average to a new target average within the same innings. Given Data / Assumptions: - Before the current innings, the batsman has completed 11 innings with an average of 14 runs. - While playing his twelfth innings, after scoring 26 runs, his average becomes 15 runs per innings. - We assume 26 is the partial score so far in this twelfth innings when the average first reaches 15. - We need the total score in this same twelfth innings that will raise his average to 19 runs per innings. Concept / Approach: Batting average is defined as total runs scored divided by number of innings played. First, we use the given information about the average changing from 14 to 15 to find how many innings he has already completed. Then we use the required final average of 19 over the same number of innings (still 12) to find the total runs needed in this innings, which directly gives the required total score, not just extra runs beyond 26. Step-by-Step Solution: Step 1: Let the number of innings before this match be n. His total runs before this innings are 14 * n. Step 2: After scoring 26 runs in this twelfth innings, the new average becomes 15, so (14 * n + 26) / (n + 1) = 15. Step 3: Solving 14n + 26 = 15n + 15 gives n = 11, so this is his twelfth innings. Step 4: Total runs after 12 innings for an average of 15 are 15 * 12 = 180. Step 5: To have an average of 19 over 12 innings, his total runs must be 19 * 12 = 228. Step 6: Therefore, his total score in this twelfth innings must be 228 - (total runs before this innings), that is 228 - 14 * 11 = 228 - 154 = 74. Verification / Alternative check: If he scores 74 runs in the twelfth innings, his total runs become 154 + 74 = 228. The new average is 228 / 12 = 19 runs per innings, exactly as required. Thus, the answer is consistent. Why Other Options Are Wrong: - Option 79: With 79 runs, total runs are 154 + 79 = 233, giving an average of 233 / 12 which is greater than 19. - Option 72: With 72 runs, total runs are 154 + 72 = 226, giving an average below 19. - Option 60: With 60 runs, the total is 214, and the average is much less than 19. Common Pitfalls: A common error is to treat 26 as the final score in the innings instead of a partial score at which the average first becomes 15. Another mistake is to try to first calculate an intermediate average and then guess extra runs. The safest approach is always to write clear equations for total runs before and after the innings and use the definition of average systematically. Final Answer: The batsman must score a total of 74 runs in this twelfth innings to raise his average to 19.

Question 19

All the students of a school are made to stand in rows of 54 students each and 30 such rows are formed. If instead they are arranged in rows of 45 students each, how many rows will be formed?

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Introduction / Context: This problem uses the basic concept of average and multiplication to count how many students there are in total and how rearranging them into different sized rows changes the number of rows. It is a typical counting and grouping puzzle from school level aptitude. Given Data / Assumptions: - When students stand in rows of 54, there are 30 such rows. - The total number of students remains the same for any arrangement. - In the second arrangement, each row contains 45 students. - We need to find how many rows of 45 students can be formed. Concept / Approach: The total number of students is equal to number of rows multiplied by students per row. First we find the total number of students using the first arrangement. Then, for the second arrangement, we divide the same total by 45 to find the new number of rows. There is no rounding because the problem is designed so that the total divides exactly. Step-by-Step Solution: Step 1: Compute the total number of students from the first arrangement: total = 54 * 30. Step 2: Calculate 54 * 30 = 1620 students. Step 3: In the second arrangement, each row has 45 students. Step 4: The number of rows formed is total students divided by students per row, that is 1620 / 45. Step 5: Compute 1620 / 45 = 36 rows. Verification / Alternative check: You can quickly check by multiplying 36 * 45. This gives 1620, the same total number of students as before. Since the total is unchanged, 36 rows of 45 students each is consistent with the original arrangement. Why Other Options Are Wrong: - Option 22: 22 * 45 = 990, which is less than the total 1620, so many students would be left out. - Option 32: 32 * 45 = 1440, still less than 1620, so it cannot be the correct number of rows. - Option 26: 26 * 45 = 1170, which again does not match the required total. Common Pitfalls: Some students confuse the two numbers 54 and 30 and mistakenly divide instead of multiplying to find the total. Another error is to try to adjust the number of rows directly without first working out the total. Always remember that when the group size changes but the total remains constant, multiply to find the total first and then divide by the new group size. Final Answer: The students will form 36 rows when arranged with 45 students in each row.

Question 20

A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the whole family?

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Introduction / Context: This problem is a classic example of combined average. Instead of every individual age being listed, you are given average ages for different subgroups within the family. The task is to combine these subgroup averages to find the overall average age for all family members together. Given Data / Assumptions: - There are two grandparents with an average age of 67 years. - There are two parents with an average age of 35 years. - There are three grandchildren with an average age of 6 years. - The total number of family members is 2 + 2 + 3 = 7. Concept / Approach: To combine averages, we convert each subgroup average back into a total sum for that subgroup by multiplying by the number of people in that subgroup. Then we add all these subgroup sums to get the total sum of ages for the entire family. Finally, we divide this total by the total number of people, which in this case is seven, to get the overall average age. Step-by-Step Solution: Step 1: Compute the total age of the grandparents: 2 * 67 = 134 years. Step 2: Compute the total age of the parents: 2 * 35 = 70 years. Step 3: Compute the total age of the grandchildren: 3 * 6 = 18 years. Step 4: Add all these totals to get the family total age: 134 + 70 + 18 = 222 years. Step 5: The total number of family members is 7, so the overall average age is 222 / 7 years. Step 6: Divide 222 by 7 to get 31 with a remainder of 5, so the average age is 31 5/7 years. Verification / Alternative check: You can check the mixed fraction form: 7 * 31 = 217 and 217 + 5 = 222, which agrees with the total family age. So 31 5/7 is correctly expressed. Why Other Options Are Wrong: - Option 32 6/7: This corresponds to 32 + 6/7 = 32.857..., which is higher than the actual overall average. - Option 34 4/7: This is even larger, corresponding to 34.571..., which is inconsistent with the given subgroup data. - Option 35 4/9: This value is around 35.44, far too high given that three of the family members are only 6 years old. Common Pitfalls: A common mistake is to simply take the average of the three subgroup averages (67, 35 and 6) without weighting them by the number of people in each group. This ignores the fact that there are different numbers of grandparents, parents and grandchildren. Always convert averages to totals, sum them, and then divide by the overall count. Final Answer: The average age of the entire family is 31 5/7 years.

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