Average – mcqpedia.in

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[Concept: Basic Average & Sum] The average of 5 numbers is 40. What is the sum of these 5 numbers?

Explanation: Book Formula: Sum of the observation = Average × total number of observations. Sum = 40 × 5 = 200.

[Concept: Basic Average & Sum] The average of 10 items is 25. If one item whose value is 40 is removed, what is the sum of the remaining 9 items?

Explanation: Total sum of 10 items = 10 × 25 = 250. If the item with value 40 is removed, the sum of the remaining 9 items = 250 – 40 = 210.

[Concept: Basic Average & Sum] The average of three numbers A, B, and C is 15. If the sum of A and B is 25, find the value of C.

Explanation: Total sum of A, B, and C = 3 × 15 = 45. Given A + B = 25. Therefore, C = 45 – (A + B) = 45 – 25 = 20.

[Concept: Deviation Method] Find the average of the following numbers using the deviation method: 45, 47, 50, 52, 56.

Explanation: Book Method: Let the assumed average be 50. The deviations are -5, -3, 0, +2, +6. Sum of deviations = (-5) + (-3) + 0 + 2 + 6 = 0. Net deviation = 0/5 = 0. Actual Average = 50 + 0 = 50.

[Concept: Deviation Method] The average of 6 numbers is 30. Five of these numbers are 28, 32, 25, 35, and 27. What is the value of the sixth number?

Explanation: Taking 30 as the base, deviations of the 5 numbers are: -2, +2, -5, +5, -3. The sum of these deviations is -3. For the overall average to be exactly 30, the net deviation must be 0. Thus, the 6th number must have a deviation of +3. Sixth number = 30 + 3 = 33.

[Concept: Deviation Method] The average of 4 numbers is 100. Three of the numbers are 90, 105, and 110. Find the fourth number.

Explanation: Deviations from 100 are: -10, +5, +10. The sum of these deviations is +5. To balance the average at 100, the fourth number must have a deviation of -5. Fourth number = 100 – 5 = 95.

[Concept: Weighted/Combined Average] In a class, 20 boys have an average score of 50, and 30 girls have an average score of 60. What is the combined average score of the class?

Explanation: Book Formula: Combined Average = (n1*a1 + n2*a2) / (n1 + n2). Combined Avg = (20*50 + 30*60) / (20 + 30) = (1000 + 1800) / 50 = 2800 / 50 = 56.

[Concept: Weighted/Combined Average] Section A has 40 students with an average weight of 70 kg. Section B has 60 students with an average weight of 80 kg. Find the average weight of all 100 students.

76 kg
75 kg
74 kg
78 kg

Explanation: Combined Avg = (40*70 + 60*80) / (40 + 60) = (2800 + 4800) / 100 = 7600 / 100 = 76 kg.

[Concept: Weighted/Combined Average] The ratio of workers in Department A and B is 3:2. The average salary of Dept A is Rs 4000 and Dept B is Rs 5000. Find the combined average salary.

Rs 4400
Rs 4500
Rs 4600
Rs 4200

Explanation: We can use the ratio directly as the number of workers (n1=3, n2=2). Combined Avg = (3*4000 + 2*5000) / (3 + 2) = (12000 + 10000) / 5 = 22000 / 5 = Rs 4400.

[Concept: 1st ‘n’ Natural Numbers] What is the average of the first 99 consecutive natural numbers?

50
49.5
50.5
49

Explanation: Book Formula: Average of 1st ‘n’ natural numbers = (n + 1) / 2. Here n = 99. Average = (99 + 1) / 2 = 100 / 2 = 50.

[Concept: Squares of ‘n’ Natural Numbers] What is the average of the squares of the first 7 consecutive natural numbers?

Explanation: Book Formula: Average of squares of 1st ‘n’ natural numbers = (n + 1)(2n + 1) / 6. Here n = 7. Avg = (7 + 1)(2*7 + 1) / 6 = (8 * 15) / 6 = 120 / 6 = 20.

[Concept: Cubes of ‘n’ Natural Numbers] Find the average of the cubes of the first 4 consecutive natural numbers.

Explanation: Book Formula: Average of cubes of 1st ‘n’ natural numbers = n(n + 1)² / 4. Here n = 4. Avg = 4 * (4 + 1)² / 4 = 4 * 25 / 4 = 25.

[Concept: 1st ‘n’ Even/Odd Numbers] What is the average of the first 50 consecutive even natural numbers?

51
50
25.5
26

Explanation: Book Formula: Average of 1st ‘n’ consecutive even natural numbers = (n + 1). Here n = 50. Average = 50 + 1 = 51.

[Concept: 1st ‘n’ Even/Odd Numbers] What is the average of the first 100 consecutive odd natural numbers?

