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Try this beautiful problem on balls based on Number theory, useful for ISI B.Stat Entrance.

A bag contains colored balls of which at least 90% are red. Balls are drawn from the bag one by one and their color noted. It is found that 49 of the first 50 balls drawn are red. Thereafter 7 out of every 8 balls are red. The number of balls in the bag can not be

- (a) \(170\)
- (b) \(210\)
- (c) \(250 \)
- (d) \(194\)

Number theory

Percentage

Inequility

But try the problem first...

Answer: (b) \(210\)

Source

Suggested Reading

TOMATO, Problem 128

Challenges and Thrills in Pre College Mathematics

First hint

Let the number of balls in the bag is n.Let, m number of times 8 balls are drawn. Therefore, \(n = 50 + 8m\). Red balls =\( 49 + 7m\)

Can you now finish the problem ..........

Final Step

Percentage of red balls = \(\frac{(49 + 7m)}{(50 + 8m)}≥ \frac{90}{100}\)

\(\Rightarrow \frac{(49 + 7m)}{(50 + 8m)}≥0.9\)

\(\Rightarrow 49 + 7m ≥ 45 + 7.2m\)

\(\Rightarrow 0.2m ≤ 4\)

\(\Rightarrow m ≤ 20\)

\(\Rightarrow n ≤ 50 + 8 \times 20 = 210\)

Therefore we can say that at most \(210\) balls can be drawn. So there must be at most \(210\) balls in the bag, and so there cannot be \(250\) balls in the bag (because \(250>210\)).

Therefore option (C) is the correct answer

- https://www.cheenta.com/divisibility-problem-from-amc-10a-2003-problem-25/
- https://www.youtube.com/watch?v=XOrePzJWFiE

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