In a mixture of three varieties of tea ratio of there weights is 4:5:8. If 5 kg tea of first variety, 10 kg tea of the second variety and some quantity of tea of the third variety are added to the mixture the ratio of the weights of the three varieties becomes 5:7:9. In the final mixture the quantity (in kg) of the third variety of tea was?

• 5 Kg.
  
  If original quantity(in Kg.) of three mixtures are 4x, 5x and 8x
  After adding 5 Kg. in first, 10 Kg. is second and 'y' Kg. in third mixture, ratio becomes 5: 7: 9
  Comparing first two weights (4x+5)/(5x+10) =5/7, so x=5
  Comparing weight of second & third mixture (5x+10)/(8x+y)= 7/9, On substituting x=5, we have y=5
• 8 years agoHelpfull: Yes(1) No(0)
• The ratio of the three mxture of the tea is = 4:5:8
  Lets the 1st mixture is 4x kg
  the 2nd mixture is 5x kg
  and the 3rd mixture is 8x kg
  
  now as per question 5kg and 10kg tea are added with 1st and 2nd type tea
  respectively.
  So new weight of 1st type of tea = 4x+5 kg
  new weight of 2nd type of tea = 5x+10 kg
  Now lets Y g of the third type of tea is mixed with the 3rd tea to get new nixture
  So total weight of 3rd type of tea is = 8x+Y kg
  
  So as per the problem,
  (4x+5):(5x+10):(8x+Y) = 5:7:9
  
  From here we can write, (4x+5):(5x+10) = 5:7
  solving this we get x=5
  
  now, similarly, (5x+10):(8x+Y) = 7:9
  or, (25+10):(8x+Y) = 7:9
  or, 35:(8x+Y) = 7:9
  or, Y = 5
  So 5kg of tea of third variety are added to this mixture.
• 5 years agoHelpfull: Yes(0) No(0)