Need help with a problem!

in CCNA & CCENT
My sister got this problem for homework. I am a CCENT and I was not able to figure it out! Can anyone solve it and explain to me how it is solved?
Rewrite this forwarding table using the a.b.c.d/x notation instead of the binary string notation.
Destination Address Range | Link Interface
Destination Address Range: 11001000 00010000 00000000 through 11001000 00010111 00010111 11111111 Link Interface: 0
Destination Address Range: 11001000 00010111 00011000 00000000 through 11001000 00010111 00011000 11111111 Link Interface: 1
Destination Address Range: 11001000 00010111 00011001 00000000 through 11001000 00010111 00011001 00000000 through 11001000 00010111 0001111 11111111 Link Interface: 2
Destination Address Range: otherwise Link Interace: 3
Rewrite this forwarding table using the a.b.c.d/x notation instead of the binary string notation.
Destination Address Range | Link Interface
Destination Address Range: 11001000 00010000 00000000 through 11001000 00010111 00010111 11111111 Link Interface: 0
Destination Address Range: 11001000 00010111 00011000 00000000 through 11001000 00010111 00011000 11111111 Link Interface: 1
Destination Address Range: 11001000 00010111 00011001 00000000 through 11001000 00010111 00011001 00000000 through 11001000 00010111 0001111 11111111 Link Interface: 2
Destination Address Range: otherwise Link Interace: 3
Comments
Convert the binary values to decimal then use that to derive the subnet mask, start there
To make life easier, create a table with the following columns.
128 - 64 -32 -16 - 8 - 4 - 2 - 1
Use this to help you determine the value of each binary string. Once you have that figured out it should be easy.
http://yuri.easytospell.net
Ok, what do I do after this? (This is for the first one).
11000000 10101000 00000000 00000000 - 11000000 10101000 00000011 11111111 in decimal this is 192.168.0.0 - 192.168.3.255.
Take a look at the third octet where the increment is 4, so the mask would be 255.255.252.0 or /22.
So the network address would be 192.168.0.0/22
Hope this makes sense.