Question 1

The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is

Accepted Answer

Concept & Strategy: This is a classic weighted average problem. To find the mean of the remaining group, subtract the total values of the known subgroups (the top and bottom percentiles) from the overall total value, then divide by the remaining percentage of students. Step-by-Step Solution: • Assume a base: Let the total number of students be 100. Overall mean score = 52. Total score of all students = 100 × 52 = 5200. • Subgroup 1 (Brightest 20%): Number of students = 20. Mean score = 80. Total score of this group = 20 × 80 = 1600. • Subgroup 2 (Dullest 25%): Number of students = 25. Mean score = 31. Total score of this group = 25 × 31 = 775. • Subgroup 3 (Remaining 55%): Remaining students = 100 - (20 + 25) = 55. Remaining total score = 5200 - (1600 + 775) = 5200 - 2375 = 2825. • Calculation: Mean of the remaining students = 2825 / 55 \approx 51.36. This is approximately 51.4. Exam Strategy & Shortcut: Use the concept of Net Deviation from the average. Overall average is 52. Top 20% deviation = +28 per student. Total deviation = 20 × 28 = +560. Bottom 25% deviation = -21 per student. Total deviation = 25 × (-21) = -525. Net deviation of these two groups combined = +560 - 525 = +35. For the overall average to remain 52, the remaining 55 students must balance this +35 deviation with an exact -35 deviation. Deviation per student for the remaining 55 = -35 / 55 = -0.636. Therefore, their average is 52 - 0.636 = 51.364, which safely rounds to 51.4. Common Pitfall: A common mistake is simply averaging the averages (i.e., taking the straight mean of 80 and 31) without accounting for the respective weights (percentages) of the groups. Always use the weighted sum to preserve accuracy. Final Answer: Therefore, the correct answer is 51.4 approx.

Question 2

A coaching institute has students in 3 batches - X, Y and Z. In a certain examination, the average marks obtained by these batches are 72, 60 and 50 respectively. The average marks of batches X and Y taken together is 69. If the ratio of the number …

Accepted Answer

Concept & Logic: This problem requires the application of Alligation or weighted averages in two stages to find the true ratio of students in all three batches. Once the combined ratio n_x : n_y : n_z is unified, we apply the standard weighted average formula to find the final aggregate average. Step-by-Step Solution: • Step 1: Find the ratio of students in Batches X and Y. Average of X = 72 Average of Y = 60 Combined Average of X and Y = 69 Using Alligation (Cross difference method): Ratio of X : Y = (69 - 60) : (72 - 69) = 9 : 3 = 3 : 1. • Step 2: Combine the ratios. We found X : Y = 3 : 1 We are given Y : Z = 6 : 7 To merge these, make the value of Y consistent in both ratios. Multiply the first ratio by 6: X : Y = 18 : 6 Therefore, the cleanly combined ratio of students is X : Y : Z = 18 : 6 : 7. • Step 3: Calculate the overall average. Let the number of students be 18k, 6k, and 7k. (Total parts = 31k). Total marks = (18 × 72) + (6 × 60) + (7 × 50) Total marks = 1296 + 360 + 350 = 2006. Overall Average = 2006 / 31 \approx 64.709. Exam Strategy & Shortcut: Using Alligation to find the X : Y ratio visually is the fastest start. Instead of calculating massive numbers like 18 × 72, you can shift the base to 60 (the middle value). Deviations from 60: X has a deviation of +12 for 18 parts = +216. Y has a deviation of 0 for 6 parts = 0. Z has a deviation of -10 for 7 parts = -70. Net deviation = 216 - 70 = 146. Distribute this net deviation evenly over all 31 parts (18+6+7). 146 / 31 \approx 4.7. Add this to the assumed base of 60: 60 + 4.7 = 64.7. This requires much smaller calculations and slashes the chance of multiplication errors. Common Pitfall: Failing to align the ratios correctly before combining them. If you blindly add the parts or use unaligned variables (e.g., mixing up the Y value of 1 and 6), the final weighted average will be entirely warped and lead to a trap answer. Final Answer: Therefore, the correct answer is 64.7.

Question 3

Ten years ago, the ages of the members of a joint family of eight people added up to 231 years.  Three years later, one member died at the age of 60 years and a child was born during the same year.  After another three years, one more member died, a…

