Question 1

In an examination, the average marks obtained by John in English, Maths, Hindi and Drawing together are 50. His average marks in Maths, Science, Social Studies and Craft are 70. If the average marks in all seven subjects taken together is 58, what w…

Accepted Answer

Introduction / Context: This problem uses overlapping averages of different subject groups to find John's marks in a particular subject, Maths. By converting each given average into a total and carefully combining these totals, we can solve for the unknown score. Given Data / Assumptions: • Subjects: English (E), Maths (M), Hindi (H), Drawing (D), Science (Sc), Social Studies (SS), Craft (C).• Average of E, M, H, D = 50.• Average of M, Sc, SS, C = 70.• Average of all seven subjects = 58.• All averages are computed over John's marks in those subjects. Concept / Approach: The plan is to convert each average into a total sum. The total of all seven subjects equals the sum of the first four subjects plus the sum of the last three subjects. Since one of the groups with average 70 includes Maths, we express the last three subjects in terms of M and substitute into the total for all seven. This yields an equation in M which we solve to find the Maths marks. Step-by-Step Solution: Step 1: From the first average, E + M + H + D = 4 * 50 = 200.Step 2: From the second average, M + Sc + SS + C = 4 * 70 = 280.Step 3: Let T be the total marks in all seven subjects. From the overall average, T = 7 * 58 = 406.Step 4: T can also be written as (E + M + H + D) + (Sc + SS + C).Step 5: From Step 1, E + M + H + D = 200.Step 6: From Step 2, Sc + SS + C = 280 - M.Step 7: Substitute into T: T = 200 + (280 - M) = 480 - M.Step 8: Using T = 406, we have 480 - M = 406.Step 9: Rearranging, M = 480 - 406 = 74. Verification / Alternative check: With M = 74, we have Sc + SS + C = 280 - 74 = 206. The total of all seven subjects is then 200 + 206 = 406. The overall average is 406 / 7 = 58, which matches the given condition. All relationships are satisfied, confirming that John scored 74 in Maths. Why Other Options Are Wrong: 50, 52, 60: Substituting any of these into the equation T = 480 - M gives totals that do not equal 406. The resulting overall average would not be 58, so these values violate the given conditions. Common Pitfalls: Students may incorrectly assume that the Maths marks are the simple average of the two given group averages or may forget to subtract M properly when expressing Sc + SS + C. Carefully keeping track of the overlapping subject (Maths) and using algebra to express totals helps avoid such mistakes. Final Answer: John's marks in Maths were 74.

Question 2

Mahesh buys 3 shirts at an average price of Rs 1250 per shirt. He then buys 2 more shirts at an average price of Rs 1450 per shirt. What is the average price (in Rs) of all 5 shirts taken together?

Accepted Answer

Introduction / Context: This question is about combining two groups with different averages to find a combined average. Mahesh buys some shirts at one average price and then buys more shirts at a different average price. You must determine the overall average price per shirt for all of his purchases. Given Data / Assumptions: • Number of shirts in the first purchase = 3.• Average price in the first purchase = Rs 1250 per shirt.• Number of shirts in the second purchase = 2.• Average price in the second purchase = Rs 1450 per shirt.• All prices are in rupees. Concept / Approach: The combined average is found by calculating the total cost of all shirts and dividing by the total number of shirts. We first convert each group's average into a total cost, then sum these totals, and finally divide by the total quantity to get the required average. Step-by-Step Solution: Step 1: Total cost of the first 3 shirts = 3 * 1250 = Rs 3750.Step 2: Total cost of the next 2 shirts = 2 * 1450 = Rs 2900.Step 3: Total cost of all 5 shirts = 3750 + 2900 = Rs 6650.Step 4: Total number of shirts = 3 + 2 = 5.Step 5: Combined average price = total cost / total number of shirts = 6650 / 5.Step 6: 6650 / 5 = Rs 1330. Verification / Alternative check: Note that the second group of shirts is more expensive than the first, so the combined average must be between 1250 and 1450, but closer to 1250 because there are more shirts from the cheaper batch. The calculated value of 1330 lies in this reasonable range and therefore makes sense. Why Other Options Are Wrong: 1370, 1390: These are too high given that only 2 out of 5 shirts are from the more expensive batch.1310: This is lower than the balanced value determined from the exact ratio of cheaper and costlier shirts. Common Pitfalls: Students may incorrectly average the two given averages directly without weighting by quantity, for example taking (1250 + 1450) / 2. This approach ignores that the number of shirts in each group is different. Always convert to totals first and then divide by the total quantity. Final Answer: The average price of all 5 shirts is Rs 1330.

Question 3

In a one day cricket match of 50 overs, Team A bats first and scores at a run rate of 6.1 runs per over, so that its total is fixed. Team B is now batting. With 10 overs left to play, the required run rate for Team B to tie the match is 6.5 runs per…

Accepted Answer

Introduction / Context: This question applies average rate concepts to cricket scoring. Team A has already set a target with a known run rate. Team B is partway through its innings, and you are given the required run rate for the remaining overs to tie the match. From this information, you need to determine Team B's current score. Given Data / Assumptions: • Total overs in the innings = 50.• Team A's run rate = 6.1 runs per over for all 50 overs.• Team B is chasing Team A's total.• There are 10 overs left for Team B, so they have already played 40 overs.• Required run rate for the remaining 10 overs to tie the match = 6.5 runs per over.• We assume no extra balls or interruptions; overs are full and complete. Concept / Approach: First, we calculate Team A's total runs from its run rate and overs. This total is also the target Team B must reach to tie. Then, using the required run rate and remaining overs, we compute how many runs Team B still needs. Subtracting this from the target total gives the current score of Team B. Step-by-Step Solution: Step 1: Team A's total runs = run rate * overs = 6.1 * 50.Step 2: Compute 6.1 * 50 = 305 runs.Step 3: To tie the match, Team B must score exactly 305 runs.Step 4: There are 10 overs left, and the required run rate is 6.5 runs per over.Step 5: Runs needed in remaining 10 overs = 6.5 * 10 = 65 runs.Step 6: Let S be Team B's current score.Step 7: To tie, S + 65 must equal 305.Step 8: So S = 305 - 65 = 240. Verification / Alternative check: If Team B currently has 240 runs, they need 305 - 240 = 65 more runs. With 10 overs remaining, scoring 65 runs means an exact run rate of 65 / 10 = 6.5 runs per over, which matches the required run rate stated in the question. Thus the value S = 240 is fully consistent. Why Other Options Are Wrong: 235: This would require 70 runs in 10 overs, implying a required run rate of 7.0, not 6.5.230: This would lead to a required run rate of 7.5, again inconsistent with the given value.225: This would require 80 runs in 10 overs, an even higher required rate of 8.0 runs per over. Common Pitfalls: Some students incorrectly compute Team A's total or misinterpret the required run rate as the current run rate. Others might forget that the required rate applies only to the remaining overs, not to the entire innings. Carefully separating completed overs from remaining overs and applying the definitions of run rate prevents such mistakes. Final Answer: Team B's current score is 240 runs.

