3/5th of the balls in a bag are red, the rest are green. If 3/4th of the red balls and 5/7th of the green balls are defective, find the total number of balls in the bag given that the number of non-defective balls is 111.
(1) 516 (2) 432 (3) 420 (4) 504

• let x balls present initially...
  3/5x balls red, 2/5x balls green
  3/4(3/5x)+5/7(2/5x)--defective
  x-(9/20x+10/35x)=111
  x-(63/140x+40/140x)=111
  x-103/140x=111
  37/140x=111
  1/140x=3
  x=420
• 4 years agoHelpfull: Yes(8) No(2)
• Let 5x are the balls
  Red Green =3x,2x
  Defective=3x*3/4,2x*5/7=9x/4,10x/7
  Total defective =103x/28
  Non-deffective =5x-103x/28=(140-103)x/28
  37x/28=111=>Total balls 5x=5*111*28/37
  =420
• 4 years agoHelpfull: Yes(4) No(1)
• Let Total no of balls is X.
  3/5 X Balls are Red
  2/5 X Balls are green
  (3/5) (3/4) X + (2/5) (5/7) X = X - 111
  9/40 X + 10/35 X = X - 111
  
  Solving we Get, X = 420
  
  Ans is C.
• 3 years agoHelpfull: Yes(0) No(0)