100
101
50
50.5

Explanation: Book Formula: Average of 1st ‘n’ consecutive odd natural numbers = n. Here n = 100. Average = 100.

[Concept: 1st ‘n’ Even/Odd Numbers] Find the average of all even numbers from 1 to 40.

21
20
20.5
41

Explanation: Book Formula: Average of 1 to n even numbers = (Last even no. + 2) / 2. Here the last even number is 40. Avg = (40 + 2) / 2 = 42 / 2 = 21.

[Concept: Consecutive Numbers & Multiples] What is the average of the consecutive numbers from 7 to 45?

26
25
27
25.5

Explanation: Book Formula: The average of consecutive numbers = (First no. + Last no.) / 2. Avg = (7 + 45) / 2 = 52 / 2 = 26.

[Concept: Consecutive Numbers & Multiples] What is the average of the first 15 multiples of 4?

Explanation: Book Formula: Average of 1st ‘n’ multiples of a number ‘x’ = x(1 + n) / 2. Here x = 4, n = 15. Avg = 4(1 + 15) / 2 = 4 * 16 / 2 = 64 / 2 = 32.

[Concept: Consecutive Numbers & Multiples] Find the average of consecutive even numbers from 10 to 60.

Explanation: Book Formula: For any sequence with a common difference (like even numbers), Average = (First Term + Last Term) / 2. Avg = (10 + 60) / 2 = 70 / 2 = 35.

[Concept: Average Speed – 2 Distances] A man goes from P to Q at a speed of 30 km/h and returns from Q to P at 20 km/h. What is his average speed for the whole journey?

24 km/h
25 km/h
26 km/h
22 km/h

Explanation: Book Formula: Average speed = 2xy / (x + y). Here x=30, y=20. Avg Speed = (2 * 30 * 20) / (30 + 20) = 1200 / 50 = 24 km/h.

[Concept: Average Speed – 3 Distances] A car travels an equilateral triangular path. It covers the three equal sides at speeds of 10 km/h, 15 km/h, and 30 km/h. Find the average speed.

15 km/h
18 km/h
20 km/h
16 km/h

Explanation: Book Formula for 3 equal distances: Avg Speed = 3abc / (ab + bc + ca). Here a=10, b=15, c=30. Avg Speed = (3 * 10 * 15 * 30) / (150 + 450 + 300) = 13500 / 900 = 15 km/h.

[Concept: Average Speed – 2 Distances] A boy travels half of his total journey at 40 km/h and the remaining half at 60 km/h. What is his average speed?

48 km/h
50 km/h
45 km/h
52 km/h

Explanation: Since the distances are equal (half and half), use the formula 2xy / (x + y). Avg Speed = (2 * 40 * 60) / (40 + 60) = 4800 / 100 = 48 km/h.

[Concept: Replacement of Person] The average weight of 10 people increases by 2 kg when a person weighing 40 kg is replaced by a new person. What is the weight of the new person?

60 kg
50 kg
58 kg
62 kg

Explanation: Book Formula: Age/Weight of new member = (Age/Weight of replaced member) + (x * n). Here, Replaced = 40 kg, x (increase) = +2, n = 10. New person = 40 + (2 * 10) = 40 + 20 = 60 kg.

[Concept: Replacement of Person] In a group of 8 men, the average weight drops by 1.5 kg when a man weighing 50 kg is replaced by a new man. Find the weight of the new man.

38 kg
40 kg
35 kg
42 kg

Explanation: Book Formula: New member = Replaced member + (x * n). Here, x is a decrease, so x = -1.5. New = 50 + (-1.5 * 8) = 50 – 12 = 38 kg.

[Concept: Replacement of Person] The average age of 15 students increases by 1 year when a 20-year-old student is replaced by a new student. What is the age of the new student?

35 years
30 years
25 years
40 years

Explanation: Formula: New member = Replaced member + (x * n). New student = 20 + (1 * 15) = 20 + 15 = 35 years.

[Concept: Addition of a Person] The average age of 11 players is 22 years. When the coach’s age is included, the average increases by 1 year. What is the age of the coach?

34 years
33 years
40 years
36 years

Explanation: Book Formula: Age of added member = Old Average + x(n + 1). Here Old Avg = 22, x (increase) = +1, n = 11. Coach’s age = 22 + 1(11 + 1) = 22 + 12 = 34 years.

[Concept: Addition of a Person] The average age of 20 students is 15 years. If the teacher’s age is added, the average increases by 2 years. Find the teacher’s age.

57 years
55 years
45 years
60 years

Explanation: Formula: Added member = Old Avg + x(n + 1). Teacher’s age = 15 + 2(20 + 1) = 15 + 2(21) = 15 + 42 = 57 years.

