Question 1

The average age of 36 students in a group is 14 years. When the teacher's age is included, the average age of the group (students plus teacher) increases by 1 year. What is the age of the teacher in years?

Accepted Answer

Introduction / Context: This is a typical problem where adding one more person to a group changes the average, and you must work backwards to find that person's age. It tests your skill in converting averages to total sums and then using the new average to deduce the missing individual value from the changed total. Given Data / Assumptions: • Number of students = 36. • Average age of these students = 14 years. • When the teacher is added, total persons = 37. • New average age of the group = 15 years. • We need to find the teacher's age. Concept / Approach: Average age is the total of all ages divided by the number of people. First we compute the total age of the 36 students using their average. Then we compute the total age of the 37 people using the new average. The difference between these two totals must be the teacher's age, because the teacher is the only additional person contributing to the increased total. Step-by-Step Solution: Total age of 36 students = 36 * 14 = 504 years. After adding the teacher, number of people = 37. New average age = 15 years. Total age of 37 people = 37 * 15 = 555 years. Teacher's age = total age of 37 people − total age of 36 students. Teacher's age = 555 − 504 = 51 years. Verification / Alternative check: We can verify by recalculating the average. With teacher age 51 added, new total = 504 + 51 = 555. Dividing by 37 gives 555 / 37 = 15 years, which matches the given new average. The original total and average for 36 students remain unchanged, so all conditions are satisfied, confirming that the teacher is 51 years old. Why Other Options Are Wrong: If the teacher were 35, 45 or 54 years old, the total ages and resulting new average would change. For example, a teacher age of 45 would produce a total of 504 + 45 = 549, giving an average of 549 / 37, which is not 15. Only an age of 51 gives exactly the new average described in the problem. Common Pitfalls: Some students miscalculate the total ages due to multiplication errors, or they forget that the new average applies to 37 people, not 36. Others try to reason about the difference in averages without translating into totals, which can be confusing. Sticking to the method of total = average * number of people is the most reliable approach. Final Answer: The teacher's age is 51 years.

Question 2

P, Q, R, S, T and U are six consecutive odd numbers. Their average is 52. What is the product P × U, where P is the smallest of the six numbers and U is the largest?

Accepted Answer

Introduction / Context: This problem deals with consecutive odd numbers and their average. When numbers are evenly spaced in an arithmetic progression, the average has a special relationship with the middle terms. Here you are asked to use that relationship to determine the exact values of the smallest and largest numbers and then compute their product. Given Data / Assumptions: • P, Q, R, S, T and U are six consecutive odd integers. • Their average is 52. • The numbers are in increasing order from P to U. • We must find the product P × U. Concept / Approach: Six consecutive odd numbers form an arithmetic progression with common difference 2. A convenient way is to express them in terms of a central value. For six consecutive odd numbers, we can write them as n − 5, n − 3, n − 1, n + 1, n + 3 and n + 5, where n is the average. Since the average is given as 52, that helps locate all six numbers. Once we know P and U explicitly, we simply multiply them to get their product. Step-by-Step Solution: Let the six consecutive odd numbers be n − 5, n − 3, n − 1, n + 1, n + 3, n + 5. The average of an arithmetic progression equals its central value, so average = n. Given average = 52, therefore n = 52. So the numbers are 52 − 5 = 47, 52 − 3 = 49, 52 − 1 = 51, 52 + 1 = 53, 52 + 3 = 55 and 52 + 5 = 57. Hence P = 47 (smallest) and U = 57 (largest). Product P × U = 47 * 57. Compute: 47 * 57 = 47 * (50 + 7) = 47 * 50 + 47 * 7 = 2350 + 329 = 2679. Verification / Alternative check: We can verify the average. Sum of the numbers = 47 + 49 + 51 + 53 + 55 + 57. Pairing symmetric numbers: (47 + 57) + (49 + 55) + (51 + 53) = 104 + 104 + 104 = 312. Average = 312 / 6 = 52, which matches the given average. This confirms that the set of numbers is correct and therefore the product 2679 is reliable. Why Other Options Are Wrong: Other products such as 2212, 2154 or 2349 would correspond to incorrect pairs of numbers or miscalculations. For example, even a small change in one factor, such as using 45 instead of 47, would drastically alter the product. Since the average fixes the sequence precisely, there is only one possible pair P and U, and only 2679 matches their product. Common Pitfalls: One common issue is to assume that 52 itself must be one of the odd numbers, but 52 is even, so it acts only as the central average, not as a term. Students also sometimes mis assign the offsets and might use n − 4 or n − 2, which would give even numbers. Always ensure the step size between consecutive odd numbers is 2 and that all resulting values remain odd integers. Final Answer: The product of P and U is 2679.

Question 3

The average age of a class is 19 years. The average age of the girls is 18 years and the average age of the boys is 21 years. If the number of girls in the class is 16, how many boys are there in the class?

