Cidr

Question

An internet service provider (ISP) has the following chunk of IP address available with it: 192.248.128.0/22. The ISP wants to give half of this chunk to organization A and one fourth of remaining half to organization B and organization C then what is the valid allocation of address to A, B and C?
(A) 192.248.128./23, 192.248.128.0/23, 192.24S.128.0/25
(B) 192.248.128.0/23, 192.248.128.0/25, 192.248.128.0/25
(C) 192.248.128.0/22, 192.248.128.0/23, 192.248.128.0/24
(D) None
https://gateoverflow.in/?qa=blob&qa_blobid=12308793692587589147

Answer 1

192.248.128.0/22
Total host possible = 32 - 22 = 10 = > 2^10 = 1024

so half of chunk to A means 512 host = 2^9 = $2^{32-23}$

i.e. 192.248.128.0/23 or 192.248.130.0/23

now one fourth of rmaining half to B and C
means 128 host to each = 2^9 = $2^{32-25}

i.e B 192.248.130.0/25  or 192.248.128.0/25
    C 192.248.130.128/25  or 192.248.128.128/25

Here no matching combination in options
So option D.