Question 1

A square plot of land has an area of 289 square feet. What will be the cost of building a fence around this square plot if the price per foot of fencing is Rs. 58?

Accepted Answer

Introduction / Context: This question requires converting area of a square into side length, then using that to compute the perimeter and finally the fencing cost. It brings together basic geometry and simple arithmetic cost calculation. Given Data / Assumptions: - Area of square plot = 289 square feet. - Cost of fencing = Rs. 58 per running foot. - We assume fencing is required on all four sides of the square. Concept / Approach: For a square, if the side length is s, then: Area = s^2Perimeter = 4 * s We first find the side from the area, then the perimeter, and finally multiply the perimeter by the cost per foot to get the total cost of fencing. Step-by-Step Solution: Step 1: Let side of square = s feet.Step 2: Given s^2 = 289.Step 3: Since 17 * 17 = 289, s = 17 feet.Step 4: Perimeter of the square = 4 * s = 4 * 17 = 68 feet.Step 5: Cost per foot = Rs. 58.Step 6: Total cost = 68 * 58 rupees.Step 7: Compute 68 * 58 = (68 * 60) - (68 * 2) = 4080 - 136 = 3944. Verification / Alternative check: You can also multiply 58 * 68 directly in parts: 58 * 70 = 4060, then subtract 58 * 2 = 116 to get 4060 - 116 = 3944. Both methods confirm the same total cost. Why Other Options Are Wrong: - 3748, 3847 and 3154 result from incorrect multiplication or from using the wrong side length or perimeter. None equals 68 * 58. Common Pitfalls: - Taking the square root of 289 incorrectly, for example using 19 instead of 17. - Forgetting that fencing cost is based on perimeter, not area, and mistakenly multiplying area by cost per foot. Final Answer: The cost of building the fence around the plot is Rs. 3944.

Question 2

It costs Rs. p to send each of the first 1000 messages and Rs. q to send each message after the first 1000. If r is greater than 1000, what is the total cost in rupees of sending r messages in terms of p, q and r?

Accepted Answer

Introduction / Context: This algebraic word problem uses a piecewise pricing scheme. The cost per message changes after a threshold, which is a typical model in telecommunications and bulk pricing scenarios. We must express the total cost symbolically in terms of p, q and r. Given Data / Assumptions: - Cost per message for the first 1000 messages = Rs. p. - Cost per message for each additional message beyond 1000 = Rs. q. - Total number of messages sent = r, with r greater than 1000. - We assume the pricing scheme is exactly as stated with no fixed charges. Concept / Approach: The total cost is the sum of two parts: 1. Cost of the first 1000 messages.2. Cost of the remaining (r - 1000) messages. Then we simplify the expression and match it with the given options. Step-by-Step Solution: Step 1: Cost of the first 1000 messages = 1000 * p.Step 2: Number of additional messages beyond 1000 = r - 1000.Step 3: Cost of additional messages = (r - 1000) * q.Step 4: Total cost = 1000p + (r - 1000)q.Step 5: Expand (r - 1000)q = rq - 1000q.Step 6: Total cost = 1000p + rq - 1000q.Step 7: Factor 1000 from the first and last terms: 1000p - 1000q = 1000(p - q).Step 8: So total cost = 1000(p - q) + qr. Verification / Alternative check: Write the expression as 1000p + (r - 1000)q and check with an example. Suppose p = 2, q = 1 and r = 1500. Then cost should be 1000 * 2 + 500 * 1 = 2000 + 500 = 2500. Using the simplified form 1000(p - q) + qr gives 1000(2 - 1) + 1 * 1500 = 1000 + 1500 = 2500, which matches. Why Other Options Are Wrong: - 1000 (r - p) + pq incorrectly treats r - p as a count of messages and pq as an added term. - 1000p + qr includes all r messages at cost q without subtracting the first 1000 at cost p. - 1000(r - q) + pr mixes units incorrectly and does not separate the threshold at 1000 messages. Common Pitfalls: - Forgetting to subtract 1000 from r for the additional messages. - Treating p and q as counts instead of rates per message. - Not simplifying the expression carefully, which can cause a mismatch with the correct option. Final Answer: The total cost of sending r messages is 1000(p - q) + qr.

Question 3

Three maths classes A, B and C take an algebra test. The average score of class A is 83, of class B is 76 and of class C is 85. The combined average score of classes A and B together is 79, and that of classes B and C together is 81. What is the ave…

Accepted Answer

Introduction / Context: This problem involves multiple averages for different subsets of three classes. From the given information, we infer the relative sizes of the classes and then use a weighted average to find the overall average score. Given Data / Assumptions: - Average score of class A = 83. - Average score of class B = 76. - Average score of class C = 85. - Average of classes A and B together = 79. - Average of classes B and C together = 81. - We assume each class has a certain number of students but not necessarily the same size. Concept / Approach: Let the number of students in classes A, B and C be a, b and c respectively. Then: Total score of A = 83a, of B = 76b, of C = 85c. The combined averages give two equations: (83a + 76b) / (a + b) = 79(76b + 85c) / (b + c) = 81 We use these equations to find the ratios between a, b and c, then compute the overall average. Step-by-Step Solution: Step 1: From (83a + 76b) / (a + b) = 79, cross multiply:83a + 76b = 79a + 79b.Step 2: Rearrange: 83a - 79a = 79b - 76b, so 4a = 3b.Step 3: Hence a : b = 3 : 4.Step 4: From (76b + 85c) / (b + c) = 81, cross multiply:76b + 85c = 81b + 81c.Step 5: Rearrange: 85c - 81c = 81b - 76b, so 4c = 5b.Step 6: Hence b : c = 4 : 5.Step 7: Combining the two ratios, we get a : b : c = 3 : 4 : 5.Step 8: Let a = 3k, b = 4k, c = 5k.Step 9: Total score = 83a + 76b + 85c = 83 * 3k + 76 * 4k + 85 * 5k.Step 10: Compute: 83 * 3 = 249, 76 * 4 = 304, 85 * 5 = 425, so total score = (249 + 304 + 425)k = 978k.Step 11: Total number of students = a + b + c = 3k + 4k + 5k = 12k.Step 12: Overall average = 978k / 12k = 978 / 12 = 81.5. Verification / Alternative check: You can also check consistency with each pair of classes. Using the ratios, confirm that the computed totals give the stated combined averages of 79 and 81, which they do, so the ratios and overall average are consistent. Why Other Options Are Wrong: - 81, 78 and 80.5 do not satisfy the weighted combination of the three class averages given the derived class size ratios 3 : 4 : 5. They would imply incorrect total scores or inconsistent pair averages. Common Pitfalls: - Assuming equal class sizes and taking the simple average of 83, 76 and 85, which would give a different result. - Not setting up the pairwise average equations correctly, which leads to incorrect ratios and hence an incorrect overall average. Final Answer: The average score of all three classes A, B and C together is 81.5.

Question 4

The average of marks in three subjects is 224. The marks in the first subject are twice the marks in the second subject, and the marks in the second subject are twice the marks in the third subject. What are the marks in the second subject?