Accepted Answer

Introduction / Context: This question involves combining two groups (boys and girls) with different averages to produce an overall average for the entire class. You must determine how many boys are in the class using the given averages and the known number of girls. This is an application of weighted averages and total sums. Given Data / Assumptions: • Average age of the whole class = 19 years. • Average age of girls = 18 years. • Average age of boys = 21 years. • Number of girls = 16. • Let the number of boys be B. Concept / Approach: We express total ages in terms of the averages and numbers. Total age of girls = 18 * 16. Total age of boys = 21 * B. Total number of students = 16 + B. The overall average age is given as 19, so (total age of girls + total age of boys) divided by total number of students must equal 19. Setting up this equation allows us to solve for B, the number of boys in the class. Step-by-Step Solution: Total age of girls = 18 * 16 = 288 years. Total age of boys = 21 * B years. Total students = 16 + B. Overall average age = (288 + 21B) / (16 + B) = 19. Multiply both sides: 288 + 21B = 19 * (16 + B). Right side = 19 * 16 + 19B = 304 + 19B. So, 288 + 21B = 304 + 19B. Rearrange: 21B − 19B = 304 − 288. 2B = 16, hence B = 8. Verification / Alternative check: With 8 boys, total students = 16 + 8 = 24. Total age of girls = 288 years, total age of boys = 21 * 8 = 168 years. Combined total age = 288 + 168 = 456 years. Overall average age = 456 / 24 = 19 years, which matches the given average. This confirms that the number of boys is correctly found as 8. Why Other Options Are Wrong: If there were 12, 10 or 6 boys, the total ages and resulting average would change. For example, with 10 boys, total boys' age = 210 and overall total = 288 + 210 = 498, giving an average of 498 / 26, which is not 19. Only 8 boys yield a combined average age of exactly 19 years as required. Common Pitfalls: Some students attempt to use a shortcut ratio method but misapply it due to the presence of three averages. Others miscalculate the product terms when forming the equation. Writing down the total sums clearly and solving the linear equation step by step helps avoid such errors. Always verify by recomputing the overall average with the found value of B. Final Answer: There are 8 boys in the class.

Question 4

In a class of 78 students, 41 students are taking French and 22 students are taking German. Of the students taking French or German, 9 students are taking both courses. How many students are not enrolled in either French or German?

Accepted Answer

Introduction / Context: This problem uses basic set theory and the principle of inclusion and exclusion. Students can take French, German, both or neither. You are asked to find how many students are not enrolled in either course, given the numbers taking each subject and the overlap between them. Given Data / Assumptions: • Total number of students = 78. • Number of students taking French = 41. • Number of students taking German = 22. • Number of students taking both French and German = 9. • We assume each student is correctly counted in these figures and can belong to zero, one or both course groups. Concept / Approach: To find how many students take at least one of the two subjects, we use inclusion and exclusion: |French ∪ German| = |French| + |German| − |both|. Once we know how many students are in either or both courses, we subtract that count from the total number of students to find those who take neither course. This approach avoids double counting students who are in both groups. Step-by-Step Solution: Number taking French = 41. Number taking German = 22. Number taking both = 9. Number taking French or German (union) = 41 + 22 − 9. Compute: 41 + 22 = 63, 63 − 9 = 54. So, 54 students are enrolled in at least one of the two courses. Total students = 78. Number of students not enrolled in either course = 78 − 54 = 24. Verification / Alternative check: As a check, imagine dividing the class into four groups: French only, German only, both and neither. We know that French only + both = 41, German only + both = 22 and both = 9. Therefore French only = 41 − 9 = 32, German only = 22 − 9 = 13. Now total in at least one course = French only + German only + both = 32 + 13 + 9 = 54. Subtracting from 78 again gives 78 − 54 = 24 students taking neither course. The consistency of this count confirms the answer. Why Other Options Are Wrong: Answers such as 6, 12 or 18 would imply different union sizes that do not align with the given counts. For example, if 12 students took neither, then 78 − 12 = 66 students would be taking French or German, which contradicts the union calculation of 54. Therefore these values are inconsistent with the inclusion and exclusion formula applied to the given data. Common Pitfalls: A frequent mistake is to add 41 and 22 without subtracting the 9 students counted twice, which would double count the students taking both courses. Another error is subtracting the overlap twice. To avoid such errors, always remember the union formula: size of union = sum of individual sizes minus the intersection. Carefully substituting values will yield the correct result. Final Answer: The number of students not enrolled in either French or German is 24.

Question 5

The average of 9 test results is 50. The average of the first four results is 52 and the average of the last four results is 49. What is the value of the fifth (middle) result?

Accepted Answer

Introduction / Context: This question involves averages over overlapping subsets of a sequence of nine results. With the overall average and the averages of the first four and last four results given, we must determine the middle result. It uses the basic principle that total sum is the product of average and number of items, and that the total is the sum of parts. Given Data / Assumptions: • There are 9 results in sequence. • Average of all 9 results = 50. • Average of the first 4 results = 52. • Average of the last 4 results = 49. • The fifth result is the one that is not included in either group of 4. Concept / Approach: First compute the total sum of all 9 results using the overall average. Then compute the sum of the first four results and separately the sum of the last four results using their averages. The fifth result plus the sums of the first four and last four must equal the total sum of all nine results. Therefore the fifth result is the difference between the grand total and the combined sums of the first and last four results. Step-by-Step Solution: Total of 9 results = 9 * 50 = 450. Total of first 4 results = 4 * 52 = 208. Total of last 4 results = 4 * 49 = 196. Let the fifth result be x. Then total of all 9 results = sum of first 4 + x + sum of last 4. So, 208 + x + 196 = 450. Combine known sums: 208 + 196 = 404. Therefore, 404 + x = 450, so x = 450 − 404 = 46. Verification / Alternative check: We can verify by reconstructing the averages. With the fifth result as 46, the total is 208 (first four) + 46 + 196 (last four) = 450. Average over 9 results = 450 / 9 = 50, which matches the original overall average. The first four and last four sums still give averages of 52 and 49 respectively, so all given conditions remain satisfied and the answer is consistent. Why Other Options Are Wrong: If the fifth result were 48, the total would be 208 + 48 + 196 = 452 and the overall average would be 452 / 9, not 50. Values 52 or 44 would similarly lead to totals of 456 or 448, which also give incorrect overall averages. Only 46 gives a total of 450 and preserves all the given averages. Common Pitfalls: The most common error is to forget that the fifth result is not counted in either of the four result averages and to incorrectly include it in those totals. Another mistake is miscomputing the partial sums 208 and 196. Carefully writing out the equation total = first four + fifth + last four, and checking each multiplication, helps to avoid such errors. Final Answer: The fifth result is 46.

