Marcel is taking a trip, driving at a constant speed of X miles per ho

Average speed over 5 hours

• Total travel: 2x + 3y
• Total time: 5
• Avg speed: \(\frac{2x+3y}{5}\)

Average speed over 5X miles

• Total travel: 5x
• Total time: \(2 + \frac{5x-2x}{y} = 2 + \frac{3x}{y}\)
• Avg speed:

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Let's plug in numbers:

X = 3 km/h
Y = 2 Km/h

Average speed = total distance / total time.

Marcel drove the first two hours at X, and the last 3 at Y, then in the two first hours Marcel drove 6 km, and for the last 3 hours, 6 km.
(You can do it with mental math, or just apply the formula D = V x T)

Average speed = 12/5 km/h

Option 1 - 3Y/5

3x2/5 is diferrent from 12/5

Option 2 - (2X + 3Y)/5

(2x3 + 3x2)/5 = 12/5.

So we have our first option conclued, let's move on to second.




The distance Marcel traveled was 5X, hence 5x3 = 15 km.

Consider he always drive the first 2 hours at 3km/h and the remaining at 2km/h

Now it's just logic and convertion.

In two hours he drove 6 km, there's 9 left to run at 2 km/h, hence 8 km in 4 hours + 1 km in 0.5 hours.
(Or just apply the formula T = D/V for the second part of the trip: T = 9/2, Total time = 4,5 + 2 from the first parte of the trip.)

So he drove 15 miles in 6.5 hours, then his average speed is 15/6.5 or 30/13.

Let's check the options, we already know the correct answer for the second part can't be the two first ones.

Option 3 - (3X - 2Y)/5

Don't even calculate, we know it yields a smaller number than we need.

Look at option 5, it also yields a smaller number, hence we only have the fourth option as the correct one.



Want to check it out? Ok.

Option 4 - 5YX/(2Y + 3X)

5x2x3/(2x2 + 3x3) = 30/13.