Question 4

A batsman scores 98 runs in the 17th match of his career, and as a result his average runs per match increases by 2.5 runs. What was his average runs per match before playing this 17th match?

Accepted Answer

Introduction / Context: This problem tests the concept of average in the context of cricket statistics. We are given the score in a particular match and the change in the overall average after that match. From this, we need to work backwards to find the original average runs per match before the latest innings. Such questions are common in aptitude exams and help in understanding how averages behave when a new value is added. Given Data / Assumptions: Total matches before the latest innings = 16 Runs scored in the 17th match = 98 Increase in average after the 17th match = 2.5 runs Let the old average before the 17th match be a runs per match Concept / Approach: The key idea is that: average = total runs / number of matches. Before the 17th match, the batsman has 16a total runs. After scoring 98 runs in the 17th match, his total runs become 16a + 98 and the number of matches becomes 17. We are told that the new average is 2.5 runs more than the old average, so we can form an equation involving a and solve it. Step-by-Step Solution: Let the old average = a runs per match. Total runs before 17th match = 16 * a. After 17th match, total runs = 16 * a + 98. Number of matches after 17th match = 17. New average = (16 * a + 98) / 17. Given that new average = old average + 2.5, so: (16 * a + 98) / 17 = a + 2.5. Multiply both sides by 17: 16 * a + 98 = 17 * a + 42.5. Rearrange: 98 - 42.5 = 17 * a - 16 * a. 55.5 = a. Therefore, the old average = 55.5 runs per match. Verification / Alternative check: Total runs before 17th match = 16 * 55.5 = 888 runs. After scoring 98 runs, total = 888 + 98 = 986 runs. New average = 986 / 17 = 58 runs per match. Increase in average = 58 - 55.5 = 2.5 runs, which matches the given condition. Why Other Options Are Wrong: Option A (58): This is actually the new average after the 17th match, not the old one. Option B (60.5): Gives a different increase in average and does not satisfy the equation. Option C (63): Also fails when substituted, leading to an incorrect new average. Option D (55.5): Correct, as it satisfies all the given conditions. Common Pitfalls: A common mistake is to directly average 98 with some guessed old average instead of using the total runs approach. Another error is to assume that the increase in average is equal to the difference between 98 and the old average, which is not true because the average involves many matches. Students sometimes miscount the number of matches before and after the innings, leading to a wrong equation. Final Answer: The batsman's average runs per match before the 17th match was 55.5 runs.

Question 5

What is the average of all integers between 100 and 200 (both limits exclusive) that are exactly divisible by 13?

Accepted Answer

Introduction / Context: This question checks understanding of averages of terms in an arithmetic progression, specifically numbers divisible by a given integer within a certain range. Instead of listing every number and summing them, we can use properties of arithmetic sequences to work efficiently, which is crucial in competitive exams. Given Data / Assumptions: Range of interest: greater than or equal to 100 and less than or equal to 200. We need numbers divisible by 13. We seek the average of all such numbers. Average of an arithmetic sequence = (first term + last term) / 2. Concept / Approach: Numbers divisible by 13 form an arithmetic progression with common difference 13. Within the given range, we find the smallest and largest multiples of 13 that lie between 100 and 200. Once we know the first and last terms of this sequence, the average of all such terms is simply the mean of these two extremes, because the terms are equally spaced. Step-by-Step Solution: Find smallest multiple of 13 greater than or equal to 100. 13 * 7 = 91 (too small), 13 * 8 = 104 (acceptable). So, first term = 104. Find largest multiple of 13 less than or equal to 200. 13 * 15 = 195 (acceptable), 13 * 16 = 208 (too large). So, last term = 195. Average of all terms in this arithmetic sequence = (first term + last term) / 2. Average = (104 + 195) / 2 = 299 / 2 = 149.5. Verification / Alternative check: List the multiples: 104, 117, 130, 143, 156, 169, 182, 195. There are 8 numbers. Sum = (104 + 195) * 8 / 2 = 299 * 4 = 1196. Average = 1196 / 8 = 149.5, which confirms the earlier result. Why Other Options Are Wrong: Option A (147.5): Slightly less than the correct value, possibly from arithmetic error. Option B (145.5): Does not match the mean of the first and last multiples. Option C (143.5): Too far from the center of the range of multiples. Option D (149.5): Correct average based on the arithmetic progression formula. Common Pitfalls: Students sometimes include 91 or 208 by mistake, which lie outside the given interval. Another error is averaging 100 and 200 directly instead of focusing on numbers divisible by 13. Forgetting that the average of equally spaced numbers equals the mean of the first and last terms can also lead to a long and error prone approach. Final Answer: The required average of all numbers between 100 and 200 divisible by 13 is 149.5.

Question 6

The average marks of 18 students in an examination was first calculated as 60, but it was later found that one student's marks had been wrongly entered as 63 instead of 36. What is the correct average marks of the class after this correction?

