In counting m coloured balls, some blue and some white, it was found that 49 of the first 50 counted were blue. Thereafter 7 out of every 8 counted were blue. If, in all, 90% or more of the balls counted were blue, the maximum value of m is

Out of first 50 balls, number of blue balls = 49

From now, for each $$x$$ balls, number of blue balls = $$\frac{7x}{8}$$

Total number of balls = $$m=(x+50)$$

According to ques,

=> $$\frac{\frac{7x}{8}+49}{x+50}\geq \frac{9}{10}$$

=> $$\frac{70x}{8}+490\geq9x+450$$

=> $$40\geq\frac{x}{4}$$

=> $$x\leq160$$

$$\therefore$$ To maximize $$m$$, we need to maximize $$x$$, which has a maximum value of 160

=> Maximum number of balls = $$160+50=210$$

=> Ans - (C)