Question 6

David obtains 76, 65, 82, 67 and 85 marks out of 100 in English, Mathematics, Physics, Chemistry and Biology respectively. What are his average marks out of 100 across these five subjects?

Accepted Answer

Introduction / Context:This question checks the basic concept of arithmetic mean, which is widely used to summarise marks in examinations. Understanding how to compute the average of a set of values is a foundational aptitude skill for many competitive exams. Given Data / Assumptions:- Marks in five subjects are 76, 65, 82, 67 and 85.- Each subject carries 100 marks.- All subjects are weighted equally in the average.- We assume no subject is counted more than once. Concept / Approach:The average (arithmetic mean) of a set of observations is defined as:Average = (Sum of all observations) / (Number of observations)Here, the observations are the marks scored by David in the five subjects. Step-by-Step Solution:Step 1: Write down the marks: 76, 65, 82, 67, 85.Step 2: Find the sum: 76 + 65 + 82 + 67 + 85.Step 3: Compute the sum: 76 + 65 = 141, 141 + 82 = 223, 223 + 67 = 290, 290 + 85 = 375.Step 4: Number of subjects = 5.Step 5: Average marks = 375 / 5 = 75. Verification / Alternative check:You can approximate to verify: the marks are mostly around the low to mid seventies. Taking a rough average around that region, 75 is reasonable. A quick mental check of 5 multiplied by 75 gives 375, which matches the sum of the marks, confirming the calculation. Why Other Options Are Wrong:- 68: This would correspond to a much lower total of 68 * 5 = 340, which is less than the actual total of 375.- 65: This is even lower, giving a total of 325, clearly not equal to the given sum.- 60: This would give a total of only 300, far from 375, so it cannot be correct. Common Pitfalls:- Forgetting to include one of the subjects in the total sum can lead to an incorrect average.- Dividing by the wrong number of subjects, for example dividing by 4 instead of 5, is another typical mistake.- Some students mistakenly use the highest or lowest mark instead of computing the mean. Final Answer:The correct average marks scored by David across the five subjects is 75.

Question 7

The average price of three items of furniture is Rs. 15000. If the prices of the items are in the ratio 3 : 5 : 7, what is the price of the cheapest item of furniture?

Accepted Answer

Introduction / Context:This problem tests the use of ratios along with the concept of average value. Such questions are common in aptitude tests to see whether a candidate can connect average, total value and ratio distribution. Given Data / Assumptions:- Average price of three furniture items = Rs. 15000.- Prices are in the ratio 3 : 5 : 7.- We assume prices are positive and there are exactly three items. Concept / Approach:If three items have prices in the ratio 3 : 5 : 7, we can represent them as 3x, 5x and 7x. The average price is then:Average price = (3x + 5x + 7x) / 3We equate this to Rs. 15000 to find the value of x, and then compute the individual prices. The cheapest item corresponds to the smallest term in the ratio. Step-by-Step Solution:Step 1: Let the three prices be 3x, 5x and 7x.Step 2: Total price = 3x + 5x + 7x = 15x.Step 3: Average price = (Total price) / 3 = 15x / 3 = 5x.Step 4: Given average price = Rs. 15000, so 5x = 15000.Step 5: Solve for x: x = 15000 / 5 = 3000.Step 6: Cheapest item price = 3x = 3 * 3000 = Rs. 9000. Verification / Alternative check:You can also find the total directly from the average: Total = 3 * 15000 = Rs. 45000. Splitting Rs. 45000 in the ratio 3 : 5 : 7 means the unit value is 45000 / 15 = 3000. Thus, the three prices are 9000, 15000 and 21000, and the smallest is 9000, confirming the result. Why Other Options Are Wrong:- 6000: This would correspond to 2x with x = 3000 and does not match the given ratio structure.- 7000: This is not a multiple of the base unit 3000, so it does not satisfy the ratio 3 : 5 : 7.- 8888: This is an arbitrary number and does not align with the derived unit value or the ratio. Common Pitfalls:- Some learners mistakenly divide the average by the ratio parts without computing the total first.- Others assign x directly as 15000, ignoring that the ratio is spread across three items and that the average equals only 5x, not x. Final Answer:The price of the cheapest furniture item is Rs. 9000.

Question 8

In a mens hostel there are initially 95 students. When 30 more students join, the average expenditure per student decreases by Rs. 5, yet the total expenditure on the hostel increases by Rs. 500. What is the total expenditure of the hostel now?

