| The speed of boat A is 2 km/h less than the speed of the boat B. The time taken by boat A to travel a distance of 20 km downstream is 30 min more than time taken by B to travel the same distance downstream. If the speed of the current is one-third of the speed of the boat A, then what is the speed of boat B? [LIC (AAO) 2014] |
A) 4 km/h
B) 6 km/h
C) 12 km/h
D) 10 km/h
E) 8 km/h
Correct Answer: E
Solution :
| Let speed of boat \[B=x\,\,km/h\] |
| and speed of boat \[A=(x-2)\,km/h\] |
| \[\therefore \] Speed of current \[=\left( \frac{x-2}{3} \right)\,\,km/h\] |
| Now, according to the question, |
| \[\frac{20}{(x-2)+\frac{(x-2)}{3}}\,\,=\,\,\frac{20}{x+\frac{x-2}{3}}\,\,+\,\frac{30}{60}\] |
| \[\Rightarrow \] \[\frac{20\times 3}{3x-6+x-2}\,\,=\,\,\frac{20\times 3}{3x+x-2}+\frac{1}{2}\] |
| \[\Rightarrow \]\[\frac{60}{4x-8}-\frac{60}{4x-2}=\frac{1}{2}\]\[\Rightarrow \]\[\frac{60}{4\,(x-2)}-\frac{60}{2\,(2x-1)}=\frac{1}{2}\] |
| \[\Rightarrow \] \[\frac{15}{x-2}-\frac{30}{2x-1}=\frac{1}{2}\] |
| \[\Rightarrow \] \[\frac{30x-15-30x+60}{(x-2)\,(2x-1)}=\frac{1}{2}\] |
| \[\Rightarrow \] \[\frac{45}{(x-2)(2x-1)}=\frac{1}{2}\] |
| \[\Rightarrow \] \[(x-2)(2x-1)=90\] |
| \[\Rightarrow \] \[2{{x}^{2}}-x-4x+2=90\] |
| \[\Rightarrow \] \[2{{x}^{2}}-5x+2-90=0\] |
| \[\Rightarrow \] \[2{{x}^{2}}-5x-88=0\] |
| \[\Rightarrow \] \[2{{x}^{2}}-16x+11x-88=0\] |
| \[\Rightarrow \] \[2x\,(x-8)+11\,(x-8)=0\] |
| \[\Rightarrow \] \[(x-8)(2x+11)=0\] |
| \[\Rightarrow \] \[2x+11=0\] and \[x-8=0\] |
| \[\Rightarrow \] \[x=-\frac{11}{2}\]and \[x=8\] |
| [speed cannot be negative] |
| Speed of boat B = 8 km/ h |
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