hi guys,

im planning to take the exam v soon & was just needed help with the following practise q:

''The plans for a new office requiere 300 network connections, which of the following network sizes should be allocated to the new office?,,

- a: /23

- b: /25

- c: /24

- d: /30

apparently the answer is A but can somebody kindly explain how this is so?

**0**

# Subnet Q Help Plz

Started by
z_ahmed
, Oct 14 2011 07:00 AM

3 replies to this topic

### #1

Posted 14 October 2011 - 07:00 AM

### #2

Posted 14 October 2011 - 08:51 AM

QUOTE (z_ahmed @ Oct 14 2011, 11:00 AM) <{POST_SNAPBACK}>

hi guys,

im planning to take the exam v soon & was just needed help with the following practise q:

''The plans for a new office requiere 300 network connections, which of the following network sizes should be allocated to the new office?,,

- a: /23

- b: /25

- c: /24

- d: /30

apparently the answer is A but can somebody kindly explain how this is so?

im planning to take the exam v soon & was just needed help with the following practise q:

''The plans for a new office requiere 300 network connections, which of the following network sizes should be allocated to the new office?,,

- a: /23

- b: /25

- c: /24

- d: /30

apparently the answer is A but can somebody kindly explain how this is so?

Look at it this way. You need 300 hosts, so you need atleast 9 host bits.

2^9 = 512-2 = 510 usable host addresses.

If you used 8 host bits, then it would have been

2^8 = 256-2 = 254 usable host bits.

By that logic, subtract 9 host bits from 32 bits of IP address,

32-9 = 23.

So your Subnet mask is then,

32-9 = 23 = 255.255.254.0 or FFFF.FFFF.FFFE.0000 or 11111111.11111111.11111110.00000000.

Does it help? If not, ask away. But seriously, if you are not comfortable with this kind of subnet problems, may be you should consider reading some reference material on subnetting. Such problems are going to pop-up in real life scenarios, if you work in networking.

**Petite et accipietis**### #3

Posted 16 October 2011 - 12:43 PM

QUOTE (timtom22 @ Oct 14 2011, 05:51 PM) <{POST_SNAPBACK}>

Look at it this way. You need 300 hosts, so you need atleast 9 host bits.

2^9 = 512-2 = 510 usable host addresses.

If you used 8 host bits, then it would have been

2^8 = 256-2 = 254 usable host bits.

By that logic, subtract 9 host bits from 32 bits of IP address,

32-9 = 23.

So your Subnet mask is then,

32-9 = 23 = 255.255.254.0 or FFFF.FFFF.FFFE.0000 or 11111111.11111111.11111110.00000000.

Does it help? If not, ask away. But seriously, if you are not comfortable with this kind of subnet problems, may be you should consider reading some reference material on subnetting. Such problems are going to pop-up in real life scenarios, if you work in networking.

2^9 = 512-2 = 510 usable host addresses.

If you used 8 host bits, then it would have been

2^8 = 256-2 = 254 usable host bits.

By that logic, subtract 9 host bits from 32 bits of IP address,

32-9 = 23.

So your Subnet mask is then,

32-9 = 23 = 255.255.254.0 or FFFF.FFFF.FFFE.0000 or 11111111.11111111.11111110.00000000.

Does it help? If not, ask away. But seriously, if you are not comfortable with this kind of subnet problems, may be you should consider reading some reference material on subnetting. Such problems are going to pop-up in real life scenarios, if you work in networking.

Thank u & much appreciated

i was wondering is a calculator provided in the exam?

### #4

Posted 16 October 2011 - 06:52 PM

QUOTE (z_ahmed @ Oct 16 2011, 04:43 PM) <{POST_SNAPBACK}>

Thank u & much appreciated

i was wondering is a calculator provided in the exam?

i was wondering is a calculator provided in the exam?

Sorry , you are out of luck. No calculators are allowed in the exam.

However, they give you a marker, eraser, and a small board to scribble your notes on, calculations etc.

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