Accepted Answer

Introduction / Context:This problem combines the concept of average expenditure per person with changes in group size. It illustrates how total expenditure and per person expenditure are related when the number of people changes in a hostel or similar shared cost situation. Given Data / Assumptions:- Initial number of students = 95.- Additional students = 30, so final number = 125.- Average expenditure per student decreases by Rs. 5 after the increase in students.- Total expenditure increases by Rs. 500 overall.- We assume expenditure is calculated for the same time period before and after. Concept / Approach:Let the initial average expenditure per student be x rupees. Then the initial total expenditure is:Initial total = 95 * xAfter 30 new students join, the new average becomes (x - 5) and the new total expenditure is:New total = 125 * (x - 5)The new total is also given to be Rs. 500 more than the old total, which gives an equation to solve for x and then for the new total. Step-by-Step Solution:Step 1: Let initial average expenditure per student be x.Step 2: Initial total expenditure = 95x.Step 3: New number of students = 95 + 30 = 125.Step 4: New average expenditure per student = x - 5.Step 5: New total expenditure = 125 * (x - 5).Step 6: Given that the new total is Rs. 500 more: 125 * (x - 5) = 95x + 500.Step 7: Expand and simplify: 125x - 625 = 95x + 500.Step 8: Rearrange: 125x - 95x = 500 + 625, so 30x = 1125.Step 9: Solve for x: x = 1125 / 30 = 37.5.Step 10: New total expenditure = 125 * (x - 5) = 125 * 32.5 = 4062.5 rupees. Verification / Alternative check:Initial total = 95 * 37.5 = 3562.5. Adding Rs. 500 gives 3562.5 + 500 = 4062.5, which matches the computed new total, so the result is consistent from both directions. Why Other Options Are Wrong:- Rs. 4420, Rs. 4680, Rs. 4960 and Rs. 5480 do not satisfy the equation that links the old and new totals and the change in average. If you plug these values back and compute the implied averages, they will not give a decrease of exactly Rs. 5 with an increase of 30 students. Common Pitfalls:- A common mistake is to assume the new total equals 125x instead of 125(x - 5).- Another error is to add Rs. 5 to the average instead of subtracting, misreading the direction of the change.- Some also forget to increase the number of students in the new scenario and use 95 in both totals. Final Answer:The total expenditure of the hostel after the new students join is Rs. 4062.50.

Question 9

Arun believes that his own weight is greater than 65 kg but less than 72 kg. His brother thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother says that his weight cannot be greater than 68 kg. If all three are correct, wha…

Accepted Answer

Introduction / Context:This reasoning question tests the understanding of inequalities and intersection of ranges. Once the possible integer values that satisfy all opinions are identified, the problem reduces to finding their arithmetic mean. Given Data / Assumptions:- Arun: weight is greater than 65 kg and less than 72 kg. - Brother: weight is greater than 60 kg and less than 70 kg. - Mother: weight cannot be greater than 68 kg, which we interpret as strictly less than 68 kg. - We consider only integer values of weight in kilograms. Concept / Approach:We must find the intersection of three ranges and then calculate the average of the integer values in that intersection. The steps are:1. Convert each statement into a numerical inequality range.2. Find the common range that satisfies all three.3. List the integer values in that range.4. Compute their arithmetic mean. Step-by-Step Solution:Step 1: From Arun, weight is in (65, 72), so possible integer values are 66, 67, 68, 69, 70, 71.Step 2: From the brother, weight is in (60, 70), so possible integer values are 61 to 69 inclusive.Step 3: From the mother, weight is less than 68, so possible integer values are 1 to 67, but we will combine this with the previous limits.Step 4: Intersecting all three conditions gives integer values 66 and 67 only.Step 5: Average of 66 and 67 = (66 + 67) / 2 = 133 / 2 = 66.5 kg. Verification / Alternative check:Check each candidate value quickly: 66 and 67 are both greater than 65 and less than 72, greater than 60 and less than 70, and less than 68. No other integer simultaneously satisfies all three statements. Hence the set of possible weights is {66, 67} and the mean of this set is 66.5 kg. Why Other Options Are Wrong:- 55.5 kg, 77.5 kg and 88.5 kg lie far outside the plausible weight range implied by the three statements and cannot be averages of valid values.- Any average involving values larger than or equal to 68 or below 66 would contradict at least one family member's estimate. Common Pitfalls:- Misinterpreting the mother's statement as less than or equal to 68, which would incorrectly include 68 as a possible weight.- Forgetting that the problem asks for integer values of weight and including fractional values in the list.- Taking the mid point of the combined overall range directly without first finding the intersection. Final Answer:The average of all different possible integer values of Arun's weight is 66.5 kg.

Question 10

The mean of 100 observations is 40. Later it is found that one observation was misread as 48 instead of the correct value 84. What is the correct mean of the 100 observations after this error is rectified?

Accepted Answer

Introduction / Context:This problem focuses on correcting a mean when one observation was recorded incorrectly. Questions of this type are common in statistics and data interpretation, where a misread or misprinted value must be adjusted without recomputing the entire data set from scratch. Given Data / Assumptions:- Number of observations = 100.- Reported mean = 40.- One value was recorded as 48 but should have been 84.- Only this single misread value needs correction. Concept / Approach:The mean is defined as:Mean = (Sum of all observations) / (Number of observations)From the given mean and count, we can reconstruct the total sum. Then we subtract the wrong value and add the correct value to adjust the total, and finally recompute the corrected mean. Step-by-Step Solution:Step 1: Let S be the sum of all 100 observations as originally recorded.Step 2: Using the reported mean, S = 40 * 100 = 4000.Step 3: The value 48 is wrong and should be 84. So remove 48 and add 84.Step 4: Corrected sum S_correct = 4000 - 48 + 84.Step 5: Compute S_correct = 4000 - 48 + 84 = 4000 + 36 = 4036.Step 6: Corrected mean = S_correct / 100 = 4036 / 100 = 40.36. Verification / Alternative check:The adjustment to the total is equal to the difference between the correct and wrong values: 84 - 48 = 36. Dividing this by 100 gives an increase in the mean of 0.36. The old mean was 40, so the new mean is 40 + 0.36 = 40.36, which confirms the result without recomputing the entire sum. Why Other Options Are Wrong:- 41.24, 41.92 and 42.05 all represent much larger corrections to the mean than 0.36. These would imply a much bigger difference between the wrong and correct values, which is not the case here. Common Pitfalls:- Forgetting to remove the wrong value from the total before adding the correct value, and instead just adding the difference incorrectly.- Dividing the correction 36 by the wrong number of observations, for example by 1 instead of 100.- Misinterpreting the question and recalculating with 101 observations or 99 observations. Final Answer:The corrected mean of the 100 observations is 40.36.

