The average of 5 consecutive numbers is k. If the next 3 numbers are included, the new average will be?

Answer: k+1.5. For natural consecutive numbers, adding 'n' next numbers increases the average by n/2. Here n=3, so average increases by 1.5.

Average of 20 numbers is 0. At most how many of them can be greater than zero?

Answer: 19. To have an average of 0, the sum must be 0. We can have 19 large positive numbers, and a single massive negative number to balance them perfectly.

Average of first 50 natural numbers is?

Answer: 25.5. Use AP formula: $\frac{First + Last}{2} = \frac{1+50}{2} = 25.5$.

If 11a + 11b = 33, what is the average of a and b?

Answer: 1.5. $11(a+b) = 33 \implies a+b = 3$. Average of two numbers is their sum divided by 2. $3/2 = 1.5$.

Average of x numbers is y, and average of y numbers is x. Total Average of all numbers?

Answer: 2xy / (x+y). Sum 1 = $x \times y = xy$. Sum 2 = $y \times x = yx$. Total Sum = 2xy. Total Count = x+y. Avg = $2xy/(x+y)$.

Can the average of a data set be smaller than the smallest number in the set?

Answer: No, never. By definition, the average is the center point. It must strictly lie between the minimum and maximum values of the dataset.

A car covers a distance at 60 km/h and returns at 40 km/h. Its average speed is?

Answer: 48 km/h. Harmonic Mean: $\frac{2xy}{x+y} = \frac{2(60)(40)}{100} = 48$ km/h.

For a bowler in cricket, does a higher bowling average indicate a better performance?

Answer: No, lower is better. Bowling Ave = $\frac{\text{Runs Given}}{\text{Wickets}}$. Giving away fewer runs for a wicket is better, so a lower number is superior.

Average of x, x+2, x+4, x+6, x+8 is 11. Find the average of the last three.

Answer: 13. Since it is an AP series, the initial Avg is the middle term: $x+4 = 11$. The last three terms are $x+4, x+6, x+8$. Their avg is the middle: $x+6$. Wait $x+4=11 implies x+6=13$.

Average of squares of first 10 natural numbers?

Answer: 38.5. Formula: $\frac{(n+1)(2n+1)}{6} = \frac{(11)(21)}{6} = \frac{231}{6} = 38.5$.

Find Average of: 693, 699, 706, 708, 715.

Answer: 704.2. Assume 700. Deviations: -7, -1, +6, +8, +15. Net = +21. Avg = $700 + (21/5) = 704.2$.

Average of 10 numbers is 40. A number 35 is wrongly read as 53. Correct average?

Answer: 38.2. Difference = Correct - Wrong = $35 - 53 = -18$. Avg Change = $-18/10 = -1.8$. New Avg = $40 - 1.8 = 38.2$.

Average weight of 8 men increases by 2.5 kg when a new man replaces one weighing 65 kg. Weight of new man?

Answer: 85 kg. New = Old + (Total Increase). New = $65 + (8 \times 2.5) = 65 + 20 = 85$ kg.

Avg age of 24 students and 1 teacher is 15 years. If teacher leaves, avg drops by 1 year. Teacher's age?

Answer: 39. Teacher took away their own 15 years PLUS 1 year from each remaining student (24). Age = $15 + 24(1) = 39$.

A cricketer has avg 28 in 10 innings. To raise avg to 30 in next inning, he must score?

Answer: 50. Needs 30 for the new inning PLUS 2 extra for each of the old 10 innings. Score = $30 + (10 \times 2) = 50$.

In a class, avg of 20 boys is 60kg, avg of 30 girls is 50kg. Class average?

Answer: 54. Ratio Boys:Girls = 2:3. Gap is 10. Avg = $50 + (2 \times 10)/5 = 54$.

Avg temp Mon-Wed is 40°C. Tue-Thu is 41°C. If Thu is 42°C, find Mon.

