PS Arithmetic

Expert replies

sud21: Master | Next Rank: 500 Posts; Posts: 344; Joined: Sat Nov 12, 2011 3:21 amFollowed by: 2 members; Thanked: 1 times

by sud21 » Sat Sep 19, 2015 5:16 am

Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

x minutes

x/2 minutes

2x/3 minutes

3x/2 minutes

2x minutes

Like this post

Source: Beat The GMAT — Problem Solving | Sat Sep 19, 2015 5:16 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Sep 19, 2015 7:08 am

sud21 wrote:Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

x minutes

x/2 minutes

2x/3 minutes

3x/2 minutes

2x minutes

Let x = 60 minutes.
In 60 minutes -- in other words, in 1 hour -- Jessica travels 30 kilometers, while Nicole travels 20 kilometers.
Implication:
When Jessica stops, Nicole must travel 10 kilometers to catch up to Jessica.
At a rate of 20 kilometers per hour, the time for Nicole to travel 10 kilometers = d/r = 10/20 = 1/2 hour = 30 minutes. This is our target.

Now plug x=60 into the answer choices to see which yields our target of 30 minutes.
Only B works:
x/2 = 60/2 = 30 minutes.

The correct answer is B.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Like this post

Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Thu Sep 24, 2015 11:16 am

I'd pick numbers here.

Suppose x = 6. Then Jessica bikes 30*6 = 180 km and Nicole bikes 20*6 = 120 km.

So Nicole still has 60 km left to bike. Her rate is 20km per hour, so this will take her THREE hours.

Hence Jessica has to wait for 3 hours, which is just (x/2), or (6/2).

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Sep 24, 2015 11:41 am

Matt@VeritasPrep wrote:I'd pick numbers here.

Suppose x = 6. Then Jessica bikes 30*6 = 180 km and Nicole bikes 20*6 = 120 km.

So Nicole still has 60 km left to bike. Her rate is 20km per hour, so this will take her THREE hours.

Hence Jessica has to wait for 3 hours, which is just (x/2), or (6/2).

Matt,

This solution presumes that Jessica travels for 6 HOURS after passing Nicole.
But the prompt states that Jessica travels for x MINUTES after passing Nicole.
As a result, the values in red do not accurately represent the distances traveled.

Like this post

Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Thu Sep 24, 2015 11:50 pm

GMATGuruNY wrote:
Matt@VeritasPrep wrote:I'd pick numbers here.

Suppose x = 6. Then Jessica bikes 30*6 = 180 km and Nicole bikes 20*6 = 120 km.

So Nicole still has 60 km left to bike. Her rate is 20km per hour, so this will take her THREE hours.

Hence Jessica has to wait for 3 hours, which is just (x/2), or (6/2).
Matt,

This solution presumes that Jessica travels for 6 HOURS after passing Nicole.
But the prompt states that Jessica travels for x MINUTES after passing Nicole.
As a result, the values in red do not accurately represent the distances traveled.

Sure, but that's trivial - the ratio of speeds is constant, irrespective of the unit of time; we only need the time traveled so far and the time required for Nicole to catch Jessica to be the same unit (here minutes). Given any amount of time t, Jessica travels (3/2) Nicole's distance. Suppose Nicole's distance traveled at the time Jessica stops = d. Then Jessica has traveled (3/2)d, so Nicole will need to make up (1/2)d. She'll do this in half the time she has already spent traveling, i.e. t/2. and the answer follows.

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Sep 25, 2015 2:47 pm

Matt@VeritasPrep wrote:
GMATGuruNY wrote:
Matt@VeritasPrep wrote:I'd pick numbers here.

Suppose x = 6. Then Jessica bikes 30*6 = 180 km and Nicole bikes 20*6 = 120 km.

So Nicole still has 60 km left to bike. Her rate is 20km per hour, so this will take her THREE hours.

Hence Jessica has to wait for 3 hours, which is just (x/2), or (6/2).
Matt,

This solution presumes that Jessica travels for 6 HOURS after passing Nicole.
But the prompt states that Jessica travels for x MINUTES after passing Nicole.
As a result, the values in red do not accurately represent the distances traveled.
Sure, but that's trivial - the ratio of speeds is constant, irrespective of the unit of time; we only need the time traveled so far and the time required for Nicole to catch Jessica to be the same unit (here minutes). Given any amount of time t, Jessica travels (3/2) Nicole's distance. Suppose Nicole's distance traveled at the time Jessica stops = d. Then Jessica has traveled (3/2)d, so Nicole will need to make up (1/2)d. She'll do this in half the time she has already spent traveling, i.e. t/2. and the answer follows.

My concern is that your initial post does not clearly indicate that x=6 represents hours and that the switch from minutes to hours is mathematically justified.
As a result, a reader might incorrectly construe that a time in terms of minutes can be multiplied by a rate in terms of hours.

Like this post

Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

by Jeff@TargetTestPrep » Mon Dec 11, 2017 7:11 am

sud21 wrote:Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

x minutes

x/2 minutes

2x/3 minutes

3x/2 minutes

2x minutes

We can classify this problem as a "catch-up" rate problem, for which we use the fact that distance travelled by Nicole = distance travelled by Jessica.

We are given that Nicole cycles at a constant rate of 20 kilometers per hour and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. We are also given that Jessica cycles at her constant rate for x minutes after passing Nicole, and then stops to wait for her. We can let t be the waiting time in minutes; that is, Nicole has to cycle x + t minutes to catch up to Jessica, who has to cycle x minutes only.

Since distance = rate x time, we can calculate each person's distance in terms of t and x. However, the rate is in km per hour, but the time is in minutes. Thus we need to convert the time from minutes to hours. Recall that there are 60 minutes in an hour. Thus, x minutes = x/60 hours and x + t minutes = (x + t)/60 hours.

Nicole's distance = 20[(x + t)/60] = (x + t)/3

Jessica's distance = 30(x/60) = x/2

We can equate the two distances and determine t in terms of x.

(x + t)/3 = x/2

2(x + t) = 3x

2x + 2t = 3x

2t = x

t = x/2

Answer: B

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review