On what subnet can the address 210.17.23.200 /27 be found?
Remember the steps: First, convert the IP address and mask to
binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
210.17.23.200 11010010 00010001 00010111 11001000 Subnet Mask
255.255.255.224 (/27) 11111111 11111111 11111111 11100000 Boolean AND Result 11010010 00010001 00010111 11000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 1 0 1 0 0 1 0 210 Second Octet 0 0 0 1 0 0 0 1 17 Third Octet 0 0 0 1 0 1 1 1 23 Fourth Octet 1 1 0 0 0 0 0 0 192 The subnet on which this IP address is found is 210.17.23.192 /27.
On what subnet can the IP address 24.194.34.12 /10 be found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
24.194.34.12 00011000 11000010 00100010 00001100 Subnet Mask
255.192.0.0 (/10) 11111111 11000000 00000000 00000000 Boolean AND Result 00011000 11000000 00000000 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 0 0 0 1 1 0 0 0 24 Second Octet 1 1 0 0 0 0 0 0 192 Third Octet 0 0 0 0 0 0 0 0 0 Fourth Octet 0 0 0 0 0 0 0 0 0
Chris Bryant, CCIE #12933 www.thebryantadvantage.com
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© 2005 The Bryant Advantage
The subnet is 24.192.0.0 /10.
On what subnet can the IP address 190.17.69.175 /22 be found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
190.17.69.175 10111110 00010001 01000101 10101111 Subnet Mask
255.255.252.0 (/22) 11111111 11111111 11111100 00000000 Boolean AND Result 10111110 00010001 01000100 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 0 1 1 1 1 1 0 190 Second Octet 0 0 0 1 0 0 0 1 17 Third Octet 0 1 0 0 0 1 0 0 68 Fourth Octet 0 0 0 0 0 0 0 0 0 The subnet is 190.17.68.0 /22.
On what subnet can the IP address 111.11.126.5 255.255.128.0 be found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
111.11.126.5 01101111 00001011 01111110 00000101 Subnet Mask
255.255.128.0 (/17) 11111111 11111111 10000000 00000000 Boolean AND Result 01101111 00001011 00000000 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 0 1 1 0 1 1 1 1 111 Second Octet 0 0 0 0 1 0 1 1 11 Third Octet 0 0 0 0 0 0 0 0 0 Fourth Octet 0 0 0 0 0 0 0 0 0 Chris Bryant, CCIE #12933
www.thebryantadvantage.com
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© 2005 The Bryant Advantage
The subnet is 111.11.0.0 255.255.128.0, or 111.11.0.0 /17.
On what subnet can the IP address 210.12.23.45 255.255.255.248 be found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
210.12.23.45 11010010 00001100 00010111 00101101 Subnet Mask
255.255.255.248 (/29) 11111111 11111111 11111111 11111000 Boolean AND Result 11010010 00001100 00010111 00101000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 1 0 1 0 0 1 0 210 Second Octet 0 0 0 0 1 1 0 0 12 Third Octet 0 0 0 1 0 1 1 1 23 Fourth Octet 0 0 1 0 1 0 0 0 40 The subnet is 210.12.23.40 255.255.255.248.
On what subnet is the IP address 222.22.11.199 /28 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
222.22.11.199 11011110 00010110 00001011 11000111 Subnet Mask
255.255.255.240 (/28) 11111111 11111111 11111111 11110000 Boolean AND Result 11011110 00010110 00001011 11000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 1 0 1 1 1 1 0 222 Second Octet 0 0 0 1 0 1 1 0 22 Third Octet 0 0 0 0 1 0 1 1 11 Fourth Octet 1 1 0 0 0 0 0 0 192
The subnet is 222.22.11.192 /28, or 222.22.11.192 255.255.255.240.
Chris Bryant, CCIE #12933 www.thebryantadvantage.com
49
© 2005 The Bryant Advantage
On what subnet is the IP address 111.9.100.7 /17 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
111.9.100.7 01101111 00001001 01100100 00000111 Subnet Mask
255.255.128.0 (/17) 11111111 11111111 10000000 00000000 Boolean AND Result 01101111 00001001 00000000 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 0 1 1 0 1 1 1 1 111 Second Octet 0 0 0 0 1 0 0 1 9 Third Octet 0 0 0 0 0 0 0 0 0 Fourth Octet 0 0 0 0 0 0 0 0 0 The subnet is 111.9.0.0 /17, or 111.9.0.0 255.255.128.0.
On what subnet is the IP address 122.240.19.23 /10 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
122.240.19.23 01111010 11110000 00010011 00010111 Subnet Mask
255.192.0.0 (/10) 11111111 11000000 00000000 00000000 Boolean AND Result 01111010 11000000 00000000 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 0 1 1 1 1 0 1 0 122 Second Octet 1 1 0 0 0 0 0 0 192 Third Octet 0 0 0 0 0 0 0 0 0 Fourth Octet 0 0 0 0 0 0 0 0 0 The subnet is 122.192.0.0 /10, or 122.192.0.0 255.192.0.0.
