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10 Jun 2019, 01:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
81% (02:21) correct
19%
(02:37)
wrong
based on 181
sessions
History
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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
(A) \(\frac{11v}{3}\)
(B) \(\frac{10v}{3}\)
(C) 3v
(D) \(\frac{5v}{2}\)
(E) 2v
10 Jun 2019, 01:15
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Bunuel
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
Archit3110
Major Poster
10 Jun 2019, 11:24
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Bunuel
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
