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FORMULA USED:

Average = Sum of numbers/Total numbers

CALCULATION:

Total age of the class (Only students) = 40 × 12 = 480 years

And

Total age of the class (With teacher) = 41 × 13 = 533 years

Hence,

Age of the teacher = 533 - 480 = 53 years

Given:

Average weight of A, B, and C = 70 kg

Average weight of A, B, C, and D = 60 kg

Weight of E = Weight of D + 5 kg

Average weight of B, C, D and E = 59 kg

Concept used:

Sum of weight = Average × Number of observation

Calculation:

A + B + C = 70 × 3

⇒ A + B + C = 210 kg               .............(1)

A + B + C + D = 60 × 4

⇒ A + B + C + D = 240 kg        .............(2)

B + C + D + E = 59 × 4

⇒ B + C + D + E = 236 kg         ............ (3)

Subtracrt (1) in (2),

(A + B + C + D) – (A + B + C) = 240 – 210

⇒ D = 30 kg

Weight of E = Weight of D + 5 kg

⇒ E = 30 + 5

⇒ E = 35 kg

B + C + D + E = 236                 [Using eqn (3)]

⇒ B + C + 30 + 35 = 236

⇒ B + C = 236 – 65

⇒ B + C = 171 kg                ------------(4)

Subtract (4) in (1)

(A + B + C) – (B + C) = 210 – 171

⇒ A = 39 kg

∴ The weight of A is 39 kg.

Total age of 39 persons = (39 x 15) years = 585 years. 

Average age of 40 persons= 15 yrs 3 months = 61/4 years. 

Total age of 40 persons = ((61/4 )x 40) years= 610 years. 

∴ Age of the teacher = (610 - 585) years=25 years.

Given:

Average weight of Aman, Vistara, and Damini = 70 kg

Average weight of Aman, Vistara, Damini, and Vishwa = 66 kg

Average weight of Vistara, Damini, Vishwa, and Tara = 75 kg

Formula used:

Average = (Total weight) / Number of people

Calculation:

Total weight of Aman, Vistara, and Damini = 70 × 3

⇒ 210 kg

Total weight of Aman, Vistara, Damini, and Vishwa = 66 × 4

⇒ 264 kg   ----(i)

Weight of Vishwa = 264 – 210

⇒ 54 kg

Weight of Tara = 54 + 6 = 60 kg

Total weight of Vistara, Damini, Vishwa, and Tara = 75 × 4

⇒ 300 kg   ----(ii)

Subtracting (i) from (ii)

Tara – Aman = 300 – 264

⇒ Weight of Aman = 60 – 36

⇒ Weight of Aman = 24 kg

Given:

The average weight of the group = 44 kg

Formula used:

\(Average\ =\ {The\ sum\ of\ the\ all\ items \over The\ total\ number\ of\ items}\)

Calculation:

Let us assume the total boys in the group be X

⇒ The sum of the weight of the all-boys = 44X

⇒ 42X = 44X - 50 + 40

⇒ 42X = 44X - 10

⇒ 44X - 42X = 10

⇒ 2X = 10

⇒ X = 5

∴ The required result will be 5.

Given:

A new student of weight 45 kg came in place of the old student. The average weight of 20 students in the class increased by 1.5 kg.

Concepts used:

Average weight = (Total weight of all students)/(Number of students)

Calculation:

Let the original average weight of students be x kg.

Original total weight of all students = 20 × x = 20x

The average weight of 20 students in the class increased by 1.5 kg.

So, New average weight = x + 1.5 kg

A new student of weight 45 kg came in place of the old student.

Let the weight of the old student be y kg.

New total weight of 20 students = Original total weight of all students + Weight of new student – weight of old student

⇒ New total weight of 20 students = 20x + 45 – y

New average weight = x + 1.5 kg

Average weight = Total weight of all students/Number of students

⇒ x + 1.5 = (20x + 45 – y)/20

⇒ 20 × (x + 1.5) = 20x + 45 – y

⇒ 20x + 30 = 20x + 45 – y

⇒ y = 45 – 30 kg = 15 kg

∴ The weight of old student was 15 kg.

Given:

There are total 15 students in a class.

An old student of weight 25 kg was replaced by the new student.

New average weight is more than the old average weight by 2 kg. 

Concepts used:

Average weight = (Total weight of all students)/(Number of students)

Calculation:

Let the original average weight of students be x kg.

So, Total weight of all students = 15x

According to the question

New average weight = x + 2 kg

Let the weight of the new student be y kg.