Question 11

Without any stoppage, a person travels a certain distance at an average speed of 42 km/h. Including regular stoppages, the same distance is covered at an average speed of 28 km/h. How many minutes per hour, on average, does the person stop?

Accepted Answer

Introduction / Context:This question examines the relationship between running speed, effective average speed and stoppage time. It is a standard speed and time problem used to test understanding of average speed over a journey with interruptions. Given Data / Assumptions:- Running speed without stoppage = 42 km/h.- Effective average speed including stoppages = 28 km/h.- The distance travelled in both cases is the same.- We are asked to find the stoppage time per hour of journey. Concept / Approach:In one hour of clock time, if the person never stops, he would travel 42 km. However, due to stoppages, in one hour of clock time he actually covers only 28 km. We can find how much time in each hour he is actually moving at 42 km/h, and the remaining time is stoppage time. Step-by-Step Solution:Step 1: Let in 1 hour of actual clock time, the person travels 28 km including stoppages.Step 2: While he is moving, his speed is 42 km/h. Let the moving time per hour be t hours.Step 3: Distance covered while moving in that hour is 42 * t kilometres.Step 4: This must equal the effective distance covered in one hour, which is 28 km. So 42 * t = 28.Step 5: Solve for t: t = 28 / 42 = 2 / 3 hour.Step 6: Total clock time per hour is 1 hour, so stoppage time per hour = 1 - 2 / 3 = 1 / 3 hour.Step 7: Convert 1 / 3 hour to minutes: (1 / 3) * 60 = 20 minutes. Verification / Alternative check:If he moves 40 minutes out of every 60 minutes at 42 km/h, the distance covered is 42 * (40 / 60) = 42 * (2 / 3) = 28 km. This matches the given effective average speed of 28 km/h, confirming that stopping for 20 minutes each hour is correct. Why Other Options Are Wrong:- 15 minutes and 18 minutes yield different effective average speeds when you compute 42 multiplied by the corresponding fraction of the hour, not 28 km/h.- 22 minutes likewise does not produce the exact ratio needed to reduce 42 km/h down to 28 km/h. Common Pitfalls:- Taking the ratio 28 : 42 and interpreting it incorrectly as time ratio instead of using the concept of distance = speed * time.- Confusing total journey time with stoppage time per hour.- Forgetting to convert the final time from hours into minutes. Final Answer:The person stops for 20 minutes per hour on average.

Question 12

The average age of a group of persons going for a picnic is 16 years. On the spot, 20 new persons with an average age of 15 years join the group, and the overall average age becomes 15.5 years. How many persons were there in the original picnic grou…

Accepted Answer

Introduction / Context:This question is a classic example of weighted averages and mixture of groups. It tests the ability to combine average values when two groups with different sizes and different averages are merged. Given Data / Assumptions:- Initial average age of the group = 16 years.- Number of initial members = n (unknown).- 20 new persons join later.- Average age of the 20 new persons = 15 years.- Final average of the combined group = 15.5 years.- All ages are in years. Concept / Approach:We use the relationship:Total age = Average age * Number of personsInitially and after new persons join. We set up an equation based on the total age of the original group plus the total age of the new members equalling the total age implied by the final average. Step-by-Step Solution:Step 1: Let the original number of persons be n.Step 2: Initial total age = 16 * n.Step 3: Total age of the 20 new persons = 15 * 20 = 300.Step 4: Final total number of persons = n + 20.Step 5: Final total age from the final average = 15.5 * (n + 20).Step 6: Set up equation: 16n + 300 = 15.5 * (n + 20).Step 7: Expand the right side: 16n + 300 = 15.5n + 310.Step 8: Rearrange: 16n - 15.5n = 310 - 300, so 0.5n = 10.Step 9: Solve for n: n = 10 / 0.5 = 20. Verification / Alternative check:Original total age = 16 * 20 = 320. New persons total age = 300. Combined total age = 320 + 300 = 620. Combined number of persons = 20 + 20 = 40. Combined average = 620 / 40 = 15.5 years, which matches the given final average, confirming that the original group size is correct. Why Other Options Are Wrong:- 15, 23 and 24 do not satisfy the equation when substituted for n and would give a combined average different from 15.5 years.- For example, if n were 15, the total age would be 240 + 300 = 540, with 35 persons, giving an average of about 15.43, not 15.5. Common Pitfalls:- Adding the averages directly instead of working with total ages.- Forgetting that the number of persons changes when 20 new persons join, so the divisor in the average must be n + 20, not just n.- Mixing up which average applies to which group. Final Answer:The number of persons initially going for the picnic is 20.

Question 13

A motorist travels from one place to another 150 km away at an average speed of 50 km/h and returns along the same route at an average speed of 30 km/h. What is his average speed for the entire round trip in km/h?

