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Annie took a trip consisting of three segments...

Expert replies
by BTGmoderatorLU » Thu Jan 04, 2018 8:44 am
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?

$$A.\ \frac{11v}{3}$$
$$B.\ \frac{10v}{3}$$
$$C.\ 3v$$
$$D.\ \frac{5v}{2}$$
$$E.\ 2v$$

The OA is D.

I'm confused with this PS question. Experts, any suggestion about the solution? Thanks in advance.
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Source: — Problem Solving |

by DavidG@VeritasPrep » Thu Jan 04, 2018 9:36 am
LUANDATO wrote: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?

$$A.\ \frac{11v}{3}$$
$$B.\ \frac{10v}{3}$$
$$C.\ 3v$$
$$D.\ \frac{5v}{2}$$
$$E.\ 2v$$

The OA is D.

I'm confused with this PS question. Experts, any suggestion about the solution? Thanks in advance.
Set D = 6 and v = 1

First leg Distance: 5D = 30
First leg speed = 2v = 2
First leg Time: 30/2 = 15

Second leg Distance: 4D = 24
Second leg speed: 3v = 3
Second leg Time: 8

Third leg Distance: 6
Third leg speed: 6v = 6
Third leg Time: 6/6 = 1

Total distance: 30 + 24 + 6 = 60
Total Time: 15 + 8 + 1 = 24
Avg speed = 60/24 = 5/2

Now plug v =1 into your answer choices and see what gives you 5/2. D is the only option that works.
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by Scott@TargetTestPrep » Mon Sep 02, 2019 5:59 pm
BTGmoderatorLU wrote: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?

$$A.\ \frac{11v}{3}$$
$$B.\ \frac{10v}{3}$$
$$C.\ 3v$$
$$D.\ \frac{5v}{2}$$
$$E.\ 2v$$

The OA is D.

I'm confused with this PS question. Experts, any suggestion about the solution? Thanks in advance.
Using the formula: average speed = total distance/total time, we have average speed equal to:

(5D + 4D + D)/(5D/2v + 4D/3v + D/6v)

(10D)/(5D/2v + 4D/3v + D/6v)

Multiplying by 6v/6v, we have:

60Dv/(15D + 8D + D)

60Dv/24D = 10v/4 = 5v/2

Answer: D

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