Question

A train travels a distance at a speed of 40 km/h and returns at of 60 km/h. What is the average speed of the train?
(i)12km/h
(ii)24km/h
(iii)16km/h
(iv) 48km/h​

Answer 1

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

Provided that:

• A train travels a distance at a speed of 40 km/h and returns at of 60 km/h. What is the average speed of the train?

Solution:

• Average speed = 48 kmph

Using concepts:

• 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 as a

Required solution:

~ Firstly finding total distance!

→ As it travel and return too and take same distance or we can say say path. Therefore,

• Total distance = 2a

~ Now let us find out the time taken

$:\implies \sf Time \: taken \: = \dfrac{Distance}{Speed} \\ \\ :\implies \sf t \: = \dfrac{s}{v} \\ \\ :\implies \sf t \: = \dfrac{a}{40} + \dfrac{a}{60} \\ \\ \leadsto \sf Taking \: LCM \: of \: 40 \: and \: 60 \\ \\ :\implies \sf t \: = \dfrac{3 \times a + 2 \times a}{120} \\ \\ :\implies \sf t \: = \dfrac{3a + 2a}{120} \\ \\ :\implies \sf t \: = \dfrac{5a}{120} \\ \\ :\implies \sf Time \: = \dfrac{5a}{120} \: hour$

~ Now let us calculate the average speed!

$:\implies \sf Average \: speed \: = \dfrac{Total \: distance}{Total \: time} \\ \\ :\implies \sf v \: = \dfrac{s}{t} \\ \\ :\implies \sf v \: = \dfrac{\dfrac{2a}{5a}}{120} \\ \\ :\implies \sf v \: = \dfrac{2a \times 120}{5a} \\ \\ :\implies \sf v \: = \dfrac{2\not{a} \times 120}{5\not{a}} \\ \\ :\implies \sf v \: = \dfrac{2 \times 120}{5} \\ \\ :\implies \sf v \: = \dfrac{240}{5} \\ \\ :\implies \sf v \: = \cancel{\dfrac{240}{5}} \: (Cancelling) \\ \\ :\implies \sf v \: = 48 \: kmh^{-1}$

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

Provided that:

• A train travels a distance at a speed of 40 km/h and returns at of 60 km/h. What is the average speed of the train?

Solution:

Average speed = 48 kmph

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:

$:\implies \sf v \: = \dfrac{2 \: v_1 \: v_2}{v_1 \: + v_2} \\ \\ :\implies \sf v \: = \dfrac{2 \times 40 \times 60}{40 + 60} \\ \\ :\implies \sf v \: = \dfrac{2 \times 2400}{100} \\ \\ :\implies \sf v \: = \dfrac{2 \times 24}{1} \\ \\ :\implies \sf v \: = 48 \: kmh^{-1}$

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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! :)