Solution:
Since the month begins with a Sunday, to there will be five Sundays in the month.
$$\eqalign{ \therefore {\text{Required}}\,{\text{average}} \cr = {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \cr = \frac{{8550}}{{30}} \cr = 285 \cr} $$

Solution:
Sum of the weights of the students after replacement
= [(52 × 45) - (48 × 5) + (54 × 5)] kg
= 2370 kg
∴ New average
= $$\left( {\frac{{2370}}{{45}}} \right)$$ kg
= $$52\frac{2}{3}$$ kg

Solution:
Sum of the present ages of A and B
= (18 × 2 + 4 × 2) years
= 44 years
Sum of the present ages of A, B and C
= (24 × 3) years
= 72 years
C's present age
= (72 - 44) years
= 28 years
∴ C's age after 8 years
= (28 + 8) years
= 36 years

Solution:
Let a, b, c, d are four number
∵ Average of first three number a, b, c = 16
Total of (a + b + c) = 16 × 3 = 48.....(i)
Again
∴ Average of last 3 numbers b, c, d = 15
⇒ Total of (b + c + d) = 15 × 3 = 45.....(ii)
⇒ Subtracting equation (i) - (ii), we get
⇒ (a + b + c) - (b + c + d) = 48 - 45
⇒ a - d = 3
⇒ a = - 20 = 3       [Given d = 20]
⇒ a = 23
⇒ Therefore, first number a = 23

Solution:
According to the question,
Average = $$\frac{1 + 2 + 2 + 3 + 3 + 3 + 4 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 6 + 6 + 7 + 7 + 7 + 7 + 7 + 7 + 7}{28}$$
Average = $$\frac{1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 + 7 × 7}{28}$$
Average = $$\frac{1 + 4 + 9 + 16 + 25 + 36 + 49}{28}$$
Average = 5

Solution:
Let the eight number be x
Then, sixth number = (x - 7)
Seventh number = (x - 7) + 4 = (x - 3)
So,
$$ \Leftrightarrow \left( {2 \times 15\frac{1}{2}} \right) + \left( {3 \times 21\frac{1}{2}} \right)$$     $$ + \left( {x - 7} \right)$$   $$ + \left( {x - 3} \right)$$   $$ + \,x = 8 \times 20$$
$$\eqalign{ \Leftrightarrow 31 + 64 + \left( {3x - 10} \right) = 160 \cr \Leftrightarrow 3x = 75 \cr \Leftrightarrow x = 25 \cr} $$

Solution:
Sum of numbers
= (41.4 × 15)
= 621