CCNA 640-802 Exam

QUESTION 21:
The corporate LAN shown in the Certkiller network uses IP network 172.28.4.0/22 for all departments. All workstations use 172.28.4.1 as a default gateway address. Network administrators have recently become concerned that excessive broadcasts are slowing network performance. Which change is most likely to reduce broadcast traffic on the corporate LAN?
Exhibit:


A. Configure an access control list on the router to prevent broadcast forwarding.
B. Configure each NIC and switch port to operate at full duplex.
C. Change the router-to-switch connection from Fast Ethernet to Gigabit Ethernet.
D. Implement VLANs after creating IP subnets for each department.
E. Increase the number of switches in the network closet of each department.

Explanation:
Switches using VLANs create the same division of the network into separate broadcast domains but do not have the latency problems of a router. Switches are also a more cost-effective solution.
There are several benefits to using VLANs, including:
1. Increased performance
2. Improved manageability
3. Network tuning and simplification of software configurations
4. Physical topology independence
5. Increased security options Increased performance
Switched networks by nature will increase performance over shared media devices in use today, primarily by reducing the size of collision domains. Grouping users into logical networks will also increase performance by limiting broadcast traffic to users performing similar functions or within individual workgroups. Additionally, less traffic will need to be routed, and the latency added by routers will be reduced.

QUESTION 22:
DRAG DROP



Answer:


QUESTION 23:
Two buildings on the San Jose campus of the Certkiller network must be connected to use Ethernet with a bandwidth of at least 100 Mbps. The company is concerned about possible problems from voltage differences between the two buildings. Which media type should be used for the connection?
Exhibit:


A. STP cable
B. Coaxial cable
C. Fiber optic cable
D. UTP cable
E. None of the above

Explanation:
Since fiber optic cable does not carry electrical charges, all electrical cable problems disappear. When fiber optic cable (outdoor quality) is used to link buildings, grounding problems, ground loops, and voltage spikes are eliminated and fiber-optic cable is all so immune to electronic eavesdropping. The other options (STP, Coax, and UTP) are all copper based and prone to electrical interferences.

QUESTION 24:
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

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 25:
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

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 (Cisco Press, ISBN
1-58720-083-X) Page 559

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.

QUESTION 26:
Which one of the binary number ranges shown below corresponds to the value of the first octet in Class B address range?
A. 10000000-11101111
B. 11000000-11101111
C. 10000000-10111111
D. 10000000-11111111
E. 11000000-10111111

Explanation:
Class B addresses are in the range 128.0.0.0 through 191.255.255.255.
In binary, the first octet (128 through 191) equates to 10000000-10111111

Incorrect Answers:
A. Binary 10000000 does equate to 128 but binary 11101111 equates to 239
B. Binary 11000000 equates to 192 and binary 11101111 equates to 239
D. Binary 10000000 does equate to 128 but binary 11011111 equates to 223
E. Binary 11000000 equates to 192 but binary 10111111 does equate to 191

QUESTION 27:
How would the number 231 be expressed as a binary number?
A. 11011011
B. 11110011
C. 11100111
D. 11111001
E. 11010011

Explanation
Decimal number 231 equates to 11100111 in binary (128+64+32+0+0+4+2+1)

Incorrect Answers:
A: Binary 11011011 equates to 219 (128+64+0+16+8+0+2+1)
B: Binary 11110011 equates to 243 (128+64+32+16+0+0+2+1)
D: Binary 11101011 equates to 249 (128+64+32+16+8+0+0+1)
E: Binary 11010011 equates to 211 (128+64+0+16+0+0+2+1)

QUESTION 28:
How would the number 172 be expressed in binary form?
A. 10010010
B. 10011001
C. 10101100
D. 10101110

Explanation:
10101100= 128 + 0 + 32 + 0 + 8 + 4 + 0 + 0 = 172

Incorrect Answers:
A. Binary 10010010 = 128+0+0+16+0+0+2+0=146
B. Binary 10011001 = 128+0+0+16+8+0+0+1=153
D. Binary 10101110 = 128+0+32+0+8+4+2+0= 174

QUESTION 29:
The MAC address for your PC NIC is: C9-3F-32-B4-DC-19. What is the address of the OUI portion of this NIC card, expressed as a binary number?
A. 11001100-00111111-00011000
B. 11000110-11000000-00011111
C. 11001110-00011111-01100000
D. 11001001-00111111-00110010
E. 11001100-01111000-00011000
F. 11111000-01100111-00011001

Explanation:
The first half of the address identifies the manufacturer of the card. This code, which is
assigned to each manufacturer by the IEEE, is called the organizationally unique
identifier (OUI). In this example, the OUI is C9-3F-32. If we take this number and
convert it to decimal form we have:
C9 = (12x16) + 9 = 201
3F = (3x16) + 15 = 63
32 = (3x16) + 2 = 50
So, in decimal we have 201.63.50. If we then convert this to binary, we have:
201 = 11001001
63 = 00111111
50 = 00110010
So the correct answer is D: 11001001-00111111-00110010

QUESTION 30:
How do you express the binary number 10110011 in decimal form?
A. 91
B. 155
C. 179
D. 180
E. 201
F. 227

Explanation:
If you arrange the binary number 10110011, against the place value and multiply the values, and add them up, you get the correct answer.
1 0 1 1 0 0 1 1
128 64 32 16 8 4 2 1
128 + 0 + 32 +16 + 0 + 0 +2 +1 = 179