How many nos. are there between 100 and 200 both inclusive and divisible by 2 or 3?

a) 67 b) 68 c) 84 d) 100

• well, there are 51 divisible by 2. 50 from 101 to 200 (= 200-100)/2 + 100 itself.
  there are 33 divisible by 3 (200-100)/3 rounded down
  there are 16 divisible by 6 (200-100)/6 rounded down,
  which are counted in both sets above.
  51 + 33 - 16 = 72
• 11 years agoHelpfull: Yes(7) No(19)
• There will be (200-100)/2+1 = 51 numbers divisible by 2.There will be Floor[(200 - 100)/3] = 33 numbers divisible by 3.Multiples of 6 from the 17th through the 33rd will be counted twice by the sum of these numbers (51+33). There are 17 of them.That means the count of numbers you are interested in is 51+33-17 = 67.
• 10 years agoHelpfull: Yes(4) No(1)
• ans is 67................................
• 11 years agoHelpfull: Yes(3) No(3)
• 51 are divisible by 2
  33 are divisible by 3
  51+33 = 84
  now the number divisible by both are counted twice so we will hv to substract them once to find the actual result
  for that
  17 of them are divisible by 6 i.e (3*2) so
  req ans is 84-17 = 67
  
• 10 years agoHelpfull: Yes(3) No(0)
• There will be (200-100)/2+1 = 51 numbers divisible by 2.
  There will be Floor[(200 - 100)/3] = 33 numbers divisible by 3.
  Multiples of 6 from the 17th through the 33rd will be counted twice by the sum of these numbers (51+33). There are 17 of them.
  so the answer will be 67.(A)
• 10 years agoHelpfull: Yes(1) No(1)
• @DEVAMANI SINGH how the answer is 67?
• 10 years agoHelpfull: Yes(0) No(0)