Question

1. Abdul, while driving to school, computes the average speed for his trip to be 20 km hl. On his return trip along the same route, there is less traffic and the average speed is 30 km hl. What is the average speed for Abdul's trip?
full explanation please​

Answer 1

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${\large{\pmb{\sf{\bigstar \:{\underline{RequirEd \: Solution...}}}}}}$

$\: \: \: \: \: \:{\large{\pmb{\sf{\bigstar \:{\underline{By \: method \: first...}}}}}}$

ProvidEd that:

⋆ Average speed by going to a trip is equal to 20 kilometres per hour.

⋆ Average speed when returning from a trip is equal to 30 kilometres per hour.

To calculaTe:

⋆ Average speed for Abdul's trip

SolutioN:

⋆ Average speed for Abdul's trip = 24 kilometres per hour

Using concepts:

⋆ Dimension to calculate total distance

⋆ Average speed formula

⋆ Formula to calculate time taken

Using formulas:

• Total distance is given by,

• ${\small{\underline{\boxed{\sf{Total \: distance \: = 1st \: distance \: + 2nd \: distance}}}}}$

• Average speed is given by,

• ${\small{\underline{\boxed{\sf{v \: = \dfrac{s}{t}}}}}}$

Where, v denotes average speed, s denotes total distance and t denotes total time taken.

• Time is given by,

• ${\small{\underline{\boxed{\sf{t \: = \dfrac{s}{v}}}}}}$

Where, t denotes time taken, s denotes distance and v denotes speed

Assumption:

• Let the distance travelled by Abdul as a

Required solution:

~ Firstly finding total distance!

→ As he go to school and return back again from a distance,a therefore, the total distance becames

• Total distance = 2a

~ Now let us find out the time taken

$\begin{gathered}:\implies \sf Time \: taken \: = \dfrac{Distance}{Speed} \\ \\ :\implies \sf t \: = \dfrac{s}{v} \\ \\ :\implies \sf t \: = \dfrac{a}{20} + \dfrac{a}{30} \\ \\ \leadsto \sf Taking \: LCM \: of \: 20 \: and \: 30 \\ \\ :\implies \sf t \: = \dfrac{3 \times a + 2 \times a}{60} \\ \\ :\implies \sf t \: = \dfrac{3a + 2a}{60} \\ \\ :\implies \sf t \: = \dfrac{5a}{60} \\ \\ :\implies \sf Time \: = \dfrac{5a}{60} \: hour\end{gathered}$

~ Now let us calculate the average speed!

$\begin{gathered}:\implies \sf Average \: speed \: = \dfrac{Total \: distance}{Total \: time} \\ \\ :\implies \sf v \: = \dfrac{s}{t} \\ \\ :\implies \sf v \: = \dfrac{\dfrac{2a}{5a}}{60} \\ \\ :\implies \sf v \: = \dfrac{2a \times 60}{5a} \\ \\ :\implies \sf v \: = \dfrac{2\not{a} \times 60}{5\not{a}} \\ \\ :\implies \sf v \: = \dfrac{2 \times 60}{5} \\ \\ :\implies \sf v \: = \dfrac{120}{5} \\ \\ :\implies \sf v \: = \cancel{\dfrac{120}{5}} \: (Cancelling) \\ \\ :\implies \sf v \: = 24 \: kmh^{-1}\end{gathered}$

Therefore, average speed = 24 km/h

___________________________

$\: \: \: \: \: \:{\large{\pmb{\sf{\bigstar \:{\underline{By \: method \: second...}}}}}}$

ProvidEd that:

⋆ Average speed by going to a trip is equal to 20 kilometres per hour.

⋆ Average speed when returning from a trip is equal to 30 kilometres per hour.

To calculaTe:

⋆ Average speed for Abdul's trip

SolutioN:

⋆ Average speed for Abdul's trip = 24 kilometres per hour

Using concept:

⋆ Average speed formula

Using formula:

${\small{\underline{\boxed{\sf{v \: = \dfrac{2 \: v_1 \: v_2}{v_1 \: + v_2}}}}}}$

Where, v denotes average speed, v_1 denotes speed first and v_2 denotes speed second.

Required solution:

$\begin{gathered}:\implies \sf v \: = \dfrac{2 \: v_1 \: v_2}{v_1 \: + v_2} \\ \\ :\implies \sf v \: = \dfrac{2 \times 20 \times 30}{20 + 30} \\ \\ :\implies \sf v \: = \dfrac{2 \times 600}{50} \\ \\ :\implies \sf v \: = \dfrac{1200}{50} \\ \\ :\implies \sf v \: = \cancel{\dfrac{1200}{50}} \: (Cancelling) \\ \\ :\implies \sf v \: = 24 \: kmh^{-1}\end{gathered}$

Average speed = 24 km/h

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Apply the second method if you are in standard 11 or more as this formula will start from class 11th. If you wanna apply this formula and you are in 11th less then please ask your teacher about this formula that you are getting marking or not. Thanks for understanding! :)