Average (SSC)

Original number of students (N) = 25

Original average weight (A) = 28 kg

Number of new students (n) = 10 

Difference between the average weights (d) = 2 kg

Now, apply this shortcut formula:

Average weight of new students = A + (N/n + 1)d

= 28 + (25/10 + 1)2

= 35 kg

So, the average weight of 10 new students is 35 kg.

Let the average of the remaining 3 quantities be x.

Then, as per the question,

Average of 9 quantities is 21.

So, 21 = (6 * 22 + 3 * x)/9

189 = 132 + 3x

x = 57/3

x = 19

So, required average = 19

Number of labourers of factory A (m) = 21

Number of labourers of factory B (N) = 24

Average salary of labourers of factory A (a) = Rs. 5520 

And average salary of labourers of factory B (b) = Rs. 6530 

Now, to find the average salary of labourers of both the factories, use this shortcut formula:

Average salary of labourers of both the factories = (ma + na)/(m + n)

= (5520 * 21 + 6530 * 21)/(21 + 24)

= 6058.67

So, the average salary of labourers of both the factories is Rs. 6058.67.

Using the shortcut formula:

Weight of the new man - weight of the removed man = NX 

Where, N = number of men in the group and X = increase or decrease in average 

Weight of the new man - 75 = 20(-2)

Weight of the new man = 75 - 40

= 35 kg

Original number of teachers (N) = 25 

Original average salary (A) = Rs. 6050 

Number of teachers who left the school (n) = 5

Difference between the average salaries (d) = Rs. 250 

Now, apply the following shortcut formula to find the average salary of teachers who left the school:

Average salary of the teachers who left the school = A - (1 - N/n)d

= 6050 - (1 - 25/5) * 250 

= Rs. 7050 

Now, total salary of 5 teachers = 5 * 7050

= Rs. 35250

Sum of quantities = average of quantities x number of quantities

Then, let the average price of a copy be Rs. x.

So, 32 * 155 + 40 * y = 6960

4960 + 40y = 6960

y = 50

Therefore, the average price of copies is Rs. 50.

Original average weight of (A) = 46 kg

Original number of candidates (N) = 31

Increase in average weight (d) = 0.25 kg

Now, apply the following shortcut formula:

Weight of the teacher = A + (N + 1 )d

= 46 + (31 + 1) * 0.25 

= 48 kg

Original average age (A) = 15 years

Original number of students (N) = 41

Difference in average ages (d) = 0.05 years

Now, apply the shortcut formula:

Age of the boy who left the class = A - (1 - N)d 

= 15 - (1 - 41) * 0.05 

= 17 years 

So, the age of the boy who left the class was 17 years.

Average of 15 numbers = 26.2

Then, sum of all 15 numbers = 26.2 * 15

= 393 

Now, average of first 8 numbers = 27.875 

So, sum of first 8 numbers = 27.875 * 8

= 223

And average of last 8 numbers = 25.125 

Then, sum of last 8 numbers = 25.125 * 8

= 201 

So, 8^th number = 203 + 201 - 393

= 424 - 393

= 31

Sum of quantities = average of quantities x number of quantities

So, average marks of class 9 = (22 * 52 + 23 * 50 + 25 * 48)/(22 + 23 + 25)

= 49.91

Average of A and B = 24 years

So, (A + B)/2 = 24

A + B = 48    -------------(1)

On replacing A with C,

(C + B)/2 = 23.5

C + B = 47   -------------(2)

On putting the value of B in equation (2), we get

C + 48 - A = 47

C - A = -1    ---------------(3)

Now, replacing B with C

(A + C)/2 = 22.5

A + C = 45 --------------(4) 

Now, on solving equations (3) and (4), we get

C = 22 years

Present average age of the family of 9 members is 26 years.

So, present sum of the ages of all the 9 members = 26 * 9 = 234 years

Now, sum of the ages at the time of birth of the youngest member = 234 - (8 * 10 + 10)

= 144 years

So, average age of 8 members at the time of birth of the youngest member = 144/8 = 18 years

Let the first number be x.

Then, second number = x/16 and third number = x/4 

Now, average of 3 numbers = 28

(x + x/16 + x/4)/3 = 28

x = 64

So, the first number is 64.

Then, second number = 64/16 = 4

Let his original average runs be x runs.

So, sum of runs of 12 innings = 12x 

New sum of runs after 13 innings = 12x + 69

So, new average = (12x + 69)/13

x - 2 = (12x + 69) /13

x = 95 

Therefore, original average runs = 95

The average of x, y, z and w is a.

So, a = (x + y + z + w)/4    --------------(1)

Now, average of x, y and a = (x + y + a)/3

On putting the value of (x + y) from equation (1) 

= (4a - z - w + a)/3

= [5a - (z + w)]/3