Given :   $P = 4.4$ kW             $t = 4$ h

Electrical energy consumed by an electric appliance is equal to product of its power rating and time for which it is used $E = P \times t$.

Electricity bill is measured in units where 1 unit is equal to kWh which is the unit of energy. Hence, when we pay for our electricity bill, we pay for the amount of electrical energy consumed by us.

Energy consumed, $E =2.5 \times 2 = 5\ kWh $
Cost, $C = 5 \times 0.60 = Rs\ 3 $

For bulbs :

$Energy\,\,=\,power\times time$

Total hours in 30 days $5\times 30=150hours$

So, $ energy=0.1\times 150=15J $

 $  $Power of 4 bulbs

 $ =4\times 15=60kW $

 $  $For LED:

Total hours in 30 days $=150hours$

$E=0.008\times 150=1.2J$

So, power of 4 LEDs $=1.2\times 4=4.8hours$

Saved energy 

  $ =\,\,60-4.8 $

 $ =55.2units $

Energy consumed $=10\times 50\times 10\times 30\times 3600 J$
$[1 Kwh=3600\times 1000J]$
$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$

Daily usage of electricity   $E _1 = 2.5 \ kw\times 8 = 20 \ kwh = 20 \ units$

Power P is proportional to V^2. If V decreases by 2.5%, V_new = 0.975 V_old. P_new is proportional to (0.975)^2 = 0.9506. The decrease is approximately 5%.