Accepted Answer

Introduction / Context: This question uses ratio relationships between three subject marks combined with an average condition. It is a straightforward algebra problem where we express all marks in terms of one variable and use the given average to solve. Given Data / Assumptions: - Average of marks in three subjects = 224. - Let the marks in the three subjects be M1, M2 and M3. - M1 = 2 * M2. - M2 = 2 * M3. - All marks are non negative numbers. Concept / Approach: First, we express all three marks in terms of the third subject. Then we use the average condition to determine their actual values. Finally, we extract the value of the second subject. Step-by-Step Solution: Step 1: Let M3 = x.Step 2: Then M2 = 2x (since the second is twice the third).Step 3: M1 = 2 * M2 = 2 * 2x = 4x.Step 4: Sum of marks = M1 + M2 + M3 = 4x + 2x + x = 7x.Step 5: Average of the three subjects = 224, so 7x / 3 = 224.Step 6: Multiply both sides by 3: 7x = 224 * 3 = 672.Step 7: Solve for x: x = 672 / 7 = 96.Step 8: M3 = 96, M2 = 2 * 96 = 192, M1 = 4 * 96 = 384.Step 9: The second subject marks are therefore 192. Verification / Alternative check: The sum of the marks is 384 + 192 + 96 = 672. The average is 672 / 3 = 224, which matches the given average, confirming that the derived values are consistent. Why Other Options Are Wrong: - 384 is the mark in the first subject, not the second. - 96 is the mark in the third subject. - 206 is not consistent with the required ratio relations and would not produce an average of 224 when used with matching values for the other two subjects. Common Pitfalls: - Mixing up the order of relationships, for example taking the third subject as twice the second rather than the reverse. - Using the average directly as one of the subject marks without considering the ratio. Final Answer: The marks in the second subject are 192.

Question 5

How many separate window coverings are required to completely span a row of 50 windows, if each single window covering can continuously cover 15 windows in a line?

Accepted Answer

Introduction / Context: This question is a basic application of averages and division in a real life style context. We are told that one long window covering can span several windows in a row and we must work out how many such coverings are needed to span all the windows in the line. The core idea is to see how many groups of a fixed size can fit into a total and then recognise that if there is any remainder, we still need one extra covering to complete the span. Given Data / Assumptions: Total number of windows to be covered = 50. Each window covering can span 15 windows in a straight line. We assume that window coverings cannot be cut or partially used; each covering is used as a whole piece. We must find the minimum number of coverings needed so that all 50 windows are covered. Concept / Approach: The main concept is simple division with a remainder. To see how many coverings are needed, we divide the total number of windows by the number of windows that one covering can span. If the total is not a perfect multiple, the remainder indicates that we need one additional full covering. Mathematically, this is the ceiling of 50 divided by 15. Step-by-Step Solution: Step 1: Write the basic division: number of coverings = 50 / 15. Step 2: Compute 50 / 15 = 3.333..., which means 3 complete groups of 15 windows and some windows left over. Step 3: Each full group of 15 windows uses one covering, so 3 complete groups use 3 coverings. Step 4: Number of windows covered by 3 coverings = 3 * 15 = 45 windows. Step 5: Remaining windows to be covered = 50 - 45 = 5 windows. Step 6: Even though only 5 windows are left, we still need one whole extra covering to span these remaining windows. Step 7: Therefore, the total number of coverings required = 3 + 1 = 4. Verification / Alternative check: We can quickly check by multiplying the answer back. If we use 4 coverings with a capacity of 15 windows each, the maximum span is 4 * 15 = 60 windows. Since 60 is more than 50, we definitely have enough length to cover all windows. With only 3 coverings, we would cover at most 45 windows, leaving 5 uncovered, which is not acceptable. Therefore 4 coverings is the smallest number that works. Why Other Options Are Wrong: Three coverings (option 3) can cover only 45 windows, so 5 windows remain uncovered. Fifteen coverings or 50 coverings are far more than necessary and do not reflect the capacity of each covering. The option 5 coverings is not minimal because 4 coverings already suffice to cover all 50 windows. Common Pitfalls: A common mistake is to take only the integer part of 50 / 15 and answer 3 without thinking about the remainder. Another typical error is to assume that a partially used covering somehow does not count as a full covering. In reality, if you need even part of another covering, you must count that entire piece. Remember that for these types of questions, if there is any remainder after division, you must take the next whole number. Final Answer: The number of window coverings required to span all 50 windows is 4.

Question 6

Sixteen children are divided into two groups, A and B, with 10 children in group A and 6 children in group B. The average percentage marks of group A is 75, and the overall average percentage marks of all 16 children is 76. What is the average perce…

Accepted Answer

Introduction / Context: This question is a classic weighted average problem involving two subgroups within a larger group. We know the average marks of one subgroup and the overall average, and we must use this information to find the average of the remaining subgroup. Such questions are very common in aptitude tests and help you understand how averages behave when combining or splitting groups. Given Data / Assumptions: Total number of children = 16. Group A has 10 children with average percentage marks = 75. Group B has 6 children with unknown average percentage marks. The overall average percentage marks of all 16 children = 76. We assume percentage marks are computed out of the same maximum and can be added normally. Concept / Approach: The key idea is that the total marks for the whole group is the sum of the total marks for each subgroup. The average of a group is defined as total marks divided by number of children. So we first compute the total marks of all 16 children using the overall average. Then we compute the total marks of group A using its average. The difference between these totals gives the total marks of group B. Finally, we divide by the number of children in group B to find their average. Step-by-Step Solution: Step 1: Compute total marks of all 16 children using the overall average. Total marks (all) = 16 * 76 = 1216. Step 2: Compute total marks of group A using its average. Total marks (A) = 10 * 75 = 750. Step 3: Compute total marks of group B. Total marks (B) = Total marks (all) - Total marks (A) = 1216 - 750 = 466. Step 4: Number of children in group B = 6. Step 5: Average marks of group B = Total marks (B) / Number of children in B = 466 / 6. Step 6: Simplify 466 / 6 by dividing numerator and denominator by 2: 466 / 6 = 233 / 3. Step 7: Thus the average percentage marks of group B is 233/3, which is approximately 77.67 percent. Verification / Alternative check: We can verify by substituting the average of group B back into the weighted average formula. Total marks from group B is (233 / 3) * 6 = 233 * 2 = 466. Combined with group A total of 750, we get 750 + 466 = 1216. Divide by the total number of children 16 to get 1216 / 16 = 76, which matches the given overall average. This confirms that our computed average for group B is correct. Why Other Options Are Wrong: The value 243/4 corresponds to 60.75 percent, which would bring the overall average below 76. Similarly, 254/5 equals 50.8 percent, also too low. The fraction 345/7 is approximately 49.29 percent, again inconsistent with a higher overall average. Simply using 76 (the overall average) for group B is not justified because group A already has a lower average (75), so group B must be higher than 76 to pull the combined average to 76. Only 233/3 satisfies the weighted average relationship exactly. Common Pitfalls: A frequent mistake is to average the averages directly, for example taking (75 + 76) / 2, which is incorrect because the groups have different sizes. Another common error is to forget to multiply the average by the number of children to obtain total marks. Some students also try to solve the problem using guesswork instead of a clear algebraic approach, which can be risky with fractional answers. Always remember that average = total / number, and that total is additive across groups. Final Answer: The average percentage marks of the children in group B is 233/3.