Accepted Answer

Introduction / Context: This problem focuses on correcting an average when one of the data values was entered wrongly. Such questions test your understanding of how a single incorrect entry affects the overall mean and how to adjust the total sum and recompute the average once the error is identified. Given Data / Assumptions: Number of students = 18. Originally calculated average = 60 marks. One student's marks were entered as 63 instead of the correct value 36. We need to find the corrected average after replacing 63 with 36. Concept / Approach: Average is defined as total sum of observations divided by the number of observations. First, we compute the incorrect total using the wrong average. Then we adjust this total by removing the wrong value and adding the correct one. Finally, we divide the corrected total by the same number of students to obtain the correct average. Step-by-Step Solution: Incorrect average = 60 marks. Number of students = 18. Incorrect total marks = 60 * 18 = 1080. Wrong entry used = 63 marks. Correct value should be = 36 marks. Corrected total marks = 1080 - 63 + 36. Corrected total marks = 1080 - 27 = 1053. Correct average = corrected total / number of students. Correct average = 1053 / 18 = 58.5. Verification / Alternative check: Difference between wrong and correct marks = 63 - 36 = 27. So the original total was 27 marks too high. Thus, average must decrease by 27 / 18 = 1.5 marks. Original average = 60, so correct average = 60 - 1.5 = 58.5, which matches the earlier result. Why Other Options Are Wrong: Option A (59): Assumes a smaller correction than required. Option B (59.5): Does not match the exact decrease needed from the wrong entry. Option C (58): Overshoots the necessary reduction in average. Option D (58.5): Exactly reflects the corrected total and is therefore correct. Common Pitfalls: A common mistake is to adjust the average directly without first correcting the total. Some students forget to subtract the wrong entry before adding the correct one, effectively double counting. Another error is to divide the change in marks by the wrong number of students, which gives an incorrect adjustment to the average. Final Answer: The correct average marks of the class after correction is 58.5.

Question 7

The average of 44 consecutive odd numbers is 144. What is the largest odd number in this list of 44 consecutive odd integers?

Accepted Answer

Introduction / Context: Here we are dealing with consecutive odd numbers, which form an arithmetic progression with a common difference of 2. The question gives us the average of 44 such numbers and asks for the largest number. This tests understanding of averages and the structure of arithmetic sequences, especially when the count of terms is even. Given Data / Assumptions: There are 44 consecutive odd numbers. Their average is 144. We need to find the largest number in this sequence. The numbers are in increasing order and differ by 2 between consecutive terms. Concept / Approach: For any arithmetic sequence, the average is equal to the value of the middle term (if the number of terms is odd) or the average of the two middle terms (if the number of terms is even). Here, 44 is an even number, so the mean is the average of the 22nd and 23rd terms. Let the smallest odd number be a. Then the nth term is a + (n - 1) * 2. We use the mean relation to link the first and last terms. Step-by-Step Solution: Let the first term (smallest odd number) be a. Number of terms n = 44. Common difference d = 2. Last term L = a + (44 - 1) * 2 = a + 86. Average of an arithmetic sequence = (first term + last term) / 2. Given average = 144, so: (a + (a + 86)) / 2 = 144. (2a + 86) / 2 = 144. 2a + 86 = 288. 2a = 202. a = 101. Therefore, largest term L = a + 86 = 101 + 86 = 187. Verification / Alternative check: First term = 101, last term = 187. Check average: (101 + 187) / 2 = 288 / 2 = 144, which matches the given value. Thus our computed largest number is consistent. Why Other Options Are Wrong: Option A (189): Would give average (a + 189) / 2 greater than 144, contradicting the information. Option B (191): Even further from the required mean. Option C (187): Correct largest term obtained from the arithmetic sequence formula. Option D (193): Too large and incompatible with average 144. Common Pitfalls: Some students mistakenly assume that the given average is itself one of the sequence terms, which is not always true when there is an even number of terms. Others may forget that the common difference is 2 for odd numbers and use incorrect expressions for the last term. Errors also occur if the formula for the mean of an arithmetic sequence is not applied correctly. Final Answer: The largest odd number in the list is 187.

Question 8

A batsman scores 100 runs in the 25th match of his career, which increases his average runs per match by 1.4 runs. What was his average runs per match before playing this 25th match?

Accepted Answer

Introduction / Context: This question again uses the concept of average in a sports setting. By knowing the score in a particular match and the change in the overall average, we can work backwards to determine the previous average performance. This is a standard pattern in aptitude tests and helps strengthen understanding of averages and algebraic equations. Given Data / Assumptions: Before the 25th match, the batsman had already played 24 matches. Runs scored in the 25th match = 100. Average runs per match increases by 1.4 after this match. Let the old average before 25th match be a runs per match. Concept / Approach: Average is total runs divided by number of matches. Before the 25th match, the total runs are 24 * a. After scoring 100 runs, total runs become 24 * a + 100 and the number of matches becomes 25. This new average is given to be a + 1.4. We form an equation based on this relationship and solve for a. Step-by-Step Solution: Let old average = a runs per match. Total runs before 25th match = 24 * a. After 25th match, total runs = 24 * a + 100. Number of matches after 25th match = 25. New average = (24 * a + 100) / 25. Given new average = a + 1.4. So, (24 * a + 100) / 25 = a + 1.4. Multiply both sides by 25: 24 * a + 100 = 25 * a + 35. Rearrange terms: 24 * a + 100 - 35 = 25 * a. 24 * a + 65 = 25 * a. 65 = a. Therefore, old average = 65 runs per match. Verification / Alternative check: Total runs before 25th match = 24 * 65 = 1560. After scoring 100, total runs = 1560 + 100 = 1660. New average = 1660 / 25 = 66.4 runs per match. Increase in average = 66.4 - 65 = 1.4 runs, which matches the given condition. Why Other Options Are Wrong: Option B (55): Leads to a different new average that does not match the given increase. Option C (75): Produces an incorrect change in average when used in the equation. Option D (45): Far from the actual value and does not satisfy the relation. Option A (65): Correct, since it satisfies the equation and the given increase in average. Common Pitfalls: Confusing the number of matches before and after the innings can lead to wrong equations. Some students attempt to simply average 100 with a guess for the previous average, which is incorrect. Others may forget that the increase in average applies to all matches, not just the latest one. Final Answer: The batsman's average runs per match before the 25th match was 65 runs.

