Question

a man travelled from his house to market on a straight road at 10 kilometre per hour and immediately returned at 15 kilometre per hour average speed of the man is​

Answer 1

Answer:

$\underline{\green{Given:}}$

$At\:first\:speed\:of\:the\:man=10\:km/hr$

$After\:speed\:of\:the\:man=15\:km/hr$

$\underline{\red{To\:find}}$

$Average\:speed\:of\:the\:man=??$

$\underline{\pink{Taken}}$

$Average\:speed=\frac{Total\:distance\:travelled}{Total\:time\:taken}$

$\underline{\blue{Concept}}$

▪︎First do sum of total distance travelled and total time taken . Then divide both of them .

$\underline{\orange{Solution}}$

$\rightarrow{Average\:speed=\frac{10\:km+15\:km}{1\:hr+1\:hr}}$

$\rightarrow{Average\:speed=\frac{25\:km}{2\:hr}}$

$\therefore\boxed{Average\:speed=14.5\:km/hr}$

$\underline{\purple{Conclusion}}$

So , the average speed is 14.5 km per hour.

HOPE IT HELPS YOU

Answer 2

$\color{red}\large\underline{\underline{Question:}}$

$\text{A man travelled from his house to market on a}$ $\text{straight road at 10 kilometre per hour and }$$\text{immediately returned at 15 kilometre per hour }$$\text{average speed of the man is.....}$

______________________________________

$\color{purple}\large\underline{\underline{To\:Find:}}$

$\underline{\color{purple}{The\:Average\:speed\:of\:the\:man}}$

______________________________________

$\color{navy}\large\underline{\underline{Given:}}$

$Speed = s_{1} = 10\:km\:hr^{-1}$

$Speed = s_{2} = 15\:km\:hr^{-1}$

______________________________________

$\color{orange}\large\underline{\underline{We\:Know:}}$

$\rightarrow Average\:speed = \color{magenta}{\dfrac{Total\:distance\:travelled}{Total\:time\:taken}}$

$\rightarrow speed = \color{purple}{\dfrac{distance\:travelled}{time\:taken}}$

$\therefore time = \color{blue}{\dfrac{distance\:travelled}{speed}}$

______________________________________

$\color{violet}\large\underline{\underline{Taken:}}$

$\text{Let the distance be x}$

______________________________________

$\color{blue}\large\underline{\underline{Concept:}}$

• $\text{Here the distance is same in both conditions}$
• $\text{By the given distance and time we can find it's time}$

______________________________________

$\color{magenta}\large\underline{\underline{Solution:}}$

$\underline{First\:let's\:find\:its\:time}$

$Let\:the\:time\:be\:t_{1}\:in\:first\:case$

so,

$t_{1} = \dfrac{x}{10}\:hr$

$Let\:the\:time\:be\:t_{2}\:in\:second\:case$

so,

$t_{2} = \dfrac{x}{15}\:hr$

______________________________________

$\underline{Average\:speed}$

$\Rightarrow Average\:speed = {\dfrac{x + x}{\dfrac{x}{10} + \dfrac{x}{15}}}$

$\Rightarrow Average\:speed = {\dfrac{2x}{\dfrac{3x + 2x}{30}}}$

$\Rightarrow Average\:speed = {\dfrac{2x}{\dfrac{5x}{30}}}$

$\Rightarrow Average\:speed = \dfrac{2x}{5x} \times 30$

$\Rightarrow Average\:speed = \dfrac{2}{5} \times 30$

$\Rightarrow Average\:speed = \dfrac{2}{1} \times 6$

$\Rightarrow Average\:speed = 12\:km\:hr^{-1}$

______________________________________

$\red{\boxed{\therefore Average\:speed = 12\:km\:hr^{-1}}}$