Juan goes out cycling outdoors. He travels at an average...

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BTGmoderatorLU: Moderator; Posts: 2550; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Fri Mar 09, 2018 3:04 pm

Juan goes out cycling outdoors. He travels at an average speed of 15km/h, to the top of the hill where the midpoint of the trip is. Going down hill, Juan travels at an average speed of 20km/h. Which of the following is the closest approximation of Juan's average speed, in kilometers per hour, for the round trip?

A. 15.0
B. 17.1
C. 17.5
D. 17.9
E. 20.0

The OA is B.

I'm confused by this PS question. Is there some direct formula to solve it? Experts, any suggestion? Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Fri Mar 09, 2018 3:04 pm

mbawisdom: Senior | Next Rank: 100 Posts; Posts: 94; Joined: Tue Dec 16, 2014 9:50 amFollowed by: 4 members; Location: London, UK; Thanked: 2 times; GMAT Score:770

by mbawisdom » Sat Mar 10, 2018 5:35 am

LUANDATO wrote:Juan goes out cycling outdoors. He travels at an average speed of 15km/h, to the top of the hill where the midpoint of the trip is. Going down hill, Juan travels at an average speed of 20km/h. Which of the following is the closest approximation of Juan's average speed, in kilometers per hour, for the round trip?

A. 15.0
B. 17.1
C. 17.5
D. 17.9
E. 20.0

The OA is B.

I'm confused by this PS question. Is there some direct formula to solve it? Experts, any suggestion? Thanks in advance.

Where did you get up to?

I would approach it as follows:

Speed = Distance/Time, lets call it S = D/T
Divide the journey up into two parts with equal distance but different times and average speeds.

Part 1: 15 =D/T1 --> T1 = D/15
Part 2: 20 = D/T2 --> T2 = D/20

Total Time = T1 + T2 = D/15 + D/20 = (15D + 20D/300) = (35D/300) = 7D/60

Speed = 2D/(7D/60) = 120D/7D = 120/7

Long division of 7 into 120 gives us 17 and 1/7 which is closest to 17.1 B

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by [email protected] » Sat Mar 10, 2018 4:37 pm

Hi AAPL,

We're told that Juan cycles at an average speed of 15 km/h, to the top of the hill (where the MIDPOINT of the trip is), then travels at an average speed of 20 km/h downhill. We're asked which of the following is the closest approximation of Juan's average speed, in kilometers per hour, for the round trip. This question can be approached in a number of different ways. Here's how you can solve it with almost no math at all - just a little logic (and using the 'spread' of the Answer choices to your advantage):

Juan cycled 15 km/hour in one direction and then drove back (the same distance) at 20 km/hour. Driving the same distance at a faster speed takes LESS time, so Juan spent MORE time traveling 15 km/hour than he spent cycling 20 km/hour. This is ultimately a 'weighted average' - meaning that the average speed for the entire trip will be CLOSER to 15 than it is to 20. Looking at the Answers, we can immediately eliminate Answer C, D and E. Since the average speed has to be BETWEEN 15 and 20, we can also eliminate Answer A (it's clearly not between those numbers). There's only one answer left...

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Jun 09, 2019 6:00 pm

BTGmoderatorLU wrote:Juan goes out cycling outdoors. He travels at an average speed of 15km/h, to the top of the hill where the midpoint of the trip is. Going down hill, Juan travels at an average speed of 20km/h. Which of the following is the closest approximation of Juan's average speed, in kilometers per hour, for the round trip?

A. 15.0
B. 17.1
C. 17.5
D. 17.9
E. 20.0

We can let the one-way trip be 60 miles; therefore, we have:

Average Speed = (Total Distance)/(Total Time)

Average Speed = (60 + 60)/(60/15 + 60/20)

Average Speed = 120/7

Average Speed â‰ˆ 17.1

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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