[Concept: Addition of a Person] The average weight of 4 boys is 50 kg. A 5th boy joins, and the average drops by 1 kg. What is the weight of the 5th boy?

45 kg
46 kg
44 kg
49 kg

Explanation: Formula: Added member = Old Avg + x(n + 1). Here x is a decrease (-1). Weight = 50 + (-1)(4 + 1) = 50 – 5 = 45 kg.

[Concept: Removal of a Person] The average age of 6 members is 30 years. If one member leaves, the average drops to 28 years (a decrease of 2). What is the age of the member who left?

40 years
38 years
42 years
36 years

Explanation: Book Formula: Age of left member = Old Average – x(n – 1). Here Old Avg = 30, x (change in avg) = -2 (decrease). Left member = 30 – (-2)(6 – 1) = 30 – (-2)(5) = 30 + 10 = 40 years. (Or Total Age: 180 – 140 = 40).

[Concept: Removal of a Person] The average age of 10 workers is 45 years. One worker leaves, and the average rises by 1 year. What is the age of the worker who left?

36 years
35 years
40 years
34 years

Explanation: Formula: Left member = Old Average – x(n – 1). Here x (increase) = +1. Left worker = 45 – (+1)(10 – 1) = 45 – 9 = 36 years.

[Concept: Removal of a Person] The average weight of 5 friends is 60 kg. One friend leaves, and the new average becomes 62 kg (rises by 2). What is the weight of the friend who left?

52 kg
50 kg
54 kg
56 kg

Explanation: Formula: Left member = Old Average – x(n – 1). Here x = +2. Weight = 60 – (+2)(5 – 1) = 60 – 8 = 52 kg.

[Concept: Misread Data (1 Error)] The average of 10 numbers was calculated as 50. Later, it was found that the number 40 was mistakenly read as 60. What is the correct average?

Explanation: Book Formula: Correct Average = m + (a – b) / n, where a is correct value and b is wrong value. Correct Avg = 50 + (40 – 60) / 10 = 50 + (-20) / 10 = 50 – 2 = 48.

[Concept: Misread Data (1 Error)] The average of 20 numbers is 35. It was later discovered that a number 85 was misread as 45. Find the correct average.

37
36
38
35.5

Explanation: Correct Avg = m + (a – b) / n. Correct Avg = 35 + (85 – 45) / 20 = 35 + 40 / 20 = 35 + 2 = 37.

[Concept: Misread Data (2 Errors)] The average of 50 numbers is 60. Later it was found that two numbers 45 and 55 were misread as 65 and 85 respectively. What is the true average?

59
58
61
59.5

Explanation: Book Formula: Correct Avg = m + (a + b – p – q) / n. True sum of errors = (45 + 55) – (65 + 85) = 100 – 150 = -50. Correct Avg = 60 + (-50 / 50) = 60 – 1 = 59.

[Concept: Pass/Fail Mixture] In a class of 100 students, the overall average is 50. The passed students’ average is 60 and the failed students’ average is 20. How many students passed?

Explanation: Book Formula: Passed students = n(a – y) / (x – y), where n=100, a=50 (overall), x=60 (passed), y=20 (failed). Passed = 100(50 – 20) / (60 – 20) = 100(30) / 40 = 3000 / 40 = 75.

[Concept: Pass/Fail Mixture] A class has 60 students with an overall average of 45. The passed average is 55 and the failed average is 35. Find the number of passed students.

Explanation: Formula: Passed students = n(a – y) / (x – y). Passed = 60(45 – 35) / (55 – 35) = 60(10) / 20 = 600 / 20 = 30.

[Concept: Pass/Fail Mixture] Out of 80 students, the overall average is 40. The average of passed students is 50, and that of failed students is 10. Find the number of FAILED students.

Explanation: First find passed students using n(a – y) / (x – y). Passed = 80(40 – 10) / (50 – 10) = 80(30) / 40 = 60. Therefore, Failed students = Total (80) – Passed (60) = 20.

[Concept: Batting & Bowling Average] A batsman has scored 450 runs in 10 innings. What is his batting average?

Explanation: Book Formula: Batting Average = Total runs scored / Total number of innings played. Batting Average = 450 / 10 = 45.

[Concept: Batting & Bowling Average] A bowler has conceded 600 runs and taken 30 wickets so far. What is his bowling average?

Explanation: Book Formula: Bowling Average = Total runs given / Total wickets taken. Bowling Average = 600 / 30 = 20.

[Concept: Batting & Bowling Average] A batsman has an average of 40 runs in his first 5 innings. If he scores 100 runs in his 6th inning, what will be his new batting average?

Explanation: Total runs in first 5 innings = 5 × 40 = 200. Runs in 6th inning = 100. Total runs in 6 innings = 200 + 100 = 300. New Batting Average = 300 / 6 = 50.