A bowl contains raspberries and blueberries. Natacha added \(10\) more raspberries to the bowl, then determined that \(\frac{3}{10}\) of the berries in the bowl were raspberries. Jing then added \(20\) more blueberries to the bowl, then determined that \(\frac{1}{4}\) of the berries in the bowl were raspberries.
What fraction of the original berries in the bowl were raspberries?
Let \(r\) represent the number of raspberries originally in the bowl, and let \(b\) represent the number of blueberries originally in the bowl.
After Natacha added \(10\) more raspberries, there were \(r+10\) raspberries and \(b\) blueberries. Since \(\frac{3}{10}\) of the berries in the bowl were raspberries, \[\begin{align} \frac{r+10}{r+10+b} &= \frac{3}{10}\\ 10r + 100 &= 3r + 30 + 3b\\ 7r - 3b + 70 &= 0 \tag{1} \end{align}\] After Jing added \(20\) more blueberries, there were \(r+10\) raspberries and \(b+20\) blueberries. Since \(\frac{1}{4}\) of the berries in the bowl were raspberries, \[\begin{align} \frac{r+10}{r+10+b+20} &= \frac{1}{4}\\ 4r + 40 &= r + b + 30\\ 3r + 10 &= b \tag{2} \end{align}\] Substituting equation \((2)\) into equation \((1)\), \[\begin{align} 7r - 3(3r+10) + 70 &= 0\\ 7r - 9r - 30 + 70 &= 0\\ -2r+40 &= 0\\ r &= 20 \end{align}\] Then \(b=3(20)+10=70\). Therefore, there were originally \(20\) raspberries and \(70\) blueberries in the bowl. Thus, the fraction of the original berries in the bowl that were raspberries is \(\frac{20}{20+70}=\frac{20}{90}=\frac{2}{9}\).