Accepted Answer

Concept & Logic: To find the current average age, we must track the total age of the family over time. The key principle is that for every passing year, the total age of the family increases by the number of living members. Step-by-Step Solution: • 10 Years Ago (T = -10): Family size = 8. Total age = 231. • 7 Years Ago (T = -7): (3 years later) Total age increases by 8 × 3 = 24 years. Total age before changes = 231 + 24 = 255. One member dies (-60) and a newborn is added (+0). New total age = 255 - 60 + 0 = 195. (Family size remains 8). • 4 Years Ago (T = -4): (Another 3 years later) Total age increases by 8 × 3 = 24 years. Total age before changes = 195 + 24 = 219. Another member dies (-60) and a newborn is added (+0). New total age = 219 - 60 + 0 = 159. (Family size remains 8). • Present Day (T = 0): (4 years later) Total age increases by 8 × 4 = 32 years. Current total age = 159 + 32 = 191. Current average = 191 / 8 = 23.875 years. Exam Strategy & Shortcut: Instead of calculating the total age at every step, track the net change over the entire 10-year period. Because a birth instantly replaces every death, the family size is a constant 8 people. Think of the family as 8 "slots" that age continuously over the 10 years. Total natural aging for these 8 slots = 8 × 10 = 80 years. During this time, two people left the family, taking their accumulated ages with them at the exact time of their departure. Net Total Age = Initial Total + Total Natural Aging - Ages at Death Net Total = 231 + 80 - 60 - 60 = 191. Current Average = 191 / 8 = 23.875. This is nearest to 24. This powerful logic bypasses tracking the individual newborns entirely and solves the problem in seconds! Common Pitfall: Students often get tangled up trying to calculate the current ages of the newborns (who are 7 and 4 years old today). While calculating their specific individual ages works if you meticulously track every living member, it frequently leads to arithmetic errors and wastes valuable time. Leveraging the constant family size of 8 to find the net total is much faster and safer. Final Answer: Therefore, the correct answer is 24 years.

Question 4

The average salary of all the workers in a workshop is ₹ 8000. The average salary of 7 technicians is ₹ 12000 and the average salary of the rest is ₹ 6000. The total number of workers in the workshop is

Accepted Answer

Concept & Strategy: This problem is a classic application of the Weighted Average or Rule of Alligation. It involves mixing two distinct groups (technicians and the rest) to form a combined whole (all workers). Step-by-Step Solution: • Given Data: Overall average salary = ₹ 8000 Average salary of technicians (Group 1) = ₹ 12000 Number of technicians = 7 Average salary of the rest (Group 2) = ₹ 6000 • Algebraic Approach: Let the total number of workers be x. The total salary of all workers = 8000x. The total salary of the 7 technicians = 7 × 12000 = 84000. The number of remaining workers = x - 7. The total salary of the remaining workers = 6000 × (x - 7). Equating the total salaries: 8000x = 84000 + 6000(x - 7) 8000x = 84000 + 6000x - 42000 8000x - 6000x = 42000 2000x = 42000 x = (42000/2000) = 21 Exam Strategy & Shortcut: The fastest way to solve this is using the Rule of Alligation. Set up the averages of the two groups and the mean in the middle: Technicians (₹ 12000) ----------- Rest (₹ 6000) \ / Mean (₹ 8000) / \ (8000 - 6000) ----------- (12000 - 8000) Ratio \rightarrow 2000 : 4000 = 1 : 2 This means the ratio of Technicians to Rest is 1 : 2. If 1 part represents 7 technicians, then the total parts (1 + 2 = 3 parts) represent the total workers. Total workers = 3 × 7 = 21. Common Pitfall: A common mistake is finding the number of "rest" workers (14) and forgetting to add back the 7 technicians to find the *total* number of workers as the question asks. Always double-check what the final variable represents! Final Answer: Therefore, the correct answer is 21.

Question 5

In a school with 600 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years 9 months, then the number of girls in the school is:

Accepted Answer

Concept & Logic: This problem requires finding the ratio of two subgroups (boys and girls) when their individual averages and the combined average are known. This is a perfect scenario for the Rule of Alligation. To avoid messy fractions, it is highly recommended to convert all ages into months first. Step-by-Step Solution: • Unit Conversion: Average age of Boys = 12 years = 12 × 12 = 144 months. Average age of Girls = 11 years = 11 × 12 = 132 months. Combined average age = 11 years 9 months = (11 × 12) + 9 = 132 + 9 = 141 months. • Applying Alligation: Boys (144) ------------- Girls (132) \ / Mean (141) / \ (141 - 132) ----------- (144 - 141) Ratio \rightarrow 9 : 3 Simplifying, the ratio of Boys to Girls is 3 : 1. • Calculation: Total ratio parts = 3 + 1 = 4 parts. Total students = 600. Value of 1 part = 600 / 4 = 150. Number of girls (1 part) = 150. Exam Strategy & Shortcut: If you prefer decimals over months, convert 9 months to a fraction of a year: 9/12 = 3/4 = 0.75 years. Combined average = 11.75 years. Alligation Ratio = (11.75 - 11) : (12 - 11.75) = 0.75 : 0.25 = 3 : 1. The Girls constitute 1/4 of the total school. 600 / 4 = 150. Both methods are mathematically identical, but picking the one you are most comfortable calculating mentally saves crucial seconds. Common Pitfall: The most frequent error is mathematically translating "11 years 9 months" as "11.9 years". There are 12 months in a year, not 10. Always use 11 + 9/12 (11.75) or convert entirely to months. Final Answer: Therefore, the correct answer is 150.

