Answer:
Given, a jar contains red, blue and orange balls. Let the number of red balls \[=x\] and the number of blue balls \[=y\] Number of orange bails \[=10\] Then, total number of balls \[=x+y+10\] Let P be the probability of selecting a red ball from the jar \[P=\frac{x}{x+y+10}\] But P (a red balls) \[=\frac{1}{4}\] (Given) \[\therefore \frac{1}{4}=\frac{x}{x+y+10}\] \[x+y+10=4x\] \[3x-y=10\] ?(i) Similarly, P (a blue ball) \[=\frac{y}{x+y+10}\] But P (a blue ball) \[=\frac{1}{3}\] \[\therefore \frac{1}{3}=\frac{y}{x+y+10}\] \[x+y+10=3y\] \[x-2y=-10\] ?(ii) On multiplying equation (ii) by 3, we get \[3x-6y=-30\] ...(iii) On subtracting equation (iii) from (i) \[_{\begin{smallmatrix} 3x\,\,-\,\,6y\,\,=\,\,-30 \\ -\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,+ \\ \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,5y\,\,=\,\,40\,\,\,\,\,\,\,} \end{smallmatrix}}^{3x\,\,-\,\,y\,\,=\,\,10}\] \[y=8\] On putting the value of y in (iii), we get \[3x-6\times 8=-30\] \[3x=-30+48\] \[x=\frac{18}{3}\] \[x=6\] Total number of balls \[=x+y+10\] \[=6+8+10\] \[=24\] Hence, total number of balls in the jar is 24.
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