A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of 24. If three…

Let number of red balls be x & number of blue balls be y A.T.Q. \[\frac{10 \mathrm{x}}{100}+\frac{20 \mathrm{y}}{100}=24 \] or \(\displaystyle \mathrm{x}+2 \mathrm{y}=240\) Also, \(\displaystyle 3 \mathrm{x}-\mathrm{y}=20\) Solving (i) and (ii) , we get \[x=40, y=100 \] ∴ Number of red balls = $\displaystyle 40$ and Number of blue balls = $\displaystyle 100$