A car completes a 200 km journey at an average speed of x km/h. The car completes the return journey of 200 km at an average speed of (x + 10) km/h. Show that the difference betwee... A car completes a 200 km journey at an average speed of x km/h. The car completes the return journey of 200 km at an average speed of (x + 10) km/h. Show that the difference between the time taken for each of the two journeys is equal to 2000 / (x(x + 10)) hours.

Understand the Problem

The question states that a car travels 200km at speed x km/h, then returns along the same 200km at a new speed of (x+10) km/h. We need to show that the difference in time of the outbound and return journey is equal to (\frac{2000}{x(x+10)}). To show this, we need to find an expression for the time taken for each part of the journey, and subtract them.

Answer

$\frac{2000}{x(x+10)}$

Steps to Solve

Calculate the time taken for the outbound journey.

We know that time = distance / speed. The distance of the outbound journey is 200 km and the speed is $x$ km/h. Therefore, the time taken for the outbound journey is: $$ \text{Time}_\text{outbound} = \frac{200}{x} $$

Calculate the time taken for the return journey.

The distance of the return journey is 200 km and the speed is $(x+10)$ km/h. Therefore, the time taken for the return journey is: $$ \text{Time}_\text{return} = \frac{200}{x+10} $$

Find the difference in time between the two journeys.

We are asked to find the difference in time, which is the time taken for the outbound journey minus the time taken for the return journey: $$ \text{Time}\text{difference} = \text{Time}\text{outbound} - \text{Time}_\text{return} = \frac{200}{x} - \frac{200}{x+10} $$

Simplify the expression.

To subtract the two fractions, we need a common denominator, which is $x(x+10)$. $$ \text{Time}_\text{difference} = \frac{200(x+10)}{x(x+10)} - \frac{200x}{x(x+10)} = \frac{200x + 2000 - 200x}{x(x+10)} = \frac{2000}{x(x+10)} $$

State the final answer.

Therefore, the difference in time between the two journeys is $\frac{2000}{x(x+10)}$ hours, as required.

The difference in time is $\frac{2000}{x(x+10)}$ hours.

More Information

This question is related to distance, speed and time calculations, with a bit of algebraic manipulation thrown in. It demonstrates how changes in speed affect travel time, and how to represent these relationships algebraically.

Tips

A common mistake would be to subtract the fractions incorrectly, either by not finding a common denominator or by making an error when expanding the brackets. Also, some might subtract the wrong way around (return - outbound), but since we are not given any additional info, both directions are correct since the question asks for the difference between the time.

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