If a car traveled the first third of the distance at 80 km/h, the

Here's an algebraic approach:

Let d = 1/3 of the total distance traveled
So, the car traveled d km at 80 kmh, then the car traveled d km at 24 kmh, and then d km at 48 kmh

Average speed = (total distance traveled)/(total travel time)

Total distance traveled = d + d + d = 3d

time = distance/speed
Total time spent traveling = (time spent traveling 80kmh) + (time spent traveling 24 kmh) + (time spent traveling 48 kmh)
= d/80 + d/24 + d/48 [let's rewrite this with a common denominator]
= 3d/240 + 10d/240 + 5d/240
= 18d/240
= 9d/120
= 3d/40

Average speed = (3d)/(3d/40)
= 120d/3d
= 40

Answer: B

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We can let the total distance = 3d and create the equation:

average = 3d/(d/80 + d/24 + d/48)

average = 3d/(d/80 + d/24 + d/48)

average = 3d/(3d/240 + 10d/240 + 5d/240)

average = 3d/(18d/240)

average = 720d/18d = 40

Answer: B

Such questions are best done by taking smart numbers, because whatever be the distance the answer will remain the same.

What can be a smart number here. It has to be something to do with the LCM of three speeds.
LCM(80,24,48) = 240.
So you can take each \(\frac{1}{3}^{rd}\) portion as 240 or 480.

Let us take the distance for each part as 480.

A car went the first third of the distance at 80 kmh, so time taken => \(\frac{480}{80}=6\)
The second third at 24 kmh, so time taken => \(\frac{480}{24}=20\)
The last third at 48 kmh, so time taken => \(\frac{480}{48}= 10\)

Total time = \(6+20+10=36\)\(\)
Distance =\(3*480\)

The average speed of the car for the entire trip = \(\frac{3*480}{36}=40\)kmph

B

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