A) 2500
B) 1250
C) 950
D) 122
Correct Answer: C
Solution :
\[x+y+z=50;\text{ }xyz=3750\] \[\therefore \] \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz+zx+xy}{xyz}=\frac{31}{150}\] \[\Rightarrow \]\[xy+yz+zx=\frac{31}{150}xyz\] \[=\frac{31}{150}\times 3750=775\] \[\therefore \]\[{{(x+y+z)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(xy+yz+zx)\] \[\Rightarrow \] \[{{(50)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\times 775\] \[\Rightarrow \] \[2500={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+1550\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=2500-1550=950\]
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