A) 0.549
B) 0.200
C) 0.786
D) 0.478
Correct Answer: D
Solution :
\[{{P}_{T}}=P_{p}^{0}{{x}_{p}}+P_{h}^{0}{{x}_{h}}\]= \[440\times \frac{1}{5}+120\times \frac{4}{5}\] = \[88+96\]=\[184\]; \[P_{p}^{0}{{x}_{p}}={{y}_{p}}{{P}_{T}}\]; \[\frac{88}{184}={{y}_{p}}\] \[{{y}_{p}}=0.478\]
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