PS Arithmetic

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.

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@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.

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.

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.

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