CCNA FOR YOU

QUESTION NO: 12
Which Layer 1 devices can be used to enlarge the area covered by a single LAN
segment? (Select two)
A. Switch
B. Router
C. NIC
D. hub
E. Repeater
F. RJ-45 transceiver
Answer: D, E
Explanation:
Both hub, Repeater, Router and Switch repeat the packet. But only hub and Repeater do
not segment the network. Repeaters and Hubs are contained in layer one of the OSI model
(Physical layer) while a switch lies in layer two and a router is in layer 3.

QUESTION NO: 13
CDP is running between two devices. What information is supplied by
CDP? (Select three)
A. Device identifiers
B. Capabilities list
C. Platform
D. Route identifier
E. Neighbor traffic data
Answer: A, B, C

QUESTION NO: 14
If a host on a network has the address 172.16.45.14/30, what is the address of the
subnetwork to which this host belongs?
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.18
Answer: D
Explanation:
The last octet in binary form is 00001110. Only 6 bits of this octet belong to the subnet
mask. Hence, the subnetwork is 172.16.45.12.

How many broadcast domains are shown in the graphic assuming only the default
VLAN is configured on the switches?
A. One
B. Two
C. Six
D. Twelve
Answer: A
Explanation:
There is only one broadcast domain because switches and hubs do not segment the
broadcast domains when only a single VLAN is configured. Only layer 3 devices can
segment the broadcast domains, or VLAN-capable switches where multiple VLANs are
configured.

QUESTION:
You have the binary number 10011101. Convert it to its decimal and hexadecimal
equivalents. (Select two answer choices)
A. 158
B. 0x9D
C. 156
D. 157
E. 0x19
F. 0x9F
Answer: B, D
Explanation:
10011101 = 128+0+0+16+8+4+0+1 = 157
For hexadecimal, we break up the binary number 10011101 into the 2 parts:
1001 = 9 and 1101 = 13, this is D in hexadecimal, so the number is 0x9D. We can further
verify by taking the hex number 9D and converting it to decimal by taking 16 times 9,
and then adding 13 for D (0x9D = (16x9)+13 = 157).

QUESTION NO:
The subnet mask on the serial interface of a router is expressed in binary as
11111000 for the last octet. How do you express the binary number 11111000 in
decimal?
A. 210
B. 224
C. 240
D. 248
E. 252
Answer: D
Explanation:
128 + 64+32+16+8 = 248. Since this is the last octet of the interface, the subnet mask
would be expressed as a /29.
Reference:
CCNA Self-Study CCNA ICND exam certification Guide

Incorrect Answers:
A. The number 210 would be 11010010 in binary.
B. The number 224 would be 11100000 in binary.
C. The number 240 would be 11110000 in binary
E. The number 252 would be 11111100 in binary. This is known as a /30 and is used
often in point-point links, since there are only 2 available addresses for use in this subnet.