There is a number which is greater than the aggregate of its third. tenth and the twelfth parts by 58.Can u guess the number?

• Let the number be x
  It is given that number is greater than the aggregate of its third, tenth and twelvth parts by 58
  
  X-(X/3+ X/10 + X/12) =58
  
  take LCM
  X-(20X + 6X + 5X)/60 = 58
  (60X- 31X)/60 = 58
  29X/60 =58
  
  Now, X = (58 * 60)/29 = 60 *2 = 120.
  
  Thus, the number is 120.
  
• 11 years agoHelpfull: Yes(18) No(2)
• let the no. is x
  X-(X/3-X/10 - X/12) =58
  X*(1-1/3-1/10-1/12)=58
  x*(60-20-6-5)/60 = 58
  x*(60-31)/60=58
  x*29/60=58
  x=58*29/60
  x=2*60
  x=120
• 6 years agoHelpfull: Yes(2) No(0)