# Mixed Real World Math Problems

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The following is a set of word problems that are practical problems, meaning a reasonable person can apply them to the real world one way or another. The solutions are also shown beneath each question, so you can check your answer.

## Example 1[edit | edit source]

At the Simpsons’ home in Springfield, Homer Simpson has a generally stable router that works well far more often than not. However, once in a great while, he will be faced with a router-related difficulty or problem that requires him to contact Springfield Tech Support. On average, he will face one problem like this every 18 months. The occurrence of one problem requiring assistance from Springfield Tech Support does not affect the probability of it occurring again; in other words, each Tech Support problem is independent of previous occurrences, and the probability of it happening in any 12-hour interval is constant and memoryless. Find the expected number of times Homer will need to contact Springfield Tech Support in the next 9 years (108 months).

### Solution[edit | edit source]

Since the probability of a Tech Support problem is constant and thus independent of previous occurrences and previous gaps between occurrences, and no time interval is “favored,” this is a Poisson distribution. This information is not needed to answer this question. The expected number of Tech Support problems in 108 months is simply 108 x (1/18) = 6.

## Example 2[edit | edit source]

A follow-up to the previous question. What is the probability that there will be exactly three Tech Support problems in the next three years (36 months)?

### Solution[edit | edit source]

Again, as mentioned before, this is a Poisson distribution. Recall that the formula for the probability of observing an event X times is λ^{X} • e^{-λ} / X! where λ is the mean or expected number of occurrences of the event. Since Tech Support problems occur at a monthly probability of 1/18 (about 5.56%), we should expect, on average, two such problems in a period of three years or 36 months, so λ = 2. The probability that there will be *three* problems that the Simpson family can’t solve without Tech Support is 2^{3} • e^{-2} / 3! = 8(0.1353)/6 = 0.1804 = 18.04%.

## Example 3[edit | edit source]

John has a Zoom meeting with his teacher Jeff once a week. Under their current weekly schedule, they always meet for two hours on Wednesday nights from 8:00 pm until 10:00 pm. The computer that John usually uses to connect to Zoom is his bedroom computer. Unfortunately, it is prone to system updates, which happen every fifteen days from 7:30 pm until 9:00 pm. John always uses his bedroom computer to attend the weekly Zoom meeting as long as he is able to do so. One Wednesday night, his computer was installing updates, so John was unable to use that computer to attend the meeting and instead had to use a different computer to do so. How many days will pass before this happens again?

### Solution[edit | edit source]

There are seven days in a week, so the Zoom meeting, which takes place on and only on Wednesdays, takes place every seven days. The computer performs a system update once every fifteen days. The length of the compound cycle of the two regular events is the lowest common multiple (LCM) of 7 and 15, which is 105 because LCM(7,15) = 105. Since 7 and 15 are relatively prime, meaning they share no common divisors greater than one, denoted as GCF(7,15) = 1, the two individual cycles are independent. The answer is 105 days.

## Example 4[edit | edit source]

In November 2018, MandJTV published a video to his YouTube channel counting down his top ten least favorite Pokémon. There were 809 Pokémon in existence at that time. Delphox, a Fire/Psychic starter Pokémon introduced in Pokémon X and Y, took the number 5 spot on that list. In other words, Delphox is his fifth least favorite Pokémon, or the fifth Pokémon on his list if he ranks all Pokémon from least favorite to favorite. Had he made a video ranking all 809 Pokémon from favorite to least favorite, Delphox would rank at what number?

- A) 5
- B) 600
- C) 804
- D) 805
- E) 809

### Solution[edit | edit source]

This is a hard question. In the unlikely event it is used on an official SAT math section, it would be number 19 or number 20, near the end of the section. Finding the correct answer to this question must be harder than simply subtracting 5 from 809 to get 804. It is a Joe Bloggs answer. Eliminate C. Answer choices A and E simply repeat numbers from the question, so they are also very unlikely to be correct. Eliminate A and E. If you are stuck, you can guess with coin-flip odds between B and D. Of course, if you know how to solve the problem directly, and you are confident you can do it in reasonable time, do so. To find the correct answer, subtract 5 from 809 to get 804, and then add one more: 804 + 1 = 805.

