a bag contains 2/3rd blue balls and rest are pink.then there are some part of blue balls are defective and some part of black balls are defective(i don't recall the exact value)n tatal non defective balls are 146.then total no. of balls in bag.?
ANS:432

• a bag contains 2/3rd blue balls and rest are pink.5/9 blue balls are defective and 7/8 pink balls are defective.total non defective balls are 146.then total no. of balls in bag.?
  
  sol:let balls in bag be x
  blue=2x/3 ,pink=x/3
  non defective blue balls=(4/9)*(2x/3)=8x/27,
  ....do........ pink =(1/8)*(x/3)=x/24
  so (8x/27)+(x/24)=146
  on solving x=432.So total balls=432
• 12 years agoHelpfull: Yes(37) No(0)
• a bag contains 2/3rd blue balls and rest are pink.t5/9 blue balls are defective and 7/8 black balls are defective.total non defective balls are 146.then total no. of balls in bag.?
  
  sol: 2/3,5/9,7/8
  No. divisible by 3,9,8 is 72
  72*3=216 so that it is divisible by 9
  2/3(216)=144 blue
  5/9(144)=80 defective blue
  pink balls=216-144=72
  7/8(72)=63 defective pink
  total no.of defective balls=80+63=143
  total no. of non defective balls=216-143=73
  bt given 146 non defective balls which is exactly twice of what v got(73*2=146)
  so 216*2=432 is the total no.of balls
• 12 years agoHelpfull: Yes(3) No(5)