A 3-gallon mixture contains one part S and two parts R. In order to change it to a mixture containing 25% S, how much R should be added?
(a) 1/2 gallon
(b) 2/3 gallon
(c) 3/4 gallon
(d) 1 gallon
(e) 1 1/2 gallon

• S:R = 1:2 in order to make S:R = 25:75 = 1:3 we need to "add 1 gallon".
  
  as S's share is (1/1+3)*100 = 1/4*100 = 25%
  
  Thus ans is 1 gallon.
• 12 years agoHelpfull: Yes(16) No(4)
• 3 gallon mixture;1gallon S and 2 gallons of R.If the new mixture should contain 25% of S then it must contain 75% of R.Let x gallons of R be added so,
  2+x=(75/100)(3+x)
  200+100x=225+75x
  25x=25
  x=1.
• 12 years agoHelpfull: Yes(6) No(0)
• A 3-gallon mixture contains one part S and two parts R.
  means 2 gallon R and 1 gallon S.
  Now if S is 25% of a mixture, then total vol of mixture is 4 gallon.
  
  so 1 gallon of R should be added to 3 gallon mixture.
• 12 years agoHelpfull: Yes(5) No(2)
• let assume the R to be added is x.
  
  according to the current situation.
  in 3 gallon , 1 gallon is S, 2 gallon is R.
  
  we can write it as : 100/3 + 200/3 =100 % mixture.
  
  when x gallon R is added to make part S 25% ,
  
  the equation becomes: [100/(3+x)] + [(2+x)*100/(3+x)]= 100% mixture.
  
  here, [100/(3+x)] part is 25% S.
  
  so, [100/(3+x)]=25
  solving this we get x= 1 gallon.
  
  ANS IS : (d) 1 gallon.
  
• 12 years agoHelpfull: Yes(4) No(1)
• ans is 1 gallon
• 12 years agoHelpfull: Yes(0) No(0)