If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
A. 36 kmh
B. 40 kmh
C. 42 kmh
D. 44 kmh
E. 50 kmh
[spoiler]OA=B[/spoiler].
What is the formula that helps me to solve this PS question? I need some help here. Thanks in advance.
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If a car went the first third of the distance at 80 kmh
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Gmat_mission
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Vincen
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Hello Gmat_mission.
If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
Let call d the total distance. The average speed is equal to: (total distance )/ (total time). We have
$$\frac{d}{3}\cdot\left(\frac{1}{80}\right)$$ $$\frac{d}{3}\cdot\left(\frac{1}{24}\right)$$ $$\frac{d}{3}\cdot\left(\frac{1}{48}\right)$$ Therefore, $$Average\ speed=\frac{d}{3}\left(\frac{1}{80}+\frac{1}{24}+\frac{1}{48}\right)=\frac{d}{3}\left(\frac{3}{40}\right)=\frac{d}{40}.$$ this implies that the correct answer is the option B.
I hope it helps.
If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
Let call d the total distance. The average speed is equal to: (total distance )/ (total time). We have
$$\frac{d}{3}\cdot\left(\frac{1}{80}\right)$$ $$\frac{d}{3}\cdot\left(\frac{1}{24}\right)$$ $$\frac{d}{3}\cdot\left(\frac{1}{48}\right)$$ Therefore, $$Average\ speed=\frac{d}{3}\left(\frac{1}{80}+\frac{1}{24}+\frac{1}{48}\right)=\frac{d}{3}\left(\frac{3}{40}\right)=\frac{d}{40}.$$ this implies that the correct answer is the option B.
I hope it helps.
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Here's an algebraic approach:Gmat_mission wrote:If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
A. 36 kmh
B. 40 kmh
C. 42 kmh
D. 44 kmh
E. 50 kmh
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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Let each third of the distance = 240 km, implying that the total distance = 3*240 = 720 km.Gmat_mission wrote:If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
A. 36 kmh
B. 40 kmh
C. 42 kmh
D. 44 kmh
E. 50 km
Since the rate for the first third = 80 kmh, the time to travel the first third = d/r = 240/80 = 3 hours.
Since the rate for the second third = 24 kmh, the time to travel the second third = d/r = 240/24 = 10 hours.
Since the rate for the last third = 48 kmh, the time to travel the last third = d/r = 240/48 = 5 hours.
Average speed for the entire trip = (total distance)/(total time) = 720/(3+10+5) = 720/18 = 40 kmh.
The correct answer is B.
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We can let the total distance = 3d and create the equation:Gmat_mission wrote:If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
A. 36 kmh
B. 40 kmh
C. 42 kmh
D. 44 kmh
E. 50 kmh
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
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