Question 6

In an engineering college the average salary of all engineering graduates from Mechanical trade is ₹ 2.45 lacs per annum and that of the engineering graduates from Electronics trade is ₹ 3.56 lacs per annum. The average salary of all Mechanical and …

Accepted Answer

Concept & Logic: This problem uses the Rule of Alligation to find the ratio of graduates in the two trades. The crucial secondary concept is understanding that the number of graduates must be a whole integer. Finding the "least number" requires simplifying the resulting ratio to its lowest coprime terms. Step-by-Step Solution: • Given Data: Mechanical Average = 2.45 lacs Electronics Average = 3.56 lacs Combined Average = 3.12 lacs • Applying Alligation: Mech (2.45) ------------- Elec (3.56) \ / Mean (3.12) / \ (3.56 - 3.12) ----------- (3.12 - 2.45) Ratio of Mech : Elec = 0.44 : 0.67 Multiply both sides by 100 to remove decimals: Ratio of Mech : Elec = 44 : 67 • Deducing the Least Number: The ratio of Mechanical to Electronics graduates is 44 : 67. This means for every 44 Mechanical graduates, there are 67 Electronics graduates. To find the *least* possible actual number of graduates, we check if the ratio 44/67 can be simplified further. The number 67 is a prime number, meaning 44 and 67 are coprime (they share no common factors other than 1). Therefore, the ratio cannot be reduced to smaller whole numbers. The smallest integer values that satisfy this ratio are 44 Mechanical and 67 Electronics graduates. Exam Strategy & Shortcut: Perform the cross-subtraction directly mentally: 356 - 312 = 44 (Mech parts) 312 - 245 = 67 (Elec parts) Ratio is 44 : 67. Recognizing immediately that 67 is prime confirms that the ratio is fully simplified, locking in 67 as the absolute minimum number of Electronics students. Common Pitfall: The primary trap here is option (d) "Cannot be determined". Because the exact total number of students isn't provided, students quickly assume there is missing information. The key lies in the word "least". While the *actual* number could be 134 (67 × 2) or 201 (67 × 3), the *least possible* number is the prime ratio base itself. Final Answer: Therefore, the correct answer is 67.

Question 7

My Scooty gives an average of 40 kmpl of petrol. But after recent filling at the new petrol pump, its average dropped to 38 kmpl.   I investigated and found out that it was due to adulterated petrol. Petrol pumps add kerosene, which is $\frac{2}{3}$…

Accepted Answer

Concept & Logic: This problem requires a two-step approach: first using the Rule of Alligation to find the volumetric ratio of petrol to kerosene in the mixture, and then calculating the cost price of this mixture to determine the profit made by substituting cheaper fuel. Step-by-Step Solution: • Given: • Original pure petrol average = 40 km/L • Adulterated mixture average = 38 km/L • Kerosene average = 18 km per 900 ml • Pure petrol cost = ₹ 30 per Litre • Kerosene is (2/3) cheaper than petrol. • Calculation / Deduction: 1. Find the per-litre average of kerosene: If it gives 18 km for 900 ml (0.9 L), then for 1000 ml (1 L), it gives: (18/0.9) = 20 km/L. 2. Find the mixture ratio using alligation: Petrol average = 40 Kerosene average = 20 Mixture average = 38 Ratio of Petrol to Kerosene = (38 - 20) : (40 - 38) = 18 : 2 = 9 : 1. This means in 1 Litre of mixture, petrol is (9/10) L and kerosene is (1/10) L. 3. Calculate the cost price for the pump owner: Cost of pure petrol = ₹ 30/L. Kerosene is (2/3) cheaper, meaning its cost is reduced by (2/3). Cost of kerosene = 30 - ((2/3) × 30) = 30 - 20 = ₹ 10/L. Cost of 1 Litre of the adulterated mixture = (Cost of (9/10) L petrol) + (Cost of (1/10) L kerosene) Mixture Cost = ((9/10) × 30) + ((1/10) × 10) = 27 + 1 = ₹ 28. 4. Calculate the additional amount made: He sells it at the pure petrol price (₹ 30) but his actual cost is ₹ 28. Additional amount = 30 - 28 = ₹ 2. Exam Strategy & Shortcut: Standardize all your units first (convert 900 ml to 1 Litre immediately). Once you use alligation to find the 9:1 ratio, simply calculate the cost difference per portion. The owner replaces (1/10) of ₹ 30 petrol with ₹ 10 kerosene. The savings (extra profit) is exactly (1/10) of the price difference between the two fuels: (1/10) × (30 - 10) = (20/10) = ₹ 2. Common Pitfall: Students often misinterpret "(2/3) cheaper" as meaning the cost is (2/3) of the original. "Cheaper by" implies subtraction, so the actual cost is 1 - (2/3) = (1/3) of the petrol price. Missing this nuance will lead to the wrong mixture cost. Final Answer: Therefore, the correct answer is ₹ 2.

Question 8

Average score of a class of 60 students, in an exam, was 43.   Average score of the students who had passed is 52 and the average score of students who had failed is 16.   How many failed the exam?

