Latest TCS Aptitude Question SOLUTION: 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 t

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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

Asked In TCS atul patel (12 years ago)

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Sutanu Paul  (12 years ago)

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

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Divya Kumari.C.D  (12 years ago)

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

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