10th Class Mathematics Solved Paper - Mathematics-2018

  • question_answer
    A motor boat whose speed is 18 km/hr. in still water takes 1 hr. more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
    A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr. more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?


    Given, speed of motor boat in still water = 18 km/hr.
    Let speed of stream \[=x\,km/hr\].
    \[\therefore \] Speed of boat downstream \[=(18+x)\text{ }km/hr\]
    And speed of boat upstream \[=(18-x)\text{ }km/hr\].
    Time of the upstream journey =\[\frac{24}{(18-x)}\]
    Time of the downstream journey \[\frac{24}{(18+x)}\]
    According to the question,
               \[\frac{24\times 18+24x-24\times 18+24x}{324-{{x}^{2}}}=1\]
                \[\Rightarrow \]           \[{{x}^{2}}+48x-324=0\]
                \[\Rightarrow \]   \[{{x}^{2}}+54x-6x-324=0\]
                \[\Rightarrow \]   \[x(x+54)-6(x+54)=0\]
                \[\Rightarrow \]             \[(x+54)(x-6)=0\]
                Either                         \[x+54=0\]
    Rejected, as speed cannot be negative
    Or                     \[x-6=0\]
    Thus, the speed of the stream is 6 km/hr.
    Let original average speed of train be x km/hr.
    \[\therefore \] Increased speed of train \[=(x+6)\text{ }km/hr\].
    Time taken to cover 63 km with average speed \[=\frac{63}{x}hr.\]
    Time taken to cover 72 km with increased speed \[=\frac{72}{(x+6)}hr.\]
    According to the question,
    \[\Rightarrow \]       \[\frac{63(x+6)+72(x)}{(x)(x+6)}=3\]
    \[\Rightarrow \]      \[\frac{63x+378+72x}{{{x}^{2}}+6x}=3\]
    \[\Rightarrow \]   \[135x+378=3({{x}^{2}}+6x)\]
    \[\Rightarrow \]   \[135x+378=3{{x}^{2}}+18x\]
    \[\Rightarrow \]   \[3{{x}^{2}}+18x-135x-378=0\]
    \[\Rightarrow \]   \[3{{x}^{2}}-117x-378=0\]
    \[\Rightarrow \]   \[3({{x}^{2}}-39x-126)=0\]
    \[\Rightarrow \]       \[{{x}^{2}}-39x-126=0\]
    \[\Rightarrow \]   \[{{x}^{2}}-42x+3x-126=0\]
    \[\Rightarrow \]   \[x(x-42)+3(x-42)=0\]
    \[\Rightarrow \]   \[(x-42)(x+3)=0\]
    Either                \[x-42=0\]
    Or                     \[x+3=0\]
    Rejected (as speed cannot be negative)
    Thus, average speed of train is 42 km/hr.


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