1. There are two plausible interpretations of the "lifetime" of a pair of shoes; it could mean (a) the amount of use I get out of the shoes (measured by how much time I can spend walking in them until they wear out), or (b) how long I can own the shoes until they wear out.
If wearing the shoes every third day doubles the lifetime of the shoes in sense (a), then it's a wise thing to do, but if wearing the shoes every third day doubles the lifetime of the shoes in sense (b), then it's not so smart, because the extra longevity of the shoes (by a factor of 2) doesn't compensate for the fact that I can less use out of the shoes during that period (by a factor of 3).
To illustrate this, suppose a pair of shoes would last a year if I wore them every day. That's 365 days of use. Under interpretation (a), the salesman is saying that if I wear the shoes only every third day, I'll get 2x365 = 730 days of use out of the shoes. That's more value for the price of the shoes than 365 days of use. On the other hand, under interpretation (b), the shoes will last 730 days if I wear them every third day, but then I'm getting only 730/3 or about 243 days of use out of the shoes. That's less value for my money than 365 days of use.
A more detailed analysis takes into account the fact that if you're going to wear a pair of shoes only every third day, you're going to have to buy more shoes to wear the other days. To keep things simple, let's assume that a pair of shoes will last one year if we wear the shoes every day. So, if we wear the shoes every day, we'll have to buy another pair in a year, and another pair the year after that. So, for the price of three pairs of shoes, we'll get THREE YEARS of use of the shoes. Now, suppose we decide to buy the three pairs right at the start, and wear each pair every third day, in rotation. How long will it take until we need to buy new shoes? Under the first interpretation of the salesman's claim, each pair of shoes will give me 365x2 = 730 days of use, so all three pairs taken together will give me 365x2x3 = 2190 days of use. So it'll be SIX YEARS until I need to buy shoes again: a good deal! On the other hand, under the second interpretation of the salesman's claim, each pair of shoes will now last two years instead of one year, so it'll be just TWO YEARS until I need to buy shoes again: a bad deal!
2. This problem requires sifting through the story in search of the relevant facts, which (in chronological order) are:
(a) Cecille met William when she was eighteen.
(b) Cecille married William in 1928.
(c) Cecille was 99 years old on October 30, 2005.
To answer the question "How much time elapsed between (b) and (c)?", we need to make use of (a) to determine when Cecille was born.
When might Cecille have been born?: On Oct. 30, 2005, Cecille was 99 years old. So on Oct. 30, 1906, she was 0 years old. That is, she was born between Oct. 31, 1905 and Oct. 30, 1906 (inclusive).
When might Cecille have met William?: She was eighteen at the time. If she was born on Oct. 31, 1905, she would have been eighteen from Oct. 31, 1923 to Oct. 30, 1924. But if she was born on Oct. 30, 1906, she would have been eighteen from Oct. 30, 1924 to Oct. 29, 1925. So all we can say for sure is that she met William sometime between Oct. 31, 1923 and Oct. 29, 1925.
When might Cecille have married William?: Sometime between Jan. 1, 1928 and Dec. 31, 1928.
Now we can answer the original question. If Cecille and William met late (Oct. 29, 1925) and married early (Jan. 1, 1928), slightly more than two years elapsed. If they met early (Oct. 31, 1923) and married late (Dec. 31, 1928), slightly more than five years elapsed. So all we can say for sure is "More than two years and less than six years."
3. Since Day 1 is on July 22, Day 25 (which is 25‒1=24 days later than Day 1) is on "July 22+24" or August 22+24‒31=15, and Day 29 (if she has a 29-day period) is on August 22+28‒31=19. (Check: Day 29 is four days later than Day 25, and August 19 is four days later than August 15.) So Day 1 of her second cycle could be anywhere between August 16 and August 20, inclusive.
If Rhonda's second cycle starts early (on August 16) and ends quickly (after 25 days), her third cycle will start on September 16+25‒31=10. But if her second cycle starts late (on August 20) and ends slowly (after 29 days), her third cycle will start on September 20+29‒31=18. So Day 1 of her third cycle could be anywhere between September 10 and September 18, inclusive.
If Rhonda's third cycle starts early (on September 10) and she ovulates early (on Day 12, which is 11 days after Day 1), she will ovulate on September 10+11=21. If Rhonda's third cycle starts late (on September 18) and she ovulates late (on Day 14, which is 13 days after Day 1), she will ovulate on September 18+13=31, aka October 1. So Rhonda's ovulation could occur anytime between September 21 and October 1.
Should Roger plan to take a business trip in the last half of September?
Probably not, especially in view of the fact that the assumptions
underlying the preceding analysis were empirical and hence possibly
flawed. E.g., the problem said that Rhonda's menstrual cycle "typically
lasts 25 to 29 days", implying that it might on rare occasions last
fewer than 25 days or more than 29. A realistic version of this problem
might weight such factors as the probability of an unusually long or
unusually short period (or an unusually early or unusually late
ovulation, for that matter), how urgently Roger and Rhonda want to
start a family, and what the financial benefits of Roger's trip might
be. Since we are not given enough data to do a risk-benefit analysis,
all we can say is that it is likely that Rhonda's late-September
ovulation will occur between September 21 and October 1, so the
days in late September when Roger must be sure to be at home are
September 21 through September 30. (I will also accept
"September 21 through October 1" as an answer.)
Here is a different approach. We first solve a version of the problem in which all the uncertainty has been removed: Rhonda's cycles always last exactly 27 days (the midpoint of the interval from 25 to 29), and ovulation always occurs on Day 13 of her cycle (the midpoint of the interval from Day 12 to Day 14). Then her second cycle starts on August 22+27‒31=18, and her third cycle starts on September 18+27‒31=14, and she ovulates on September 14+12=26. (Note that we add 12, not 13, because Day 13 is twelve days later than Day 1). Recall that this is the version I solved in class.
Now we add in the effects of uncertainty. Since the duration of Rhonda's first cycle is not 27 days exactly but rather 27±2 days, the start-date of Rhonda's second cycle should now be August 18±2 (instead of August 18). Since the duration of Rhonda's second cycle is also 27±2 days, the combined duration of Rhonda's first two periods is 54±4 days and the start-date of Rhonda's third cycle should now be September 14±4 days. And since the day of ovulation is not necessarily Day 13 but rather Day 13±1, Rhonda's September ovulation should now be on September 26±5. Note that this agrees with our previous answer. The amount of uncertainty in the final answer, namely, ±5, comes from combining ±2 (the uncertainty for the duration of Rhonda's first period), ±2 (the uncertainty for the duration of Rhonda's second period), and ±1 (the uncertainty for the number of days between the start of Rhonda's third period and the day of Rhonda's third ovulation).
Of course, once Rhonda has her next period, she and Roger will have
more precise information, and it may be possible for them to use
that information in planning his business trip. But until that
information becomes available, Roger must plan to be home by
September 21 and must stay home through the end of the month.
Here is yet another approach: Rhonda's first period could be as short as 25 days, her second period could be as short as 25 days, and her day of ovulation in her third cycle could be as early as the 12th day of her cycle. So she could be ovulating as early as 25+25+11 days after July 22; likewise, she could be ovulating as late as 29+29+13 days after July 22. (Note that we add 11 and 13, instead of 12 and 14, since the first day of a period is counted as Day 1, not Day 0; beware of fence-post error!) July 22+25+25+11=83 is September 83‒31‒31=21, and July 22+29+29+13=93 is October 93‒31‒31‒30=1. Roger must plan to be home by September 21 and be prepared to stay home for the rest of the month if need be.
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