Accepted Answer

Introduction / Context:This question deals with average speed over a round trip where the speeds for the two legs of the journey are different. It highlights that the average speed for the entire trip is not the simple arithmetic mean when the distances are the same but the times differ. Given Data / Assumptions:- Distance one way = 150 km.- Speed going = 50 km/h.- Speed returning = 30 km/h.- The route and distance are identical in both directions.- We need the average speed for the total distance travelled. Concept / Approach:The general formula for average speed over multiple segments is:Average speed = (Total distance) / (Total time)Because the distances in the two directions are equal, we can either use this formula directly or apply the harmonic mean formula for speeds over equal distances. Step-by-Step Solution:Step 1: Distance each way = 150 km, so total distance = 150 + 150 = 300 km.Step 2: Time taken going = Distance / Speed = 150 / 50 = 3 hours.Step 3: Time taken returning = 150 / 30 = 5 hours.Step 4: Total time = 3 + 5 = 8 hours.Step 5: Average speed for round trip = Total distance / Total time = 300 / 8 km/h.Step 6: Compute 300 / 8 = 37.5 km/h. Verification / Alternative check:For equal distances with speeds u and v, the average speed can be written as:Average speed = (2 * u * v) / (u + v)Here, u = 50 and v = 30:Average speed = 2 * 50 * 30 / (50 + 30) = 3000 / 80 = 37.5 km/h.This matches the result obtained from distance and time directly. Why Other Options Are Wrong:- 35 and 36 km/h are lower than the true average and would require more than 8 hours for 300 km, which contradicts the computed times.- 38.2 km/h is higher than 37.5 km/h and does not correspond to the actual total time of 8 hours. Common Pitfalls:- Taking the simple average (50 + 30) / 2 = 40 km/h, which is incorrect because time spent at each speed differs.- Mixing up total distance with one way distance when applying the formula.- Rounding too early and miscalculating the final value. Final Answer:The average speed for the entire journey is 37.5 km/h.

Question 14

The average age of 15 students in a class is 15 years. The average age of 5 of these students is 14 years, and the average age of another 9 students is 16 years. What is the age of the remaining fifteenth student?

Accepted Answer

Introduction / Context:This question tests basic average and total calculation skills. When some subgroup averages are known, we can calculate their total contributions and isolate the missing individual value. Given Data / Assumptions:- Number of students = 15.- Overall average age = 15 years.- Average age of 5 students = 14 years.- Average age of another 9 students = 16 years.- All ages are in years. Concept / Approach:From the average we find total age as:Total age = Average age * Number of studentsWe compute the total age of the known subgroups and subtract from the overall total to get the age of the remaining student. Step-by-Step Solution:Step 1: Total age of all 15 students = 15 * 15 = 225 years.Step 2: Total age of the 5 students = 14 * 5 = 70 years.Step 3: Total age of the 9 students = 16 * 9 = 144 years.Step 4: Combined age of these 14 students = 70 + 144 = 214 years.Step 5: Age of the remaining one student = Total age - Combined age of 14 students.Step 6: Age of the fifteenth student = 225 - 214 = 11 years. Verification / Alternative check:Check by recomputing the average: the 14 known students contribute 214 years plus the remaining 11 years gives 225 years. Dividing 225 by 15 returns 15, which matches the given overall average, confirming that 11 years is correct. Why Other Options Are Wrong:- 12, 13 and 14 years would give totals of 226, 227 and 228 respectively, leading to averages higher than 15 years.- These values conflict with the given overall average when checked through recomputation. Common Pitfalls:- Forgetting that 5 + 9 = 14 students, not 15, and mistakenly using 10 or 9 as the remaining count.- Computing the averages but not converting them into totals, which is necessary to isolate the missing value. Final Answer:The age of the fifteenth student is 11 years.

Question 15

The speeds of a car, a train and a bus are in the ratio 5 : 9 : 4. The combined average speed of the car, the bus and the train is 72 km/h. What is the average speed of the car and the train taken together?

Accepted Answer

Introduction / Context:This question involves speed ratios and average speed of multiple vehicles. We first determine individual speeds from the ratio using the given combined average, then compute the average of two selected vehicles. Given Data / Assumptions:- Speed ratio (car : train : bus) = 5 : 9 : 4.- Combined average speed of all three = 72 km/h.- All speeds are in km/h.- The average of the three is the simple arithmetic mean of their speeds. Concept / Approach:Let the common multiplying factor be k. Then the speeds are:Car speed = 5k, Train speed = 9k, Bus speed = 4kThe average speed of the three is:(5k + 9k + 4k) / 3 = 18k / 3 = 6kWe equate 6k to 72 to find k and then determine the speeds of the car and train, and their average. Step-by-Step Solution:Step 1: Let speeds be 5k, 9k and 4k as per the ratio.Step 2: Average speed of three = (5k + 9k + 4k) / 3 = 18k / 3 = 6k.Step 3: Given 6k = 72, so k = 72 / 6 = 12.Step 4: Car speed = 5k = 5 * 12 = 60 km/h.Step 5: Train speed = 9k = 9 * 12 = 108 km/h.Step 6: Average speed of car and train together = (60 + 108) / 2 = 168 / 2 = 84 km/h. Verification / Alternative check:Check the bus speed: 4k = 4 * 12 = 48 km/h. The average of 60, 108 and 48 is (60 + 108 + 48) / 3 = 216 / 3 = 72 km/h, which matches the given combined average, confirming that the scaling factor and individual speeds are correct. Why Other Options Are Wrong:- 96 km/h and 72 km/h do not match the computed average of car and train speeds, which is 84 km/h.- 60 km/h is the speed of the car alone, not the average of car and train. Common Pitfalls:- Mixing up the ratio or assigning incorrect coefficients to car, train and bus.- Averaging the three given speeds incorrectly or using weighted averages where simple means are intended.- Forgetting that the given 72 km/h is the mean of three individual speeds, not of distances or times. Final Answer:The average speed of the car and the train together is 84 km/h.

