問題 1
Question#1Topic1
Refer to the exhibit. Which type of route does R1 use to reach host 10.10.13.10/32?
正解を表示
正解
B
Cisco · 200-301
200-301 試験練習問題
Cisco Certified Network Associate (CCNA)
200-301 の PDF 練習問題。資料の言語は 英語です。購入はこの試験のみで、自動更新はありません。
試験概要
200-301 は「Cisco Certified Network Associate (CCNA)」の認定試験です。試験形式、時間、対応言語は以下をご確認ください。最新の出題範囲や受験条件は公式サイトをご参照ください。
- 試験コード
- 200-301
- 公式の試験・資格名
- Cisco Certified Network Associate (CCNA)
- 試験形式
- Technical assessment; consult the official exam topics and booking details for item formats
- 試験時間
- 120 分
- 受験可能な言語
- 英語、日本語
公式試験の対応言語と練習問題ファイルの言語は異なります。試験の版と言語は公式の予約ページでご確認ください。
公式情報： 公式サイト
情報確認日：2026-10-20
対応言語
- 英語
- 簡体字中国語
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問題 2
Question#2Topic1
Refer to the exhibit. Which prefix does Router1 use for traffic to Host A?
正解を表示
正解
D
問題 3
Question#3Topic1
DRAG DROP -
Drag and drop the descriptions of file-transfer protocols from the left onto the correct protocols on the right.
Select and Place:
正解を表示
正解
DAE, CBF
問題 4
Question#4Topic1
A frame that enters a switch fails the Frame Check Sequence. Which two interface counters are incremented? (Choose two.)
正解を表示
正解
AD
問題 5
Question#5Topic1
DRAG DROP -
Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right.
Select and Place:
正解と解説を表示
正解
CBDEA
解説
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet 172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -
> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The 3rd octet of the network address is 228
(because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or
172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is 172.28.229.254.
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -
> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The 3rd octet of the network address is 228
(because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or
172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is 172.28.229.254.
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