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300-410 Practice Question: Router R1 and R2 are running EIGRP as the IGP,…

Router R1 and R2 are running EIGRP as the IGP, and R1 is redistributing a connected subnet 10.1.1.0/24 into EIGRP. R2 also runs BGP with an external peer, and BGP is redistributing the same prefix 10.1.1.0/24 into EIGRP with a route-map that sets the administrative distance to 100. On R3, a downstream EIGRP router, 'show ip route 10.1.1.0' shows the route via R2. What is the most likely cause of suboptimal routing?

Answer choices

Why each option matters

Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.

Correct answer & explanation

The internal EIGRP route is not present in R3's routing table due to a distribute-list inbound from R1, so the redistributed route with AD 100 is the only path.

EIGRP internal routes have AD 90, external routes AD 170. Redistribution from BGP into EIGRP creates external routes (AD 170) unless a route-map changes the distance. Setting AD to 100 makes the redistributed route preferred over the original internal route (AD 90) because 100 > 90, so the internal route should be preferred. However, if the internal route is not present due to a filter, or if the distance is set lower than 90, the redistributed route is chosen. The correct answer is that the route-map set distance 100 is higher than 90, so the internal route (AD 90) is still preferred; the issue is that the internal route is being suppressed by a distribute-list on R3.

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

  • The route-map set distance 100 overrides the default AD, making the redistributed route preferred over the internal route (AD 90).

    Why it's wrong here

    100 is higher than 90, so internal route is still preferred.

  • The internal EIGRP route is not present in R3's routing table due to a distribute-list inbound from R1, so the redistributed route with AD 100 is the only path.

    Why this is correct

    If the internal route is filtered, the redistributed route becomes the best path, causing suboptimal routing.

  • The redistributed route has AD 170 by default, and the route-map is ignored because redistribution from BGP always uses AD 170.

    Why it's wrong here

    Route-map can change AD.

  • R3 prefers routes with lower metric, not AD, and the redistributed route has a better metric.

    Why it's wrong here

    AD is checked first; if AD is equal, metric is compared.

Visual reference

192.168.1.0 /24 256 addresses (254 usable) 192.168.1.0 /25 Subnet A 128 addr (126 usable) 192.168.1.128 /25 Subnet B 128 addr (126 usable) Borrowing 1 bit from host portion creates 2 subnets (/25)

Quick reference

Routing Protocol Comparison

ProtocolMetricMax HopsAlgorithmType
RIP v2Hop count15Bellman-FordDistance vector
OSPFCost (bandwidth)UnlimitedDijkstra (SPF)Link state
EIGRPComposite metricUnlimitedDUALHybrid
IS-ISCostUnlimitedDijkstraLink state
BGPPolicy / attributesUnlimitedPath vectorPath vector

RIP's 15-hop limit makes it unsuitable for large networks. OSPF and EIGRP dominate modern enterprise deployments.

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JA

Written by Johnson Ajibi, MSc IT Security

Senior Network & Security Engineer · founder of Courseiva

This 300-410 practice question is part of Courseiva's free Cisco certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 300-410 exam.