New total weight of 15 students = Original total weight of all students + Weight of new student - weight of old student

⇒ (15x + y – 25) kg

Average weight = Total weight of all students/Number of students

⇒ x + 2 = (15x + y – 25)/15

⇒ 15 × (x + 2) = 15x + y – 25

⇒ 15x + 30 = 15x + y – 25

⇒ y = 30 + 25 = 55 kg

∴ The weight of new student is 55 kg.

Given:

Average of 62 innings is 42 runs.

Highest score – lowest score = 196

New average excluding highest and lowest innings = 40 runs

Formula used:

Average = Sum of observation/Total number of observation

Calculation:

Let the highest score be H and the lowest score be L.

Total run scored in 62 innings = 62 × 42 = 2604

Now, If his highest and lowest innings are removed then the average is decreased by 2

Total run excluding H and L = 60 × 40 = 2400

⇒ H + L = 2604 – 2400 = 204

⇒ H – L = 196

⇒ 2H = 400

⇒ H = 200

∴ The highest score scored by the batsman is 200 runs

Given:

Average after 10^th innings = 45

Highest score – lowest score = 100

New average = 45 + 4 = 49

Score of 11^th innings = (Highest score + lowest score)/2

Formula used:

Average = Total sum/Total number

Total sum = Average × Total number

Calculation:

Total score after 10 innings = 45 × 10 = 450

Let, lowest score = l and highest score = h

Score of 8 innings + l + h = 450     ---- (1)

New average = 49

Total score after 11 innings = 49 × 11 = 539

Score of 11^th inning = (l + h)/2

Score of 8 innings + (l + h) + (l + h)/2 = 539     ---- (2)

Now, eqn. (2) – (1)

⇒ (l + h)/2 = 539 – 450 = 89

⇒ l + h = 89 × 2 = 178     ---- (a)

⇒ h – l = 100     ---- (b)

Solving eqn. (a) and (b)

h = 139 and l = 39

∴ The lowest score is 39.

Given:

The average weight of the family = x

The weight of the guest = 40

Then average increases by = 2 kg

Weight of the guest = 28

Average decreases by = 2 kg

Calculations:

Let the number of family members be y.

∴ Total weight = xy

Equation for the first scenario:

⇒ (y + 1) (x + 2) = xy + 40

⇒ xy + x + 2y + 2 = xy + 40

⇒ x + 2y = 38      ----(i)

Equation for the second scenario:

⇒ (y + 1) (x - 2) = xy + 28

⇒ xy + x - 2y - 2 = xy + 28

⇒ x - 2y = 30      ----(ii)

Solving equation (i) and (ii), we get

x = 34.

The value of x is 34.

Given:

Average runs of player in 20 matches = 35

Average runs of player in first 12 matches = 45

Formula:

Average = (Sum of all observations)/(Total number of all observations)

Calculation:

Sum of runs of player in 20 matches = 35 × 20 = 700

Sum of runs of player in first 12 matches = 12 × 45 = 540

Sum of runs of player in last 8 matches = 700 - 540 = 160

∴ Average runs of player in last 8 matches = 160/8 = 20

GIVEN:

 the average score of twenty-five matches = 75 runs

 In the next five-matches score = 240.

FORMULA USED:

Average = Sum of numbers/Total numbers

CALCULATION:

Total runs till twenty-five matches = 25 × 75 = 1875

So,

New average = [1875 + 240]/30 = 2115/30

New average = 70.5

Given:

Total number of matches = 5

The average of the first 3 = t

Calculations:

Let the three scores be a – 1, a, a + 1.

∴ average of 3 scores

⇒ (a – 1 + a + a + 1)/3

⇒ 3a/3

⇒ a

∴ a = t

Similarly, the average of 5 score is:

⇒ (a – 1 + a + a + 1 + a + 2 + a + 3)/5

⇒ (5a + 5)/5

⇒ a + 1

⇒ t + 1

The average of the five score is t + 1.

Calculation:

The Sum of integers is

⇒ 30 × 5 = 150

According to the question, 20 integers not exceed 5, and asked the maximum average

So, we have to find the minimum value of the remaining value of 10 integers

Which should be greater than 5.

Sum of 10 integers whose value of 6 is

⇒ 6 × 10 = 60

Remaining sum

⇒ 150 - 60 = 90

Average of 20 integers.

⇒ 90/20 = 4.5

∴ The highest average of 20 integers is 4.5.

