A man buys 12 lts of liquid which contains 20% of the liquid and the rest is water. He then mixes it with 10 lts of another mixture with 30% of liquid. What is the % of water in the new mixture?

• 20% in 12 lts is 2.4. So water = 12-2.4 = 9.6 lts. 30% of 10 lts = 3. So water in 2nd mixture = 10-3 = 7 lts. Now total quantity = 12 + 10 = 22 lts. Total water in it will be 9.6 + 7 = 16.6 lts. % of water = (100*16.6)/22 = 75.45
• 12 years agoHelpfull: Yes(59) No(0)
• liquid after mixing = 12/5 + 3 = 5.4ltr
  
  % of liquid = (5.4/22)*100
  = 24.5%
  
  % of water = 100 - 24.5
  
  =75.5%
• 10 years agoHelpfull: Yes(7) No(0)
• firstly ,in 12 litre
  ratio of liquid and water is 1:4 -> so part of water in this -> 4/5*12=48/5 litre
  
  now in 10 litre
  ratio of liquid and water is 3:7 -> so part of water in this -> 7/10*10= 7litre
  After adding both we get --> (7+(48/5)) = 83/5 litre water and after adding mixture has become =22 litre
  so percent of water in 22 litre is --> ((83/5*22)*100) = 75.45% Ans.
• 7 years agoHelpfull: Yes(2) No(0)
• % of Liquid in Solution A = 20%
  
  ⇒ % of Water in Solution A = 80%
  
  Volume of Solution A = 12 Litres
  
  ⇒ Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres
  
  % of Liquid in Solution B = 30%
  
  ⇒ % of Water in Solution A = 70%
  
  Volume of Solution B = 10 Litres
  
  ⇒ Volume of Water in Solution B = (70/100) × 10 = 7 Litres
  
  If Solution A is mixed with Solution B
  
  ⇒ Net Total Volume of Solution = (12 + 10) = 22 Ltrs
  
  ⇒ Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs
  
  % of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100
  
  % of Water in New Mixture = (16.6 / 22) × 100 = 75.45%
• 6 years agoHelpfull: Yes(1) No(0)