average — BSSC Quantitative Aptitude

Given, Correct Marks = 86 and Mistaken Marks = 44. • Reduction in total marks = 86 - 44 = 42. • Decrease in average = 1.5. • Number of students = Reduction in total marks / Decrease in average = 42 / 1.5 = 28. • Exam Trick: Total Change = Number of items × Change in Average. So, 42 = n × 1.5, which gives n = 28.

The prime numbers between 32 and 69 are: 37, 41, 43, 47, 53, 59, 61, and 67. • Number of prime numbers (n) = 8. • Sum of these prime numbers = 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 = 408. • Average = Sum / n = 408 / 8 = 51.

Let the 15 consecutive even numbers be represented as an Arithmetic Progression (A.P.). • First term (smallest) = a, Last term (largest) = l. • Common difference (d) = 2. • Total terms (n) = 15. • Difference between largest and smallest term = (n - 1) × d = (15 - 1) × 2 = 28. • Exam Trick: For any 'n' consecutive even or odd numbers, the difference between the largest and smallest number is always 2 × (n - 1). Here, 2 × (15 - 1) = 28.

Let the 11 consecutive integers be a, a+1, ..., a+10. • The average of these 11 numbers is the middle term, which is the 6th term: a + 5 = d. • If the last two numbers (a+9 and a+10) are removed, we are left with 9 consecutive integers: a, a+1, ..., a+8. • The new average of these 9 numbers is their middle term, which is the 5th term: a + 4. • New Average = d - 1. Thus, the average decreases by 1. • Exam Trick: Removing the largest 'k' terms from consecutive integers always decreases the average by k/2. Here, removing 2 terms decreases the average by 2/2 = 1.

Number of girls (n1) = 22, Average weight of girls (A1) = 46 kg. • Total students (n) = 37, Average weight of class (A) = 50 kg. • Number of boys (n2) = 37 - 22 = 15. Let average weight of boys be A2. • Total weight of class = Total weight of girls + Total weight of boys. • 37 × 50 = 22 × 46 + 15 × A2 => 1850 = 1012 + 15 × A2 => 15 × A2 = 838 => A2 = 55.87 kg. • Exam Trick: Use deviation method. The girls' average is 4 kg below the class average. Total deficiency for 22 girls = 22 × (-4) = -88 kg. This deficiency must be compensated by the 15 boys. So, boys' average = 50 + 88/15 = 50 + 5.87 = 55.87 kg.

Let the average weight of all 14 persons be A kg. • Sum of weights of the first 13 persons = 13 × 78 = 1014 kg. • Weight of the 14th person = A + 39 kg. • Setting up the equation: 1014 + (A + 39) = 14 × A => 1053 + A = 14A => 13A = 1053 => A = 81 kg. • Weight of the 14th person = A + 39 = 81 + 39 = 120 kg. • Exam Trick: The 14th person brings 39 kg more than the average. This extra 39 kg is distributed equally among the first 13 persons to raise them to the global average. Increase per person = 39 / 13 = 3 kg. Therefore, overall average = 78 + 3 = 81 kg. Weight of 14th person = 81 + 39 = 120 kg.

Sum of ages of the 3 initial family members = 44 × 3 = 132 years. • When the daughter-in-law joins, there are 4 members and the average age is 39 years. • Sum of ages of all 4 members = 39 × 4 = 156 years. • Age of daughter-in-law = 156 - 132 = 24 years. • Exam Trick: Think in terms of deviation. The new average is 39, which is 5 years less than the old average of 44. The reduction for the 3 original members is 3 × 5 = 15 years. So, the new member's age must be 15 years less than the new average: 39 - 15 = 24 years.

Initial sum of marks in 6 subjects = 6 × 110 = 660. • Reduction in marks in one subject = 148 - 112 = 36 marks. • New sum of marks = 660 - 36 = 624. • New average marks = 624 / 6 = 104. • Exam Trick: Calculate the decrease in average directly: Total reduction / Number of subjects = 36 / 6 = 6. New average = Old average - 6 = 110 - 6 = 104.

Given, 24x + 24y = 192. • Factor out 24: 24(x + y) = 192. • Solve for the sum: x + y = 192 / 24 = 8. • Average of x and y = (x + y) / 2 = 8 / 2 = 4. • Exam Trick: The sum is simply the constant divided by the coefficient (192 / 24 = 8). Since we want the average of two numbers, we divide this sum by 2 to get 4.

The first 6 multiples of 10 are: 10, 20, 30, 40, 50, and 60. • Since the common difference is constant (d = 10), the terms form an Arithmetic Progression (A.P.). • Average of an A.P. = (First term + Last term) / 2 = (10 + 60) / 2 = 70 / 2 = 35. • Exam Trick: For an even number of terms in an A.P., the average is the mean of the two middle terms. The middle terms of 10, 20, 30, 40, 50, 60 are 30 and 40. Their average is (30 + 40) / 2 = 35.