Question 7

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 / Context: This question is designed to test your understanding of how averages work when combining two subgroups into one larger group. We know the average ages of boys and girls separately, and we are asked to find the average age of the whole class. However, the problem contains a subtle but important missing piece of information, which makes it a test of logical reasoning as well as basic arithmetic. Given Data / Assumptions: The average age of the boys in the class is 16 years. The average age of the girls in the class is 15 years. The total number of students in the class is not given. The numbers of boys and girls are also not given. We are asked to find the average age of all students taken together. Concept / Approach: The overall average of a combined group depends on both the subgroup averages and the sizes of those subgroups. In mathematical terms, if there are b boys with average 16 and g girls with average 15, then the overall average A is given by: A = (16b + 15g) / (b + g). Without knowing either b or g, we cannot simplify this expression to a single unique number. Different values of b and g will generally produce different overall averages, even though the subgroup averages remain the same. Step-by-Step Solution: Step 1: Assume the number of boys is b and the number of girls is g. Step 2: Total age of all boys = 16b years. Step 3: Total age of all girls = 15g years. Step 4: Total age of the whole class = 16b + 15g. Step 5: Total number of students = b + g. Step 6: Overall average age A = (16b + 15g) / (b + g). Step 7: Without a specific relation between b and g, A cannot be simplified to a single fixed number. Step 8: Choose two different combinations, for example b = 1, g = 1 and b = 10, g = 1, to see that the resulting overall averages are different. Verification / Alternative check: Take b = 1, g = 1. The overall average = (16 * 1 + 15 * 1) / (1 + 1) = 31 / 2 = 15.5 years. Now take b = 10, g = 1. The overall average = (16 * 10 + 15 * 1) / 11 = (160 + 15) / 11 = 175 / 11 ≈ 15.91 years. Since both situations respect the given subgroup averages but produce different class averages, there is no unique answer for the average age of the whole class. Why Other Options Are Wrong: The value 15 years would be the overall average only if there were no boys, which contradicts the given condition that there are boys with average 16 years. The values 20 and 21 years are clearly impossible because they exceed the maximum subgroup average of 16 years. The value 16 would be possible only if there were no girls, which again contradicts the given information. Therefore none of these numerical options can be guaranteed as the class average. Common Pitfalls: A very common mistake is to simply average the two subgroup averages, for example taking (16 + 15) / 2 = 15.5, without considering the relative sizes of the subgroups. This is incorrect unless the number of boys and girls is the same, which is not stated in the problem. Another mistake is to assume that the overall average must be somewhere between 15 and 16 and then to guess a value, which is not logically justified. Always remember that to compute a combined average you must know subgroup sizes or have enough information to deduce them. Final Answer: Since the numbers of boys and girls are not known, the overall average age of the whole class cannot be uniquely determined. The correct response is data inadequate.

Question 8

What is the average (arithmetic mean) of the decimal numbers four tenths and five thousandths?

Accepted Answer

Introduction / Context: This problem checks your comfort with decimal notation and basic averages. You are given two numbers expressed in words as decimal fractions and asked to find their average. The key skills are to correctly convert the verbal expressions into numerical decimals and then apply the standard formula for the arithmetic mean of two numbers. Given Data / Assumptions: First number is four tenths. Second number is five thousandths. We must compute the average of these two decimal numbers. We assume standard decimal place value: tenths, hundredths, thousandths and so on. Concept / Approach: The average (or arithmetic mean) of two numbers a and b is given by: Average = (a + b) / 2. We first translate the words into decimal form: four tenths becomes 0.4 and five thousandths becomes 0.005. Then we add these decimals and divide by 2. Careful handling of decimal places ensures we get the exact correct value. Step-by-Step Solution: Step 1: Convert four tenths into decimal form. Four tenths = 4 / 10 = 0.4. Step 2: Convert five thousandths into decimal form. Five thousandths = 5 / 1000 = 0.005. Step 3: Add the two decimals: 0.4 + 0.005. Step 4: Write 0.4 as 0.400 to make the place values line up: 0.400 + 0.005 = 0.405. Step 5: Now compute the average: (0.4 + 0.005) / 2 = 0.405 / 2. Step 6: Divide 0.405 by 2: 0.405 / 2 = 0.2025. Step 7: Therefore, the average of four tenths and five thousandths is 0.2025. Verification / Alternative check: We can verify by working with fractions instead of decimals. Four tenths = 4 / 10 and five thousandths = 5 / 1000. Convert both to a common denominator of 1000: 4 / 10 = 400 / 1000. Then the sum is (400 / 1000) + (5 / 1000) = 405 / 1000. The average is (405 / 1000) / 2 = 405 / 2000. Dividing numerator and denominator by 5 gives 81 / 400, which in decimal form is 0.2025. This matches the earlier decimal calculation exactly. Why Other Options Are Wrong: The options 25002 and 2502 clearly ignore the decimal place values and look like misinterpretations of the numbers as whole integers. The option 0.225 would be correct if the sum was 0.45, which would come from misreading five thousandths as 0.05 instead of 0.005. The option 'none of these' is not correct because 0.2025 is already present among the choices. Therefore, 0.2025 is the only correct answer. Common Pitfalls: A common mistake is confusing five thousandths (0.005) with five hundredths (0.05). Another frequent error is to add the decimals incorrectly, for example writing 0.4 + 0.005 = 0.45. Students also sometimes forget to divide by 2 after adding the two numbers. Paying attention to place values and following the operations in sequence helps avoid these errors. Final Answer: The average of four tenths and five thousandths is 0.2025.

Question 9

The average of any 5 consecutive odd natural numbers is k. If two more consecutive odd natural numbers, immediately following those 5 numbers, are added to the set, what will be the new average in terms of k?

