How can I find the time when two people agree to meet if both of their watches have an offset?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

Jenny and Vincent agreed to meet at the library at $6\,p.m$. Both synchronized their watches at midnight ($\textrm{0 hours}$). We know that Vincent's watch is not functioning correctly and gets ahead of the real time $\textrm{50 seconds}$ each hour and Jenny's watch gets delayed from the real time $\textrm{50 seconds}$ each hour. Vincent arrived to the library $\textrm{15 minutes}$ before the agreed time accoring to his watch and Jenny $\textrm{15 minutes}$ late by looking at her watch. Using this information, find how long did Vincent waited Jenny?

The alternatives according to my book are as follows:

$\begin{array}{ll}
1.&\textrm{15 min}\\
2.&\textrm{0 min}\\
3.&\textrm{25 min}\\
4.&\textrm{60 min}\\
\end{array}$

This part I'm stuck at exactly how should I make up an equation which can relate both offsets. Can someone help me here? I'm assuming that by $\textrm{6 p.m}$ the time which would be seen by Vincent will be:

$18\times 50= 900\,s$

which would be $60$ minutes

But Vincent seen in his watch was:

$\textrm{5:45 pm}$

hence until 5 pm would be:

$17\times 50=850\,s$

$56\frac{2}{3}$ minutes

and Vincent's watch would have seen:

$\textrm{5:56:40 pm}$

At this point I could try guessing reducing the number of minutes until adjusting the time which will be seen by Vincent and this would be same for Jenny, but it doesn't seem something which can be effective. Can someone help me here? How exactly can I find what is being requested?
 

skeeter

Math Team
Jul 2011
3,360
1,850
Texas
50 seconds an hour = 5 minutes in 6 hours

Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.

Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.

The only answer choice that makes sense is the 60 minutes Vincent waited.

I worked out the problem using proportions, where T represents the actual time

for Jenny ...
T/18.25 = 72/71

for Vincent ...
T/17.75 = 72/73
 
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Jun 2017
399
6
Lima, Peru
50 seconds an hour = 5 minutes in 6 hours

Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.

Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.

The only answer choice that makes sense is the 60 minutes Vincent waited.

I worked out the problem using proportions, where T represents the actual time

for Jenny ...
T/18.25 = 72/71

for Vincent ...
T/17.75 = 72/73
This is the major problem which I had with this question. It mentioned that until his time was 17:45, not 18:00. How would be the difference between those fractions? Can you help me with that part?