Question 9

A factory buys 10 machines in total: 2 of type A, 3 of type B and the remaining 5 of type C. The prices of machines A, B and C are Rs 95,000, Rs 60,000 and Rs 50,000 respectively. What is the average cost (in rupees) of all these machines together?

Accepted Answer

Introduction / Context: This question focuses on finding the average cost of a group of items when different quantities of each type are purchased at different prices. It is a weighted average problem, where we multiply each price by the number of items of that type, add these products, and then divide by the total number of items. Given Data / Assumptions: Total number of machines = 10. Number of type A machines = 2, price per machine = Rs 95,000. Number of type B machines = 3, price per machine = Rs 60,000. Remaining machines are type C, so number of type C machines = 10 - 2 - 3 = 5, price per machine = Rs 50,000. Required: average cost of all 10 machines. Concept / Approach: Average cost is computed as total cost divided by total number of machines. Since each type has a different cost and count, we compute the total cost by summing individual costs: (number of A * price of A) + (number of B * price of B) + (number of C * price of C). Then we divide this total by 10 to obtain the average cost per machine. Step-by-Step Solution: Total cost of type A machines = 2 * 95000 = Rs 190000. Total cost of type B machines = 3 * 60000 = Rs 180000. Total cost of type C machines = 5 * 50000 = Rs 250000. Total cost of all machines = 190000 + 180000 + 250000. Total cost = Rs 620000. Total number of machines = 10. Average cost per machine = total cost / total number. Average cost = 620000 / 10 = Rs 62000. Verification / Alternative check: Compute approximate average by observation: prices range from Rs 50000 to Rs 95000. Most machines (3 + 5 = 8 machines) cost Rs 60000 or Rs 50000, while only 2 machines cost Rs 95000. So the overall average must be somewhat above Rs 55000 but clearly less than Rs 70000. The value Rs 62000 fits well in this reasonable range, supporting our detailed calculation. Why Other Options Are Wrong: Option B (68333): Too high given that only 2 machines are costly; it overestimates total cost. Option C (74666): Even higher and impossible given the mixture of cheaper machines. Option D (60500): Closer but still not equal to the exact computed average of 62000. Option A (62000): Matches the correct calculation of total cost divided by number of machines. Common Pitfalls: Some students try to simply average the three prices without considering the different quantities of each type. Others may miscount the number of type C machines by not subtracting correctly. Calculation slips in multiplying or adding large numbers may also lead to incorrect answers. Final Answer: The average cost of all 10 machines is Rs 62000.

Question 10

A straight line cuts the x-axis at the point (3, 0) and the y-axis at the point (0, 6). What is the equation of this line in the form y = mx + c?

Accepted Answer

Introduction / Context: This problem tests understanding of the equation of a straight line given its intercepts on the coordinate axes. The line is specified by two points, one on the x-axis and one on the y-axis, and we are required to determine its equation, ideally in slope intercept form y = m x + c. Given Data / Assumptions: The line passes through point A(3, 0) on the x-axis. The line also passes through point B(0, 6) on the y-axis. We assume a standard Cartesian coordinate system. We need to find the equation of the line in the form y = m x + c. Concept / Approach: The equation of a line through two given points can be found using the slope formula. First, we calculate the slope m = (y2 - y1) / (x2 - x1). Then we substitute one of the points into the equation y = m x + c to solve for the y intercept c. Finally, we write the equation in the required slope intercept form and compare with the options. Step-by-Step Solution: Let A(3, 0) and B(0, 6) be the two points. Compute the slope m of the line: m = (6 - 0) / (0 - 3). m = 6 / -3 = -2. So the line has the form y = -2x + c. Use point B(0, 6) to find c: Substitute x = 0, y = 6 into y = -2x + c. 6 = -2 * 0 + c, so c = 6. Thus, the equation of the line is y = -2x + 6. Verification / Alternative check: Check with point A(3, 0): y = -2 * 3 + 6 = -6 + 6 = 0, which matches the given point. Check with point B(0, 6): y = -2 * 0 + 6 = 6, which again matches. Therefore, y = -2x + 6 is consistent with both points. Why Other Options Are Wrong: Option B (y = 2x - 6): Has positive slope. Through x = 3, y = 0 would give 0 = 6 - 6 = 0, but for x = 0, y = -6, not 6. Option C (y = 2x + 6): Intercepts and slope are incorrect; it does not pass through (3, 0). Option D (y = -2x - 6): Gives y intercept -6 instead of 6, so it fails to pass through (0, 6). Option A (y = -2x + 6): Correct equation, passes through both given points. Common Pitfalls: A frequent error is to swap x and y changes when computing the slope, changing its sign. Some students also mix up the intercepts and write equations like x = 2y + 6, which represents a different line. Another mistake is to assume c equals the x intercept, whereas in y = m x + c, c is the y intercept. Final Answer: The correct equation of the line is y = -2x + 6.

Question 11

The average weight of P, Q and R together is 71 kg. The average weight of P and Q is 66 kg, while the average weight of Q and R is 76.5 kg. What is the weight (in kg) of Q?