## Example 5[edit | edit source]

Jaycee Dugard was kidnapped from a bus stop on June 10, 1991. For the next 18 years, 2 months, and 16 days, she was held captive in the backyard owned by her two kidnappers and sealed off from the rest of the world. Her captivity ended on August 26, 2009 when she was rescued by police and went home to reunite with her family. Person XYZ was born during Jaycee’s captivity. Person XYZ was old enough to vote by the time Jaycee’s captivity ended. Which of the following could be Person XYZ’s date of birth?

- A) May 1, 1990
- B) June 5, 1991
- C) July 25, 1991
- D) August 31, 1991
- E) January 1, 1992

### Solution[edit | edit source]

Jaycee Dugard’s captivity lasted from June 10, 1991 until August 26, 2009. The countless gory details of her captivity are not relevant to answering this question. Answer choices A and B are incorrect because they occurred before the captivity began. Answer choices D and E are incorrect because their eighteenth anniversaries (August 31, 2009 and January 1, 2010) happened after she escaped from captivity and regained her freedom. This leaves only C) July 25, 1991 as a viable answer choice. A person born on that date was born 45 days into the captivity, and turned 18 over a month before August 26, 2009, the day the captivity ended.

## Example 6[edit | edit source]

In North America, daylight saving time ends on the first Sunday of November. Which of the following is true?

- A) Daylight saving time cannot end on November 2.
- B) Daylight saving time always ends exactly 21 days before the last Sunday in November.
- C) Daylight saving time cannot end on November 6.
- D) Daylight saving time cannot end on the same date two years in a row.
- E) Daylight saving time sometimes ends later than November 11.

### Solution[edit | edit source]

November always has 30 days. If the 2nd falls on a Sunday, then so do the 9th, 16th, 23rd, and 30th. Therefore, daylight saving time can end on November 2, and if it does it is not 21 days before the last Sunday of the month, but 28 days. Eliminate A and B. If November begins on a Tuesday, its Sundays are the 6th, 13th, 20th, and 27th, so the end of daylight saving time *can* fall on November 6, so answer choice C is incorrect. Eliminate C. With choices A, B, and C out of the way, two answer choices remain. Answer choice E is incorrect because any given period of seven days contains a Sunday, so the latest date that daylight saving time can end is November 7. Even when November 11 falls on a Sunday, it is not the first Sunday of the month, because the 4th falls on a Sunday as well. Eliminate E. This leaves answer choice D, which is correct because a date can never fall on the same day of the week twice in a row because 365 days is not a multiple of 7 days, and 366 isn’t either. Therefore, daylight saving time never ends on the same date two years in a row. The answer is D.

## Example 7[edit | edit source]

On April 6, 2003, residents of the United States “lost” one hour of sleep as daylight saving time began in all parts of the country except Arizona and Hawaii. When local time was about to reach 2:00 am, local time jumped forward one hour, so that 01:59:59 was immediately followed by 03:00:00. If a random point was chosen uniformly from April 6, 2003, what is the probability that it is before 12:00:00 noon?

### Solution[edit | edit source]

Be very careful. Before you bask in the warm glow of success, the answer is not one-half. On 363 days of the year, noon is indeed the midpoint of the day. The day that daylight saving time begins is one of two annual exceptions, the other being the day that daylight saving time ends. Because local time skipped an hour, the day lasted only 23 hours instead of the usual 24 hours. Likewise, since the period between midnight and noon includes the moment the clocks changed by one hour, it too was shortened one hour from 12 hours to only 11 hours. Since any given point in the day is equally likely to be chosen, the answer is 11/23, or approximately 47.83%.