Q. In a party there are women and cats.There are 18 heads and 72 legs altogether. How many women and cats?

• let women=w cat=c
  so w+c=18
  2w+4c=72
  solving the eqn
  we get w=0 c=18
• 11 years agoHelpfull: Yes(36) No(0)
• Let No of cats be x
  So no of women = 18-x
  (4*x)+(2*(18-x))=72
  2x+36=72
  x=18
  So no of cats=18
  no of women=0
  
• 11 years agoHelpfull: Yes(15) No(2)
• let the number of women is x and number of cats is y
  according to question.
  x+y=18
  2x+4y=72
  by solving these two equations
  number of womens=8 and no. of cats =14
  
• 11 years agoHelpfull: Yes(14) No(62)
• let women=w,cat=c;
  accoding to Q. w+c=18
  2w+4c=71(2 legs for women and 4legs for cat.)
  after solve above both equation we have,
  w=0 and c=18 .
• 10 years agoHelpfull: Yes(5) No(0)
• assume all are cats then heads=18 and legs=4*18=72
  so all are cats only
  if you assume all are women mens head=18 and legs=2*18=36 will not satisfy so answer is all are cats only
• 10 years agoHelpfull: Yes(0) No(0)
• If we cosider that the 18 heads are collective some of both cat and woman then answer will be 18 cats and zero womans . If we consider that the heads which is visible belongs only to womens then we get there are 18 womens and 9 cats are there present in the party.
• 7 years agoHelpfull: Yes(0) No(0)