Accepted Answer

Introduction / Context: This question uses averages of overlapping groups to determine individual values. Three people P, Q and R have certain average weights in different pairs and as a whole. From these overlapping averages, we need to deduce the exact weight of one person, Q. This type of problem relies on forming and solving simple linear equations. Given Data / Assumptions: Average weight of P, Q and R together = 71 kg. Average weight of P and Q = 66 kg. Average weight of Q and R = 76.5 kg. We are asked to find the weight of Q. Concept / Approach: If the average of some people is known, we can multiply by the number of people to get the total weight. Let the weights of P, Q and R be p, q and r respectively. From the given averages, we form three equations: p + q + r, p + q and q + r. These equations can be combined to eliminate p and r, leaving us with q. This is a classic system of linear equations problem. Step-by-Step Solution: Let weights be: P = p, Q = q, R = r. Average of P, Q, R = 71 kg. So, p + q + r = 71 * 3 = 213. (1) Average of P and Q = 66 kg. So, p + q = 66 * 2 = 132. (2) Average of Q and R = 76.5 kg. So, q + r = 76.5 * 2 = 153. (3) Add equations (2) and (3): (p + q) + (q + r) = 132 + 153. p + 2q + r = 285. (4) From (1), p + q + r = 213. Subtract (1) from (4): (p + 2q + r) - (p + q + r) = 285 - 213. q = 72. Verification / Alternative check: If q = 72, then from (2): p + 72 = 132, so p = 60. From (3): 72 + r = 153, so r = 81. Check total: p + q + r = 60 + 72 + 81 = 213. Average = 213 / 3 = 71, which matches the given value. Why Other Options Are Wrong: Option A (60): This is actually the weight of P, not Q. Option C (81): This is the weight of R after solving, not Q. Option D (75): Does not satisfy all three average conditions. Option B (72): Correct, as shown by the equations and verification. Common Pitfalls: Students may confuse which person corresponds to which equation and misassign values. Some try to average the given averages directly rather than forming proper equations. Arithmetic mistakes in multiplying averages by the number of persons can also lead to wrong totals and thus wrong answers. Final Answer: The weight of Q is 72 kg.

Question 12

Rita buys 5 sarees at an average cost of Rs 2250 each. She then buys 3 more sarees at an average cost of Rs 2750 each. What is the overall average cost (in rupees) of all 8 sarees that she buys?

Accepted Answer

Introduction / Context: This question involves combining two groups with different averages to find a new overall average. Rita buys sarees in two batches at different average prices. We need to compute the total cost from these averages and then determine the combined average. This is a straightforward weighted average problem. Given Data / Assumptions: First batch: 5 sarees, average cost = Rs 2250 each. Second batch: 3 sarees, average cost = Rs 2750 each. Total sarees purchased = 5 + 3 = 8. We need the overall average cost per saree. Concept / Approach: Average cost is defined as total cost divided by total number of items. When averages for subgroups are known, we can recover the total cost of each group by multiplying the group average by the group size. Then we sum these totals and divide by the total number of sarees to get the combined average. Step-by-Step Solution: First batch: number of sarees = 5, average cost = Rs 2250. Total cost of first batch = 5 * 2250 = Rs 11250. Second batch: number of sarees = 3, average cost = Rs 2750. Total cost of second batch = 3 * 2750 = Rs 8250. Total number of sarees = 5 + 3 = 8. Total cost of all sarees = 11250 + 8250 = Rs 19500. Overall average cost = total cost / total number. Overall average = 19500 / 8 = Rs 2437.5. Verification / Alternative check: We can approximate: since most sarees are bought at Rs 2250 and some at Rs 2750, the final average should lie between these two values. 2437.5 is indeed between 2250 and 2750, and closer to 2250, which is reasonable because more sarees were bought at the lower price. Why Other Options Are Wrong: Option B (2500): Slightly higher than the correct average, ignoring the fact that more sarees were bought at the cheaper price. Option C (2450): Close but does not match the exact calculation. Option D (2332.5): Too low and does not reflect the effect of higher cost sarees. Option A (2437.5): Matches the exact computed overall average, so it is correct. Common Pitfalls: A common error is to simply take the average of 2250 and 2750 without considering how many sarees were bought in each batch. Some students mistakenly divide the sum of the two averages by 2, which ignores the different group sizes. Calculation mistakes when multiplying or adding large numbers can also cause wrong answers. Final Answer: The overall average cost of all 8 sarees is Rs 2437.5.

Question 13

In a one day cricket match of 50 overs, Team A scores at a run rate of 5.3 runs per over for their entire innings. Team B is chasing this target. With 5 overs remaining, Team B needs a run rate of 7.2 runs per over to tie the match. What is Team B's…

Accepted Answer

Introduction / Context: This problem involves cricket scoring and run rates. Team A has already completed its innings, and Team B is chasing the target. We are given Team A's run rate and the required run rate for Team B with a certain number of overs left to tie the match. From this, we need to determine Team B's current score. This is a direct application of rate and total concepts. Given Data / Assumptions: Total overs in the match for each team = 50. Team A's run rate = 5.3 runs per over. Team B has 5 overs remaining. Required run rate for Team B to tie = 7.2 runs per over for the last 5 overs. We assume Team B has played 45 overs already (50 - 5). Concept / Approach: First, we find the total score made by Team A using run rate * overs. That score is the target Team B must match to tie. Next, we use the required run rate for the remaining overs to figure out how many more runs Team B needs. Subtracting these required runs from the target gives Team B's current score after 45 overs. Step-by-Step Solution: Team A's total score = run rate * overs. Team A's total = 5.3 * 50 = 265 runs. To tie, Team B must also score 265 runs. Remaining overs for Team B = 5. Required run rate for last 5 overs = 7.2 runs per over. Runs needed in last 5 overs = 7.2 * 5 = 36 runs. Therefore, current score of Team B = target - runs still needed. Current score = 265 - 36 = 229 runs. Verification / Alternative check: If Team B is on 229 after 45 overs, then to reach 265, they need 36 more runs. Required run rate = 36 / 5 = 7.2 runs per over, which matches the condition given in the question. Why Other Options Are Wrong: Option A (265): This would mean Team B already tied the match, so required run rate would be 0, not 7.2. Option B (238): Would make remaining runs = 27, giving a required run rate of 27 / 5 = 5.4, not 7.2. Option C (254): Would leave only 11 runs, resulting in a required rate of 11 / 5 = 2.2, which is incorrect. Option D (229): Correct, since it leads to a required rate of 7.2 runs per over to reach 265. Common Pitfalls: Students may mistakenly treat 7.2 as the average for the whole innings instead of just the last 5 overs. Another common mistake is to add 7.2 * 5 to Team A's score instead of subtracting it. Some may forget that the run rate given for Team A applies to all 50 overs and must be used to compute total runs first. Final Answer: Team B's score at that point is 229 runs.

