Question

A boy goes from his house to school by bus at a speed of 20 km/h and returns back through the same route at a speed of 30 km/h . What is the average speed of his journey?

Answer 1

Hi.

Here is your answer------

Let the distance between the school and the house is x km

speed of the boy from house to school = 20 km/hr.

Using formula, Time taken  = distance/speed

So, time taken = x/20 hours.

Also,
Speed of the boy from school to the house = 30 km/hr

so, time taken = x/30 hours.

Total time taken = x/20 + x/30
                          = x/12 hours

Also total distance covered by boy = x +x 
                                                        = 2 x km.
Using formula,
Average speed = Total distance/ Total time
                         = 2 x/(x/12)
                         = 24 km/hr.
Thus, average speed of the boy's journey is 24 km/hr.

Hope this will help u.

Have a nice day.

Please mark this answer as Brainliest.

Answer 2

Given:

Speed from the house to school $\[ = 20{\text{ }}km/h\]$

Speed from the school to the house $\[ = 30{\text{ }}km/h\]$

To Find: Average speed

Solution:

The average speed is given as:

$\[{\text{Average speed = }}\frac{{{\text{total distance}}}}{{{\text{total time taken}}}}\]$

Suppose the distance between the house and the school is $d$.

Therefore,

Time taken to travel from house to school will be

${t_1} = \frac{d}{{20{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}$

Time taken to travel from school to house will be

${t_2} = \frac{d}{{30{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}$

Thus, the total time to travel from house to school and school to house can be given as:

$t = {t_1} + {t_2} = \frac{d}{{20{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}} + \frac{d}{{30{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}$

Therefore, the average speed can be calculated as:

$\[\begin{gathered} {V_{avg}} = \frac{{2d}}{{\frac{d}{{20{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}} + \frac{d}{{30{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}}} \\ {V_{avg}} = \frac{2}{{\frac{1}{{20{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}} + \frac{1}{{30{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}}} \\ \end{gathered} \]$

$\[{V_{avg}} = \frac{2}{{\frac{{20{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h} + 30{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}{{20{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h} \times 30{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}}}\]$

$\[{V_{avg}} = \frac{{2 \times 600{{{{k{m^2}} \mathord{\left/ {\vphantom {{k{m^2}} h}} \right. \kern-\nulldelimiterspace} h}}^2}}}{{50{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}}}\]$

Thus,

$\[{V_{avg}} = 24{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}\]$

Hence, the average speed of the boy is $\[24{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h}\]$.

#SPJ2