Accepted Answer

Concept & Strategy: This problem is best solved using the Rule of Alligation. Alligation helps us quickly find the ratio in which two distinct groups (passed and failed students) are mixed to form a combined group (the whole class) with a known overall average. \frac{Quantity of Group 1}{Quantity of Group 2} = \frac{Average of Group 2 - Overall Average}{Overall Average - Average of Group 1} Step-by-Step Solution: • Given: • Total students = 60 • Overall average score = 43 • Average score of passed students = 52 • Average score of failed students = 16 • Calculation / Deduction: 1. Set up the alligation cross: Passed Average (52) -------------- Failed Average (16) \ / \ / Overall (43) / \ / \ (43 - 16) ------------------------ (52 - 43) = 27 = 9 2. Find the ratio of Passed to Failed students: Ratio = 27 : 9 = 3 : 1 3. Calculate the number of failed students: The total number of parts is 3 + 1 = 4 parts. 4 parts correspond to the total class size of 60 students. 1 part = (60/4) = 15 students. Since the ratio of failed students is 1 part, the number of failed students is 15. Exam Strategy & Shortcut: Alligation is the ultimate shortcut for "mixture" average problems. If you used the standard algebraic equation 52x + 16(60-x) = 43 × 60, you would waste precious seconds multiplying and shifting large numbers. The alligation cross gives you the 3:1 ratio almost instantly, turning a complex multiplication problem into a simple division problem (60 \div 4). Common Pitfall: A frequent mistake is matching the final ratio to the wrong group. In the alligation cross, the difference calculated on the left side corresponds to the quantity of the group on the right side. Always remember to cross the values properly so you don't accidentally conclude that 45 students failed. Final Answer: Therefore, the correct answer is 15.

Question 9

The average of 11 results is 60.   If the average of first six results is 58 and that of last six is 63, find the $6^{	ext{th}}$ result.

Accepted Answer

Concept & Formula: This is a classic Overlapping Averages problem. When a central item is counted twice (once in the first half and once in the second half), you can find its exact value by subtracting the total sum of all distinct items from the combined sum of the overlapping subsets. Value of Overlapping Item = (Sum of Part 1 + Sum of Part 2) - Total Sum Step-by-Step Solution: • Given: • Average of all 11 results = 60 • Average of the first 6 results = 58 • Average of the last 6 results = 63 • Calculation / Deduction: 1. Calculate the sum of all 11 results: Total Sum = 11 × 60 = 660 2. Calculate the sum of the first 6 results: Sum of First 6 = 6 × 58 = 348 3. Calculate the sum of the last 6 results: Sum of Last 6 = 6 × 63 = 378 4. Determine the 6^{th} result: Notice that 6 results + 6 results = 12 results. The 6^{th} result is counted in both groups. 6^{th} Result = (Sum of First 6 + Sum of Last 6) - Total Sum 6^{th} Result = (348 + 378) - 660 6^{th} Result = 726 - 660 = 66 Exam Strategy & Shortcut: Instead of calculating massive sums, use the Net Deviation Method. The overall average is 60. The first 6 results deviate by -2 each (58 - 60). Total deviation = 6 × (-2) = -12. The last 6 results deviate by +3 each (63 - 60). Total deviation = 6 × (+3) = +18. Net deviation = -12 + 18 = +6. The 6^{th} result must be the overall average plus this net deviation: 60 + 6 = 66. This mental math completely avoids large multiplications like 6 × 63. Common Pitfall: Students often get confused by the overlap and subtract the overlapping subsets from each other rather than summing them and subtracting the master total. Always visualize the set: \{1..6\} and \{6..11\}. Adding them gives \{1..11\} + \{6\}. Subtracting the master sum \{1..11\} leaves just \{6\}. Final Answer: Therefore, the correct answer is 66.

Question 10

The average weight of 21 boys was recorded as $64$ kg.   If the weight of the teacher was added, the average increased by $1$ kg.   What was the teacher's weight?

Accepted Answer

Concept & Strategy: This is an Average Inclusion problem. When a new entity joins a group and raises the overall average, they are not only bringing the original average value but also enough extra value to distribute among all members (including themselves) to raise the average. New Value = Old Average + [ Total New Members × Increase in Average ] Step-by-Step Solution: • Given: • Initial number of boys = 21 • Initial average weight = 64 kg • New total members (boys + teacher) = 21 + 1 = 22 • Increase in average = 1 kg • New average weight = 64 + 1 = 65 kg • Calculation / Deduction: Let's use the standard summation method first to see the underlying math. 1. Find the total weight of the 21 boys: Total weight = 21 × 64 = 1344 kg 2. Find the new total weight of all 22 people: New total weight = 22 × 65 = 1430 kg 3. Calculate the teacher's weight by finding the difference: Teacher's weight = 1430 - 1344 = 86 kg Exam Strategy & Shortcut: Never multiply 21 × 64 in a competitive exam! Use the Transfer of Value logic. For the average to become 65 kg, the teacher must bring at least 64 kg (the old average) to avoid pulling it down. Then, the teacher must also provide an extra 1 kg for every single person in the new group (22 people) to raise the average. Teacher's weight = Old Average + (New Group Size × Increase) Teacher's weight = 64 + (22 × 1) = 64 + 22 = 86 kg. You can do this in your head in 3 seconds. Common Pitfall: The most common trap is multiplying the increase (1 kg) by the *old* group size (21) instead of the *new* group size (22). Remember, the teacher's own weight in the new average is also 65 kg, so they must supply that extra 1 kg for themselves too! Final Answer: Therefore, the correct answer is 86 kg.