Question 14

Find the average of all even numbers between 104 and 148 (both inclusive). These even integers form a simple arithmetic sequence. What is their average value?

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Introduction / Context: This question asks for the average of even numbers within a given inclusive range. Since consecutive even numbers differ by 2, they form an arithmetic sequence. For such a sequence, we can use a shortcut to find the average rather than individually listing or summing all terms. Given Data / Assumptions: Range considered: from 104 to 148 inclusive. Only even numbers are taken. 104 and 148 are both even, so they are part of the sequence. We need the average of all these even numbers. Concept / Approach: Consecutive even numbers form an arithmetic progression with common difference 2. For any arithmetic sequence, the average is equal to (first term + last term) / 2. This is true regardless of how many terms there are, as long as the spacing between terms is uniform. So we simply identify the first even number and the last even number in the range and compute their mean. Step-by-Step Solution: First even number in the range = 104. Last even number in the range = 148. These numbers form an arithmetic sequence: 104, 106, 108, ..., 148. Average of an arithmetic sequence = (first term + last term) / 2. Average = (104 + 148) / 2. Average = 252 / 2 = 126. Verification / Alternative check: Observe that the even numbers are symmetric around their average. Pairs like (104, 148), (106, 146), (108, 144) all sum to 252. Since each pair has the same sum and there are equally many such pairs, the overall average must be 252 / 2 = 126, confirming the result. Why Other Options Are Wrong: Option A (128): Slightly higher than the correct mean; may arise if someone miscomputes the sum. Option B (130): Too high and not equal to the midpoint of 104 and 148. Option C (124): Too low and again not matching the midpoint. Option D (126): Correct, since it is exactly the average of the first and last even numbers. Common Pitfalls: Some students mistakenly average 104 and 148 with 100 and 150 or perform incorrect arithmetic. Others may think they must list all even numbers and individually sum them, which is time consuming and error prone. Forgetting that the average of an arithmetic sequence is simply the mean of the first and last terms is another common oversight. Final Answer: The average of all even numbers between 104 and 148 inclusive is 126.

Question 15

The average of 6 consecutive integers is 15/2 (that is, 7.5). If these six integers are in increasing order, what is the average of the last three integers in this sequence?

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Introduction / Context: This question involves consecutive integers and their averages. We are given the average of 6 consecutive integers and asked to find the average of the last three numbers. Consecutive integers form an arithmetic progression with common difference 1, which allows us to represent them algebraically and compute required averages quickly. Given Data / Assumptions: There are 6 consecutive integers. The average of these 6 integers is 15/2 = 7.5. We need the average of the last three integers in this ordered list. Concept / Approach: Let the six consecutive integers be represented in terms of a variable. Because they are consecutive, they can be written as n, n + 1, n + 2, n + 3, n + 4, n + 5. Their average will be the middle value between the third and fourth integer, which is n + 2.5. Setting this equal to the given average allows us to solve for n and thereby obtain all six integers. Then we directly compute the average of the last three integers. Step-by-Step Solution: Let the six consecutive integers be: n, n + 1, n + 2, n + 3, n + 4, n + 5. Average of these integers = (n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + 5)) / 6. Sum = 6n + 15, so average = (6n + 15) / 6 = n + 2.5. Given average = 15/2 = 7.5. So, n + 2.5 = 7.5. n = 7.5 - 2.5 = 5. Therefore, the six integers are: 5, 6, 7, 8, 9, 10. The last three integers are: 8, 9, 10. Average of last three integers = (8 + 9 + 10) / 3 = 27 / 3 = 9. Verification / Alternative check: Check the original average: (5 + 6 + 7 + 8 + 9 + 10) / 6. Sum = 45, average = 45 / 6 = 7.5 = 15/2, which confirms our list of integers. Thus, the computed average of last three integers is consistent with the given condition. Why Other Options Are Wrong: Option A (8): This would be the middle of 7, 8, 9 and not of 8, 9, 10. Option C (17/2 = 8.5): Does not match the average of 8, 9 and 10. Option D (7): Too small and corresponds to the lower part of the sequence. Option B (9): Correct, as it equals the average of the last three integers 8, 9 and 10. Common Pitfalls: Some students mistakenly think the given average is one of the integers in the sequence, which is not true here because the average is 7.5, a non integer. Others may forget that six numbers will not have a single middle integer but rather a pair of central values. Miswriting the consecutive integers or making arithmetic errors while adding them can also lead to incorrect answers. Final Answer: The average of the last three integers is 9.

Question 16

The average revenues of a company over 9 consecutive years is Rs 80 lakhs. The average revenue for the first 5 years is Rs 75 lakhs, and the average revenue for the last 5 years is Rs 87 lakhs. What is the revenue (in lakhs of rupees) in the 5th yea…

