A company will use 1212 plates to number all of the houses in a new de

The question is confusingly worded but I get:

9 houses with 1-digit plates (uses 9 plates)
90 houses with 2-digit plates (uses 180 plates)
341 houses with 3-digit plates (uses 1023 plates)

This uses 1212 plates for 440 houses.
B

You know that the answer is going to be three digits, and it's easy to figure out how many 1 and 2 digit numbers (9 and 90), so the real question is how many 3 digit number are there?

99+(1212-9-90*2)/3=440 B

My first attempt at this forum and I proved myself inadequate. These are the questions that bug me GMAT prep tests.
Quite clever question and cleverer responders. kudos ...

Just now I got the question . I misunderstood the plates part. It one digit one plate. I thought each plate must have a few digits.

I was confused on reading the question but seems simple to solve.
Thanks for posting

good question....

Bunuel Can you help here?

The first 9 houses will require 9 plates. The next 90 houses will require \(2*90 = 180\) plates. The rest \(x - 99\) houses will require \((x - 99)*3\) plates. The total number of plates will be \(9 + 180 + 3x - 297 = 1212\). From this equation \(x = 440\).


Answer: B
_________________

Thanks Bunuel

Keeping in mind digits plates only what number from the choice leads to three times 1212?
Why three times, as order is for thousands we need one less since repetition is allowed
440 is close

Sent from my iPhone

Hi All,

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:

GMAT assassins aren't born, they're made,
Rich

Its saying each house will be numbered using digit plates only... So, for a house with number 10 requires 2 Plates (One for 1 and the other for 0) . So, first 9 house will need 9 plates, next 90 houses will need \(2*90 = 18\)0 plates.. now, the equation will be \(9+180+(x-99)*3 = 1212\)...\(x = 341\) So, totatl will be \(9+180+x.. 440\)

House no. 1 to 9 will use 1 plate each (9 houses)
House no. 10 to 99 will use 2 plates each (90 houses)
House no. x to y will use 3 plates each where x and y are 3 digit numbers (z houses)

Total plates are 1212.
Hence 1212 = 1*9 + 2*90 + 3*z
z = 341

There are 341 houses which have 3 digit numbers,
So total number of houses = 9 + 90 + 341 = 440