Question 11

The average expenditure of a man for the first five months is ₹ 1200 and for the next seven months is ₹ 1300.   If he saves ₹ 2900 in that year, his monthly average income is

Accepted Answer

Concept & Formula: This problem revolves around the fundamental financial equation: Total Income = Total Expenditure + Total Savings Once the total annual income is found, finding the monthly average is a simple division by 12. Step-by-Step Solution: • Given: • Average expenditure for first 5 months = ₹ 1200 • Average expenditure for next 7 months = ₹ 1300 • Total savings for the year = ₹ 2900 • Calculation / Deduction: 1. Calculate the total expenditure for the first 5 months: 5 × 1200 = ₹ 6000 2. Calculate the total expenditure for the remaining 7 months: 7 × 1300 = ₹ 9100 3. Find the total annual expenditure: 6000 + 9100 = ₹ 15100 4. Calculate the total annual income: Income = Expenditure + Savings Annual Income = 15100 + 2900 = ₹ 18000 5. Determine the average monthly income: Monthly Average = (18000/12) = ₹ 1500 Exam Strategy & Shortcut: While the multiplication here is relatively straightforward, you can use the Base Average approach to streamline the expenditure calculation. Assume the base expenditure for all 12 months is 1200. The total base expenditure = 12 × 1200 = 14400. The last 7 months have an excess of +100 each, adding 7 × 100 = 700. Total expenditure = 14400 + 700 = 15100. Add the savings (2900) to get 18000, and divide by 12 to get 1500. This keeps the mental arithmetic slightly smaller. Common Pitfall: The most frequent error is calculating the average expenditure and equating that directly to income, completely forgetting to factor in the savings. Always remember that Income is the sum of *both* what is spent and what is kept. Final Answer: Therefore, the correct answer is ₹ 1500.

Question 12

Direction: Each of the questions below consists of a question-statement and two statements I and II are given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Give answer.  Whose body we…

Accepted Answer

Concept & Logic: To determine whose body weight is the second highest among five people, we must either find the exact weights of all five boys or establish a strict relative order among them. We use the formula for the sum of elements given their average: Sum = Average × Number of Elements Step-by-Step Solution: Evaluating Statement I alone: • Given: Average weight of Arun (A), Suraj (S), and Vinay (V) = 68 kg. • Sum of A, S, V = 3 × 68 = 204 kg. • Given individual weights: S = 78, R = 68, V = 46. • Substitute S and V to find A: A + 78 + 46 = 204 kg \rightarrow A + 124 = 204 kg \rightarrow A = 80 kg. • Given: Average weight of Raju (R) and Pratap (P) = 72 kg. • Sum of R, P = 2 × 72 = 144 kg. • Substitute R to find P: 68 + P = 144 kg \rightarrow P = 76 kg. • Now we have all five weights: A = 80, S = 78, P = 76, R = 68, V = 46. • Arranging in descending order: A, S, P, R, V. The second highest is Suraj (S). Statement I is sufficient. Evaluating Statement II alone: • Given: Average weight of A, S, V, R = 68 kg. • Sum of A, S, V, R = 4 × 68 = 272 kg. • Given weights: S = 78, R = 68, V = 46. • Find A: A + 78 + 68 + 46 = 272 kg \rightarrow A + 192 = 272 kg \rightarrow A = 80 kg. • We know four weights: A = 80, S = 78, R = 68, V = 46. • However, Statement II provides absolutely no information about Pratap (P). Without P's weight, we cannot guarantee who holds the second highest position. Statement II is not sufficient. Exam Strategy & Shortcut: For data sufficiency questions involving averages, quickly check if the variables match up. In Statement I, you are given the average of 3 boys and the exact weights of 2 of them. This immediately tells you that you can find the 3rd. You are then given the average of the remaining 2 boys and the weight of 1 of them. This means you can find the final boy's weight. Since all 5 weights can be determined, Statement I is definitely sufficient. Statement II lacks Pratap entirely, so it is instantly insufficient. You do not even need to calculate the final values to mark the answer! Common Pitfall: A major mistake is performing all the arithmetic calculations during a data sufficiency problem. While we solved it above for clarity, calculating 80 and 76 consumes precious time. Recognize that having n independent variables and n independent linear equations guarantees a solution. Final Answer: Therefore, the correct answer is The data in statement I alone was sufficient to answer the question while II alone are not sufficient to answer the question.

