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.

OR

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?

Answer:

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,

\[\frac{24}{(18-x)}-\frac{24}{(18+x)}=1\]

\[\frac{24(18+x)-24(18-x)}{(18-x)(18+x)}=1\]

\[\frac{24\times 18+24x-24\times 18+24x}{324-{{x}^{2}}}=1\]

\[\frac{48x}{324-{{x}^{2}}}=1\]

\[48x=324-{{x}^{2}}\]

\[\Rightarrow \]           \[{{x}^{2}}+48x-324=0\]

\[\Rightarrow \]   \[{{x}^{2}}+54x-6x-324=0\]

\[\Rightarrow \]   \[x(x+54)-6(x+54)=0\]

\[\Rightarrow \]             \[(x+54)(x-6)=0\]

Either                         \[x+54=0\]

\[x=-54\]

Rejected, as speed cannot be negative

Or                     \[x-6=0\]

\[x=6\]

Thus, the speed of the stream is 6 km/hr.

OR

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,

\[\frac{63}{x}+\frac{72}{x+6}=3\]

\[\Rightarrow \]       \[\frac{63(x+6)+72(x)}{(x)(x+6)}=3\]

\[\Rightarrow \]      \[\frac{63x+378+72x}{{{x}^{2}}+6x}=3\]

\[\Rightarrow \]   \[135x+378=3({{x}^{2}}+6x)\]

\[\Rightarrow \]   \[135x+378=3{{x}^{2}}+18x\]

\[\Rightarrow \]   \[3{{x}^{2}}+18x-135x-378=0\]

\[\Rightarrow \]   \[3{{x}^{2}}-117x-378=0\]

\[\Rightarrow \]   \[3({{x}^{2}}-39x-126)=0\]

\[\Rightarrow \]       \[{{x}^{2}}-39x-126=0\]

\[\Rightarrow \]   \[{{x}^{2}}-42x+3x-126=0\]

\[\Rightarrow \]   \[x(x-42)+3(x-42)=0\]

\[\Rightarrow \]   \[(x-42)(x+3)=0\]

Either                \[x-42=0\]

\[x=42\]

Or                     \[x+3=0\]

\[x=-3\]

Rejected (as speed cannot be negative)

Thus, average speed of train is 42 km/hr.