Hi utkalnayak,
Mitch's approach for this type of situation is right-on, but if you don't immediately "see" that approach, there's still a way to estimate the solution. You just have to do 'enough' work to spot the pattern.
We're looking for the number of 3-digit integers that are divisibly by BOTH 3 and 4.
Starting with the first 3-digit integer....
100 is divisibly by 4 but NOT 3
We can work "up" by adding 4 (since that would give us the "next" multiple of 4)....
104 is divisible by 4 but NOT 3
108 is divisible by 4 AND by 3
Notice the pattern so far...."miss", "miss", "hit"......
112 by 4 but NOT 3
116 by 4 but NOT 3
120 by 4 AND by 3
This also fits the pattern: "miss", "miss", "hit"....
It stands to reason that this pattern will continue, so we can leapfrog the misses and find the "hits" (notice that each is 12 greater than the prior one); here are the first several....
108, 120, 132, 144, 156, 168, 180, 192.....
So we have 8 multiples in the range of 100 - 200. Given this approximate pattern, there will probably be 8 or 9 terms in every set of 100 3-digit numbers. There are 9 groups of 100 from 100 to 999, so (approximately 8 per set)(9 sets) = about 72 multiples. There's only one answer that's close....
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
