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
OA:[spoiler]B
[/spoiler]
A construction company wants to number
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- fiza gupta
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Plates required for 1 digit plates = 1-9 = 9
Plates required for 2 digit plates = 10-99 = 90
Plates required for 3 digit plates = 100- up = needs to find.
1,212 = plates required for single digit + plates required for two digits + plates required for three digit
1,212 = 9 (n) + 90 (2*n) + x (3*n)
x = (1,212 - 9 - 180)/3 (since n = 1)
x = 341
no of houses = 9 + 90 + 341 = 440.
Plates required for 2 digit plates = 10-99 = 90
Plates required for 3 digit plates = 100- up = needs to find.
1,212 = plates required for single digit + plates required for two digits + plates required for three digit
1,212 = 9 (n) + 90 (2*n) + x (3*n)
x = (1,212 - 9 - 180)/3 (since n = 1)
x = 341
no of houses = 9 + 90 + 341 = 440.
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Hi fiza gupta,
This question ultimately comes down to doing some basic arithmetic and staying organized.
Each of the 1-digit house numbers (numbers 1 - 9, inclusive) requires one plate, so that's 9 houses and 9 plates
Each of the 2-digit house numbers (numbers 10 - 99, inclusive) requires two plates, so that's 90 houses and 180 plates
So far, we've used 9 + 180 = 189 of the 1212 total plates available, which leaves us with 1212 - 189 = 1023 plates
Each of the 3-digit house numbers (numbers 100 - 999, inclusive) requires three plates...
With 1023 plates remaining, we can put numbers on 1023/3 = 341 additional houses
Total houses = 9 + 90 + 341 = 440
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
This question ultimately comes down to doing some basic arithmetic and staying organized.
Each of the 1-digit house numbers (numbers 1 - 9, inclusive) requires one plate, so that's 9 houses and 9 plates
Each of the 2-digit house numbers (numbers 10 - 99, inclusive) requires two plates, so that's 90 houses and 180 plates
So far, we've used 9 + 180 = 189 of the 1212 total plates available, which leaves us with 1212 - 189 = 1023 plates
Each of the 3-digit house numbers (numbers 100 - 999, inclusive) requires three plates...
With 1023 plates remaining, we can put numbers on 1023/3 = 341 additional houses
Total houses = 9 + 90 + 341 = 440
Final Answer: B
GMAT assassins aren't born, they're made,
Rich