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GATE 2026 CS1 Q.56: solution and answer

GATE 2026 · CS1Q.56 · 2 marks · MSQ

An ISP having an address block 202.16.0.0/15 assigns a block of 6000 IP addresses to a client, using the classless internet domain routing (CIDR) super-netting approach. Which of the following address blocks can be assigned by the ISP?

  1. (A)

    202.16.0.0/19

  2. (B)

    202.17.64.0/19

  3. (C)

    202.16.32.0/19

  4. (D)

    202.17.24.0/19

See the question in the paper
GATE 2026 CS1 question 56Tap to open full size

Answer and solution

ANSWER

The ISP can assign (A) 202.16.0.0/19, (B) 202.17.64.0/19 and (C) 202.16.32.0/19. Option (D) is not a valid /19 block.

Official answer: A, B, CSource: Official answer key, GATE 2026 CS1 (IIT Guwahati)

  1. The ISP block 202.16.0.0/15 has host bits, so it holds addresses. A /15 fixes the first 15 bits, so the second octet may differ only in its last bit. The block therefore runs from 202.16.0.0 to 202.17.255.255.

  2. A block of 6000 addresses needs host bits, because . The client's block is therefore a /19 (), which holds 8192 addresses.

  3. A /19 block must be aligned. Its network address has all 13 host bits equal to zero. A /19 fixes 3 bits of the third octet, so the third octet must be a multiple of , that is 0, 32, 64, 96, and so on.

  4. A candidate must also lie inside the ISP's range, 202.16.0.0 to 202.17.255.255.

  5. (A) 202.16.0.0/19: the third octet 0 is a multiple of 32. The block runs from 202.16.0.0 to 202.16.31.255, inside the ISP range. Valid.

  6. (B) 202.17.64.0/19: the third octet 64 is a multiple of 32. The block runs from 202.17.64.0 to 202.17.95.255, inside the ISP range. Valid.

  7. (C) 202.16.32.0/19: the third octet 32 is a multiple of 32. The block runs from 202.16.32.0 to 202.16.63.255, inside the ISP range. Valid.

  8. (D) 202.17.24.0/19: 24 is not a multiple of 32, so the host bits are not all zero. The /19 that contains this address is 202.17.0.0/19, so 202.17.24.0/19 is not a valid block. Invalid.

  9. The blocks that can be assigned are A, B and C.

KEY IDEA

6000 addresses need 13 host bits, so the client gets a /19 of 8192 addresses. A valid /19 starts on a multiple of 32 in the third octet and lies inside the ISP's /15, which spans 202.16.0.0 to 202.17.255.255.

WATCH OUT FOR

Checking only that the block lies inside the ISP's range and forgetting alignment. 202.17.24.0 is inside the range, but a /19 must start on a multiple of 32 in the third octet, and 24 is not one. Another slip is rounding 6000 down to a /20 (4096 addresses), which is too small.

From the GATE 2026 CS1 paper. Every question of this paper · Download the full paper · All 20 IPv4 questions, with notes

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