Accepted Answer

Introduction / Context: This problem involves averages of consecutive odd natural numbers and how the average changes when more terms are added. The question is algebraic in nature and is intended to test your ability to represent patterns using variables, compute sums of sequences and manipulate averages symbolically. Recognising the symmetry of consecutive odd numbers around their middle term is the key insight. Given Data / Assumptions: We have 5 consecutive odd natural numbers. The average (mean) of these 5 numbers is k. Two more consecutive odd numbers, immediately after these 5, are added to the set. We must find the new average of all 7 numbers in terms of k. All numbers are positive odd integers. Concept / Approach: Five consecutive odd numbers can be written in terms of their middle number. If the middle number is n, then the five numbers are n - 4, n - 2, n, n + 2 and n + 4. The average of these five numbers is exactly the middle one, n. We are told that this average is k, so n = k. When we add the next two consecutive odds, n + 6 and n + 8, we can compute the new sum and average using algebra. Step-by-Step Solution: Step 1: Let the five consecutive odd numbers be n - 4, n - 2, n, n + 2 and n + 4. Step 2: Their average is the middle term n, and we are given that this average is k, so n = k. Step 3: The next two consecutive odd numbers after n + 4 are n + 6 and n + 8. Step 4: Now the full set of 7 numbers is: n - 4, n - 2, n, n + 2, n + 4, n + 6, n + 8. Step 5: Sum of these 7 numbers = 7n + ( -4 - 2 + 0 + 2 + 4 + 6 + 8 ). Step 6: Compute the constant part: -4 - 2 + 0 + 2 + 4 + 6 + 8 = 14, so sum = 7n + 14. Step 7: New average = (7n + 14) / 7 = n + 2. Step 8: Replace n with k, since n = k, to get new average = k + 2. Verification / Alternative check: Use a concrete example to confirm the algebra. Suppose the 5 consecutive odd numbers are 3, 5, 7, 9 and 11. Their average is (3 + 5 + 7 + 9 + 11) / 5 = 35 / 5 = 7, so k = 7. The next two odd numbers are 13 and 15. Now the 7 numbers are 3, 5, 7, 9, 11, 13 and 15. The new average is (3 + 5 + 7 + 9 + 11 + 13 + 15) / 7 = 63 / 7 = 9. This equals k + 2 = 7 + 2, which verifies our general result. Why Other Options Are Wrong: Options such as 2(k + 1), 2k - 3 and 2k + 1 incorrectly assume that the average doubles or follows some other incorrect pattern and do not match the actual algebraic derivation. For instance, 2(k + 1) would give 16 for k = 7, which is much larger than the correct new average 9 in our example. The option k - 2 suggests a decrease in the average, which is impossible because we are adding larger numbers to the original set. Common Pitfalls: A common mistake is to think that the new average is simply k plus the average of the two added numbers divided by some factor, without recalculating the total sum. Others may misrepresent the five consecutive odd numbers or incorrectly compute the constant part of the sum. Some students also forget that adding larger numbers to a set will generally increase the average. Writing the sequence clearly and using algebraic expressions prevents these errors. Final Answer: The new average after adding the next two consecutive odd numbers is k + 2.

Question 10

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

Accepted Answer

Introduction / Context: This question is about the relationship between the overall average of a set of numbers and the averages of two equal sized subgroups within that set. It is a simple algebraic problem that tests whether you can convert verbal statements about averages into equations using sums and then derive the correct relationship between x, y and z. Given Data / Assumptions: There are 6 numbers in total. The average of all 6 numbers is x. The average of 3 of these numbers is y. The average of the remaining 3 numbers is z. We must find the correct equation relating x, y and z. Concept / Approach: Average is defined as total sum divided by number of items. If the average of 6 numbers is x, then their total sum is 6x. If the average of one group of 3 numbers is y, then the sum of those 3 numbers is 3y. Similarly, the sum of the remaining 3 numbers is 3z. Since the 6 numbers consist exactly of these two subgroups, the total sum 6x must equal 3y + 3z. We then simplify this equation to get a direct relation between x, y and z. Step-by-Step Solution: Step 1: Let S be the total sum of all 6 numbers. Step 2: Given that the average of 6 numbers is x, we have S = 6x. Step 3: Sum of the first group of 3 numbers = 3y because their average is y. Step 4: Sum of the second group of 3 numbers = 3z because their average is z. Step 5: These two groups together make up all 6 numbers, so S = 3y + 3z. Step 6: Equate the two expressions for S: 6x = 3y + 3z. Step 7: Divide both sides by 3 to simplify: 2x = y + z. Step 8: Therefore, the correct relationship is 2x = y + z. Verification / Alternative check: Take a numerical example. Suppose the three numbers in the first group are 1, 2 and 3, and the three numbers in the second group are 7, 8 and 9. Then y = (1 + 2 + 3) / 3 = 2 and z = (7 + 8 + 9) / 3 = 8. The overall average x is (1 + 2 + 3 + 7 + 8 + 9) / 6 = 30 / 6 = 5. Now compute y + z = 2 + 8 = 10, and 2x = 2 * 5 = 10. Since 2x equals y + z for this example, this supports the derived relationship. Why Other Options Are Wrong: The equation x = y + z would suggest the overall average is larger than each subgroup average, which conflicts with the numeric example and the derived formula. The expression 3x = 2y - 2z has the wrong structure and can be rejected by substitution into a simple numeric case. The option x = (y + z) / 3 treats the overall average as a third of the average of the subgroup averages, which makes no sense dimensionally. The option 'none of these' is incorrect because we have found a specific correct relationship. Common Pitfalls: Some students mistakenly average y and z directly, writing x = (y + z) / 2, forgetting that x is the average over all 6 numbers, not just a simple mean of the subgroup averages. Others confuse sum and average and write equations like 6x = y + z, which ignore the fact that y and z are averages, not totals. Always remember that sum = average * number of items, and ensure that you use sums when combining groups. Final Answer: The correct relationship between the averages is 2x = y + z.

Question 11

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

Accepted Answer

Introduction / Context: This question involves the concept of average and how it changes when an error in one of the data points is corrected. It is a classic example of error correction problems in averages where a single incorrect entry affects the overall mean. Understanding how total sum and average are related is the key to solving such problems efficiently. Given Data / Assumptions: There are n people in the village. The original average age is 42 years. One person's age was recorded as 20 years less than the actual age. After correcting this person's age, the new average increases by 1 year, becoming 43 years. We must find the value of n. Concept / Approach: The average age is the total sum of ages divided by the number of people. If the average is 42, then the total recorded sum of ages is 42n. But one person's age was understated by 20 years, so the true total should be 20 higher than the recorded total. When we add this missing 20 years and divide by n, the average becomes 43. Setting up this relationship as an equation allows us to solve for n. Step-by-Step Solution: Step 1: Let the recorded total sum of ages be S. Since the recorded average is 42, we have S = 42n. Step 2: One person's true age is 20 years more than what was recorded. So the correct total sum is S + 20. Step 3: After correction, the new average is 43 years, so (S + 20) / n = 43. Step 4: Substitute S = 42n into this equation: (42n + 20) / n = 43. Step 5: Simplify: 42 + 20 / n = 43. Step 6: Subtract 42 from both sides: 20 / n = 1. Step 7: Multiply both sides by n: 20 = n. Step 8: Therefore, the number of people n is 20. Verification / Alternative check: Check numerically. If n = 20 and the average age is 42, the recorded total is 42 * 20 = 840. One age is 20 years too low, so the actual total should be 840 + 20 = 860. The corrected average is 860 / 20 = 43, which matches the description that the average increased by 1 year. This confirms that n = 20 is consistent with all the conditions in the problem. Why Other Options Are Wrong: If n = 21, then the original total would be 42 * 21 and adding 20 and dividing by 21 would not give exactly 43. Similarly, n = 22 or n = 19 would lead to fractional increases in the average that are not exactly 1 year. The option 'none of these' is incorrect because we have found a specific value of n that satisfies the conditions. Only n = 20 makes the increase in average exactly 1 year when 20 is added to the total. Common Pitfalls: A common mistake is to try to work directly with averages rather than converting to total sums. Some students also misinterpret the phrase '20 years less than the actual age' and add or subtract 20 incorrectly. Another error is to forget that the number of people n remains the same before and after correction. Keeping track of total sums, the number of entries and the definition of average helps avoid these mistakes. Final Answer: The total number of people in the village is 20.