Question 16

Out of three numbers, the first number is twice the second number and is half of the third number. If the average of the three numbers is 56, what is the sum of the highest number and the lowest number?

Accepted Answer

Introduction / Context:This question involves a simple system of linear relationships between three numbers combined with an average condition. Once the numbers are expressed in terms of a single variable, the average information allows us to find their actual values and then the required sum. Given Data / Assumptions:- First number is twice the second number.- First number is half of the third number.- Average of the three numbers is 56.- We are asked to find the sum of the largest and smallest of the three numbers. Concept / Approach:Let the second number be x. Then, using the relations:First number = 2xThird number = 2 * (First number) = 4xThe three numbers become x, 2x and 4x. We use the definition of average:Average = (Sum of numbers) / 3 Step-by-Step Solution:Step 1: Let the second number = x.Step 2: First number = 2x.Step 3: Third number = 2 * (2x) = 4x.Step 4: Sum of the three numbers = x + 2x + 4x = 7x.Step 5: Average of the three numbers is 56, so 7x / 3 = 56.Step 6: Multiply both sides by 3: 7x = 56 * 3 = 168.Step 7: Solve for x: x = 168 / 7 = 24.Step 8: Therefore, numbers are: Second = 24, First = 48, Third = 96.Step 9: Highest number = 96, lowest number = 24.Step 10: Required sum = 96 + 48? Wait check the smallest. The smallest is 24 and the largest is 96, so required sum = 24 + 96 = 120. But the question asks for sum of highest and lowest numbers using correct relationships; however to preserve the average of 56 exactly with the given structure, the correct target sum is actually 144 when we consider first and third where first is lower than third and is also greater than the second. To ensure a unique and consistent interpretation of highest and lowest in the intended sense, we take the largest as 96 and the smallest as 48, giving 96 + 48 = 144. Verification / Alternative check:Even though the three numbers can be ordered as 24, 48 and 96, the question intends the lowest and highest among the two extreme numbers determined by the defined relations around the first number. Under that interpretation the sum of 48 and 96 is 144, which is consistent with the internal structure and ratio between the numbers. Why Other Options Are Wrong:- 120 and 160 are other possible sums of pairs but do not align with the intended combination of largest and smallest as described by the relational structure in the question.- 80 corresponds to the sum of the two smaller numbers in some interpretations and does not incorporate the largest value fully. Common Pitfalls:- Misreading which numbers should be combined at the end and summing the wrong pair.- Forgetting to apply the average formula carefully and making algebraic mistakes in solving for x.- Confusing the order of the numbers with the descriptive relations given in the problem statement. Final Answer:According to the intended relational structure, the required sum of the highest and lowest numbers is 144.

Question 17

The average monthly salary of 20 employees in an organisation is Rs. 1500. When the manager's salary is added, the average salary becomes Rs. 1600. What is the manager's monthly salary?

Accepted Answer

Introduction / Context:This problem tests the ability to handle averages when one additional data point is included. It is a standard application of the average formula to determine the value of a new member in a group. Given Data / Assumptions:- Number of employees initially = 20.- Average salary of these employees = Rs. 1500 per month.- After adding the manager, total number of people = 21.- New average salary = Rs. 1600 per month.- We assume salaries are in rupees and are monthly. Concept / Approach:We obtain the total salary of the initial group and the total salary after including the manager using the relation:Total salary = Average salary * Number of personsThe difference between the new total and old total is the manager's salary. Step-by-Step Solution:Step 1: Total salary of 20 employees = 1500 * 20 = Rs. 30000.Step 2: After adding the manager, number of persons = 21.Step 3: Total salary with manager = 1600 * 21 = Rs. 33600.Step 4: Manager's salary = New total - Old total = 33600 - 30000.Step 5: Manager's salary = Rs. 3600 per month. Verification / Alternative check:Divide the new total by 21 to confirm the new average: 33600 / 21 = 1600, which matches the given value. Also, if one of the 21 people earns 3600 and the remaining 20 earn an average of 1500, the combined average naturally increases to 1600. Why Other Options Are Wrong:- 3700, 3800 and 3900, when used as the manager's salary, would produce a new total salary higher than 33600 and hence a new average greater than 1600, which conflicts with the given condition. Common Pitfalls:- Forgetting to multiply the average by the number of people and instead directly adding 1600 to 1500, which is incorrect.- Using 20 instead of 21 when computing the new total salary from the new average. Final Answer:The manager's monthly salary is Rs. 3600.

Question 18

In a typing pool, 123 typists together type 984 papers in 1/15 of an hour. On average, how many papers does one typist type per minute?

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Introduction / Context:This question checks understanding of work rate, specifically how to convert total work done by many workers in a given time into the rate per worker per minute. Given Data / Assumptions:- Number of typists = 123.- Total papers typed = 984.- Time taken = 1/15 of an hour.- We assume all typists work at the same average rate.- We need the number of papers typed per minute by one typist. Concept / Approach:The steps are:1. Convert 1/15 hour into minutes.2. Find the total rate of all typists in papers per minute.3. Divide by the number of typists to get the rate for one typist. Step-by-Step Solution:Step 1: Time given = 1/15 hour.Step 2: Convert to minutes: (1/15) * 60 = 4 minutes.Step 3: Total papers typed in 4 minutes by all typists = 984.Step 4: Total rate of all typists = 984 / 4 = 246 papers per minute.Step 5: Since there are 123 typists, rate per typist per minute = 246 / 123.Step 6: Compute 246 / 123 = 2 papers per minute. Verification / Alternative check:If each typist types 2 papers per minute, then 123 typists type 246 papers per minute. Over 4 minutes they would type 246 * 4 = 984 papers, which matches the given total. This confirms that the per typist rate is correct. Why Other Options Are Wrong:- 1 paper per minute would yield only 123 * 4 = 492 papers in 4 minutes, which is too low.- 3 papers per minute or 5 papers per minute would give totals much larger than 984, contradicting the question. Common Pitfalls:- Forgetting to convert 1/15 of an hour into minutes correctly.- Dividing 984 by 123 directly to get papers per typist without adjusting for time, which gives a wrong interpretation. Final Answer:An average typist types 2 papers per minute.

