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by hey_thr67 » Tue Jun 19, 2012 12:25 am
The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
(a) 246
(b) 248
(c) 492
(d) 15,128
(e) 30,256
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by gmat_and_me » Tue Jun 19, 2012 4:55 am
Is the the answer C ?

Originally the company would have had 122 * 122 codes. With
the addition of the two new codes, the total will be 124 * 124.
So, additional codes will be 122 * 2 + 2 * 124 (if you want to
look at the problem without doing complex calculation).

HTH
hey_thr67 wrote:The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
(a) 246
(b) 248
(c) 492
(d) 15,128
(e) 30,256
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by Anurag@Gurome » Tue Jun 19, 2012 7:43 am
hey_thr67 wrote:The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
If we use all the 124 signs, then each of the signs can go to any of the two positions. Hence, total number of possible codes = 124*124 = 124²

If we only 122 signs, then each of the signs can go to any of the two positions. Hence, total number of possible codes = 122*122 = 122²

Therefore, number of additional codes = (124² - 122²) = (124 - 122)(124 + 122) = 2*246 = 492

The correct answer is C.
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by dhonu121 » Tue Jun 19, 2012 8:37 am
Anurag@Gurome wrote:
hey_thr67 wrote:The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
If we use all the 124 signs, then each of the signs can go to any of the two positions. Hence, total number of possible codes = 124*124 = 124²

If we only 122 signs, then each of the signs can go to any of the two positions. Hence, total number of possible codes = 122*122 = 122²

Therefore, number of additional codes = (124² - 122²) = (124 - 122)(124 + 122) = 2*246 = 492

The correct answer is C.
Anurag: Could you please explain the problem to me ?
I am not sure what the problem is. What is meant by the statement :
If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?

How I interpreted this statement was that we have to create codes using just 2 characters since 122 out of 124 have already been used.
This can be achieved by 2*2=4 ways.

What is wrong here ?
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by Anurag@Gurome » Tue Jun 19, 2012 8:46 am
dhonu121 wrote: If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
This means the company is currently using "122 of the signs fully" and rest are unused and the question is asking "how many additional area codes can be created if the company uses all 124 signs".

Hence, the company is currently using 122*122 codes.

And if it was using all the 124 signs, it would have been using 124*124 codes.

Hence, number of additional codes = (124*124 - 122*122)

Hope that helps.
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by dhonu121 » Tue Jun 19, 2012 8:53 am
How I interpreted this statement was that we have to create codes using just 2 characters since 122 out of 124 have already been used.
This can be achieved by 2*2=4 ways.

What is wrong here ?
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by Anurag@Gurome » Tue Jun 19, 2012 8:56 am
dhonu121 wrote:How I interpreted this statement was that we have to create codes using just 2 characters since 122 out of 124 have already been used.
This can be achieved by 2*2=4 ways.

What is wrong here ?
I cannot explain the underlined part but the question clearly states "how many additional area codes can be created if the company uses all 124 signs" NOT "how many additional area codes can be created if the company uses only the unused signs"
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by sandeep_thaparianz » Tue Jun 19, 2012 8:57 am
124^2-122^2= 492
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by dhonu121 » Tue Jun 19, 2012 9:04 am
Anurag: Sorry for being a noob, but correct me what I am wrongly interpreting here.
The question says that the company used 122 out of 124 signs to create some codes of 2 digits.
Then, it asks that how many additional area codes can be created if it used all the 124 signs.
So, in effect, if we calculate the number of codes that can be created using 124-122 signs should be the answer ?

What is going wrong here ?
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by Anurag@Gurome » Tue Jun 19, 2012 10:09 am
Assume that the company had chose a special sign language containing only 3 different signs @, #, and $ to create an area code of two symbols.

Now the company has used only two signs. Say they are using only @ and #.
Hence, the company ends up with 2*2 = 4 codes (@#, @@, ##, #@)

Now if the company had used all the three signs, then they would have ended with 3*3 = 9 codes (@#, @@, ##, #@, @$, $$, $@, #$, $#)

Hence, number of additional codes = (9 - 4) = 5

But according to your understanding, we should calculate the number of two digit codes that can formed with (3 - 2) signs, i.e. only one possible code $$.

Hope it clears your confusion.
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by dhonu121 » Tue Jun 19, 2012 10:21 am
Thanks Anurag.
Got my mistake.
Basically, I was not taking into account the number of NEW 2 digit patterns that can be formed by taking the UNUSED digits with the USED digits.
I was taking the number of 2 digits patterns formed by just the 2 UNUSED digits.
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