Speeding up the heating of the heated bed Currently I am using a 12 volts, 20 amperes power supply (Model S-240-12) 



The stepper motors and the extruder need 5 amp, and the heated bed build plate needs 11 amp. 



Technically you only need to use a 12 Volts, 16 amperes power supply, but I understand that you use the one with 20 amp because pulse currents from extruders and stepper motors can be stressful to supplies loaded to the max, so for reliability and performance, it is better to use a supply rated for 25% more than you need 

In the place where I buy the spare parts for my 3d printer they also sell 12 V power supplies capable of delivering 25 amp and 30 amp and they told me that if I use those ones you are going to be able to speed up the heating of the heated bed. Is that true? I understand that the heated bed is only going to take the 11 amp that it needs so is not going to make any difference to use power supplies capable of delivering more current
Changing the PSU with one with a higher amperage will not make the bed heat up any faster unless the PSU is underrated for the amperage required and the voltage is dropping as a result of the load. This can be checked by measuring the output voltage with a multimeter (when the PSU is loaded e.g. by a heating heat bed). In this case, the PSU has a marginal higher Amperage than the printer consumes (even has some room for the over-voltage; under the assumption that it is a good working PSU). Increasing the voltage will decrease the heat up time. There is a screw next to the 12 V connectors that can change the output voltage of the PSU. Usually, it is safe to increase the voltage up to 14 V, but that depends on your whole setup (and 14 V is applied to the whole setup, increasing the current for all parts, including your printer controller board, this board must be rated for the 14 V).  Please do check the stability of the voltage during load.
Although it can be done, it is not something I used. What is an extra minute on a print of several hours?
You can do the math: say the heat bed has a resistance of 1.2 Ω. We only need two formulas:

$U=R\times I$ - potential Difference U is Resistance R times Current I
$P=U\times I=I^2\times R=\frac {U^2} R$. The power P of an item the potential difference times the current through the item.


at 12 V that will draw 10 Amps (12 V / 1.2 Ω) resulting in a 120 Watt bed: $P= 12^2 \text V \times 10^2 \text A= {10^2 \text A}\times {1.2\ \Omega}=\frac{12^2 \text V} {1.2\ \Omega} $),
at 14 V that same bed will draw 11.7 Amps (14 V / 1.2 Ω) resulting in a 163.3 Watt bed.

Use at your own risk!
What you could do to decrease time to heat the bed without changing the PSU or the voltage is to insulate the bottom of the heat bed with heat bed cotton sheets or cork (placemats from IKEA ;) ), put a sheet of cork onto the heat bed before printing and start heating the bed through the LCD panel of the printer or any attached printer controller programs over USB prior to printing.