Question 12

The average expenditure of a hotel is Rs. 60 per guest when there are 10 guests, and Rs. 40 per guest when there are 20 guests. Assuming that the hotel has some fixed daily expenses (independent of the number of guests) and the rest of the expenditu…

Accepted Answer

Introduction / Context: This question involves a mixture of fixed and variable costs and asks for the average expenditure per guest at different levels of occupancy. Such problems commonly appear in quantitative aptitude tests to check your ability to set up simple linear equations representing fixed and variable components and then use them to predict costs at other levels. Given Data / Assumptions: When there are 10 guests, the average expenditure per guest is Rs. 60. When there are 20 guests, the average expenditure per guest is Rs. 40. There is a fixed daily expenditure F (which does not depend on the number of guests). There is a variable expenditure v per guest. We must find the average expenditure per guest when there are 40 guests. Concept / Approach: Let F be the fixed cost and v be the variable cost per guest. Then for N guests total expenditure is F + vN. The average expenditure per guest is (F + vN) / N. We are given two such situations (for N = 10 and N = 20). From these two equations we can solve for F and v. Once we know F and v, we can compute the average cost when N = 40. Step-by-Step Solution: Step 1: When N = 10, average cost is 60, so (F + 10v) / 10 = 60. Step 2: Multiply both sides by 10: F + 10v = 600. Step 3: When N = 20, average cost is 40, so (F + 20v) / 20 = 40. Step 4: Multiply both sides by 20: F + 20v = 800. Step 5: Subtract the first equation from the second: (F + 20v) - (F + 10v) = 800 - 600. Step 6: This gives 10v = 200, so v = 20. Step 7: Substitute v = 20 into F + 10v = 600: F + 10 * 20 = 600 ⇒ F + 200 = 600 ⇒ F = 400. Step 8: For N = 40 guests, total expenditure = F + 40v = 400 + 40 * 20 = 400 + 800 = 1200. Step 9: Average expenditure per guest when N = 40 is 1200 / 40 = 30. Verification / Alternative check: We can cross check by computing the total expenditure for the earlier cases using F = 400 and v = 20. For 10 guests, total = 400 + 10 * 20 = 400 + 200 = 600, average = 600 / 10 = 60, which matches the given data. For 20 guests, total = 400 + 20 * 20 = 400 + 400 = 800, average = 800 / 20 = 40, also matching the problem statement. Since these match, using the same F and v for 40 guests is valid, leading to an average of 30. Why Other Options Are Wrong: An average of Rs. 25 or Rs. 20 per guest would imply a much lower variable cost or fixed cost than indicated by the given data. The option Rs. 35 does not satisfy the linear relationship defined by the two known points. Saying that the value cannot be determined ignores the fact that two equations in the two unknowns F and v are sufficient to find a unique solution. Only Rs. 30 is consistent with all the given conditions. Common Pitfalls: A common mistake is to assume that the average cost varies linearly with the number of guests and simply interpolate averages rather than expressing the total cost in terms of fixed and variable components. Another error is to forget that the fixed cost does not change with the number of guests, leading to incorrect equations. Careful identification of F and v and setting up the equations correctly ensures a straightforward solution. Final Answer: The average expenditure per guest when there are 40 guests in the hotel is Rs. 30.

Question 13

There are 44 students in a hostel. Later, 15 new students join the hostel, and as a result the total mess expense increases by Rs. 33 per day. However, the average expenditure per head decreases by Rs. 3. What was the original total daily expenditur…

Accepted Answer

Introduction / Context: This is a typical average and total cost problem in which the number of people sharing a common expense changes. When more students join the hostel, the total mess expense increases, but because more people are sharing the cost, the average expenditure per head decreases. You must use these two effects together to work out the original total daily expenditure of the mess. Given Data / Assumptions: Original number of students = 44. New number of students = 44 + 15 = 59. Original total mess expense per day = E rupees. After 15 students join, the total mess expense per day becomes E + 33 rupees. The average expenditure per head decreases by Rs. 3 after the new students join. We need to find E, the original total daily mess expenditure. Concept / Approach: The original average per student is E / 44. After 15 new students join, the new average per student becomes (E + 33) / 59. We are told that this new average is Rs. 3 less than the original average. Therefore, we can set up an equation expressing this relationship and solve for E. This is a straightforward application of the definition of average and basic algebra. Step-by-Step Solution: Step 1: Let the original total expense per day be E rupees. Step 2: Original average per head = E / 44. Step 3: After 15 new students join, total students = 59 and new total expense = E + 33. Step 4: New average per head = (E + 33) / 59. Step 5: The average decreases by Rs. 3, so (E + 33) / 59 = (E / 44) - 3. Step 6: Multiply through by 59 * 44 to clear denominators: 44(E + 33) = 59E - 3 * 59 * 44. Step 7: Expand the left side: 44E + 44 * 33 = 44E + 1452. Step 8: Compute 3 * 59 * 44 = 3 * 2596 = 7788, so the equation is 44E + 1452 = 59E - 7788. Step 9: Rearrange: 1452 + 7788 = 59E - 44E ⇒ 9240 = 15E. Step 10: Solve for E: E = 9240 / 15 = 616. Step 11: Therefore, the original total mess expense per day was Rs. 616. Verification / Alternative check: Check the averages. Original average = 616 / 44 = 14. After 15 students join, new total expense = 616 + 33 = 649 and students = 59. New average = 649 / 59 = 11. This is indeed Rs. 3 less than 14, matching the description. Also, the increase in total expense is Rs. 33 as given. This confirms that E = 616 is consistent with all parts of the problem. Why Other Options Are Wrong: If the original expense were Rs. 404, 514 or 340, then using the same reasoning would not produce an average decrease of exactly Rs. 3 when 15 students join and the total increases by Rs. 33. These values fail to satisfy the equation derived from the average relationship. The option Rs. 580 also does not satisfy the required conditions. Only Rs. 616 gives an original average of 14 and a new average of 11, exactly as required. Common Pitfalls: Some students confuse the change in total expense with the change in average and try to directly divide 33 by 3, thinking that 11 is the number of new students, which is incorrect. Others forget to use 59 in the denominator for the new average or misinterpret the direction of the change in average. Careful translation of the problem into algebraic equations is crucial to arrive at the correct answer. Final Answer: The original total daily expenditure of the mess was Rs. 616.