Given:

Average of 7 numbers = 27

Average of first two numbers = 9

Average of next two numbers = 14

Formula Used:

Average = Sum of terms/Number of terms

Calculations:

Let fifth, sixth and seventh numbers be x, y, z respectively.

As x is twice of y and y is 50% more than z.

⇒ x : y = 6 : 3, y : z = 3 : 2

⇒ x : y : z = 6 : 3 : 2

Total sum of 7 terms = 7 × 27 = 189

Sum of first 2 terms = 9 × 2 = 18

Sum of next 2 terms = 14 × 2 = 28

Total sum = Sum of first 2 terms + Sum of next 2 terms + x + y + z

⇒ 189 = 18 + 28 + x + y + z

⇒ x + y + z = 143

1 unit = 130/(x + y + z) = 143/11

⇒ 1 unit = 13

⇒ x = 6 × 13 = 78

     y = 3 × 13 = 39

     z = 2 × 13 = 26

∴ The value of seventh number is 26.

Formula Used:

Average = Sum of all items / Number of items 

Calculation:

The average of A, B and C = 27

⇒ 27 = (A + B + C)/3

⇒ 81 = (A + B + C)                   ...................(1)

The average of B, C and D = 25

⇒ 25 = (D + B + C)/3

⇒ 75 = (D + B + C)                        ....................(2)

The value of D = 4                      ....................(3)

⇒ B + C = 75 - 4 = 71      

From (1), A + 71 = 81

∴ A = 10

The correct option is 1 i.e. Both statements I and statement II are required.

Given:

The first number = 40 

The second or third number = 25% more than the first number or third number

The fourth number = 5/4 th of the third number 

The fifth number = 3/2 th  third number 

Calculations:

Let the third number be x

Second number is either 25% more than third or first number = 125% of x or 40 × 125%

⇒ (5/4) × x or 50

Then accordingly,

5x/4 = 50

⇒ x = 40

The fourth number = (5/4)x = (5/4) × 40 = 50

The fifith number = (3/2)x = (3/2) × 40 = 60

Average = [40 + 50 + 40  + 50 + 60]/5

⇒ 240/5

⇒ 48

∴ The average of the five numbers is 48.

Given:

Average votes obtained by 5 candidates in an election were 630 votes. 

Average votes obtained by other 3 candidates were 260 votes.

Concepts used:

Average votes = (Sum of votes obtained by all candidates)/(Number of candidates)

Calculations:

Sum of votes obtained by 5 candidates = 630 × 5 = 3,150 votes

Sum of votes obtained by 3 candidates = 260 × 3 = 780 votes

Total votes obtained by (5 + 3 = 8) candidates = 3150 + 780 = 3930 votes 

Average votes obtained by all candidates = 3,930/8 votes = 491.25 votes

∴ The average votes obtained by 8 candidates in an election were 491.25 votes.

Given:

The average of 5 numbers is 55.

The first number is double of the second number and three-fourth of the third number.

The sum of fourth and fifth number = 88.

Concepts used:

Average = (Sum of all observations)/(Number of observations)

Calculations:

Let the first number, second number, third number, fourth number, fifth number be a, b, c, d and e respectively.

Sum of all observations = 55 × 5 = 275

⇒ a + b + c + d + e = 275      ----(1)     

First number is double of the second number. 

⇒ a = 2b

⇒ b = a/2    

First number is 3/4th of third number.

⇒ a = 3c/4

⇒ c = 4a/3   

The sum of remaining two numbers is 88.

⇒ d + e = 88    

Putting the value of b, c, d and e in eq (1),

⇒ a + b + c + d + e = 275 

⇒ a + a/2 + 4a/3 + 88 = 275

⇒ 17a/6 = 275 - 88

⇒ 17a/6 = 187

⇒ a = 187 × 6/17 = 66

The third number = c = 4a/3 = 4 × 66/3 = 88

∴ The value of third number is 88.

GIVEN:

average of 8 students = 19 years

average of first 3 = 20 years 

average of last 4 = 17 years

FORMULA USED:
average = sum of all observations/total number of observations

CALCULATION:

sum of age of all students = 19 × 8 = 152 years

sum of age of first 3 = 20 × 3 = 60

sum of age of last 4 = 17 × 4 = 68

age of fourth student = 152 - (60 + 68) = 24 years

∴ age of 4th student is 24 years

Given:

Total numbers = 15 

By mistake, the student wrote 43 instead of 34 and 13 instead of 31.

The incorrect average = 30.