Chris Bryant, CCIE #12933 www.thebryantadvantage.com
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© 2005 The Bryant Advantage
On what subnet is the IP address 184.25.245.89 /20 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
184.25.245.89 10111000 00011001 11110101 01011001 Subnet Mask
255.255.240.0 (/20) 11111111 11111111 11110000 00000000 Boolean AND Result 10111000 00011001 11110000 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 0 1 1 1 0 0 0 184 Second Octet 0 0 0 1 1 0 0 1 25 Third Octet 1 1 1 1 0 0 0 0 240 Fourth Octet 0 0 0 0 0 0 0 0 0
The subnet is 184.25.240.0 /20, or 184.25.240.0 255.255.240.0.
On what subnet is the IP address 210.67.39.5 /30 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
210.67.39.5 11010010 01000011 00100111 00000101 Subnet Mask
255.255.255.252 (/30) 11111111 11111111 11111111 11111100 Boolean AND Result 11010010 01000011 00100111 00000100
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 1 0 1 0 0 1 0 210 Second Octet 0 1 0 0 0 0 1 1 67 Third Octet 0 0 1 0 0 1 1 1 39 Fourth Octet 0 0 0 0 0 1 0 0 4
The subnet is 210.67.39.4 255.255.255.252, or 210.67.39.4 /30.
Chris Bryant, CCIE #12933 www.thebryantadvantage.com
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© 2005 The Bryant Advantage
On what subnet is the IP address 99.140.23.143 /10 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
99.140.23.143 01100011 10001100 00010111 10001111 Subnet Mask
255.192.0.0 (/10) 11111111 11000000 00000000 00000000 Boolean AND Result 01100011 10000000 00000000 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 0 1 1 0 0 0 1 1 99 Second Octet 1 0 0 0 0 0 0 0 128 Third Octet 0 0 0 0 0 0 0 0 0 Fourth Octet 0 0 0 0 0 0 0 0 0 The subnet is 99.128.0.0 /10, or 99.128.0.0 255.192.0.0.
On what subnet is the IP address 203.27.18.161 255.255.255.224 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
203.27.18.161 11001011 00011011 00010010 10100001 Subnet Mask
255.255.255.224 (/27) 11111111 11111111 11111111 11100000 Boolean AND Result 11001011 00011011 00010010 10100000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 1 0 0 1 0 1 1 203 Second Octet 0 0 0 1 1 0 1 1 27 Third Octet 0 0 0 1 0 0 1 0 18 Fourth Octet 1 0 1 0 0 0 0 0 160
The subnet is 203.27.18.160 255.255.255.224, or 203.27.18.160 /27.
Chris Bryant, CCIE #12933 www.thebryantadvantage.com
52
© 2005 The Bryant Advantage
On what subnet is the IP address 10.191.1.1 /10 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
10.191.1.1 00001010 10111111 00000001 00000001 Subnet Mask
255.192.0.0 (/10) 11111111 11000000 00000000 00000000 Boolean AND Result 00001010 10000000 00000000 00000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 0 0 0 0 1 0 1 0 10 Second Octet 1 0 0 0 0 0 0 0 128 Third Octet 0 0 0 0 0 0 0 0 0 Fourth Octet 0 0 0 0 0 0 0 0 0 The subnet is 10.128.0.0 /10, or 10.128.0.0 255.192.0.0.
On what subnet is the IP address 187.23.191.95 /27 be found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
187.23.191.95 10111011 00010111 10111111 01011111 Subnet Mask
255.255.255.224 (/27) 11111111 11111111 11111111 11100000 Boolean AND Result 10111011 00010111 10111111 01000000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 0 1 1 1 0 1 1 187 Second Octet 0 0 0 1 0 1 1 1 23 Third Octet 1 0 1 1 1 1 1 1 191 Fourth Octet 0 1 0 0 0 0 0 0 64
The subnet is 187.23.191.64 /27, or 187.23.191.64 255.255.255.224.
Chris Bryant, CCIE #12933 www.thebryantadvantage.com
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© 2005 The Bryant Advantage
On what subnet is the IP address 222.17.32.244 /28 found?
First, convert the IP address and mask to binary. Second, perform the Boolean AND. Third, convert the Boolean AND back to dotted decimal.
Octet 1 Octet 2 Octet 3 Octet 4 IP Address
222.17.32.244 11011110 00010001 00100000 11110100 Subnet Mask
255.255.255.240 (/28) 11111111 11111111 11111111 11110000 Boolean AND Result 11011110 00010001 00100000 11110000
Converting The Boolean AND Into Dotted Decimal:
128 64 32 16 8 4 2 1 Total First Octet 1 1 0 1 1 1 1 0 222 Second Octet 0 0 0 1 0 0 0 1 17 Third Octet 0 0 1 0 0 0 0 0 32 Fourth Octet 1 1 1 1 0 0 0 0 240
The subnet is 222.17.32.240 /28, or 222.17.32.240 255.255.255.240.
Chris Bryant, CCIE #12933 www.thebryantadvantage.com
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© 2005 The Bryant Advantage
Chris Bryant, CCIE #12933 ww.thebryantadvantage.com
© 2005 The Bryant Advantage
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