Average Questions with Solutions for Competitive Exams

Explanation:
The fundamental definition of Arithmetic Mean (Average):
\text{Average} = \frac{\text{Sum of all observations}}{\text{Total number of observations}}
From this relationship, the total sum is given by:
\text{Sum of all observations} = \text{Average} \times \text{Total number of observations}

Standard Algebraic Method (Direct Formula)
Step 1: Compute the total marks across all 6 subjects:
\text{Total Sum}_{6} = 6 \times 74 = 444 \text{ marks} Step 2: Calculate the sum of marks obtained in the 5 known subjects: \text{Sum}_{5} = 68 + 72 + 85 + 63 + 76 = 364 \text{ marks} Step 3: Find the score of the 6th subject (x_6):
x_6 = \text{Total Sum}{6} - \text{Sum}{5} x_6 = 444 - 364 = 80 \text{ marks}

Explanation:
For any set of numbers in an Arithmetic Progression (AP)—where the difference between consecutive terms is constant—the Average is always equal to the Median (the middle term).
\text{Average} = \text{Middle Term} = \frac{\text{First Term} + \text{Last Term}}{2}
Let the 7 consecutive odd numbers be: x, (x+2), (x+4), (x+6), (x+8), (x+10), (x+12).
The average is:
\frac{x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10) + (x+12)}{7} = 25 \frac{7x + 42}{7} = 25 x + 6 = 25 x = 19

Explanation:
Any sequence with a constant difference between consecutive terms (like multiples of a number) forms an Arithmetic Progression (AP). For any AP, you do not need to calculate the sum of all individual terms. The average depends solely on the boundary values:
\text{Average} = \frac{\text{First Term} + \text{Last Term}}{2}
Step 1: Identify the First Term (a)
Find the smallest multiple of 9 greater than 100:\frac{100}{9} = 11.11 \implies 9 \times 12 = 108 \text{First Term} = 108
Step 2: Identify the Last Term (l)
Find the largest multiple of 9 less than 250: \frac{250}{9} = 27.77 \implies 9 \times 27 = 243 \text{Last Term} = 243
Step 3: Apply the AP Average Formula:
\text{Average} = \frac{108 + 243}{2}\text{Average} = \frac{351}{2} = 175.5
Total number of terms (n):
n = \left(\frac{\text{Last Term} - \text{First Term}}{\text{Common Difference}}\right) + 1 = \left(\frac{243 - 108}{9}\right) + 1 = \frac{135}{9} + 1 = 16 \text{ terms}
Sum of AP:
S_n = \frac{n}{2} \times (\text{First Term} + \text{Last Term}) = \frac{16}{2} \times (108 + 243) = 8 \times 351 = 2808

Average:
\text{Average} = \frac{\text{Total Sum}}{n} = \frac{2808}{16} = 175.5

Explanation:
In a replacement scenario, the total number of members in the group remains unchanged. Any change in the group’s average is entirely caused by the difference between the incoming value and the outgoing value:
\text{Net Change in Total Sum} = \text{Value of Incomer} - \text{Value of Outgoer}

Competitive Exam Shortcut: Net Deviation Formula
\text{Weight of New Member} = \text{Weight of Replaced Member} + (\text{Total Members} \times \text{Increase in Average}) Given Values:
Weight of Replaced Member = 58\text{ kg}
Total Members (n) = 12
Increase in Average (\Delta A) = +1.5\text{ kg}

Step-by-Step Calculation:
\text{Weight of New Member} = 58 + (12 \times 1.5) \text{Weight of New Member} = 58 + 18 = 76\text{ kg} Note: If the average had decreased, we would subtract the net change.

Explanation:
Step 1: Compute Sum of Correct and Incorrect Entries:
\text{Sum of Correct Values} = 48 + 62 = 110 \text{Sum of Incorrect Values} = 84 + 46 = 130 Step 2: Calculate Net Error (Difference):
\text{Net Change} = \text{Correct Sum} - \text{Incorrect Sum} = 110 - 130 = -20 Step 3: Distribute the Net Change over Total Observations (n = 40):
\text{Change in Average} = \frac{-20}{40} = -0.5 Step 4: Find the Correct Average: \text{Correct Average} = 65 + (-0.5) = 64.5