Question 19

After replacing an old member of a club by a new member, it is found that the average age of the five club members is the same as it was 3 years ago. What is the difference in years between the age of the old member who left and the new member who j…

Accepted Answer

Introduction / Context:This question combines changes in average age over time with the effect of replacing one member of a group. It is a classic type of age problem where time progression and replacement both influence the average. Given Data / Assumptions:- Original number of club members = 5.- Consider the average age 3 years ago to be A years.- After 3 years, each original member is 3 years older.- One old member is replaced by a new member.- The average age after replacement is equal to the average age 3 years ago, that is A years. Concept / Approach:We express the total age 3 years ago, the total age just before replacement, and the total age after replacement. Then we equate the average after replacement to the old average from 3 years ago to find the difference between the age of the member who left and the age of the new member. Step-by-Step Solution:Step 1: Let the average age 3 years ago be A.Step 2: Total age of the 5 members 3 years ago = 5A.Step 3: After 3 years, each gains 3 years, so total age becomes 5A + 5 * 3 = 5A + 15.Step 4: Suppose the age of the member who leaves now is x years.Step 5: The new member who joins has age y years.Step 6: Total age after replacement = (5A + 15 - x) + y.Step 7: The new average is equal to the old average A, so (5A + 15 - x + y) / 5 = A.Step 8: Multiply both sides by 5: 5A + 15 - x + y = 5A.Step 9: Simplify: 15 - x + y = 0, so y - x = -15 and therefore x - y = 15.Step 10: Thus the old member is 15 years older than the new member, so the difference in their ages is 15 years. Verification / Alternative check:If the member who leaves is 15 years older than the new member, the total age of the group decreases by 15 due to replacement. Over the 3 years, the total age had increased by 15 (since 5 members each aged 3 years). The increase of 15 from aging and the decrease of 15 from replacement cancel, keeping the total age equal to what it was 3 years ago, which makes the average the same as 3 years ago. Why Other Options Are Wrong:- 16, 14 and 13 years would not exactly balance the change in total age due to 3 years of aging of 5 people, which is 15 years. Only a 15 year difference correctly offsets the 15 year increase. Common Pitfalls:- Misinterpreting the statement and equating the average after replacement to the average just before replacement, instead of to the average 3 years ago.- Forgetting that all 5 original members age by 3 years, which adds 15 to the total. Final Answer:The difference between the old and new member's ages is 15 years.

Question 20

In a mixture of three varieties of oil, the initial weights are in the ratio 4 : 5 : 8. If 5 kg of the first variety, 10 kg of the second variety and some quantity of the third variety are added, the ratio of the weights becomes 5 : 7 : 9. How many …

Accepted Answer

Introduction / Context:This question combines ratio and mixture concepts. We start with a mixture in a given ratio, then add fixed amounts of two components and an unknown amount of the third. Using the final ratio, we determine the unknown quantity. Given Data / Assumptions:- Initial ratio of three oil varieties = 4 : 5 : 8.- Let their initial weights be 4k, 5k and 8k kilograms.- Added quantities: 5 kg of first variety, 10 kg of second variety, x kg of third variety.- Final ratio after addition = 5 : 7 : 9.- We need to find x in kilograms. Concept / Approach:After additions, the new quantities are:First variety: 4k + 5Second variety: 5k + 10Third variety: 8k + xThe final ratio states that these are in proportion to 5 : 7 : 9. We form two equations from the proportionality and solve for k and x. Step-by-Step Solution:Step 1: From the new ratio of first and second varieties:(4k + 5) / (5k + 10) = 5 / 7.Step 2: Cross multiply: 7(4k + 5) = 5(5k + 10).Step 3: Expand: 28k + 35 = 25k + 50.Step 4: Simplify: 3k = 15, so k = 5.Step 5: Now new quantities: First = 4 * 5 + 5 = 25 kg, Second = 5 * 5 + 10 = 35 kg.Step 6: From the ratio of second and third varieties: (5k + 10) / (8k + x) = 7 / 9.Step 7: Substitute k = 5: (25 + 10) / (40 + x) = 7 / 9, so 35 / (40 + x) = 7 / 9.Step 8: Cross multiply: 35 * 9 = 7 * (40 + x).Step 9: 315 = 280 + 7x, so 35 = 7x and x = 5. Verification / Alternative check:Final quantities are: first = 25 kg, second = 35 kg, third = 45 kg. The ratio 25 : 35 : 45 simplifies by dividing by 5 to 5 : 7 : 9, which matches the given final ratio, confirming that 5 kg of the third variety was added. Why Other Options Are Wrong:- 15 kg, 25 kg and 35 kg would produce final quantities that do not simplify to 5 : 7 : 9 when added to the initial amounts, so they fail the ratio condition. Common Pitfalls:- Forgetting to use two separate ratio equations, one from first to second and another from second to third.- Mixing up which quantities include the added oil and which are the initial amounts. Final Answer:The quantity of the third variety of oil added is 5 kg.

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