A) 28, 22
B) 35, 15
C) 20, 30
D) 24, 36
Correct Answer: C
Solution :
[c] \[\frac{1}{x}+\frac{1}{50-\,x}=\frac{1}{12}\] \[{{x}^{2}}-\,50x+600=0\] \[{{x}^{2}}-\,30x-\,20x+600=0\] \[x\,\,(x-30)-20\,\,(x-30)=0\] \[x=30,\,\,20\]
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