sprincej
NO

since the sum is 555 then the 3 digits number can not start with 9xx,8xx,7xx because this makes the sum > 555
also 59x,58x,57x > 555 so this too is eliminated

we are left with 54x,4xx if we list all possible combinations we get
549 >>> and 555-549=6 >> not possible
548 >>> and 555-548=7 >> not possible
547 >>> and 555-547=8 >> not possible
498 >>> and 555-498=57 >> possible
497 >>> and 555-497=58 >> possible
495 >>> and 555-495=60 >> not possible
489 >>> and 555-489=66 >> not possible
487 >>> and 555-487=60 >> not possible
485 >>> and 555-485=70 >> not possible
479 >>> and 555-479=76 >> not possible
478 >>> and 555-478=77 >> not possible
475 >>> and 555-475=80 >> not possible
459 >>> and 555-459=96 >> not possible
458 >>> and 555-458=97 >> possible
457 >>> and 555-457=98 >> possible

therefore, we have 4 possible combinations.
I hope there is a faster way to solve this problem

OA Explanation

Since the sum of these 2 numbers is 555, we cannot have any number in 5 _ _ , 7 _ _ , 8 _ _ , 9 _ _ form!
Also, the units digit of 555 is 5 and only 7 + 8 will give us the units digit as 5

4 _ _ + _ _

Case I:
4 _ 7 + _ 8
we can have 2 combinations: 497 + 58 and 457 + 98

Case II:
4 _ 8 + _ 7
we can have 2 combinations: 498 + 57 and 458 + 97

So, a total of 4 combinations
Hence, option C

This question has a very simple solution.

First note that to add up to 555, the three digit number cannot start with any of the digits except 4. (Note even the smallest three digit number starting with 5 is 547 which will give you 89 as the smallest 2-digit number).

Now lets see how many permutations are possible.

2-digit: ____ _____
3-digit: ____ _____ _____

For 3-digit's hundredths place, there is only one possibility i.e., 4.

2-digit: ____ _____
3-digit: (1 possibility) _____ _____

Now we're left with 5,7,8,9.
since the last digit in 555 is 5 so the two units digits in the 2 and 3 digit number must give 5. Only by adding 7 and 8, we get 5 in the units digit. So, the units digit of both numbers can either be 7 or 8.

2-digit: ____ (2 possibilities - 7 or 8)
3-digit: (1 possibility - 4 only) _____ (1 possibility - what ever is left from 7 and 8)

Now we're left with 5, 9. One will go in the 2-digit number and the other will go in the 3-digit number.

2-digit: (2 possibilities - 5 or 9) (2 possibilities - 7 or 8)
3-digit: (1 possibility - 4 only) (1 possibility - whatever is left from 5 and 9) (1 possibility - whatever is left from 7 and 8)

Now we multiply the possibilities.

2-digit: 2 x 2 = 4
3-digit: 1 x 1 x 1 = 1

combined possibilities
4 x 1 = 4

Answer

C

Hi,
I wanted help with these type of questions. I am unable to recognize the pattern in such questions. Where can i practice more of this type? Would be grateful for the reply.

Blitzkrieg

Just make a stroll on the forum we resources left and wide

1) This is a combinatoric problem, just click on the tag on top of the question and you countless of them https://gre.myprepclub.com/forum/search ... tag_id=241

2) The premium question bank permutation and combination https://gre.myprepclub.com/forum/gre-pr ... ml#p116136

3) The daily challenge 25th and 26th week

https://gre.myprepclub.com/forum/gre-qu ... ml#p120770
https://gre.myprepclub.com/forum/gre-qu ... ml#p120980

Thank you! Sorry for the inconvenience caused.