Question

If a bag containing red and white balls, half the number of white ball is equal to one-third the number of red balls. Thrice the total number of balls exceeds seven times the number of white balls by 6. How many balls of each colour does the bag contains?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{no}.\mathrm{of}\mathrm{red}\mathrm{balls}\mathrm{be}\mathrm{x}\mathrm{and}\mathrm{white}\mathrm{balls}\mathrm{be}\mathrm{y}. \\ \mathrm{A}.\mathrm{T}.\mathrm{Q} \\ \frac{\mathrm{y}}{2}=\frac{\mathrm{x}}{3} \\ 3\mathrm{y}=2\mathrm{x} \\ 3\mathrm{y}-2\mathrm{x}=0...\left(1\right) \\ 3\left(\mathrm{x}+\mathrm{y}\right)=7\mathrm{y}+6 \\ 3\mathrm{x}+3\mathrm{y}-7\mathrm{y}=6 \\ 3\mathrm{x}-4\mathrm{y}=6...\left(2\right) \\ \mathrm{From}\left(1\right), \\ \mathrm{x}=\frac{3\mathrm{y}}{2} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ 3\left(\frac{3\mathrm{y}}{2}\right)-4\mathrm{y}=6 \\ \frac{9\mathrm{y}}{2}-4\mathrm{y}=6 \\ \frac{9\mathrm{y}-8\mathrm{y}}{2}=6 \\ \mathrm{y}=12. \\ \mathrm{So}, \\ \mathrm{x}=\frac{3\left(12\right)}{2}=18 \\ \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{balls}\mathrm{are}18\mathrm{and}\mathrm{white}\mathrm{balls}\mathrm{are}12. \end{gathered}$$