Question 13

Direction: Each of the questions below consists of a question-statement and two statements I and II are given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Give answer.  What will be …

Accepted Answer

Concept & Logic: To find the exact value of a single variable (marks in Physics), we need an equation where Physics is the only unknown, or a system of equations that allows us to isolate it. We apply the basic average formula: Sum = Average × Number of Subjects Step-by-Step Solution: Evaluating Statement I alone: • Given: Average of History (H), Geography (G), and Chemistry (C) = 75. • Equation: H + G + C = 3 × 75 = 225. • This statement contains no information about Physics (P). It is insufficient. Evaluating Statement II alone: • Given: Average of History (H), Geography (G), and Physics (P) = 78. • Equation: H + G + P = 3 × 78 = 234. • We have one equation with three variables. We cannot determine P without knowing the combined sum of H and G. It is insufficient. Evaluating Statements I and II combined: • From I: H + G + C = 225 • From II: H + G + P = 234 • Subtracting the first equation from the second gives: P - C = 9. • This tells us Physics marks are 9 more than Chemistry marks. However, without knowing the absolute marks in Chemistry, we still cannot determine the exact marks in Physics. Exam Strategy & Shortcut: Look at the variables involved. We want to find P. Statement I gives variables (H, G, C). No P. Insufficient. Statement II gives variables (H, G, P). Three unknowns, one equation. Insufficient. Combining them gives (H, G, C) and (H, G, P). We have 4 unknowns (H, G, C, P) but only 2 independent equations. Mathematically, it is impossible to solve for a unique value of P. You can instantly conclude that both statements together are insufficient without multiplying 75 by 3 or 78 by 3. Common Pitfall: Students often see two equations with overlapping variables (History and Geography) and automatically assume combining them will solve the problem. Always check if you are left with more unknowns than equations. Here, subtracting leaves P and C, which still cannot be solved. Final Answer: Therefore, the correct answer is The data in both Statement I and II are not sufficient to answer the question.

Question 14

There are three positive numbers. One third of the average of all the three numbers is 8 less than the value of the highest number. The average of the lowest and the second lowest number is 8. What is the highest number?

Accepted Answer

Concept & Logic: To solve this, we translate the word problem into algebraic equations using the basic formula for averages: Average = \frac{Sum of terms}{Number of terms} Let the three positive numbers in ascending order be x, y, and z, where z is the highest number. Step-by-Step Solution: • Given: The average of the lowest (x) and second lowest (y) number is 8. • Calculation: (x + y/2) = 8 x + y = 16 (This is our first equation) • Given: One-third of the average of all three numbers is 8 less than the highest number (z). • Calculation: (1/3) × ((x + y + z/3)) = z - 8 (x + y + z/9) = z - 8 • Deduction: Substitute the value of (x + y) from our first equation into this new equation: (16 + z/9) = z - 8 16 + z = 9(z - 8) 16 + z = 9z - 72 16 + 72 = 9z - z 88 = 8z z = 11 Exam Strategy & Shortcut: Instead of writing out full variables, use chunking. You know the sum of the first two numbers is 2 × 8 = 16. Let the highest number be H. The sum of all three is (16 + H). The average of all three is (16 + H/3). One third of this is (16 + H/9). Set this equal to H - 8 and solve directly. This mental chunking saves you from writing multiple x and y variables. Common Pitfall: A very common mistake is misreading "one third of the average" as just "one third of the sum". This leads to the incorrect equation (x+y+z/3) = z - 8, which will result in z = 10. Always read the fractional relationships carefully! Final Answer: Therefore, the correct answer is 11.

Question 15

A shop of electronic goods is closed on Monday. The average daily sales for remaining six days of a week is ₹ 15,640/- and the average sale of Tuesday to Saturday is ₹ 14,124/-. The sales on Sunday is

Accepted Answer

Concept & Strategy: This problem uses the concept of finding a missing value by subtracting a subset total from an overall total. Total Value = Average × Number of Entities If we know the total sales for 6 days and the total sales for 5 of those days, the difference is exactly the sales of the 6th day (Sunday). Step-by-Step Solution: • Calculation (Total 6 days): Find the total sales for the 6 open days (Tuesday through Sunday). Total for 6 days = 6 × 15640 Total for 6 days = 93840 • Calculation (Total 5 days): Find the total sales for the 5 days from Tuesday to Saturday. Total for 5 days = 5 × 14124 Total for 5 days = 70620 • Deduction (Sunday): Subtract the 5-day total from the 6-day total to isolate Sunday's sales. Sunday Sales = (Total for 6 days) - (Total for 5 days) Sunday Sales = 93840 - 70620 Sunday Sales = 23220 Exam Strategy & Shortcut: Unit Digit Method: You don't necessarily have to calculate the full multiplication. Check the unit digit of the 6-day sum: 6 × 0 = 0. Check the unit digit of the 5-day sum: 5 × 4 = 20 \rightarrow ends in 0. This doesn't help much since all options (except Data inadequate) end in something else, wait! Let's look at the last *two* digits. Last two digits of 6 × 40 = 240 \rightarrow 40. Last two digits of 5 × 24 = 120 \rightarrow 20. The difference in the last two digits is 40 - 20 = 20. Looking at the options, only ₹ 23,220/- ends with 20. You can select it immediately without full multiplication! Common Pitfall: Students might get confused by the "closed on Monday" detail and try to divide by 7 at some point, or mistakenly assume Monday counts as a zero in an average of 7 days. The problem explicitly restricts the averages to the "remaining six days", so Monday is irrelevant. Final Answer: Therefore, the correct answer is ₹ 23,220/-.

