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

Question

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?.

Answer 1

Vincent's time expressed as hours $h_v$ in dependency of the real hour time $h_r$:

$h_v = \frac{3650}{3600} h_r$ where $3600$ is the number if seconds per hour.

So real time from Vincent's time could be expressed as:

$h_r = \frac{3600}{3650} h_v$

Vincent is arriving at $5:45$ p.m. of his time. So $h_v = \frac{71}{4}$ (hours).

Analogous for Jenny:

$h_r = \frac{3650}{3600} h_j$

$h_j = \frac{73}{4}$ (hours)

Know we take the difference between both real times in hours:

$\left| \frac{3600}{3650} \frac{71}{4} - \frac{3650}{3600} \frac{73}{4} \right| = \frac{20953}{21024} \approx 1$

It's kind of dissapointing that it is not exactly $1$ hour but $0.996622907 ...$

So for the multiple choice test, it gets rounded to $60$ minutes.

The text is not accurate in terms of the definition at which point in real time the clocks are differ by $50$ seconds each hour.

Edit:

The other text interpretation is: If Vincent's clock shows $1$ hour, the real time is $1$ hour $-50$ seconds.

From this interpretation it follows that:

$h_v = \frac{3600}{3600 - 50} h_r$ or respectively that $h_r = \frac{3600 - 50}{3600} h_v = \left( 1 - \frac{1}{72} \right) h_v = \frac{71}{72} h_v$

and for Julia:

$h_v = \frac{3600}{3600 + 50} h_r$ or respectively that $h_r = \frac{3600 + 50}{3600} h_j = \left( 1 + \frac{1}{72} \right) h_j = \frac{73}{72} h_j$

and the difference between both times is:

$\left| \frac{71}{72} \frac{71}{4} - \frac{73}{72} \frac{73}{4} \right| = 1$ (hour)

So, from the second interpretation, we get exactly $\mathbf{60}$ minutes which indicates that the second text interpretation would be the intended one.

Answer 2

For a multiple-choice test, you do not need a precise answer

• $50 \times 18 = 900$ seconds is $\frac{900}{60}=15$ minutes,
• so Vincent arrives about $15$ minutes before $5.45$ pm because at about $5.30$ pm his watch says $5.45$ pm
• and Jenny arrives about $15$ minutes after $6.15$ pm because at about $6.30$ pm her watch says $6.15$ pm
• making the gap between their arrival about $60$ minutes