Explanation:
Assume a base average equal to the smaller value (\text{Base} = 72):
Deviation for boys: 72 - 72 = 0
Deviation for girls: 82 - 72 = +10
Distribute the total extra score over the total ratio parts (3 + 2 = 5): \text{Total Extra Score} = (3 \times 0) + (2 \times 10) = 20 \text{Average Increase} = \frac{20}{3 + 2} = \frac{20}{5} = +4 \text{Class Average} = \text{Base Average} + \text{Average Increase} = 72 + 4 = 76

Explanation:
When combining two distinct groups with different sizes and averages, the combined average is calculated using the Weighted Average formula:
A_w = \frac{n_1 A_1 + n_2 A_2}{n_1 + n_2}
Simplify the ratio of students: 40 : 60 = 2 : 3
Substitute ratio weights (2 and 3) and their respective averages (72 and 82):
A_w = \frac{(2 \times 72) + (3 \times 82)}{2 + 3} A_w = \frac{144 + 246}{5} A_w = \frac{390}{5} = 78.0

Shortcut Technique: Deviation / Assumed Mean Method
Instead of computing large products, take the smaller average (72) as the baseline:
Deviation of Batch A (weight 2): 72 – 72 = 0
Deviation of Batch B (weight 3): 82 – 72 = +10
Total surplus marks: (2 × 0) + (3 × 10) = 30
Distribute this surplus over the total ratio parts (2 + 3 = 5):
\text{Average Increase} = \frac{30}{5} = 6 Combined Average = 72 + 6 = 78.0 marks.

Explanation:
Batting average is defined by the formula:
\text{Batting Average} = \frac{\text{Total Runs Scored}}{\text{Total Innings}}
Let the initial average for 16 innings be A.
Total runs scored in 16 innings = 16A
Runs scored in the 17th inning = 85
New average for 17 innings = A + 3
Setting up the equation using total runs:
16A + 85 = 17(A + 3)
16A + 85 = 17A + 51
17A – 16A = 85 – 51
A = 34
Old average (A) = 34
New average (A + 3) = 34 + 3 = 37

Shortcut Technique: Direct Deviation Method
The 85 runs scored in the 17th inning must cover the batsman’s own share for the 17th inning, plus supply an extra 3 runs to each of the previous 16 innings:
Total surplus distributed to previous innings: 16 × 3 = 48 runs
The remaining score becomes the New Average:
New Average = 85 – 48 = 37

Direct Formula:
\text{New Average} = \text{Runs in } n^{\text{th}}\text{ Inning} - [(n - 1) \times \text{Increase in Average}] \text{New Average} = 85 - (16 \times 3) = 37

Explanation:
A bowler’s average improves when it decreases (fewer runs given per wicket).
New overall bowling average = 24.85 – 0.85 = 24.00 runs/wicket.
In the last match, the bowler’s average was:
\text{Last Match Average} = \frac{52}{5} = 10.4 \text{ runs/wicket}
Let the initial number of wickets be W.
Total runs before last match = 24.85W
Total runs including last match = 24.85W + 52
Total wickets = W + 5
\frac{24.85W + 52}{W + 5} = 24
24.85W + 52 = 24(W + 5)
24.85W + 52 = 24W + 120
0.85W = 68
W = \frac{68}{0.85} = 80 \text{ wickets}

Shortcut Technique (Rule of Alligation):
Old Average = 24.85 | Last Match Average = 10.4
Combined Average = 24.0
Difference for Old Wickets = 24.0 – 10.4 = 13.6
Difference for New Wickets = 24.85 – 24.0 = 0.85
Ratio of Wickets = 13.6 : 0.85 = 1360 : 85 = 16 : 1
Since 1 ratio unit corresponds to 5 wickets in the last match:
Old wickets (16 units) = 16 × 5 = 80 wickets.

Explanation:
3 years ago, total age of 5 members = 5 × 24 = 120 years.
In 3 years, each of the 5 members ages by 3 years.
Present total age of original 5 members = 120 + (5 × 3) = 135 years.
Presently, there are 6 members (5 members + baby), and their average age is 24 years.
Present total age of 6 members = 6 × 24 = 144 years.
Age of the baby = 144 – 135 = 9 years…

Wait: 144 – 135 = 9 years would exceed 3 years since birth.
Re-calculating deviation: The baby needs to compensate for the aging of 5 members by 3 years each (15 years deficit relative to present average):
Age of baby = 24 – (5 × 3) = 24 – 15 = 9 years (mathematical result).>
For a realistic exam framing where age is 9 months, baby was born 3 years ago. Here, direct calculation gives:
\text{Baby's Age} = (6 \times 24) - [5 \times (24 + 3)] = 144 - 135 = 9 \text{ years}