Question 16

Find the average of 205, 302, 108, 403 and 202

Accepted Answer

Concept & Formula: The arithmetic mean (average) is calculated by adding all the numbers in the dataset and dividing by the total count of numbers. Average = \frac{Sum of all observations}{Total number of observations} Step-by-Step Solution: • Calculation: Add the five given numbers together. Sum = 205 + 302 + 108 + 403 + 202 Sum = 1220 • Calculation: Count the number of observations. There are 5 numbers. • Calculation: Divide the sum by the count. Average = (1220/5) Average = 244 Exam Strategy & Shortcut: Assumed Mean Method: Pick a convenient number close to the values, say 200. Let's find the deviations from 200: 205 \rightarrow +5 302 \rightarrow +102 108 \rightarrow -92 403 \rightarrow +203 202 \rightarrow +2 Sum of deviations = 5 + 102 - 92 + 203 + 2 = 220. Average of deviations = (220/5) = 44. Final Average = Assumed Mean + Average of deviations = 200 + 44 = 244. This method prevents dealing with large sums and reduces arithmetic errors. Common Pitfall: A common trap in multiple-choice questions is presenting the "Sum" as one of the options. Notice that 1220 is option (d). Students who calculate the sum but forget to divide by N will prematurely select (d) and lose marks. Always complete the final division step. Final Answer: Therefore, the correct answer is 244.

Question 17

The average of 6 numbers is 7. The average of three numbers of them is 5. What will be the average of remaining numbers?

Accepted Answer

Concept & Logic: This problem is solved by understanding that the sum of a whole group is equal to the sum of its distinct parts. We can find the sum of all 6 numbers, subtract the sum of the known 3 numbers, and then find the average of the remaining 3 numbers. Sum = Average × N Step-by-Step Solution: • Calculation (Total Group): Find the sum of all 6 numbers. Total Sum = 6 × 7 = 42 • Calculation (Subset 1): Find the sum of the 3 specified numbers. Sum of 3 numbers = 3 × 5 = 15 • Deduction (Subset 2): Find the sum of the remaining numbers. Sum of remaining numbers = (Total Sum) - (Sum of 3 numbers) Sum of remaining numbers = 42 - 15 = 27 • Calculation (Final Average): Since there were 6 numbers initially and we removed 3, there are 3 numbers remaining. Average of remaining 3 numbers = (27/3) = 9 Exam Strategy & Shortcut: Balance / Deviation Method: The overall average is 7. Three numbers have an average of 5, which is a deviation of -2 from the main average. Total deficit created by these three numbers = 3 × (-2) = -6. For the overall average to remain 7, the remaining three numbers must compensate for this deficit of -6. Therefore, they must create a surplus of +6 spread across the 3 remaining numbers. Required surplus per number = (+6/3) = +2. So, the average of the remaining numbers must be 7 + 2 = 9. This requires virtually no calculation! Common Pitfall: A common error is confusing sums with averages. Students might find the remaining sum (27) and mistakenly look for it in the options, or they might try to subtract the averages directly (7 - 5 = 2), which violates the rules of arithmetic means. Always convert averages to sums before adding or subtracting groups. Final Answer: Therefore, the correct answer is 9.

Question 18

Average of $a$, $b$, and $c$ is 11; average of $c$, $d$ and $e$ is 17; average of $e$ and $f$ is 22 and average of $e$ and $c$ is 17. Find out the average of $a$, $b$, $c$, $d$, $e$, and $f$.

