A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is
the fraction of the solution that was replaced?
a) ¼ b) 1/3 c) 1/2 d) 2/3

• Let X be the fraction of solution that is replaced.
  
  Then X*25% + (1-X)*40% = 35%
  
  Solving, you get X = 1/3
  
  Can someone please explain the logic behind this?
  
  
  The Amount of Solute you have is the Concentration * Volume. (Solute is what you put in to make the solution)
  
  Let's suppose your container is 1 L. If we replace X Liters, then we will have 1 - X liters left.
  
  After replacing, X liters will have a concentration of 25%, so the amount of solute is 0.25 * X. The remaining 1 - X liters will have a concentration of 40%, so the amount of solute is 0.40(1 - X). At the end, when it's all mixed, the 1 L will have a concentration of 35%, so the amount of solute is 0.35 * 1 .
  
  We can set up the equation:
  0.25X + 0.40(1 - X) = 0.35(1)
  
• 8 years agoHelpfull: Yes(21) No(1)
• Let solution be 1 ltr
  X is the certain quantity which has to be replaced
  Now,
  (40%of1-x)+(25%of1)=35%of 1
  After solving it we get ans. 3/10
• 8 years agoHelpfull: Yes(3) No(3)
• At first we take the formula for mixtures.
  then we get the mixture in the ratio of (40-35):(35-25)=1:2
  Thus we replace 1/3 of 40% solution and mix it with 25 % to get 35% solution
• 8 years agoHelpfull: Yes(2) No(0)
• (25-35):(40-35)=1:2 by simple allegation formula
• 5 years agoHelpfull: Yes(1) No(0)
• Consider a solution be 1 ltr
  X is the certain quantity which has to be replaced
  Now,
  (40%of 1-x)+(25%of 1)=35%of 1
  After solving it we get ans. 3/10
  
• 8 years agoHelpfull: Yes(0) No(2)
• someone please explain the solution..
• 8 years agoHelpfull: Yes(0) No(0)
• Ans is 1/3
• 3 years agoHelpfull: Yes(0) No(0)