Explanation:
The 6th observation is counted twice: once in the first 6 and once in the last 6.
Sum of all 11 results = 11 × 50 = 550.
Sum of first 6 results = 6 × 49 = 294.
Sum of last 6 results = 6 × 52 = 312.
Sum of (first 6 + last 6) = 294 + 312 = 606.
6th Result = (Sum of first 6 + Sum of last 6) – (Sum of 11 results) = 606 – 550 = 56.

Shortcut Technique (Net Deviation Method):
Deviation of first 6 results from 50: 6 × (49 – 50) = 6 × (-1) = -6
Deviation of last 6 results from 50: 6 × (52 – 50) = 6 × (+2) = +12
Net deviation = -6 + 12 = +6
\text{Middle (6th) Term} = \text{Base Average} + \text{Net Deviation} = 50 + 6 = 56

Explanation:
Sum of Mon + Tue + Wed = 3 × 40 = 120°C.
Sum of Tue + Wed + Thu = 3 × 41 = 123°C.
Subtracting the first equation from the second:
(Tue + Wed + Thu) – (Mon + Tue + Wed) = 123 – 120
Thu – Mon = 3°C.
Given Thursday = 42°C:
42 – Mon = 3
Mon = 42 – 3 = 39°C.

Shortcut Technique:
Difference between incoming day (Thu) and outgoing day (Mon) = Number of days × Difference in average.
Thu – Mon = 3 × (41 – 40) = 3 × 1 = 3°C.
Mon = 42 – 3 = 39°C.

Explanation:
Let the average expenditure of all 9 persons be A.
Total expenditure of all 9 persons = 9A.
Expenditure of 8 persons = 8 × 30 = Rs. 240.
Expenditure of 9th person = A + 20.
Setting up the total equation:
240 + (A + 20) = 9A
260 = 8A
A = \frac{260}{8} = 32.50
Total money spent = 9 × 32.50 = Rs. 292.50.

Shortcut Technique:
The Rs. 20 extra spent by the 9th person must be equally distributed among the other 8 persons to bring everyone to the common average:
\text{Average} = 30 + \frac{20}{8} = 30 + 2.50 = 32.50
Total expenditure = 9 × 32.50 = Rs. 292.50.

Explanation:
Using the Rule of Alligation:
Technicians Average = Rs. 12,000 | Non-technicians Average = Rs. 6,000
Overall Average = Rs. 8,000
Difference for Technicians side = 8000 – 6000 = 2000
Difference for Non-technicians side = 12000 – 8000 = 4000

Ratio of Technicians to Non-technicians:
\text{Ratio} = \frac{2000}{4000} = \frac{1}{2}
Since Technicians (1 unit) = 7:
Non-technicians (2 units) = 2 × 7 = 14 employees.
Total employees = 7 + 14 = 21 employees.

Explanation:
Simplify group count ratio: 30 : 5 = 6 : 1.
Using the Assumed Mean Method with baseline 14 years:
Deviation for first group = 6 × (14 – 14) = 0
Deviation for second group = 1 × (16 – 14) = +2
Total surplus = 2 years.
Total ratio parts = 6 + 1 = 7.
\text{Increase in Average} = \frac{2}{7} \approx 0.2857 \text{ years}
\text{New Average} = 14 + \frac{2}{7} \approx 14.28 \text{ years}

Explanation:
For n consecutive even numbers, the common difference is 2.
Smallest number = Average – (n – 1) = 47 – (8 – 1) = 47 – 7 = 40.
Largest number = Average + (n – 1) = 47 + (8 – 1) = 47 + 7 = 54.
Product = 40 × 54 = 2160.

Verification:
The numbers are 40, 42, 44, 46, 48, 50, 52, 54.
Average = (40 + 54) / 2 = 94 / 2 = 47.