Concepts used:

Average = (Sum of all observations)/(Number of observations)

Calculation:

Sum of all observations = 30 × 15 = 450

Correct sum of observations = Incorrect sum + Correct number - Incorrect number 

⇒ 450 + 34 + 31 – 43 – 13 = 459

Correct average = 459/15 = 30.6

∴ The correct average is equal to 30.6.

Given:

Average distance covered on Monday and Thursday = 110 km

Average distance covered on Tuesady, Wednesday, Friday, Saturday and Sunday = 75 km

Formula used∶

Average = Sum of all observation/Number of observation

Calculation∶

Distance travelled on Monday and Thursday = 110 × 2

⇒ 220 km

Distance travelled on Tuesday, Wednesday, Friday, Saturday and Sunday = 75 × 5

⇒ 375 km

Total distance travelled from Monday to Sunday = 220 + 375

⇒ 595 km

∴ The total distance travelled by Mr. R is 595 km.

Clearly 13th result=(sum of 25 results)-(sum of 24 results) 

=(18*25)-(14*12)+(17*12) 

=450-(168+204) 

=450-372 

=78. 

Given:

A’s weight = B’s weight + 10

B’s weight = C’s weight - 10

C’s weight = 50 kg

Concept:

Average weight = Total weight / Total persons

Solution:

C’s weight = 50 kg

B’s weight = (C’s weight - 10) ⇒ 50 - 10 ⇒ 40 kg

A’s weight = B’s weight + 10 ⇒ 40 + 10 ⇒ 50 kg

Given:

Average of the five numbers = -5

Sum of the first three numbers = 15

Formula Used:

Average of observations = Sum of all the numbers in the observations/Number of observations

Calculation:

Let the five numbers be a, b, c, d, and e.

So, we have:

a + b + c + d + e = 5 × -5 = -25      ----(i)

Also, a + b + c = 15      ----(ii)

On substituting the value of equation (i) into equation (ii), we get:

15 + d + e = -25

⇒ d + e = -40

⇒ Average of d and e = -40/2 = -20

∴ The average of the last two numbers is -20 

Given:

Average weight of A, B and C = 65 kg

Average weight of C and B = 61.5 kg

Average weight of A and C = 68.5 kg

Formula used:

Average = sum of weights/No of persons

Calculation:

Weight of A, B and C

⇒ (A + B + C)/3 = 65

⇒ A + B + C = 195

Weight of C and B

⇒ (C + B)/2 = 61.5

⇒ C + B = 123

Weight of A and C

⇒ (A + C)/2 = 68.5

⇒ A + C = 137

Weight of A

⇒ 195 - 123

⇒ 72

Weight of C

A + C = 137

⇒ C = 137 - 72

⇒ 65 Kg

∴ The weight of C is 65 Kg.

Formula used∶

Average = sum of all the observations / total number of observations

Calculations∶

Let the number of males, females and children be M, F and C respectively

M + F + C = 140

Number of males = 1 / 5 × 140 = 28

Number of females = (140 - 28) / 4 = 28

Number of children = 140 - 28 - 28 = 84

Average weight of children = 30 kg

Sum of weights of children = no. Of children × avg. Weight

⇒ 84 × 30

⇒ 2520 kg

Average weight of females = 2(30) = 60 kg

Sum of weights of females = no. Of females × avg. Weight

⇒ 28 × 60 = 1680 kg

Average weight of male = 1.5(60) = 90 kg

Sum of weights of males = no. Of males × avg. Weight

⇒ 28 × 90 = 2520 kg

Given:

10^th inning score = 70

Average increment = 1

Concept:

Average score = Total scores / Total number of innings

Solution:

Let average score before 10^th inning = p

New average score after 10^th inning = p + 1

Total score before 10^th inning = 9p

Average =

⇒ (9p + 70) / 10 = p + 1

⇒ 9p + 70 = 10p + 10

⇒ 10p - 9p = 70 - 10

⇒ p = 60

Given:

Number of boys with average age of 20 years = 15

Age of the boy who left = 23 years 

Age of the boys who joined = 18 years and 21 years

Formula Used:

Average = Sum of observations/Total number of observations

Calculation:

Sum of the ages of 15 boys = 15 × 20 = 300 years

After leaving and joining of boys,

Number of boys = 15 – 1 + 2 = 16

Sum of the ages of 16 boys = 300 – 23 + 18 + 21 = 316 years

Hence, the new average = 316/16 = 19.75 years

∴ The new average age of the group is 19 years and 9 months 

Given:

Average height of 45 sticks = 157 cm

New average height of the sticks after removing some sticks of height 163 cm = 153 cm

Formula used:

Average of n numbers = Sum of all the numbers/n

Calculation:

Total height of all the sticks initially = 45 × 157 = 7065

Let n number of sticks were removed

⇒ Total height of removed sticks = 163n

Also, according to the question:

Total height of remaining sticks = (45 - n)(153) = 6885 - 153n

Then, 7065 = 6885 - 153n + 163n

⇒ 180 = 10n

⇒ n = 18

∴ Remaining sticks = 45 - 18 = 27.

