Answer
Verified
36.9k+ views
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.
\[1{\text{ }}km{\text{ }} = {\text{ }}1000{\text{ }}m\]
\[1km/hr = \dfrac{5}{{18}}m/s\]
Recently Updated Pages
A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main
Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Other Pages
The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main
An electric bulb has a power of 500W Express it in class 11 physics JEE_Main
The cell in the circuit shown in the figure is ideal class 12 physics JEE_Main
For an electromagnet the core should have A High retentivity class 12 physics JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main