A bag contains 5 red balls and some blue balls.

Solution:

Let the number of blue balls be x.

Number of red balls = 5

Total number of balls = x + 5

$P($ getting a red ball $)=\frac{5}{x+5}$

$\mathrm{P}($ getting a blue ball $)=\frac{x}{x+5}$

Given that,

$2\left(\frac{5}{x+5}\right)=\frac{x}{x+5}$

$10(x+5)=x^{2}+5 x$

$x^{2}-5 x-50=0$

$x^{2}-10 x+5 x-50=0$

$x(x-10)+5(x-10)=0$

$(x-10)(x+5)=0$

Either $x-10=0$ or $x+5=0$

$x=10$ or $x=-5$

However, the number of balls cannot be negative.

Hence, number of blue balls = 10