Accepted Answer

Introduction / Context: This question uses averages of overlapping time periods to find the revenue in a specific year. By knowing the overall average for all 9 years, as well as averages for the first 5 and last 5 years, we can set up equations that allow us to isolate the revenue for the 5th year, which lies in the overlap between these periods. Given Data / Assumptions: Total number of years = 9. Average revenue for all 9 years = Rs 80 lakhs. Average revenue for first 5 years = Rs 75 lakhs. Average revenue for last 5 years = Rs 87 lakhs. The 5th year is common to both the first 5 and the last 5 years. We need to find the revenue in the 5th year. Concept / Approach: Let the revenues for the 9 years be R1, R2, R3, R4, R5, R6, R7, R8 and R9. We can express sums of certain groups using the given averages. The sum of all 9 years comes from the overall average. The sum of the first 5 years and the last 5 years are found from their respective averages. Since the 5th year appears in both these sums, adding them together will count R5 twice. We then use this overlap to extract R5. Step-by-Step Solution: Total revenue for 9 years = 80 * 9 = Rs 720 lakhs. Sum of first 5 years (R1 to R5) = 75 * 5 = Rs 375 lakhs. Sum of last 5 years (R5 to R9) = 87 * 5 = Rs 435 lakhs. Add these two sums: (R1 + R2 + R3 + R4 + R5) + (R5 + R6 + R7 + R8 + R9). This equals Rs 375 lakhs + Rs 435 lakhs = Rs 810 lakhs. But R1 + R2 + R3 + R4 + R5 + R6 + R7 + R8 + R9 = Rs 720 lakhs (total of 9 years), and R5 is added one extra time. So, Rs 810 lakhs = Rs 720 lakhs + R5. Therefore, R5 = 810 - 720 = Rs 90 lakhs. Verification / Alternative check: If R5 = Rs 90 lakhs, then sum of first 5 years is Rs 375 lakhs, so R1 + R2 + R3 + R4 = 375 - 90 = Rs 285 lakhs. Sum of last 5 years is Rs 435 lakhs, so R6 + R7 + R8 + R9 = 435 - 90 = Rs 345 lakhs. Total for 9 years = 285 + 90 + 345 = Rs 720 lakhs, which gives average 720 / 9 = Rs 80 lakhs, matching the data. Why Other Options Are Wrong: Option B (Rs 92 lakhs): Makes the combined sums inconsistent with the overall average. Option C (Rs 88 lakhs): Too low to reconcile both given averages. Option D (Rs 86 lakhs): Also cannot satisfy both the first 5 year and last 5 year averages simultaneously. Option A (Rs 90 lakhs): Satisfies all given conditions and is therefore correct. Common Pitfalls: Students may forget that the 5th year is counted twice when adding the sums of first 5 and last 5 years. Another common error is to mistakenly assume that the 5th year revenue is simply the average of 75 and 87, which is not correct. Some may also miscalculate the total revenue for 9 years, leading to a wrong final answer. Final Answer: The revenue in the 5th year is Rs 90 lakhs.

Question 17

The average of 31 numbers is 19. The average of the first 8 of these numbers is 24. What is the average of the remaining 23 numbers?

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Introduction / Context: This question involves splitting a data set into two parts and using averages to find the mean of the remaining subset. We are given the overall average of all 31 numbers and the average of the first 8 numbers. Using these, we must determine the average of the remaining 23 numbers. Given Data / Assumptions: Total number of numbers = 31. Average of all 31 numbers = 19. Average of first 8 numbers = 24. We need the average of the remaining 23 numbers. Concept / Approach: The basic idea is to convert averages into sums, because total sum = average * number of terms. We first compute the total sum of all 31 numbers using the overall average. Then we compute the sum of the first 8 numbers using their average. Subtracting this from the total gives the sum of the remaining 23 numbers. Dividing that sum by 23 gives the required average. Step-by-Step Solution: Total number of numbers = 31. Average of all 31 numbers = 19. Total sum of all numbers = 31 * 19. Total sum = 589. Average of first 8 numbers = 24. Sum of first 8 numbers = 8 * 24 = 192. Sum of remaining 23 numbers = total sum - sum of first 8 numbers. Sum of remaining 23 numbers = 589 - 192 = 397. Average of remaining 23 numbers = 397 / 23. Average = 397 / 23 ≈ 17.26. Verification / Alternative check: We can check that the combined effect of averages is consistent. If the remaining 23 numbers have average about 17.26, their total is about 397. Adding back the first group: 192 + 397 = 589, which divided by 31 gives 19, matching the given overall average. Why Other Options Are Wrong: Option B (23.25) and option C (26.2) are too high. If those were correct, the overall average would be much greater than 19. Option D (25.45) is even larger and clearly inconsistent with the overall average of 19, given that the first 8 numbers already have an average of 24. Option A (17.26): Correct approximate value of 397 divided by 23. Common Pitfalls: Some students mistakenly average 19 and 24 directly instead of converting to sums. Others may incorrectly divide the sum of the remaining numbers by 31 instead of 23. Calculation errors while multiplying or subtracting can also lead to wrong totals and thus wrong averages. Final Answer: The average of the remaining 23 numbers is approximately 17.26.

Question 18

The average revenues of a company over 13 consecutive years is Rs 82 lakhs. The average of the first 7 years is Rs 77 lakhs and the average of the last 7 years is Rs 89 lakhs. What is the revenue (in lakhs) in the 7th year?

Accepted Answer

Introduction / Context: This question again uses overlapping time intervals with known averages to determine a specific year's revenue. The 7th year lies in the overlap of the first 7 years and the last 7 years (years 7 to 13). By converting averages to sums, we can use the overlap to isolate the revenue for the 7th year. Given Data / Assumptions: Number of consecutive years = 13. Average revenue for all 13 years = Rs 82 lakhs. Average revenue for first 7 years = Rs 77 lakhs. Average revenue for last 7 years = Rs 89 lakhs. Year 7 is common to both the first 7 and last 7 years. We need to find the revenue in the 7th year. Concept / Approach: Let the revenues for the 13 years be R1, R2, ..., R13. The overall average gives us the total of all 13 years. The average of the first 7 years provides the sum R1 + R2 + ... + R7. The average of the last 7 years gives the sum R7 + R8 + ... + R13. Adding these two sums counts R7 twice and also includes all other years once. Using this relationship, we can solve for R7. Step-by-Step Solution: Total revenue for 13 years = 82 * 13 = Rs 1066 lakhs. Sum of first 7 years (R1 to R7) = 77 * 7 = Rs 539 lakhs. Sum of last 7 years (R7 to R13) = 89 * 7 = Rs 623 lakhs. Add these two sums: (R1 + R2 + R3 + R4 + R5 + R6 + R7) + (R7 + R8 + R9 + R10 + R11 + R12 + R13) = 539 + 623 = Rs 1162 lakhs. The total revenue for 13 years is Rs 1066 lakhs, so On the left side, all 13 years are included once, and R7 is included one extra time. Therefore, Rs 1162 lakhs = Rs 1066 lakhs + R7. R7 = 1162 - 1066 = Rs 96 lakhs. Verification / Alternative check: If R7 = Rs 96 lakhs, then sum of first 7 years is 539, so sum of R1 to R6 = 539 - 96 = Rs 443 lakhs. Sum of last 7 years is 623, so sum of R8 to R13 = 623 - 96 = Rs 527 lakhs. Total for all 13 years = 443 + 96 + 527 = Rs 1066 lakhs. Average = 1066 / 13 = 82 lakhs, consistent with the given information. Why Other Options Are Wrong: Option A (Rs 98 lakhs): Would make combined sums exceed the total implied by the overall average. Option C (Rs 94 lakhs): Too low to satisfy both subset averages simultaneously. Option D (Rs 92 lakhs): Also inconsistent with the given averages. Option B (Rs 96 lakhs): Satisfies all three average conditions and is therefore correct. Common Pitfalls: Students sometimes forget that the 7th year is counted twice when the two partial sums are added. Another mistake is to average 77 and 89 and assume that gives the 7th year revenue, which is incorrect. Errors in computing total revenue for 13 years or in basic subtraction can also cause wrong answers. Final Answer: The revenue in the 7th year is Rs 96 lakhs.

