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A construction company wants to number new houses using digit plates only. If the company puts an order for 1212 plates how many houses are to be given numbers? (The numbers of houses are consecutive and the number of the first house is 1). (A) 260 (B) 440 (C) 556 (D) 792 (E) 1200 Source: GMAT Club Tests  hardest GMAT questions Could someone provide alternative solution, the provided solution is not clear ...? thanks.



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Re: M07#37 [#permalink]
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26 Dec 2008, 17:36
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gmatclub1 wrote: A construction company wants to number new houses using digit plates only. If the company puts an order for 1212 plates how many houses are to be given numbers? (The numbers of houses are consecutive and the number of the first house is 1). 1) 260 2) 440 3) 556 4) 792 5) 1200
Could someone provide alternative solution, the provided solution is not clear ...? thanks. Plates req for 1 digit plates = 19 = 9 Plates req for 2 digit plates = 1099 = 90 Plates req for 3 digit plates = 100 up = needs to find. 1,212 = plates req for single digit + plates req for two digits + plates req for three digit 1,212 = 9 (n) + 90 (2n) + x (3n) x = (1,212  9  180)/3 (since n = 1) x = 341 no of houses = 9 + 90 + 341 = 440.
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Re: M07#37 [#permalink]
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15 Aug 2009, 00:21
Gmat tiger,
I'm trying to understand your alternative solution, but I believe the calculation is incorrect:
1,212 = plates req for single digit + plates req for two digits + plates req for three digit 1,212 = 9 (n) + 90 (2n) + x (3n) x = (1,212  9  90)/3 (since n = 1) x = 341 < this should be 371 according the the statement preceding it



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Re: M07#37 [#permalink]
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dpgxxx wrote: Gmat tiger,
I'm trying to understand your alternative solution, but I believe the calculation is incorrect:
1,212 = plates req for single digit + plates req for two digits + plates req for three digit 1,212 = 9 (n) + 90 (2n) + x (3n) x = (1,212  9  90)/3 (since n = 1) x = 341 < this should be 371 according the the statement preceding it 341 is correct but the highlighted 90 should be 180. Thanks.
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Re: M07#37 [#permalink]
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20 Jul 2010, 01:47
Is this qs too difficult? I can't get any of the solutions provided....somebody please step in and help:(



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Re: M07#37 [#permalink]
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20 Jul 2010, 04:45
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The solution is as outlined by Gmat tiger.
You can number the first 9 houses with a single digits plate each (i.e. 1st house with 1, 2nd house with 2...9nth house with 9). => You have numbered 9 houses and you have 1212  9 = 1203 digits plates left.
Then you start numbering houses 10 to 99 with two digits plates each (i.e. 11th house with 11....99th house with 99). => You have numbered 99 houses and you have 1212  9  2*90 = 1023 digits plates left.
Now you know that from now on you need 3 digits plates to number each house. Since you have 1023 digits plates left you can number 1023 / 3 = 341 additional houses
So in total you can number 9 + 90 + 341 = 440 houses.



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Re: M07#37 [#permalink]
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13 Aug 2010, 09:50
IMO B.. very good question
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Re: M07#37 [#permalink]
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13 Aug 2010, 11:42
How did the Question lead us to think that we need to determine # of houses with 3 digits? Why not 4?
I am not able to determine what is asked here.



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Re: M07#37 [#permalink]
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13 Aug 2010, 18:33
GMAT TIGER wrote: dpgxxx wrote: Gmat tiger,
I'm trying to understand your alternative solution, but I believe the calculation is incorrect:
1,212 = plates req for single digit + plates req for two digits + plates req for three digit 1,212 = 9 (n) + 90 (2n) + x (3n) x = (1,212  9  90)/3 (since n = 1) x = 341 < this should be 371 according the the statement preceding it 341 is correct but the highlighted 90 should be 180. Thanks. We need the total number of houses not plates per house90 is correct.



