Answer:
Let the speed of boat \[~x=km/hr\]. Let the speed of stream\[=y\text{ }km/hr\]. Speed of boat in upstream\[=(x-y)\text{ }km/hr\]. Speed of boat in downstream\[=(x+y)\text{ }km/hr\]. Time taken to cover 30 km upstream \[=\frac{30}{x-y}hrs.\]. Time taken to cover 40 kin downstream \[=\frac{44}{x+y}hrs.\] hrs. According to question, Total time taken \[=10\text{ }hrs\]. \[\frac{30}{x-y}+\frac{44}{x+y}=10\] ?(i) Now, Time taken to cover \[55\text{ }km\] downstream \[=\frac{55}{x+y}hrs.\] Time taken to cover \[40\text{ }km\] upstream \[=\frac{40}{x-y}hrs.\]. Total time taken \[=13\text{ }hrs\]. \[\frac{40}{x-y}+\frac{55}{x+y}=13\] ?(ii) Solving eq. (i) and eq. (ii). Let \[\frac{1}{x-y}=u,\frac{1}{x+y}=v\]. \[30u+44v=10\] \[40u+55v=13\] Or \[15u+22v=5\] ?(iii) \[8u+11v=\frac{13}{5}\] ?(iv) Multiplying eq. (iii) by 8 and eq. (iv) by 15, we get \[_{\begin{smallmatrix} 120u+165v\,\,=\,\,39 \\ -\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,- \\ \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11v\,\,=\,\,1} \end{smallmatrix}}^{120u+176v\,\,=\,\,40}\] \[v=\frac{1}{11}\] Putting the value of v in eq. (iii) \[15u+22v=5\] \[\Rightarrow \] \[15u+22\times \frac{1}{11}=5\] \[\Rightarrow \] \[15u+2=5\] \[\Rightarrow \] \[15u=3\] \[\Rightarrow \] \[u=\frac{3}{15}\] Or \[u=\frac{1}{5}\] Now, \[v=\frac{1}{11}\] \[\Rightarrow \] \[\frac{1}{x+y}=\frac{1}{11}\] \[\Rightarrow \] \[x+y=11\] ?(v) And \[u=\frac{1}{5}\] \[\Rightarrow \] \[\frac{1}{x-y}=\frac{1}{5}\] \[\Rightarrow \] \[x-y=5\] ?(vi) On solving eq. (v) and (vi). \[_{\begin{smallmatrix} x-y\,\,=\,\,5 \\ +\,\,\,\,-\,\,\,\,\,+ \\ \overline{2x\,\,\,\,\,=\,\,16} \end{smallmatrix}}^{x+y\,\,=\,\,11}\] Or \[x=8\] Put the value of x in eq. (v) \[8+y=11\] \[y=11-8\] \[y=3\] The speed of boat in still water \[=8\text{ }km/hr\]. The speed of stream \[=3\,\,km/hr\]. We learn that the speed of boat is slow in upstream and fast in downstream.
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