Question 19

A factory buys 7 machines in total: 2 machines of type A, 4 machines of type B and the remaining 1 machine of type C. The prices of machines A, B and C are Rs 56,000, Rs 45,000 and Rs 30,000 respectively. What is the average cost (in rupees) of thes…

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Introduction / Context: This is another weighted average cost question. The factory buys machines of three different types at different prices and in different quantities. We need to compute the total cost and then find the average cost per machine. Given Data / Assumptions: Total machines purchased = 7. Type A: 2 machines, each costing Rs 56,000. Type B: 4 machines, each costing Rs 45,000. Type C: 1 machine, costing Rs 30,000. We have to find the average cost of these 7 machines. Concept / Approach: Average cost is defined as total cost divided by total number of items. Since there are different quantities and prices for machine types, we compute the total cost for each type by multiplying the quantity by the unit price and then sum them up. Dividing this total by 7 gives the average cost per machine. Step-by-Step Solution: Total cost of type A machines = 2 * 56000 = Rs 112000. Total cost of type B machines = 4 * 45000 = Rs 180000. Total cost of type C machines = 1 * 30000 = Rs 30000. Total cost of all machines = 112000 + 180000 + 30000. Total cost = Rs 322000. Total number of machines = 7. Average cost per machine = total cost / number of machines. Average cost = 322000 / 7 = Rs 46000. Verification / Alternative check: We can approximate: most machines are in the mid range (Rs 45000) and a few are slightly more expensive (Rs 56000) or cheaper (Rs 30000). So the average should be around the middle of this price range, and 46000 fits this expectation very well. Why Other Options Are Wrong: Option B (43666.66): Lower than the correct average and does not match the exact total cost when multiplied by 7. Option C (41333.33): Even lower and clearly underestimates the influence of the more expensive machines. Option D (44500): Close but still not equal to the correct exact average of 46000. Option A (46000): Exactly matches the calculated average, so it is correct. Common Pitfalls: Some students average the three prices directly without considering that there are different numbers of each machine type. Others may mistakenly think that the average of 56000 and 45000 must be taken first and then adjusted incorrectly. Arithmetic errors in multiplying or adding the costs may also lead to a wrong average. Final Answer: The average cost of the 7 machines is Rs 46000.

Question 20

The average revenues of a company over 13 consecutive years is Rs 78 lakhs. The average revenue of the first 7 years is Rs 73 lakhs and the average revenue of the last 7 years is Rs 85 lakhs. What is the revenue (in lakhs) in the 7th year?

Accepted Answer

Introduction / Context: This question is similar to other overlapping average problems. We are given the overall average for 13 consecutive years and separate averages for the first 7 years and the last 7 years. Since the 7th year appears in both the first and last group, we can use this overlap to find the revenue in that specific year. Given Data / Assumptions: Number of years = 13. Average revenue for all 13 years = Rs 78 lakhs. Average revenue for first 7 years = Rs 73 lakhs. Average revenue for last 7 years = Rs 85 lakhs. The 7th year is common to both sets of 7 years. We are required to find the revenue in the 7th year. Concept / Approach: Let the revenues be R1, R2, ..., R13. The overall average gives the total of all 13 years. The average of first 7 years gives R1 + ... + R7. The average of last 7 years gives R7 + ... + R13. Adding these two sums counts all 13 years once and the 7th year twice. We can therefore write an equation where the sum of first 7 years plus the sum of last 7 years equals the total for 13 years plus the 7th year revenue. Solving that equation gives us R7. Step-by-Step Solution: Total revenue for 13 years = 78 * 13 = Rs 1014 lakhs. Sum of first 7 years = 73 * 7 = Rs 511 lakhs. Sum of last 7 years = 85 * 7 = Rs 595 lakhs. Add these sums: 511 + 595 = Rs 1106 lakhs. This combined sum includes the revenues of all 13 years once, plus the 7th year one extra time. So, 1106 = 1014 + R7. R7 = 1106 - 1014 = Rs 92 lakhs. Verification / Alternative check: If R7 = Rs 92 lakhs, then sum of first 7 years is 511, so R1 + R2 + R3 + R4 + R5 + R6 = 511 - 92 = Rs 419 lakhs. Sum of last 7 years is 595, so R8 + R9 + R10 + R11 + R12 + R13 = 595 - 92 = Rs 503 lakhs. Total for all 13 years = 419 + 92 + 503 = Rs 1014 lakhs. Average = 1014 / 13 = 78 lakhs, which matches the overall average given. Why Other Options Are Wrong: Option A (Rs 94 lakhs): Makes the combined sums inconsistent with the overall total of 1014 lakhs. Option B (Rs 90 lakhs): Too low to satisfy both subset averages at the same time. Option C (Rs 88 lakhs): Also fails to keep both partial averages correct. Option D (Rs 92 lakhs): The only value consistent with all the given averages, so it is correct. Common Pitfalls: A common error is to simply take the average of 73 and 85 and treat that as the 7th year revenue, which is not valid. Some students forget that the combined sum of the first 7 and last 7 years counts the 7th year twice. Mistakes in basic multiplication or subtraction can also lead to incorrect values for R7. Final Answer: The revenue in the 7th year is Rs 92 lakhs.

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