Question 14

On a school's Annual Day, sweets were to be distributed equally among 112 children. However, 32 children were absent on that day, so the remaining children each received 6 extra sweets. How many sweets was each child originally supposed to get?

Accepted Answer

Introduction / Context: This is a straightforward average and total quantity problem often seen in school level aptitude tests. Sweets are to be shared equally, but because some children are absent, the remaining children each receive more than planned. The main idea is to recognise that the total number of sweets remains the same, and the change in the number of children changes the per child share. Given Data / Assumptions: Total number of children planned to receive sweets = 112. Number of children absent on the day = 32. Number of children actually present = 112 - 32 = 80. Each present child receives 6 sweets more than originally planned. Total number of sweets remains fixed and is fully distributed among the present children. We must find the original number of sweets per child. Concept / Approach: Let the original number of sweets intended for each child be s. Then the total number of sweets is 112s. When only 80 children are present, these same 112s sweets are distributed equally, and each of the 80 children now receives s + 6 sweets. Therefore, the new distribution gives 80(s + 6) sweets in total. Since the total number of sweets has not changed, we can set 112s equal to 80(s + 6) and solve for s. Step-by-Step Solution: Step 1: Let s be the original number of sweets for each child. Step 2: Total sweets purchased or prepared = 112s. Step 3: Number of children present = 112 - 32 = 80. Step 4: Each present child now receives s + 6 sweets. Step 5: Total sweets distributed among present children = 80(s + 6). Step 6: Equate the totals: 112s = 80(s + 6). Step 7: Expand the right side: 112s = 80s + 480. Step 8: Subtract 80s from both sides: 32s = 480. Step 9: Divide both sides by 32: s = 480 / 32 = 15. Step 10: Therefore, each child was originally supposed to receive 15 sweets. Verification / Alternative check: Check with the computed value s = 15. Total sweets = 112 * 15 = 1680. If all 112 children had been present, each would have got 15 sweets. With only 80 present, each gets 15 + 6 = 21 sweets, for a total of 80 * 21 = 1680 sweets, which matches the original total. This confirms that the calculation is correct. Why Other Options Are Wrong: If each child originally received 14 sweets, total sweets would be 112 * 14 = 1568, and then each of the 80 present children would get 19.6 sweets, which is not a whole number. Other values like 16, 12 or 18 similarly fail to give both an integer increase of 6 sweets per child and equal distribution using whole numbers of sweets. Only 15 satisfies all the constraints nicely. Common Pitfalls: One common mistake is to try to guess the number of sweets per child without setting up the equation. Another is to confuse the number of absent children with the number of present children when writing the total distribution. Some students also mistakenly think that 6 extra sweets per child translates directly to 6 fewer children, which is not always correct. Systematically writing the equation based on fixed total sweets avoids such errors. Final Answer: Each child was originally supposed to get 15 sweets.

Question 15

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

Accepted Answer

Introduction / Context: This question involves the concept of average salary and how it changes when an additional salary is added to the group. It is a common type of aptitude problem that tests your understanding of how total sums and averages are related and how to work backwards to find an individual value from a change in the average. Given Data / Assumptions: Number of employees initially (excluding the manager) = 24. Average salary of these 24 employees = Rs. 1500 per month. When the manager is included, there are 25 people in total. The new average salary including the manager increases by Rs. 400, so the new average is Rs. 1900 per month. We must find the manager's monthly salary. Concept / Approach: Average salary is total salary divided by the number of people. We can first compute the total salary of the 24 employees using their average. Then we compute the new total salary when the manager is included using the new average for 25 people. The difference between these two totals gives the manager's salary. This is a simple but powerful approach that can be used for many average related problems. Step-by-Step Solution: Step 1: Compute the total salary of the 24 employees. Total salary (24 employees) = 24 * 1500 = 36,000. Step 2: When the manager is included, total number of people = 25. Step 3: New average salary = 1500 + 400 = 1900. Step 4: Total salary (24 employees plus manager) = 25 * 1900 = 47,500. Step 5: Manager's salary = Total salary (25) - Total salary (24) = 47,500 - 36,000. Step 6: Compute the difference: 47,500 - 36,000 = 11,500. Step 7: Therefore, the manager's monthly salary is Rs. 11,500. Verification / Alternative check: Verify by recomputing the average. Add the manager's salary to the total for 24 employees: 36,000 + 11,500 = 47,500. Divide this by 25 people to find the new average: 47,500 / 25 = 1900. Since this matches the stated new average of Rs. 1900, our calculation for the manager's salary is confirmed as correct. Why Other Options Are Wrong: If the manager's salary were Rs. 8840, the new total would be 44,840, giving an average of 44,840 / 25, which is less than 1900. Salaries such as Rs. 12,800 or Rs. 9450 or Rs. 15,000 also fail to produce the required new average of Rs. 1900 when added to the original total. Only Rs. 11,500 leads to the correct increase in average by exactly Rs. 400. Common Pitfalls: Sometimes students mistakenly divide the increase in total salary by the wrong number of people or miscalculate the new average. Others might incorrectly think the manager's salary is simply 1900 plus some difference instead of using total sums. Always remember that changes in average reflect changes in total, and the difference between new and old totals gives the contribution of the added person. Final Answer: The manager's monthly salary is Rs. 11,500.

Question 16

Sriram conducts an experiment and records 11 observations whose average is 92. The average of the first 5 observations is 89, and the average of the last 5 observations is 86. What is the value of the 6th observation?

Accepted Answer

Introduction / Context: This problem is about using averages of overlapping groups of observations to find a specific missing value. It shows how average information about different parts of a data set can be combined to reconstruct individual observations. This is a standard type of aptitude question that tests your fluency with sums and averages. Given Data / Assumptions: Total number of observations = 11. Average of all 11 observations = 92. Average of the first 5 observations = 89. Average of the last 5 observations = 86. We must find the 6th observation, which is not included entirely in either the first or last 5, but is common when we consider the whole data set. Concept / Approach: Average is total divided by count. From the overall average, we can compute the total sum of all 11 observations. From the averages of the first 5 and last 5 observations, we can compute the sums of these subsets. The key idea is that the sum of all 11 observations equals the sum of the first 5 plus the sum of the last 5 plus the 6th observation counted once, but actually the 6th observation is not included in either of the first 5 or last 5 sums. So we simply subtract the two partial sums from the overall sum to isolate the 6th observation. Step-by-Step Solution: Step 1: Compute the total of all 11 observations using the overall average. Total (all 11) = 11 * 92 = 1012. Step 2: Compute the total of the first 5 observations. Total (first 5) = 5 * 89 = 445. Step 3: Compute the total of the last 5 observations. Total (last 5) = 5 * 86 = 430. Step 4: Let the 6th observation be x. Step 5: Note that the sum of all 11 observations is the sum of the first 5, plus the 6th, plus the last 5. Step 6: Therefore, 1012 = 445 + x + 430. Step 7: Add the two known sums: 445 + 430 = 875. Step 8: So 1012 = 875 + x, so x = 1012 - 875 = 137. Step 9: Therefore, the 6th observation is 137. Verification / Alternative check: We can verify by constructing a hypothetical data set consistent with these sums, but the algebra itself is already a direct check. The combined sum of first 5 and last 5 observations is 875. Subtracting this from the total 1012 leaves 137, which must be the 6th observation. There is no other place for this missing amount to come from, so the logic is sound. Why Other Options Are Wrong: Values like 134, 139, 141 or 131 would give a different total sum for the 11 observations and would not maintain the given averages for the first 5 and last 5. For example, if the 6th observation were 134, the total would be 875 + 134 = 1009, leading to an overall average less than 92. Only 137 ensures the total stays 1012 and all averages are satisfied. Common Pitfalls: A frequent error is to think that the 6th observation is somehow the average of 89 and 86 or to attempt to average the averages. Another mistake is double counting or omitting some observations when adding partial sums. Clearly identifying the total sum and subtracting the known parts is the most reliable method. Final Answer: The 6th observation has a value of 137.

