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200-301 · Question #179

200-301 Question #179: Real Exam Question with Answer & Explanation

The correct answer is C: 192.168.25.16 255.255.255.240. The binary version of 20 is 10100. The binary version of 16 is 10000. The binary version of 24 is 11000. The binary version of 28 is 11100. The subnet mask is /28. The mask is 255.255.255.240. From the output above, EIGRP learned 4 routes and we need to find out the summary of th

Submitted by the_admin· Mar 5, 2026IP Connectivity

Question

Refer to the exhibit. Which address and mask combination represents a summary of the routes learned by EIGRP?

Options

  • A192.168.25.0 255.255.255.240
  • B192.168.25.0 255.255.255.252
  • C192.168.25.16 255.255.255.240
  • D192.168.25.16 255.255.255.252
  • E192.168.25.28 255.255.255.240
  • F192.168.25.28 255.255.255.252

Explanation

The binary version of 20 is 10100. The binary version of 16 is 10000. The binary version of 24 is 11000. The binary version of 28 is 11100. The subnet mask is /28. The mask is 255.255.255.240. From the output above, EIGRP learned 4 routes and we need to find out the summary of them: -> The increment should be: 28 - 16 = 12 but 12 is not an exponentiation of 2 so we must choose 16 (24). Therefore the subnet mask is /28 (=1111 1111.1111 1111.1111 1111.11110000) = So the best answer should be 192.168.25.16 255.255.255.240

Topics

#EIGRP#Route summarization#Subnetting

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