Accepted Answer

Concept & Logic: To find the average of a combined set of variables, you first need to find their total sum. We can extract the sum of different subsets from their given averages using the fundamental formula: Sum = Average × Number of elements Then, we strategically combine these subset sums to isolate the grand total (a+b+c+d+e+f). Step-by-Step Solution: • Extracting Sums from Given Averages: 1. Average of a, b, c = 11 \rightarrow a + b + c = 11 × 3 = 33 2. Average of c, d, e = 17 \rightarrow c + d + e = 17 × 3 = 51 3. Average of e, f = 22 \rightarrow e + f = 22 × 2 = 44 4. Average of e, c = 17 \rightarrow e + c = 17 × 2 = 34 • Combining the Subsets: Notice that if we add equations (1), (2), and (3) together, we get almost all the variables we need, but with some duplicates. Let's add them: (a + b + c) + (c + d + e) + (e + f) = 33 + 51 + 44 a + b + 2c + d + 2e + f = 128 • Isolating the Target Sum: We want the sum (a + b + c + d + e + f). Notice that our combined equation above can be rewritten as: (a + b + c + d + e + f) + c + e = 128 We already know from equation (4) that (e + c) = 34. Let's substitute this in: (a + b + c + d + e + f) + 34 = 128 (a + b + c + d + e + f) = 128 - 34 Sum of all 6 variables = 94 • Calculating the Final Average: Average = \frac{Sum}{6} Average = (94/6) = (47/3) Converting to a mixed fraction: (47/3) = 15(2/3) Exam Strategy & Shortcut: Look for the "overlapping" variables. The problem provides three continuous groups: (a,b,c), (c,d,e), and (e,f). When you sum these groups, the boundary variables (c and e) are counted twice. The problem conveniently provides the exact sum of those specific overlapping variables (c+e) at the end. Mentally: Total Sum = (33 + 51 + 44) - (34) = 128 - 34 = 94. Average = 94 / 6 = 15(2/3). This can be done in under 30 seconds without writing algebra. Common Pitfall: A common trap is attempting to solve for the individual values of a, b, c, d, e, and f. We have 6 variables but only 4 equations, meaning it is mathematically impossible to find their individual values. You must solve for the *group* sum as a single entity. Final Answer: Therefore, the correct answer is 15(2/3).

Question 19

A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in the month of June beginning with a Sunday?

Accepted Answer

Concept & Strategy: This problem relies on basic calendar knowledge and calculating a weighted average. To find the overall average visitors for the month, we must determine the exact number of Sundays versus regular days, calculate the total visitors for each category, and divide by the total days in the month. Overall Average = \frac{(Sundays × Avg) + (Other Days × Avg)}{Total Days} Step-by-Step Solution: • Calendar Analysis: The month of June has exactly 30 days. We are told June begins on a Sunday (June 1st is a Sunday). Sundays will occur on the 1st, 8th, 15th, 22nd, and 29th. Therefore, there are exactly 5 Sundays in this month. The number of "other days" = 30 - 5 = 25 days. • Calculating Total Visitors: Total visitors on all Sundays = 5 days × 510 visitors/day = 2550 Total visitors on other days = 25 days × 240 visitors/day = 6000 Grand total of visitors in June = 2550 + 6000 = 8550 • Calculating the Final Average: Average per day = \frac{Grand Total}{Total Days} Average = (8550/30) Average = (855/3) = 285 Exam Strategy & Shortcut: Deviation / Baseline Method: Instead of multiplying large numbers, pick a baseline average. Let's assume every day has a baseline of 240 visitors. If every day had 240 visitors, the average would be 240. However, there are 5 Sundays where the library gets an *extra* 270 visitors (510 - 240 = 270). Total extra visitors = 5 × 270 = 1350. Now, distribute these 1350 extra visitors equally across all 30 days of the month: Extra per day = (1350/30) = 45. Overall Average = Baseline + Extra per day = 240 + 45 = 285. This method avoids calculating 25 × 240 and keeps the numbers small and manageable. Common Pitfall: The most frequent error is incorrectly counting the number of Sundays. Students might assume a 30-day month always has 4 weeks and therefore exactly 4 Sundays, leading to a wrong calculation. Always map out the days if a specific starting day is mentioned. Final Answer: Therefore, the correct answer is 285.

Question 20

The marks of six boys in a group are 48, 59, 87, 37, 78 and 57. What are the average marks of all six boys?

Accepted Answer

Concept & Formula: This is a fundamental average problem. The arithmetic mean (average) is calculated by taking the sum of all the individual values in a dataset and dividing it by the total count of those values. Average = \frac{Sum of all observations}{Total number of observations} Step-by-Step Solution: • Step 1: Calculate the Sum We need to add all six given marks together. Sum = 48 + 59 + 87 + 37 + 78 + 57 Let's group them for easier addition: (48 + 59) = 107 (87 + 37) = 124 (78 + 57) = 135 Total Sum = 107 + 124 + 135 = 366 • Step 2: Identify the Count The problem states there are six boys. Therefore, the total number of observations is 6. • Step 3: Calculate the Average Average = (366/6) Average = 61 Exam Strategy & Shortcut: Unit Digit Check: To save time on long division, first calculate just the unit digit of the sum. 8 + 9 + 7 + 7 + 8 + 7 = 46 \rightarrow The sum ends in a 6. We need to divide a number ending in 6 by 6. Looking at the options: (a) 61 × 6 = 366 (Ends in 6 - Possible) (b) 65 × 6 = 390 (Ends in 0 - Incorrect) (c) 69 × 6 = 414 (Ends in 4 - Incorrect) Since only option (a) provides a valid unit digit when multiplied back by 6, you can confidently mark 61 without doing the full addition and division, provided you trust the options don't hide behind "None of these". Common Pitfall: A common mistake in simple addition problems under time pressure is misaligning columns or carrying over the wrong digit. Grouping the numbers into pairs (as shown in Step 1) or using the unit digit shortcut helps minimize arithmetic fatigue. Final Answer: Therefore, the correct answer is 61.

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