Question 17

The average age of a husband, wife and their child 3 years ago was 27 years. The average age of the wife and the child 5 years ago was 20 years. What is the present age of the husband?

Accepted Answer

Introduction / Context: This question tests your ability to handle age problems using averages and time shifts. Information is given about average ages some years in the past, and you must use it to deduce present ages. The key technique is to translate statements about averages into equations involving sums of ages, then adjust those sums for the passing of time. Given Data / Assumptions: Three people: husband (H), wife (W) and child (C). Three years ago, the average age of H, W and C was 27 years. Five years ago, the average age of W and C was 20 years. We must find the present age of the husband H. All ages are in years and increase linearly with time. Concept / Approach: Average age is sum of ages divided by number of people. If we know an average at a certain time in the past, we can convert it into a sum. Then we adjust that sum to the present by adding the appropriate number of years for each person. We are given two pieces of information: one about three years ago for the whole family, and one about five years ago for just the wife and child. Converting each to current sums allows us to solve for the husband's current age. Step-by-Step Solution: Step 1: Let H, W and C be the present ages of husband, wife and child respectively. Step 2: Three years ago, their ages were H - 3, W - 3 and C - 3. Step 3: The average age three years ago was 27, so the sum then was 3 * 27 = 81. Step 4: Therefore, (H - 3) + (W - 3) + (C - 3) = 81. Step 5: Simplify: H + W + C - 9 = 81 ⇒ H + W + C = 90. Step 6: Five years ago, the ages of wife and child were W - 5 and C - 5. Step 7: Their average then was 20, so the sum was 2 * 20 = 40. Step 8: Therefore, (W - 5) + (C - 5) = 40 ⇒ W + C - 10 = 40 ⇒ W + C = 50. Step 9: Substitute W + C = 50 into H + W + C = 90 to find H. Step 10: H + 50 = 90 ⇒ H = 90 - 50 = 40. Step 11: Therefore, the present age of the husband is 40 years. Verification / Alternative check: Check the conditions using these values. If H = 40 now, W + C = 50 now. Three years ago, their ages were 37, W - 3 and C - 3. The sum was 40 + 50 - 9 = 81, and the average for 3 people is 81 / 3 = 27, satisfying the first condition. Five years ago, wife and child had ages W - 5 and C - 5, with sum 50 - 10 = 40, giving an average of 40 / 2 = 20, satisfying the second condition. So all constraints hold. Why Other Options Are Wrong: If H were 30, then H + W + C would be 30 + 50 = 80, contradicting the earlier result that H + W + C = 90. Values 50 or 60 similarly give wrong total sums or fail to satisfy one of the time shifted average conditions. The value 35 also does not satisfy both equations at once. Only 40 years for the husband fits both given averages properly. Common Pitfalls: Many students attempt to guess ages or treat the averages as current rather than time shifted, which leads to inconsistent equations. Others forget to add the appropriate number of years when converting from past sums to present sums. Always write an equation for each time point and carefully adjust all ages by the same time interval when shifting between past and present. Final Answer: The present age of the husband is 40 years.

Question 18

The average value of the properties of Mittal, Ambani and Singhania is Rs. 11,111 crore. The only additional information given is that the property of Singhania is as much less than the property of Mittal as the property of Mittal is greater than th…

Accepted Answer

Introduction / Context: This question appears to be about averages and relative differences in wealth among three people: Mittal, Ambani and Singhania. While the problem provides the overall average property value and a statement about how Mittal and Singhania's properties relate to their own average, the critical issue is whether this extra statement actually adds any usable constraint. The task is to decide if the property of Ambani can be uniquely determined. Given Data / Assumptions: Let M, A and S be the property values (in crore rupees) of Mittal, Ambani and Singhania respectively. The average property value of the three is Rs. 11,111 crore. So (M + A + S) / 3 = 11,111, implying M + A + S = 33,333. The statement says: the property of Singhania is as much less than the property of Mittal as the property of Mittal is greater than the average property of Singhania and Mittal. We must decide whether A can be uniquely determined. Concept / Approach: We need to translate the given descriptive relationship between M and S into algebra. The average property of Singhania and Mittal is (M + S) / 2. The phrase can be read as: the difference M - S equals the difference between M and the average of M and S. That is: M - S = M - (M + S) / 2. We must check whether this gives a new constraint or is an identity that holds for all values of M and S, in which case it does not help determine A. Step-by-Step Solution: Step 1: Write the average of Mittal and Singhania as (M + S) / 2. Step 2: The property of Singhania is as much less than M as M is greater than this average. So write: M - S = M - (M + S) / 2. Step 3: Simplify the right side: M - (M + S) / 2 = (2M - (M + S)) / 2 = (M - S) / 2. Step 4: So the equation becomes M - S = (M - S) / 2. Step 5: Multiply both sides by 2: 2(M - S) = M - S ⇒ M - S = 0. Step 6: Therefore, M = S. Step 7: We now know only that M and S are equal, but we do not know their individual values. Step 8: From M + A + S = 33,333 and M = S, we get 2M + A = 33,333. Step 9: This is a single equation with two unknowns M and A, so there are infinitely many possible pairs (M, A) that satisfy it. Verification / Alternative check: For example, if M = S = 10,000, then A = 33,333 - 20,000 = 13,333. If M = S = 12,000, then A = 33,333 - 24,000 = 9,333. Both choices satisfy the known conditions. Because multiple different values for A are possible while still satisfying all given statements, the value of Ambani's property cannot be uniquely determined from the information provided. Why Other Options Are Wrong: Options such as Rs. 111 crore, Rs. 1111 crore or Rs. 11,111 crore are specific numerical values. To choose any of these, the data would need to pin down a single unique solution for A. However, the conditions only fix the sum M + A + S and force M = S, leaving many possibilities for M and A. The option Rs. 22,222 crore is also just a single arbitrary value without support. Therefore, no specific numeric choice is justified. Common Pitfalls: A major pitfall is interpreting the worded condition as adding meaningful information when, after simplification, it either becomes an identity or forces only a weak condition like M = S. Some students may assume symmetry and guess that Ambani's property equals the average, but this is not required by the equations. Others may over interpret the wording without checking the algebra. Always convert word conditions to equations and see how many independent equations you actually have. Final Answer: The information is not sufficient to determine Ambani's property uniquely, so the correct choice is cannot be determined.

