Marla starts running around a circular track at the same time Nick sta

Maria's rate - 32 laps per hour --> 32/60 laps/min
Nick's rate - 12 laps per hour --> 12/60 laps/min

lets set equations:

32/60*t=4 (since Maria had to run 4 laps before Nick would start)
12/60*t=0 (Hick has just started and hasn't run any lap yet)
-----------------------------------
(32/60-12/60)*t=4-0 (since Nick was chasing Maria)
t=12 min needed Maria to run 4 laps

Since Marla completes 20 more laps in 1 hour, then to complete 1/5 th of 20 laps (4 laps) she will need 1/5 th of an hour, which is 12 minutes.

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Marla completes 32laps/60min, Nick completes 12laps/60mins.
After x mins Marla would have completed 4 laps more than Nick had completed.
x((32-12)/60) = 4,
x*20/60 = 4,
x = 12 mins.

We can use the following formula:

time = change in distance/change in rate

time = 4/20 = 1/5 hour = 12 minutes

Alternate Solution:

Note that Marla’s rate is 32/60 laps per minute and Nick’s rate is 12/60 laps per minute. Let’s say after t minutes, Marla completes 4 more laps than Nick. Then, in t minutes Marla completes 32t/60 laps and Nick completes 12t/60 laps. Since the number of laps completed by Marla is 4 more than the number completed by Nick, we have:

32t/60 = 12t/60 + 4

32t/60 - 12t/60 = 4

20t/60 = 4

t/3 = 4

t = 12

Answer: C

Let the one lap distance = d
Speed of Marla \(= \frac{32}{60}\)
Speed of Nick \(= \frac{12}{60}\)
Time t would be same in both the cases

Setting up the equation

\(d = \frac{32}{60}* t\) ............ (1)

\(d-4 = \frac{12}{60} * t\)........(2)

Equating (1) & (2)

\(\frac{12}{60} * t + 4 = \frac{32}{60} * t\)

\(\frac{1}{3} * t = 4\)

t = 12

Answer = C

I also got C (12).

t= time
32 laps per hour * t = 12 laps per hour * t + 4 laps

32t = 12t + 4
20t = 4
t= 1/5 hr

1/5(60min)= 12 mins

I believe the recommended strategy is to change to minutes right away but in this situation I found it easier to do it at the end because of how the rates simplified per minute.

Given, the distance traveled by Marla in 60 minutes = 32
Distance traveled by Marla in 't' mins = \(\frac{32t}{60}\)

Also given, the distance traveled by Nick in 60 minutes = 12
Distance traveled by Nick in 't' mins = \(\frac{12t}{60}\)

Prompt says there will be a difference of 4 laps at some point in time between the above two distances.

Since Maria travels faster, the distance covered by Maria will be more than the distance covered by Nick. Therefore, subtract the distance traveled by Nick from the distance traveled by Maria in order to avoid a negative resultant.

\(\frac{32t}{60}\) - \(\frac{12t}{60}\) = 4

Solving,
\(\frac{20t}{60}\) = 4
\(\frac{t}{3}\) = 4

t= 12 mins.

Best way is to use plug in!

Always plug in C first, since this is the median value!

In 12 minutes she will have completed 32/5 rounds ( /5 because 12min is 1/5 hours) and he will have completed 12/5 rounds.

32/5 = 6 2/5
12/5 = 2 2/5

4 is the difference and hence C is th answer.



Sometimes you get lucky and plug in the right answer first! Will help you save heaps of time!

marla runs 32/60=8/15 laps per minute
nick walks 12/60=3/15 laps per minute
marla gains 8/15-3/15=1/3 laps per minute
4 laps/1/3 lap per minute=12 minutes

Marta completes 32 laps/hour or 32/60 or 8/15
Nick completes 12 laps/hour or 12/60 or 3/15

Relative speed = 8/15-3/15 = 5/15
Distance/relative speed = 4/(5/15)=4*(15/5) = 4*3=12

OR, solve by setting an equation

32t=12t+4
20t=4
t=1/5=12/60

Answer C)

The answer is C
use relative velocity to solve it

the relative velocity would be (32-12) in 1 hour.
hence, 20 laps in 1 hour
therefore time to complete four laps is 1/5hr
12 mins

Let the time taken be t
32t - 12t = 4

20t = 4
t = 4/20
1/5 * 60 = 12 minutes

Hi Dave13,

If Marla completed 4 more laps than Nick completed, then Marla's lap count is 4 greater than Nick's lap count.
In order to create an equation, we need to give Nick 4 extra laps.
If we give Marla 4 more laps, their lap counts won't be equal (i.e., we can't write an equation)

Does that help?

Bunuel - how did you get this?

Marla to complete 4 (20/5=4) more laps will need 1/5 hours, which is 12 minutes.

BrentGMATPrepNow hi Brent, how is life may i know why are you adding 4 laps to Nick , and not to Marla, mustnt Marla be 4 laps ahead of Nick? like 32t+4 i know i am wrong in this case, but this is how i think about this problem and how can you equate distances if they are different one is 4 laps ahead than another
thanks

1 hour: 32:12

30 min: 16:6
15 min: 8:3

Answer must be slightly less than 15 min.


Or algebraically:

32x/60 - 12x/60 = 4

x = 12