smc

The desired percentage of red balls is more than $90$ percent, so write an inequality. $$\frac{49+7x}{50+8x}\ge 0.9$$ Since $x>0$, the sign does not need to be swapped after multiplying both sides by $50+8x$. $$49+7x\ge 45+7.2x$$ $$4\ge 0.2x$$ $$20\ge x$$ Thus, up to $20$ batches of balls can be used, so a total of $20\cdot 8+50=210$ balls can be counted while satisfying the requirements. The answer is $\boxed{210}$.