Q. Uncle Reuben and Aunt Cynthia came to town for shopping. Reuben bought a suit and hat for $15. Cynthia paid as much as for her hat as Reuben did for his suit then she spent the rest of their money for a new dress. On the way home Cynthia called Reuben's attention to the fact that his hat cost $1 more than her dress, then she added, if we had divided our hat money differently so that we bought different hats mine costing 1 and 1/2 time cost of yours then we each would have spent the same amount of money. In that case Uncle Reuben asked "how much would my hat have cost"

• Let us use R for Reuban, C for Cynthia, H for Hat, S for Suit, and D for Dress.
  
  For Reuban, Cost of hat and Suit sums upto 15
  R(h) + R(S) = 15-----(1)
  
  Further, Cost of Cynthia's Hat = Cost of Reubena's Suit
  C(h) = R(S)-------------(2)
  
  Given, Cost of Cynthia's hat is 1 and 1/2 times that of Reubann's hat
  C(h) = 3/2 R(h)-----------(3)
  
  From (2) and (3) we get, R(S)= 3/2 R(S)-----(4)
  
  Substitute R(S) in (1) we get,
  
  R(h) + 3/2 R(h) = 15
  
  5/2 R(h) = 15
  
  R(h) = 2/5 * 15
  
  R(h) = 6
  
  Therefore, Cost of Uncle Reuben's Hat is $6.
  
  
• 10 years agoHelpfull: Yes(66) No(3)
• reuben (suit+hat)=15
  reuben suit=cynthia hat
  reuben hat=1+cynthia dress
  trial and error method 9+6=15
  rs=9
  rh=6
  cd=5
  total money=9+9+6+5=29
• 10 years agoHelpfull: Yes(5) No(2)
• r.hat+r.suit=15
  c.hat=r.suit
  c.hat=c.dress+1
  c.hat is 1 and 1/2 more than r.hat
  so,c.hat=3/2*r.hat
  since c.hat =r.suit
  therefore r.hat+(3/2*r.hat)=15
  so, r.hat=6, c.hat=9, c.dress=8,r.hat=9
  here r=reuben, c=cynthia
• 10 years agoHelpfull: Yes(3) No(0)
• in the above r.hat =c.dress+1
  c.dress=5.
  sorry for the mistake.
• 10 years agoHelpfull: Yes(2) No(0)
• Ans is 2
  Let us use r for Reuban, c for Cynthia, h for Hat, s for Suit, and d for Dress.
  1st case:-
  r(h)+r(s)=15
  c(h)=r(s) => r(h)=15-c(h)
  $ c(h)=c(d)+1 => c(d)=c(h)-1=15-r(h)-1
  => c(d)=14-r(h).....(1)
  2nd Case:-
  if, c(H)=3/2 r(h)....(2)
  {"c(H)" for different Hat.old one was "c(h)"}
  den, c(H)+c(d)=15 => 3/2r(h)+14-r(h)=15 ...(from 1&2)
  => r(h)=2
  CHEK UR ANS AS:- case 1:-
  for R: 2+13=15
  for C: 13+12=25
  case 2:-
  new c(h)=3/2.2=3
  s0,
  for R: 2+13=15
  3+12=15
  If any one thinks dis is ri8 solution den plz like
  
• 10 years agoHelpfull: Yes(1) No(8)