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Bunuel
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Took me 2mins 10 secs, here it goes :

Elements with 2 : 50,.....,98 => number of elements : (last - first)/difference + 1= 25
Elements with 3 : 51,.....,99 => number of elements : (last - first)/difference + 1= 17

Now the tricky part,

Elements with 2*3(or 6) are the ones that are counted twice, lets deduct that from the total

Elements with 6 : 54,.....,96 => number of elements : (last - first)/difference + 1= (96-54)/6+1 = 8

Total = 25+17-8=34

Cheers!
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Solution:

The total number of numbers =99-49+1 =51 (We add 1 to include both 99 and 49 in the count)

Number of multiples of 2(50,52,..98) = 98-50/2 + 1 =48/2 +1 = 25 (We add 1 to include both 50 and 98 in the count)

Number of Multiples of 3 (51,99)=99-51/3 +1 = 16+1 = 17 (We add 1 to include both 99 and 51 in the count)

The common terms that are double counted are to be eliminated. They are multiples of 6(54,..96) = 96-54 / 6 + 1

(We add 1 to include both 54 and 96 in the count)

= 7+1

=8

Thus total count =25+17-8

= 34 (option d)

Devmitra Sen
GMAT SME

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