Explanation:
Average speed for equal distances is given by the Harmonic Mean formula:
\text{Average Speed} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}}
\text{Average Speed} = \frac{3}{\frac{1}{20} + \frac{1}{30} + \frac{1}{60}} = \frac{3}{\frac{3 + 2 + 1}{60}} = \frac{3}{\frac{6}{60}} = \frac{3}{\frac{1}{10}} = 30 \text{ km/h}

Shortcut Technique (LCM Method):
Let each one-third distance be LCM(20, 30, 60) = 60 km.
Total distance = 3 × 60 = 180 km.
Time taken for part 1 = 60 / 20 = 3 hours.
Time taken for part 2 = 60 / 30 = 2 hours.
Time taken for part 3 = 60 / 60 = 1 hour.
Total time = 3 + 2 + 1 = 6 hours.
Average speed = 180 / 6 = 30 km/h.

Explanation:
Incorrect total marks read = 38 + 42 = 80.
Correct total marks = 83 + 72 = 155.
Net difference added = 155 – 80 = +75 marks.
\text{Increase in Average} = \frac{\text{Net Difference}}{\text{Total Students}} = \frac{75}{50} = 1.50
Correct Average = 64 + 1.50 = 65.50.

Explanation:
Total expenditure for first 7 months = 7 × 12,000 = Rs. 84,000.
Total expenditure for next 5 months = 5 × 15,000 = Rs. 75,000.
Total annual expenditure = 84,000 + 75,000 = Rs. 1,59,000.
Total annual savings = Rs. 41,000.
Total annual income = 1,59,000 + 41,000 = Rs. 2,00,000.
\text{Average Monthly Income} = \frac{2,00,000}{12} = \text{Rs. } 16,666.67

Explanation:
Ratio of Boys : Girls = 60% : 40% = 3 : 2.
Using Assumed Mean Method with base 55 kg:
Deviation for Boys (weight 3) = 3 × (62 – 55) = 3 × 7 = +21 kg.
Deviation for Girls (weight 2) = 2 × (55 – 55) = 0 kg.
Total ratio parts = 3 + 2 = 5.
\text{Average Increase} = \frac{21}{5} = 4.2 \text{ kg}
Average weight of the whole class = 55 + 4.2 = 59.2 kg.

Explanation:
Total runs scored in 25 innings = 25 × 56 = 1400.
Total runs after 26 innings = 1400 + 0 = 1400.
\text{New Average} = \frac{1400}{26} = \frac{700}{13} \approx 53.846 \text{ runs}
Decrease in average = 56 – 53.846 = 2.154 ≈ 2.15 runs.

Shortcut Technique:
Deviation caused by scoring 0 instead of maintaining old average (56) = -56 runs.
\text{Decrease in Average} = \frac{56}{26} = \frac{28}{13} \approx 2.15 \text{ runs}

Explanation:
When a person is replaced and the average decreases, the outgoing person is older than the incoming person by (Total Members × Decrease in Average).
Total reduction in group age = 10 × 3 = 30 years.
Age of Retired Teacher = Age of New Teacher + Total Reduction = 25 + 30 = 55 years.

Explanation:
For any Arithmetic Progression, the average is the exact midpoint of the first and last terms:
\text{Average} = \frac{\text{First Term} + \text{Last Term}}{2}
Given Average = 84
84 = \frac{\text{Smallest} + \text{Largest}}{2}
Smallest + Largest = 84 × 2 = 168.

Explanation:
Total weight of A + B + C = 3 × 84 = 252 kg.
Total weight of A + B + C + D = 4 × 80 = 320 kg.
Weight of D = 320 – 252 = 68 kg.
Weight of E = Weight of D + 3 = 68 + 3 = 71 kg.
Total weight of B + C + D + E = 4 × 79 = 316 kg.
Subtracting (B + C + D + E) from (A + B + C + D):
(A + B + C + D) – (B + C + D + E) = 320 – 316
A – E = 4 kg.
Weight of A = E + 4 = 71 + 4 = 75 kg.

Explanation:
The 7th number is common to both the first 7 and last 7 subsets.
Sum of all 13 numbers = 13 × 68 = 884.
Sum of first 7 numbers = 7 × 63 = 441.
Sum of last 7 numbers = 7 × 70 = 490.
Sum of (first 7 + last 7) = 441 + 490 = 931.
7th Number = 931 – 884 = 47.

Shortcut Technique (Deviation from Base 68):
Deviation for first 7: 7 × (63 – 68) = 7 × (-5) = -35
Deviation for last 7: 7 × (70 – 68) = 7 × (+2) = +14
Net deviation = -35 + 14 = -21
7th Number = 68 + (-21) = 47.