Given:

Average age of 12 boys = 10 years

Average age of 11 boys = 9 years

Formula used:

Sum of ages = Average age × Number of persons

Calculation:

Total age of 12 boys = 10 × 12

⇒ 120 years

Total age of 11 boys = 9 × 11

⇒ 99 years

Age of 12th boy = (120 – 99) years

⇒ 21 years

∴ The age of 12th boy is 21 years

Given:

The average area of I and II room = 400 square meters

The average area of I, II, III room = 500 square meters

Area of IV room = Area of III room + 600 square meters

The average area of II, III and IV room = 900 square meters

Formula used:

Average = Sum of all the observation/Total number of observation

Calculation:

Area of (I + II) room = 800  ….. (1)

Area of (I + II + III ) = 1500    …..(2)

Area of (II + III + IV) = 2700     ….(3)

From (1) and (2)

⇒ 800 + III = 1500

⇒ III = 1500 – 800

⇒ III = 700

Area of IV room = 700 + 600

= 1300 square metres

From (3)

⇒ II +700 + 1300 = 2700

⇒ II = 2700 – 2000

⇒ II = 700   ….. (4)

From (1) and (4)

⇒ I + 700 = 800

⇒ I = 800 – 700

⇒ I = 100

Given:

Score in 22nd innings is 210

Increase in average is 5

Formula Used:

Total score = Average × No. of Innings

Calculations:

Let the initial average be ‘k’

⇒ New average = k + 5

⇒ Total Score after 21innings = 21k

⇒ Total Score after 22 innings = 22(k + 5) = 22k + 110

⇒ 21k + 210 = 22k + 110

⇒ k = 100

∴ New Average = k + 5 = 105

Given:

First 5 odd integers = 1, 3, 5, 7 and 9

Formula used:

Mean = Sum of observations/Number of observations

Calculation:

Sum of observations (1 + 3 + 5 + 7 + 9) = 25

Number of observations = 5

Mean = 25/5 = 5

∴ The mean of first five odd integers is 5

Given:

Average age of 10 people in a village = 40 years

Formula Used:

Average Age of people in a group = Total of the age of all the people/Total number of people in the group

Calculation:

The total of age of all the 10 people in the village = 40 × 10 = 400

∵ The average age increased by 1 year, the new average age = 40 + 1 = 41

Total number of people now = 10 – 1 = 9

Hence the total age of all the remaining 9 people now = 41 × 9 = 369

So, the age of the person deceased = 400 - 369 = 31 years

∴ The age of the deceased person is 31 years

Formula Used:

Average = Sum of values/total no. of values

Calculation:

⇒ Average = (39 + 45 + 51 + 57 + 63 + 69)/6 = 324/6

∴ Average = 54

Given:

The first, second and third class third class train fare is Rs. 15, Rs. 10, and Rs. 7.50 per km respectively.

The ratio of numbers travelling in these categories is 2 : 5 : 13

Formula used:

Average = (Sum of total number)/(Total number of observation)

Using the simple concept of ratio.

Calculation:

Let the number of passenger in someday 2x, 5x and 13x.

Total fare by first fare = 15 × 2x = 30x

Total fare by first fare = 10 × 5x = 50x

Total fare by first fare = 7.5 × 13x = 97.5x

Total number of passenger = 2x + 5x + 13x = 20x

Total fare = 30x + 50x + 97.5x = 177.5x

Average = (Sum of total number)/(Total number of observation)

⇒ Average = (177.5x)/(20x)

⇒ 8.875

∴ The average fare per passenger for that day is 8.875

Given:

2 dice thrown simultaneously

Calculation:

Total number of outcomes n(T) = 6 × 6 = 36

⇒ Getting the sum divisible by 2 n(E) = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)}

 ⇒ 18 outcomes

P(E) = 18/36 = 1/2 

∴ P(E) is 1/2

Given:

Average age of 5 students = 15

The average increase by 1 when one new student join the group

Formula used:

Sum of ages = Average × Number of persons

Calculation:

