how to tackle this route summarization qs?

[ciscok1d]

well i think i got it. the answer is (b).

Same answer I got and Im confident its correct.
My method:

    8      +    8    +6(common bits)      = /22
00001010.00000000.000000|00.00000000
00001010.00000000.000000|00.00000000
00001010.00000000.000000|01.00000000
00001010.00000000.000000|10.00000000
00001010.00000000.000000|11.00000000
    10      .      0    .    0        .    0      /22
 
So the first two octets are common between all 4 and are 8 bits each so 8+8 = 16
Also the 3rd octet has 6 bits in common with all 4 addresses as shown above, 16+6 = 22
so 10.0.0.0/22 is the answer