Answer: 39. $(Tue+Wed+Thu) - (Mon+Tue+Wed) = 3(41-40) = 3$. Thu - Mon = 3. $42 - Mon = 3 \implies Mon = 39$.

Avg of 5 numbers is 27. If one is excluded, avg becomes 25. Excluded number?

Answer: 35. It took away 27, plus left a deficit of 2 for remaining 4 numbers. Value = $27 + (4 \times 2) = 35$.

3 years ago, avg age of family of 5 was 17. Baby born, avg is same today. Baby age?

Answer: 2. Today ideal total (without baby) = $85 + 15 = 100$. Actual Total (6 members) = $17 \times 6 = 102$. Baby = $102 - 100 = 2$.

Bowler avg 12.4. Takes 5 wickets for 26 runs, avg improves by 0.4. Total wickets?

Answer: 85. Old Avg: 12.4. Match Avg: 26/5 = 5.2. New Avg: 12.0. Ratio: (12.0 - 5.2) : (12.4 - 12.0) = 6.8 : 0.4 = 17:1. 1 unit = 5 wickets. Total (after match) = 17+1 = 18 units = 90? Question asks Total Wickets (before match? options says 85). Old Wickets = $17 \times 5 = 85$.

Monthly income of A&B is 5050. B&C is 6250. A&C is 5200. Income of A?

Answer: 4000. $2(A+B+C) = 10100 + 12500 + 10400 = 33000$. $A+B+C = 16500$. $A = 16500 - (B+C_{sum}) = 16500 - 12500 = 4000$.

Library visitors: Sunday 510, Other days 240. Avg in 30-day month starting Sunday?

Answer: 285. Sundays = 5. Others = 25. Avg = $\frac{5(510) + 25(240)}{30} = \frac{2550 + 6000}{30} = 285$.

Avg of 3 numbers is 60. First is 1/4th of sum of others. First number?

Answer: 36. Let others = 4x, First = x. Sum = 5x. Total Sum = $60 \times 3 = 180$. $5x = 180 \implies x = 36$.

A student finds avg of 10 two-digit numbers. If digits of one number are interchanged, avg increases by 3.6. Difference of its digits?

Answer: 4. Total Increase = $10 \times 3.6 = 36$. Property: Difference of interchanged numbers is $9(a-b)$. $9(a-b) = 36 \implies a-b = 4$.

Avg weight of A,B,C is 84. D joins, avg becomes 80. E(3kg more than D) replaces A, avg becomes 79. Weight of A?

Answer: 75. D brought avg down by 4 across 4 people. Diff = 16. D = $84-16=68$. E = $68+3=71$. When E replaces A, avg of 4 goes down by 1 (total diff 4). So A is 4kg heavier than E. $A = 71 + 4 = 75$.

Avg age of cricket team (11) is 26. Capt and Keeper excluded, avg drops by 1. Capt is 3 yrs older than Keeper. Capt age?

Answer: 32. Sum Excluded = $286 - 225 = 61$. $C + K = 61$. $C - K = 3$. $2C = 64 \implies C = 32$.

35 students in a hostel. 7 new join, exp increases by 42/day, average drops by 1. Original expenditure?

Answer: 420. $(35x + 42) / 42 = x - 1 \implies 35x + 42 = 42x - 42 \implies 7x = 84 \implies x = 12$. Original Exp = $35 \times 12 = 420$.

3 classes A,B,C take a test. Avg scores: A=83, B=76, C=85. Avg of A+B=79, B+C=81. Avg of A+B+C?

Answer: 81.5. Alligation on A&B: A:B = 3:4. Alligation on B&C: B:C = 4:5. Ratio = 3:4:5. Weighted Avg: $\frac{3(83) + 4(76) + 5(85)}{12} = \frac{978}{12} = 81.5$.

Avg of 11 numbers is 50. First 6 is 49. Last 6 is 52. The 6th number is?

