Mean weight of 7 boys = 56 kg.

Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg

= 338 kg.

Weight of the 7th boy = (total weight of 7 boys) – (total weight of 6 boys)

= (392 – 338) kg

= 54 kg.

Hence, the weight of the seventh boy is 54 kg.

Calculated average height of 30 boys = 150 cm.

Incorrect sum of the heights of 30 boys

= (150 × 30)cm

= 4500 cm.

Correct sum of the heights of 30 boys

= (incorrect sum) – (wrongly copied item) + (actual item)

= (4500 – 135 + 165) cm

= 4530 cm.

Correct mean = correct sum/number of boys

= (4530/30) cm

= 151 cm.

Hence, the correct mean height is 151 cm

Total expenditure during the year

= $[6240 × 3 + 6780 × 4 + 7236 × 5]

= $ [18720 + 27120 + 36180]

= $ 82020.

Total income during the year = $ (82020 + 7080) = $ 89100.

Average monthly income = (89100/12) = $7425.

Hence, the average monthly income of the family is $ 7425.

The sum of remaining  50 number  ( i. e 52 – 2 )

52 x 45 – 45 – 55 = 2340 – 100 =  2240

The average of remaining numbers = 2240 / 50 = 44.8

When we multiply each number by  ” X “, the average would also get multiplied by “X”.

Hence the average of new set of numbers = 15 x 12 = 180.

Take average upto 19th innings is “A”

According to given information 19A + 95 = 20 (A + 4)

So  A = 15

Now average runs after 20th innings = 15 + 4 = 19

Here take sixth number = x , then

Seventh reading =  x + 5 and 8th reading = x +7

Average of 8 readings is 24.2

Sum of 8 readings = 24.2 x 8 = 193.6

Sum of first five numbers = ( 18.5 x 2) + (21.1 x 3) = 37 + 63.6 = 100.3

Now the sum of 6th, 7th & 8th = 193.6 – 100.3 = 93.3

So  ( x + x + 5 + x + 7) = 93.3

3x = 93.3 – 12 = 81.3

x = 27.1

Here using the formula 

P =18 km , Q = 16 km , R =30 km , X = 10 km/hr, Y= 8 km/hr & Z = 6 km/hr

= 

= 80/11

The average speed of the whole distance = 80/11  =  km/hr.

Given: Average weight of batch A = 40 kg , average weight of batch B = 35 kg

1) First find the total weight of all students
– Weight of batch A = (36 x 40) = 1440
– Weight of batch B = (44 x 35) = 1540

Total weight of all students = (1440 + 1540) = 2980 kg

2) Find average weight of whole class 
(Batch A + Batch B) students = (36 + 44) = 80 students

1) Sum of observations = Average x No. of observations

= 46 x 40 = 1840

2) Correct sum = Sum of observations + (38 – 33) 

= 1840 + (5)
= 1845

The average of 6 players = X

Average increases by 10, when seventh player makes a score of 112 runs.

Therefore, average of 7 players = X + 10

Here, average = X, number of players = 6
Hence,
Sum of scores = 6X

Score of 7 players = (Score of 6 players + score of 7 player) = (6X + 112) ——- (2)
Total average = (X + 10) ——— (3)

Substitute (2) and (3), in (1)

Solving we get,

X = 42

Average of first 6 players = 42

If the weight of the man would have been 30, then the average weight would have been the same. So, the extra 100 kg that the obese man brings with him would be distributed equally amongst all of them, i.e. 100/40 = 2.5
So, the average becomes 30 + 2.5 = 32.5

The sum of salary of officers will be = 30×120 = 3600
Let the number of labourers = X.
ATQ, 3600 + 40X = 50(30+X)
2100 = 10X
X= 210

Average of 4 terms = 20
Hence, the total sum of 4 terms = 80
Let terms be A,B,C,D
So, the sum will be A+B+C+D =80
Given, 3A = B+C+D
So, 4A = 80,
A = 20

Total score for 9 innings is 75×9 = 675
Total score after 10th innings = 675 + 100 = 775
So, average = 775 / 10 = 77.5