This doesn't sound logical

I have 2 ways of solving this problem:

(1) Using physics

Get this diagram:

• In this $t_1$ represents time taken from starting point to the center whereas $t_2$ represents time taken from center to the finishing point. Moreover $x$ determines the distance travelled.

$\text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{2x}{t_1 + t_2} \color{#20A900} ------------->\text{This is equation 1}$

• Since it is uniform velocity over each sections of $x$ , therefore we can use $s = ut$ because velocity is constant for $x$ but not for $2x$ .

$\implies s = ut \\ x = 20 \times t_1 \color{#20A900} ------------->\text{This is equation 2} \\ x = 30 \times t_2 \color{#20A900} ------------->\text{This is equation 3}$

• Now substitute equation 2 and 3 in equation 1 as follows:

$\text{Average speed} = \dfrac{2x}{t_1 + t_2} \\ = \dfrac{2x}{\dfrac{x}{20} + \dfrac{x}{30}} \\ = \dfrac{2x}{x (\dfrac{1}{20} + \dfrac{1}{30})} \\ \implies \frac{2}{ (\dfrac{1}{20} + \dfrac{1}{30}) \\ \color{#69047E} = \boxed{\text{24 km/h}}}$

(2)Using mathematics

Directly use the Harmonic mean formula:

$f_{\text{HM}} = \frac{k}{\frac{1}{a_1} + \cdots + \frac{1}{a_k}}.$

Or else in our case:

$\implies \frac{2}{ \dfrac{1}{20} + \dfrac{1}{30} \\ \color{#69047E} = \boxed{\text{24 km/h}}}$

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Some common errors are:

1. You try to assume a distance. This is incorrect because time will change according to the distance.

2. You try to use the arithmetic mean which will give you 25 km/h. It may be close but not accurate enough.

Assume the distance is 120 km, then the total time taken is 3 + 2 = 5 hours => the average is 120/5 = 24 km/h

Proof that this method works for any x:

Time taken over first half: x/20 h

Time taken over second half: x/30 h

Total time: x/20 + x/30 = 5x/60 h

Total average time: total distance / total time = 2x / (5x/60) = 120/5 = 24 km/h for ANY x

So you might as well take a convenient value for x, like 60 as I did.

The average speed is $v=\frac{distance}{time}$ . The time needed to take half the distance at 20km/h is $\frac{distance/2}{20km/h}$ and for the other half of the trip $\frac{distance/2}{30km/h}$ . The total time is $time= \frac{distance/2}{20km/h}+\frac{distance/2}{20km/h}$ . Using this we get

$v=\frac{distance}{time}= \frac{distance}{\frac{distance/2}{20km/h}+\frac{distance/2}{20km/h}} = \frac{2}{\frac{1}{20}+\frac{1}{20}}=24km/h$

Let the total distance be $D$ . The time taken to travel the first half of the distance, $t_1 = \dfrac {\frac D2}{20} = \dfrac D{40} \text{ h}$ and that for the second half $t_2 = \dfrac {\frac D2}{30} = \dfrac D{60} \text{ h}$ . The average speed $v_{av} =$ total distance traveled $\div$ total time taken. $\implies v_{av} = \dfrac D{t_1+t_2} = \dfrac D{\frac D{40} + \frac D{60}} = \dfrac {120}5 = \boxed{24} \text{ km/h}$ .