Sum of ages of 5 students = (15 × 5) years

⇒ 75 years

New average after a new student joined the group = (15 + 1)

⇒ 16

Sum of ages of 6 students including new joinee = (16 × 6) years

⇒ 96 years

Age of new student = (96 – 75) years

⇒ 21 years

∴ The age of new student is 21 years

Given:

Average age of 20 persons = 18

A new person of 39 years

Formula used:

Sum of ages = Average age × Number of persons

Calculation:

Sum of age of all the 20 persons = 20 × 18 = 360 years

After a 39 years person has joined

Sum of all 21 persons = 360 + 39 = 399 years

New average of all 21 persons = 399/21 = 19 years

∴ The new average age of all persons is 19 years

Given:

Average age of 25 men = 28 years

Formula:

Average = Sum of all observations/Total number of all observation

Calculation:

Sum of age of 25 men = 28 × 25 = 700

If 5 men of average age of 25 years joined them, then

Sum of age of 5 men = 25 × 5 = 125 years

Now,

Sum of age of 30 men = 700 + 125 = 825 years

∴ Average age of 30 men = 825/30 = 27.5 years 

Given:

Average age of the first three man = 8

Average age of the last three man = 16

Average age of the 5 men = 12

Formula used:

Sum of ages = Average × Number of persons

Calculation:

Total age of 5 men = 12 × 5 years

⇒ 60 years

Total age of first three-man = 8 × 3

⇒ 24 years

Total age of last three men = 16 × 3

⇒ 48 years

Age of third man is = (24 + 48) – 60 years

⇒ 72 – 60 years

⇒ 12 years

∴ The age of third man is 12 years

Given:

Average of twelve numbers = 39

Average of last five numbers = 35

Average of first four numbers = 40

Fifth number = sixth number - 6

Fifth number = seventh number + 5

Formula used:

Average = (Sum of values)/(Number of values)

Calculations:

Let twelve numbers be x₁, x₂, x₃, x₄, x₅, x₆, x₇, x8, x9, x10, x11, x₁₂.

x5 = x6 - 6      ----(i)

x5 = x7 + 5      ----(ii)

Average of first four numbers,

(x1 + x2 + x3 + x₄)/4 = 40

⇒ (x1 + x2 + x3 + x4) = 160

Average of last five numbers,

(x₈ + x₉ + x₁₀ + x₁₁ + x₁₂)/5 = 35

⇒ (x8 + x9 + x10 + x11 + x12) = 175

Average of twelve numbers,

(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12)/12 = 39

⇒ (x1 + x2 + x3 + x4 + ​x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12) = 468

⇒ (160 + ​x5 + x6 + x7 + 175) = 468

⇒ ​x5 + x6 + x7 = 468 - 160 - 175

⇒ ​x5 + x6 + x7 = 133      ----(iii)

Putting value of x₅ from equation (i),

⇒ x6 - 6 + x6 + x7 = 133

⇒ 2x6 + x7 = 139

⇒ x7 = 139 - 2x6      ----(iv)

Putting value of x5 from equation (ii),

x7 + 5 + x6 + x7 = 133

⇒ 2x7 + x6 = 128

⇒ 2(139 - 2x6) + x6 = 128

⇒ 278 - 4x₆ + x₆ = 128

⇒ 3x₆ = 150

⇒ x₆ = 50

Putting the value of x₆ in equation (iv),

x7 = 139 - 2x6

⇒ x7 = 139 - 2(50)

⇒ x7 = 139 - 100

⇒ x7 = 39

Average of x₆ and x₇ = (50 + 39)/2

⇒ 89/2

⇒ 44.5

∴  The average of the sixth and seventh numbers is 44.5

Given:

Initial average of the group = 28 kg

Final average after three members were included = 32 kg

Weights of members included in the group = 32 kg, 34 kg and 46 kg

Formula used:

Average = Sum of observations/ Number of observations

Calculation:

Let the number of members initially in the group be x

Initial sum of observations = (28 × x)

Total number of observations when three new members joined the group

⇒ [(28 × x) + (32 + 34 + 46)]

Total number of members in the group after three members were added

⇒ (x + 3)

According to Question,

[(28 × x) + (32 + 34 + 46)]/(x + 3) = 32

⇒ [28x + 112]/(x + 3) = 32

⇒ [28x + 112] = 32 × (x + 3)

⇒ 28x + 112 = 32x + 96

⇒ 112 – 96 = 32x – 28x

⇒ 16 = 4x

⇒ x = 4

∴ The initial number of members in a group is 4.