Answer: 56. Shortcut Deviation: First 6 deviate by -1 total -6. Last 6 deviate by +2 total +12. Net = +6. Overlap implies 6th Number = Mean + Net = $50 + 6 = 56$.

Ratio of ages of Husband:Wife is 4:3. After 4 years, ratio 9:7. Age of Wife?

Answer: 24. Balance Gaps: 4:3 (diff 1). 9:7 (diff 2). Multiply first by 2 $\implies$ 8:6. Gap 8 to 9 is 1 unit. 1 unit = 4 years. Wife (6 units) = 24 years.

Home/Rajasthan Librarian Exam/Quantitative Aptitude/

Averages

Master Averages without formulas using 'The Deviation Method' and 'Weighted Balance'.

Expert Answer & Key Takeaways

Master Averages without formulas using 'The Deviation Method' and 'Weighted Balance'.

Concept 1: The 'Deviation Method' (No Formulas)

The Core Truth: Average is the point where Net Deviation is ZERO.

The Method: Instead of summing huge numbers, pick an 'Assumed Average' (A).

If numbers are 48, 52, 55. Let A = 50.

48 is -2 (below 50).
52 is +2 (above 50).
55 is +5 (above 50).
Net Deviation = +5.
Average = 50 + 5/3 = 51.66.

Example:

Q: Find avg of 693, 699, 706

Solution: Assume 700. Dev: -7, -1, +6. Net Dev: -2. Avg =

700 - 2/3 = 699.33

Concept 2: The 'Replacement Hack' (Newcomer Effect)

When a person joins a group, they bring a 'Surplus' or 'Deficit'.

New Avg = Old Avg + Total IncreaseTotal People

Example:

Q: Avg age of 10 people increases by 2 years when a new person replaces a 40-year-old.

Solution: New must cover old (40) + bring extra (10x2=20). Age =

40 + 20 = 60

Concept 3: Weighted Average (The Balance Beam)

If Group A has N1 items with Avg A1, and Group B has N2 items with Avg A2. The gap between A1 and A2 is shared in ratio N2:N1.

Weighted Avg = N1(A1) + N2(A2)N1+N2.

Example:

Q: Class A (20 boys, Avg 30), Class B (30 girls, Avg 50).

Solution: Avg =

\frac{20(30) + 30(50)}{50} = \frac{600 + 1500}{50} = 42

Concept 4: The 'AP Rule' (Visual Middle)

For any Arithmetic Progression (AP) like 12, 14, 16... or 7, 14, 21...

Average = Middle Number

Master Formula: Average = First Term + Last Term2.

Example:

Q: Average of first 50 natural numbers?

Solution: First=1, Last=50. Avg =

(1+50)/2 = 25.5

Concept 5: Average Speed (Harmonic Mean)

If distance is constant, time is variable.

Avg Speed = 2xyx+y

Example:

Q: Go at 60 kmph, return at 40 kmph.

Solution: Avg Speed =

\frac{2 \times 60 \times 40}{100} = 48

kmph.

Concept 6: The 'Cricketer's Balance'

Batting Avg: Total RunsTotal Innings [Out].

Bowling Avg: Total Runs GivenTotal Wickets Taken.

Trick: For Bowling, a LOWER avg is better.

Example:

Q: Batsman avg increases by 3 after scoring 96 in 12th inning.

Solution:

96 = \text{OldAvg} + 12(3) \implies 96 = \text{Old} + 36 \implies \text{Old} = 60

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Averages

Expert Answer & Key Takeaways

Concept 1: The 'Deviation Method' (No Formulas)

Concept 2: The 'Replacement Hack' (Newcomer Effect)

Concept 3: Weighted Average (The Balance Beam)

Concept 4: The 'AP Rule' (Visual Middle)

Concept 5: Average Speed (Harmonic Mean)

Concept 6: The 'Cricketer's Balance'

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