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Bunuel
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Annie took a trip consisting of three segments at three different spee 10 Jun 2019, 01:00
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Difficulty:

25% (medium)

Question Stats:

81% (02:21) correct 19% (02:37) wrong based on 182 sessions

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Annie took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, the average speed of Annie’s trip was

(A) \(\frac{11v}{3}\)

(B) \(\frac{10v}{3}\)

(C) 3v

(D) \(\frac{5v}{2}\)

(E) 2v
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D
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Annie took a trip consisting of three segments at three different spee 10 Jun 2019, 01:15
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Bunuel
Annie took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, the average speed of Annie’s trip was

(A) \(\frac{11v}{3}\)

(B) \(\frac{10v}{3}\)

(C) 3v

(D) \(\frac{5v}{2}\)

(E) 2v
Show more

Average Speed = \(\frac{Total Distance}{Total Time}\)

Total Distance = 5D + 4D + D = 10D
Total Time = \(\frac{5D}{2V}\) + \(\frac{4D}{3V}\) + \(\frac{D}{6V}\)
= \(\frac{24D}{6V}\)
= \(\frac{4D}{V}\)

--> Average Speed = \(\frac{10D}{4D/V}\) = 5V/2

IMO Option D
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Annie took a trip consisting of three segments at three different spee 10 Jun 2019, 11:24
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Bunuel
Annie took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, the average speed of Annie’s trip was

(A) \(\frac{11v}{3}\)

(B) \(\frac{10v}{3}\)

(C) 3v

(D) \(\frac{5v}{2}\)

(E) 2v
Show more

avg speed = total distance/ total time
total time ; 5d/2v+4d/3v+d/6v = 24d/6v; 4d/v
total distance ; 10 d
so avg speed; 5v/2
IMO D
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Annie took a trip consisting of three segments at three different spee 10 Jun 2019, 12:20
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Quote:
Annie took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, the average speed of Annie’s trip was
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Knowing principles of the "RT=D" framework, we are basically asked for total distance / total time .

From the question we know the total distance (5D+4D+D=10D), so we need to find the total time .

From the first trip segment: time=distance/rate => 5D/2v

From the second trip segment: time=distance/rate => 4D/3v

From the third trip segment: time=distance/rate => D/6v

Total time is 24D/6v=4D/v. Average speed is 10D/(4D/v)=5v/2.

Answer (D).
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Annie took a trip consisting of three segments at three different spee 04 Sep 2025, 10:38
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