A) \[P=(x+y)\]and \[Q=(x-y)\]
B) \[P=x-y\] and \[Q=x+y\]
C) \[P=\frac{1}{x+y}\] and \[Q=\frac{1}{x-y}\]
D) \[P={{x}^{2}}-{{y}^{2}}\] and \[Q={{x}^{2}}+{{y}^{2}}\]
Correct Answer: B
Solution :
(b): Let the speed of the boat in still water be x km/h and speed of the stream be y km /h. Then the speed of the boat downstream = (x + y) km/h, and the speed of the boat upstream = (x ? y) km/h Also, \[\text{time}=\frac{\text{distance}}{\text{speed}}\]; In the first case, when the boat goes 30 km upstream, let these time taken, in hour, be \[{{t}_{1}}\]. Then, \[{{t}_{1}}=\frac{30}{x-y}\] Let \[{{t}_{2}}\] be the time, in hours, taken by the boat to go 44 km downstream. Then, \[{{t}_{2}}=\frac{44}{x+y}\]. The total time taken \[{{t}_{1}}+{{t}_{2}}\], is 10 hours. Therefore, we get the equation. \[\frac{30}{x-y}+\frac{44}{x+y}=10\]?????????(2) \[\Rightarrow P=x-y\] and \[Q=x+y\].
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