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Re: M07#37 [#permalink]
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14 Aug 2010, 00:15
Quote: The numbers of houses are consecutive and the number of the first house is 1 How does the underlined text (or any other one) suggest that there is only a digit per plate? I thought consecutive numbers of houses means 1,2,3, ..., 10, 11, ..100, 101,... Also, there could be double digit plates, threedigit plates, etc... Someone please correct my erroneous statement.
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Re: M07#37 [#permalink]
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14 Aug 2010, 06:27
The construction company is using "digit plates," i.e., no S is added to digit, thus a single digit per plate.
Last edited by GaryDunn on 14 Aug 2010, 15:48, edited 2 times in total.



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Re: M07#37 [#permalink]
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17 Aug 2010, 01:32
Its B.
9 nos with 1 digit (19) & 90 nos with 2 digit (1099) and the rest are 3 digit nos!
1*9 + 2*90 + 3*(x) = 1212
So x=341
9+90+341= 440



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Re: M07#37 [#permalink]
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22 Aug 2010, 05:58
9+ 2*90 + 3*x = 1212 => 341 ==> 440 houses to be given numbers.
GMAT tiger gave a good explanation, but i usually just try to solve it practically.



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Re: M07#37 [#permalink]
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30 Aug 2010, 23:44
why isnt the 4 digits being used?



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Re: M07#37 [#permalink]
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06 Jul 2011, 13:46
I think the question may be a bit ambiguous to some. Would be helpful to mention smthing such as 1 digit equals 1 plate? What say?



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Re: M07#37 [#permalink]
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17 Aug 2011, 10:15
I used the elimination technique to find out the correct answer here. Now lets see how this works : Number of plates for numbers from 19 >9 Number of plates for number 1099 >90x2 = 180 Number of plates for numbers 100199 >100x3 =300 similarly for 200299 > 300 300399 > 300 400499 > 300 from the options which are given if we add number of plates required only for number from 100499 it should be 1200 the number of plates required are a little more than this hence our answer should be very close to 400 and hence the answer
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Re: M07#37 [#permalink]
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17 Aug 2011, 23:10
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ya really a tricky question, the trick is in the sentence "construction company wants to number new houses using digit plates only".. so it means in each board only one digit will be written. so for the first 9 houses, which wil be numbered as, 1,2,3..9, we need 9 plates for the 10th house, the house number should be 10 ( because "numbers of houses are consecutive") so for the 10th house, two plates are required ( a plate with the number 1 and a plate with the number 0) hence for the next 90 houses ( obtained as 9910), we need 90 * 2 plates = 180 plates and nw the equation comes 1*9 + 2*90 + 3*a = 1212 ==> a = 341 and number of houses = 9 +90+341 = 440 *** 4 digits are not considered because it is mentioned that "numbers of houses are consecutive"
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Re: M07#37 [#permalink]
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18 Aug 2011, 04:02
You don't have to go to 4 digits because 3*900 is greater than the number of plates (1212)
I made the same mistake; I did not take in to account the number of plates, i.e, when we use 2 digit plates, you have to multiply the various arrangements by the number of plates



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Re: M07#37 [#permalink]
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03 Sep 2011, 23:37
The answer is B. I used the answers to solve this problem. This was my approach: Firstly, irrespective of which choice the answer is: Houses 1 thru 9 require 9 platesHouses 10 thru 99 require 180 plates (90 * 2) Houses 100 and above require 3x plates If we were to now use the answer choices: (1) Too small a number. (2) 440 => 9 + 180 + (440  99)3 = 189 + (341)3 = 189 + 1023 = 1212 (3), (4) and (5) yield a number higher than 1212
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Re: M07#37 [#permalink]
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21 Aug 2012, 05:52
single numbers =9 plates doublenumbers =90 *2=180 plates triple numbers=*3=1023 plates>as total number of plates=1212 so triple numbers=341 hence total houses numbered =9+90+341=440







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