Question 19

Eight years ago, there were 5 members in Arthur's family and the average age of the family at that time was 36 years. Since then Arthur got married and a child was born. Even now, the average age of the family (now 7 members) is still 36 years. If t…

Accepted Answer

Introduction / Context: This question is a multi step age problem involving changes in family size and average age over time. Eight years ago, Arthur's family had 5 members. Since then, his wife and child have joined, and yet the average age has remained the same. We are given the present age difference between the wife and child and asked to find the wife's age at the time of the child's birth. It requires careful bookkeeping of total ages and age gaps. Given Data / Assumptions: 8 years ago: 5 family members, average age = 36 years. Present: 7 family members (the original 5 plus wife plus child). Even now, the average age is 36 years. The present difference between the wife's age and the child's age is 26 years. We must find the wife's age at the time of the child's birth. All ages increase linearly with time and there are no deaths or other new members besides the wife and child in this period. Concept / Approach: We start by computing total ages at different times using the average information. The sum of ages of the original 5 members 8 years ago is 5 * 36. Eight years later, these 5 have each aged by 8 years, so we can find their current total. We also compute the current total age of all 7 members from the current average. The difference between these totals gives the combined present ages of the wife and child. Using the stated age difference between wife and child, we solve for their individual present ages, and finally subtract the child's current age from the wife's current age to get her age at the child's birth. Step-by-Step Solution: Step 1: Eight years ago, sum of ages of 5 members = 5 * 36 = 180 years. Step 2: In the present, each of those 5 members is 8 years older, so total increase in their ages = 5 * 8 = 40 years. Step 3: Present total age of the original 5 members = 180 + 40 = 220 years. Step 4: Currently, there are 7 members and the average age is still 36 years. Step 5: Present total age of all 7 members = 7 * 36 = 252 years. Step 6: Combined present age of wife (W) and child (C) = 252 - 220 = 32 years. Step 7: Given that the present difference between wife's and child's ages is 26 years, we have W - C = 26. Step 8: Also W + C = 32 from the total just found. Step 9: Solve these two equations: adding them gives 2W = 32 + 26 = 58, so W = 29. Step 10: Then C = 32 - 29 = 3 years (present age of the child). Step 11: The child is 3 years old now, so the child was born 3 years ago. Step 12: Wife's present age is 29, so at the time of the child's birth her age was 29 - 3 = 26 years. Verification / Alternative check: We can verify the average information. Eight years ago, the sum of ages of the original 5 was 180. Now, those 5 contribute 220 years. Adding the present wife and child ages, 29 and 3, the total is 220 + 29 + 3 = 252. Dividing 252 by 7 gives an average of 36, matching the present average stated. The present age difference between wife and child is 29 - 3 = 26, which aligns with the given difference. This confirms all conditions are satisfied by these ages. Why Other Options Are Wrong: If the wife had been 25 or 20 at the time of the child's birth, her present age would be lower, which would change the combined ages and break the given average conditions. The option 'cannot be determined' is incorrect because the data lead to a unique solution for the wife and child's ages. The value 29 years is the wife's present age, not her age at the child's birth; the question specifically asks for her age at birth, which is 26. Common Pitfalls: A common mistake is to misinterpret the age difference as referring to some earlier time instead of the present. Others forget to account for the fact that the original 5 members have aged 8 years each, leading to an incorrect present total. Some students mix up the number of family members when computing totals. Keeping track of time shifts and totals systematically is essential in multi step age problems like this one. Final Answer: Arthur's wife was 26 years old at the time of the child's birth.

Question 20

The average of the five numbers 26, 29, n, 35 and 43 lies between 25 and 35. If n is an integer and must be greater than the average of these five numbers, then what is the smallest possible value of n?

Accepted Answer

Introduction / Context: This question combines inequalities with the concept of average and an additional condition on one of the numbers. We are given four fixed numbers and a variable integer n, and told that the average of the five numbers lies between 25 and 35. We are also told that n itself must be greater than this average. We must find the smallest integer n that can satisfy both conditions simultaneously. Given Data / Assumptions: The five numbers are 26, 29, n, 35 and 43. Their average lies strictly between 25 and 35. n is an integer. n must be greater than the average of the five numbers. We are asked for the smallest such integer value of n. Concept / Approach: The average of the five numbers is (26 + 29 + n + 35 + 43) / 5. First, we enforce the condition that this average lies between 25 and 35, which gives an inequality for n. Next, we apply the condition that n is greater than this average, producing another inequality. Combining these inequalities and then restricting to integer values gives a range of possible n. Finally, we choose the smallest integer in that range. Step-by-Step Solution: Step 1: Compute the sum of the four fixed numbers: 26 + 29 + 35 + 43 = 133. Step 2: The average of the five numbers is (133 + n) / 5. Step 3: The average lies between 25 and 35: 25 < (133 + n) / 5 < 35. Step 4: Multiply through by 5: 125 < 133 + n < 175. Step 5: Subtract 133: -8 < n < 42. So from this condition alone, n can be any number less than 42 and greater than -8. Step 6: Now use the condition that n is greater than the average: n > (133 + n) / 5. Step 7: Multiply both sides by 5: 5n > 133 + n. Step 8: Subtract n from both sides: 4n > 133. Step 9: Divide by 4: n > 133 / 4 = 33.25. Step 10: Combining both sets of inequalities gives 33.25 < n < 42. Step 11: Since n must be an integer, possible values are 34, 35, 36, 37, 38, 39, 40 and 41. Step 12: We are asked for the smallest possible integer n, so n = 34. Verification / Alternative check: Check n = 34. The average is (133 + 34) / 5 = 167 / 5 = 33.4. This is between 25 and 35, and n = 34 is indeed greater than 33.4. If we try n = 33, the average is (133 + 33) / 5 = 166 / 5 = 33.2; although this lies within 25 and 35, n = 33 would not be greater than the average, because 33 is slightly less than 33.2. Thus, n = 34 is the smallest integer that satisfies both conditions. Why Other Options Are Wrong: Option 33 fails the condition n > average, as shown above. The value 35 does satisfy the conditions but is not the smallest such integer. The option 'none of these' is incorrect because 34 works perfectly. Any larger value like 36 also works but is not minimal. Therefore, 34 is the only correct choice given the wording of the problem and the desire for the smallest possible n. Common Pitfalls: Some students only apply the first inequality (average between 25 and 35) and choose any value in that range, without enforcing the second condition that n must be greater than the average. Others may misinterpret 'greater than the average' as greater than some rough mental estimate and not check carefully. It is also easy to forget that n must be an integer, which is crucial when dealing with strict inequalities and averages. Final Answer: The smallest integer value of n that satisfies all conditions is 34.

Average Questions