Answer
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Hint: Speed is defined as the distance covered by any object divided by the time taken to cover that distance.
Where Distance is the total length of the path covered by that object.
\[Speed = \dfrac{{Distance}}{{Time}}\]
Complete step-by-step answer:
Given, Farmer goes from point A to B
Then,
\[\begin{array}{*{20}{l}}
{Distance{\text{ }}AB{\text{ }}\left( S \right){\text{ }} = {\text{ }}61{\text{ }}km} \\
{Total{\text{ }}Time{\text{ }}taken{\text{ }}to{\text{ }}travel{\text{ }}61{\text{ }}km{\text{ }}\left( T \right){\text{ }} = 9{\text{ }}hours} \\
{Speed{\text{ }}at{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}foot{\text{ }} = {\text{ }}4{\text{ }}km/hr} \\
{Speed{\text{ }}at{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}bicycle{\text{ }} = {\text{ }}9{\text{ }}km/hr}
\end{array}\]
Let the distance travelled on foot and bicycle be AC and BC respectively.
\[\begin{gathered}
Let{\text{ }}the{\text{ }}time{\text{ }}for{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}foot{\text{ }} = {\text{ }}x{\text{ }}hrs, \\
Then{\text{ }}time{\text{ }}for{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}bicycle{\text{ }} = {\text{ }}\left( {9 - x} \right){\text{ }}hrs \\
\end{gathered} \]
We know that,
\[Distance = Speed \times Time\]
Then, \[AC = 4km/hr \times xhr = 4x\] eqn (i)
\[BC = 9km/hr \times (9 - x)hr = 81 - 9x\] eqn (ii)
Also, \[AB = AC + BC\]
Using value of AB, AC and BC
We get,
\[\begin{gathered}
61 = 4x + 81 - 9x \\
5x = 20 \\
x = 4km \\
\end{gathered} \]
By putting value of x in eqn (i) and eqn (ii) we get,
\[\begin{array}{*{20}{l}}
{AC{\text{ }} = {\text{ }}16{\text{ }}km} \\
{BC = {\text{ }}45{\text{ }}km} \\
{\therefore Distance{\text{ }}travelled{\text{ }}on{\text{ }}foot{\text{ }}by{\text{ }}farmer{\text{ }} = {\text{ }}16{\text{ }}km}
\end{array}\]
Hence option (C) is correct
Note: Always check the unit of all the given data. For example, if the distance is given in km and speed is given in m/s then convert both the quantity into the same unit.
\[1{\text{ }}km{\text{ }} = {\text{ }}1000{\text{ }}m\]
\[1km/hr = \dfrac{5}{{18}}m/s\]
Where Distance is the total length of the path covered by that object.
\[Speed = \dfrac{{Distance}}{{Time}}\]
Complete step-by-step answer:
Given, Farmer goes from point A to B
Then,
\[\begin{array}{*{20}{l}}
{Distance{\text{ }}AB{\text{ }}\left( S \right){\text{ }} = {\text{ }}61{\text{ }}km} \\
{Total{\text{ }}Time{\text{ }}taken{\text{ }}to{\text{ }}travel{\text{ }}61{\text{ }}km{\text{ }}\left( T \right){\text{ }} = 9{\text{ }}hours} \\
{Speed{\text{ }}at{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}foot{\text{ }} = {\text{ }}4{\text{ }}km/hr} \\
{Speed{\text{ }}at{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}bicycle{\text{ }} = {\text{ }}9{\text{ }}km/hr}
\end{array}\]
Let the distance travelled on foot and bicycle be AC and BC respectively.
\[\begin{gathered}
Let{\text{ }}the{\text{ }}time{\text{ }}for{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}foot{\text{ }} = {\text{ }}x{\text{ }}hrs, \\
Then{\text{ }}time{\text{ }}for{\text{ }}which{\text{ }}farmer{\text{ }}travels{\text{ }}on{\text{ }}bicycle{\text{ }} = {\text{ }}\left( {9 - x} \right){\text{ }}hrs \\
\end{gathered} \]
We know that,
\[Distance = Speed \times Time\]
Then, \[AC = 4km/hr \times xhr = 4x\] eqn (i)
\[BC = 9km/hr \times (9 - x)hr = 81 - 9x\] eqn (ii)
Also, \[AB = AC + BC\]
Using value of AB, AC and BC
We get,
\[\begin{gathered}
61 = 4x + 81 - 9x \\
5x = 20 \\
x = 4km \\
\end{gathered} \]
By putting value of x in eqn (i) and eqn (ii) we get,
\[\begin{array}{*{20}{l}}
{AC{\text{ }} = {\text{ }}16{\text{ }}km} \\
{BC = {\text{ }}45{\text{ }}km} \\
{\therefore Distance{\text{ }}travelled{\text{ }}on{\text{ }}foot{\text{ }}by{\text{ }}farmer{\text{ }} = {\text{ }}16{\text{ }}km}
\end{array}\]
Hence option (C) is correct
Note: Always check the unit of all the given data. For example, if the distance is given in km and speed is given in m/s then convert both the quantity into the same unit.
\[1{\text{ }}km{\text{ }} = {\text{ }}1000{\text{ }}m\]
\[1km/hr = \dfrac{5}{{18}}m/s\]
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