a man has some cows and hens.if the number of heads:no of feet =12:35,find the no of hens ,if the number of heads alone is 48

• Ans: 26 hens.
  
  Ratio of head:feet = 12:35 and no. of heads=48
  Using products of means = products of extremes
  thus (35*48)/12 = 140 feets
  
  Now, as we know cow has 1 head, 4 feets and hen has 1 head, 2 feets
  Let the no. of cows be x and no. of hens be y.
  Thus x+y = 48(heads) .......(1)
  4x+2y = 140(feets) ......(2)
  
  Solving eq. (1) and (2)
  we get x=22 and y=26.
  i.e no. of cows=22 and no. of hens=26
• 7 years agoHelpfull: Yes(13) No(0)
• Given Head/feet = 12/35
  to have 48 heads alone ,lets multiply above ratios numerator and denominator with 4 , so that we get 48 heads.
  then after multiplying num. and denominator with 4, ratio head/feet = 48/140.
  
  so there r 48 heads and 140 feet
  let's x represent for cow and y for hen
  then writing the ratio in equation form, we get
  x + y = 48-------------*2
  4x + 2y = 140
  
  
  solving above equations we get
  x= 22 , y= 48-22=26
  Hence there are 26 hens.
• 7 years agoHelpfull: Yes(4) No(0)
• Ans: 26 hens.
  
  Let cows=x, hens=y....... therefore x+y=48.........(1) (bcz of total no of heads=48)
  
  so, total no of feets = 4x+2y......now as per the question the number of heads/no of feet =12/35
  therefore (x+y)/(4x+2y)=12/35.....after solving we will get x = (11y)/13
  put the value of x in equation (1).....we will get y=26 (total number of hens)
  
  I think its more easier.
• 7 years agoHelpfull: Yes(1) No(0)
• Answer: 93 marbles in 50th box
  here the factors of 50 are 1,2,5,10,25,50 so
  person with these numbers only end up keeping marbles in 50th box.
  In 50th box,
  Total marbles kept by 1st person = 1
  Total marbles kept by 2nd person = 2
  Total marbles kept by 5th person = 5
  Total marbles kept by 10th person = 10
  Total marbles kept by 25th person = 25
  Total marbles kept by 50th person = 50
  Hence total marbles in 50th box = 1+2+5+10+25+50 = 93 marbles
• 7 years agoHelpfull: Yes(0) No(0)
• 12:35=48:x where x is 140 cows have 4 legs so atmost 120 legs and remaining legs are hens so 20 hen legs i.e there are 10 hens
• 7 years agoHelpfull: Yes(0) No(2)
• Let no of hens=x
  no of cows=y
  no of heads =48, i.e. x+y=48
  again
  x+y/2x+4y=12:35
  
  solving this x=26
  y=22.
  No of hens=26
• 7 years agoHelpfull: Yes(0) No(0)
• 4x+(48-x)2=140 =>2x=44 =>x =22=head of cow
  no of hean=48-22=26
• 7 years agoHelpfull: Yes(0) No(0)
• 22*4=88; 26*2=52. Hence 26
• 7 years agoHelpfull: Yes(0) No(0)
• Given no of head:no of feet=12:35
  We know cow has 4 feet and one head, hen
  has one head and 2 feet
  Let
  No of cow =y and no of hen=z
  From ratio given say
  No of head=12x and no of feet=35x
  12x=y+z. .......(1)
  35x=4y+2z .......(2)
  Also given no of head=48=y+z
  From (2)
  35x=2(y+z)+2y=2*48+2y=96+2y
  35x=96+2y
  35*(y+z)/12=2(48+y)
  35*48/12=2(48+y)
  Y+48=70
  Y=22
  Z=48-22=26 ans
• 7 years agoHelpfull: Yes(0) No(0)
• No of heads = 48
  so 48/12=4
  so the number of feets = 35*4=140
  n=no of cows
  m no of hens
  n+m = 48
  4n+2m=140
  so m = 26
• 7 years agoHelpfull: Yes(0) No(0)
• 32 cows
  16 hens
• 4 years agoHelpfull: Yes(0) No(0)