Solution:
Required run rate
$$\eqalign{ = \frac{{282 - \left( {3.2 \times 10} \right)}}{{40}} \cr = \frac{{250}}{{40}} \cr = 6.25 \cr} $$

Solution:
$$\eqalign{ \frac{{\left[ {\left( {10 \times 50} \right) + \left( {50 \times 10} \right) + \left( {20 \times 25} \right) + \left( {25 \times 20} \right)} \right]}}{4} \cr = 500\,{\text{paise}} \cr} $$

Solution:
Sum of cubes of first $$n$$ positive consecutive numbers is
$$\eqalign{ = \frac{{{{\left\{ {n\left( {n + 1} \right)} \right\}}^2}}}{4} \cr \therefore \,{\text{Average}} = \frac{{n{{\left( {n + 1} \right)}^2}}}{4} \cr \Rightarrow n = 49 \cr \Rightarrow \frac{{49{{\left( {50} \right)}^2}}}{4} = 30625 \cr} $$

Solution:
According to the question,
First six odd numbers which are divisible by '7'
7, 21, 35, 49, 63, 77
difference in each value is 14
Sn = $$\frac{n}{2}$$ [2a + (n - 1) d]
Sn = $$\frac{6}{2}$$ [2 × 7 + (6 - 1) × 14]
Sn = 3 (14 + 70)
Sn = 252
∴ Average = $$\frac{252}{6}$$ = 42

Solution:
Average
$$ = \frac{{11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99}}{9}$$
$$ = \frac{{\left( {11 + 99} \right) + \left( {22 + 88} \right) + \left( {33 + 77} \right) + \left( {44 + 66} \right) + 55}}{9}$$
$$\eqalign{ = \frac{{4 \times 110 + 55}}{9} \cr = \frac{{495}}{9} \cr = 55 \cr} $$

Solution:
According to the question,
$$\eqalign{ \Rightarrow \frac{{{3^{30}} + {3^{60}} + {3^{90}}}}{3} \cr \Rightarrow {3^{29}} + {3^{59}} + {3^{89}} \cr} $$

Solution:
According to the question,
Annual income
= 1000 × 12
= Rs. 12000
Annual expenditure
= 1000 × 9
= Rs. 9000
Savings
= 12000 - 9000
= Rs. 3000