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?.


2 Answers 2


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.


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.

  • $\begingroup$ $\left| \frac{3550}{3600} \frac{71}{4} - \frac{3650}{3600} \frac{73}{4} \right| =1$ $\endgroup$
    – Henry
    Commented Mar 17, 2020 at 8:41
  • $\begingroup$ Yes, but nevertheless I calculated $3550$ wrong. It should be $3510$, so I deleted my edit. $\endgroup$ Commented Mar 17, 2020 at 9:02
  • $\begingroup$ @Henry So who's right? Could it be that one answer is more accurate than the other?. I am confused. $\endgroup$ Commented Mar 17, 2020 at 9:35
  • $\begingroup$ @thinkingeye I don't know what edit are you referring to? Does it mean your answer needs change? $\endgroup$ Commented Mar 17, 2020 at 9:36
  • $\begingroup$ @ChrisSteinbeckBell My problem from the text is: "Vincent's watch [...] gets ahead of the real time 50 seconds each hour". Does it mean that if it is in reality 1 hour, Vincent's watch is 1 hour +50 seconds or does it mean that if Vincent's watch shows one hour, The real time is 1 hour -50 seconds? Both interpretations of the text give different slopes. In an edit, I calculated the second slope, but I did a calculation wrong, so I deleted my edit again. $\endgroup$ Commented Mar 17, 2020 at 9:52

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
  • $\begingroup$ This part is very brief. Can you explain how did you come to that conclusion?. I could understand Vincent arrives at $5:45$ pm but why should I subtract $15$ minutes at that time if it has not elapsed sufficient time to account for that forwardness? I was thinking that when it was $6$ pm he would see $6:15$. Can you please explain this better?. $\endgroup$ Commented Mar 18, 2020 at 0:46
  • 1
    $\begingroup$ @ChrisSteinbeckBell - this is brief because multiple choice questions do not give much time for calculations and the offered answers are a large way apart - I have added an extra explanation $\endgroup$
    – Henry
    Commented Mar 19, 2020 at 4:30
  • $\begingroup$ Do you mean that by following that procedure it can work because it acts as an approximation of some sort?. I had the impression that your calculation for the time Vincent arrives is not exactly $5:35$ pm but sometime before that. Am I getting the right picture here?. $\endgroup$ Commented Mar 20, 2020 at 0:38
  • 1
    $\begingroup$ @ChrisSteinbeckBell My calculation was designed to be a fast and good enough approximation to answer the multiple choice question. Actually, based on thinkingeye's longer calculations, Vincent arrives a little after $5:30$pm (either $12.5$ seconds later or about $24.66$ seconds later depending on how you read the question) and Jenny arrives a little after $6:30$pm (either $12.5$ seconds or about $25.35$ seconds). So my approximate calculations for each arrival are accurate to half a minute or less and my approximate gap is to within a second or less. $\endgroup$
    – Henry
    Commented Mar 20, 2020 at 1:03
  • $\begingroup$ For Vincent's I'm getting $5:30:12.5$ which comes from $\frac{5}{24}$ which I believe its exactly what you calculated or mentioned in your comment, and for Jenny around the same but one hour later, so when you subtract both that little offset doesn't matter it remains being an hour. So thanks!. :) $\endgroup$ Commented Mar 20, 2020 at 1:13

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