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# A car travels the first half of a distance between two places at a speed of $30 \mathrm{Km} / \mathrm{hr}$ and the second half of the distance at $50 \mathrm{~km} / \mathrm{hr}$. The average speed of the car for the whole journey is:A) $42.5 \mathrm{~km} / \mathrm{hr}$B) $40.0 \mathrm{Km} / \mathrm{hr}$C) $37.5 \mathrm{Km} / \mathrm{hr}$D) $35.0 \mathrm{Km} / \mathrm{hr}$

Last updated date: 20th Jun 2024
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Hint: To solve this question we should know that the speed is defined as the distance to be travelled by the object divided by the total time that is taken by the object to travel the mentioned distance. We need to consider the unmentioned quantities in this question as variables and use the speed distance formula to find the answer.

Since, the distance is not mentioned in quantity, we have to assume it as a variable and develop the expressions for distance, time and speed.
So, let us consider that the total distance is 2x km.
Therefore the car travels the first half distance which is $x$ km with $30 \mathrm{~km} / \mathrm{hr}$ in the time interval $=x / 30$hr.
Now we have to find the expression for the second part of the distance that is travelled by the car. So, the expression is given as:
The car travels the second half of the distance which is $x$ km with $50 \mathrm{~km} / \mathrm{hr}$ in the time interval $=x / 50$ hr.
When the time is finally expressed, we can find the average speed. The expression is given as:
Average speed= total distance travelled divided by total time taken or:
$\text{Average speed = }\dfrac{\text{Total distance travelled}}{\text{Total time taken}}$
After we put the values, we get that: $\dfrac{2x}{\dfrac{x}{30}+\dfrac{x}{50}}$
After the evaluation, we will get that:
$\Rightarrow \dfrac{2x}{\dfrac{5x+3x}{150}}=\dfrac{2x}{\dfrac{8x}{150}}=\dfrac{150}{4}$
The above expression gives the value of $37.5 \mathrm{Km} / \mathrm{hr}$.

Hence, the correct option is option C.

Note: While answering the questions that are related to the speed, distance and time, we should be careful with the units. In this question, the units were similar so there was no conversion required to be made. But in case even one of the similar quantities were in a different unit we were required to change the unit to one common unit.