Three consecutive numbers such that twice the first, 3 times the second and 4 times the third together make 182. The numbers in question are:
1. 18, 22 and 23
2. 18, 19 and 20
3. 19, 20 and 21
4. 20, 21 and 22

• 2*x+3*x+1+4*x+2=182
  x=19
  no.s are 19,20,21
• 11 years agoHelpfull: Yes(11) No(2)
• let the first number is x.
  then consecutive numbers are x,x+1,x+2.
  so,2x+3(x+1)+4(x+2)=182
  From here, x=19.
  so, numbers are 19,20,21
• 9 years agoHelpfull: Yes(2) No(0)
• check the unit digits of the numbers given in question and solved without pen and paper
  1)6+6+2=unit digit is 4(wrong)
  2)6+7+0=unit digit is 3(wrong)
  3)8+0+4=unit digit is 2(can be)
  4)0+3+8=unit digit is 1(wrong)
  so option 3 correct answer
• 6 years agoHelpfull: Yes(0) No(0)
• Let the first no. be 'a'
  
  According to question
  
  2a+3(a+1)+4(a+2)=182
  
  2a+3a+3+4a+8=182
  
  9a=182−11
  
  9a=171
  
  a=19
• 6 years agoHelpfull: Yes(0) No(0)