Series of Question 3 - From Guide #5

[sendalot]

So, I just found out I can post questions here instead of emailing the instructor, I have looked through previous posts, but wasn't still quite sure about these.

I can't understand the solutions for these. It seems they take short-cuts??

[Series of Question 3 - From Guide #5]

P. 64 - 15 Errorenous final answer
P. 88/93 - 6 Error
P. 92 - 12 Error
P. 95 - 8 Don't know
P. 95 - 11 Don't know
P. 96 - 16 Don't know
P. 106 - 5
P. 107 - 6
P. 119 - 3
P. 125 - 6
P. 134 - 2
P. 163 - 14
P. 170 - 6 M||Calc Function usage??
P. 171 - 10 Why, 0.4, 0.4, 0.2 ??
P. 173 - 20 Why 2+2
P. 182 - 5
P. 183 - 8
P. 185 - 16
P. 185 - 17
P. 185 - 19
P. 186 - 20 hat has about no repetition??

I have posted more for different books.
Thanks.

[michael.k.bilow]

Lots and lots of tough questions here. I'll take about half of them here.

Guide 5, Page 64 #15
The worked solution on page 69 is correct, however the answer given at the top of the solution (A) is incorrect. The correct answer is B.

Guide 5, Page 88 #6
What is the median of the set {1,2,x,8} if 2 < x < 8?

The median of a set with an odd number of elements is the middle number in the set, but a set with an even number of elements has no middle element. To see this, take the numbers 1 through 10 and start crossing off the largest and smallest elements. You'll end up with 5 and 6, and you can't cross both of them off, or you'll be left with nothing. The median of a set with an even number of elements is defined to be the average of the two middle elements, so the median of the numbers 1 through 10 is (5+6)/2 = 5.5

For the set that we're given in this problem, the smallest value is clearly 1, and the middle two values are 2 and x, because both of those values are larger than 1 and smaller than 8. Then to find the median, we need to take the average of these two numbers: (x+2)/2 = x/2 + 1.

Guide 5, Page 92 #12
Variable X is nearly normally distributed, with a mean of 6 and a standard deviation of 2. Approximately what percent of the observation in X will be smaller than 4?

This problem depends on memorizing this table of values, that apply for a normally-distributed population.

Between the mean and 1st SD (standard deviation): 34% of the observation
Between 1 SD and 2SDs: 13.5% of the observation
Beyond 2SDs: 2.5% of the observation

The variable X has mean 6 and standard deviation of 2, so 4 is 1 SD below the mean. Thus the total observation that's going to be less than 4 will be the amount of observation between 1SD and 2SDs, and the amount more than 2SDs away from the mean. This amounts to 13.5% + 2.5% = 16%.

Guide 5, Page 95 #8
Matt gets a $1,000 bonus on a big sale. This commission alone raises his average commission by $150. If Matt's new average commission is $400, how many sales has Matt made?

This is a question about average commission, so let's find out how much money Matt has made in commission, and how many sales Matt has made.

Let's start by naming some variables:

# of sales Matt made before the big sale: N
# of sales Matt made after the big sale: N+1

Total amount of commission Matt made before the big sale: C
Total amount of commission Matt made after the big sale: C + 1000

Matt's average commission before the big sale: $400-$150 = $250
Matt's average commission after the big sale: $400

Now, let's set up some equations for his average commission before and after the big sale:

We know that Average = (Total Amount)/(# of Items), so before the big sale, we have

250 = C/N

And after the big sale, we have:

400 = (C+1000)/(N+1)

Let's multiply both these equations by the denominators of the respective fractions:

250N = C
400(N+1) = C + 1000

Now, let's multiply things out.

250N = C
400N + 400 = C + 1000

From which we get:
250N = C
400N = C + 600

Now, let's subtract the top equation from the bottom, which gives us:

150N = 600

N = 600/150 = 4.

Now, we have to be extremely careful, because the problem is asking for the number of sales that Matt made in total, and we've defined N to be the number of sales Matt had made before his big sale. The correct answer is going to be N+1, 5.

Guide 5, Page 95 #11
If x>0 and the range of 1,2,x,5,x^2 equals 7, what is the approximate average (mean) of the set?

This is a tricky problem, because it's hard to know what the upper and lower ends of the range should be. The elements look like they're in order, but we can't assume that, so let's look at a few cases. We know x > 0, but we don't know if x is bigger or smaller than 1. Let's test it out:

If 0 < x < 1, then 0 < x^2 < x, because multiplying by a number between zero and one shrinks numbers closer to zero. Then the smallest value in the data set would be x^2, which we know is bigger than 0. So the largest possible range we could have in this case would be 5 - 0 = 5. Uh oh, the problem told us that the range of the set was 7--so this can't be.

So we know x is bigger than 1, which means that 1 is the smallest element in the dataset. The range of the data set is 7, and the range of a data set is just the largest element in the data set minus the smallest value, so the largest value in the set must be 8, because 8-1 = 7.

Because x > 1, x^2 > x, x^2 must be the largest element of the set. Then x^2 = 8, and x = 8^(1/2). So now our whole set is:{1,2,8^(1/2),5,8}. If you add these numbers in the calculator and divide by 5 to find the average, you'll get about 3.76.

Guide 5, Page 96 #16
A is the set of the first five positive odd integers. B is the set of the first five positive even integers.

Quantity A: The standard deviation of A
Quantity B: The standard deviation of B

I'll give two ways to do this problem.

1) One of the facts about the standard deviation is that if you have a dataset S and you add any number x to each element in S to generate a new dataset T, S and T will have the exact same standard deviation. You can think of this as "shifting" the data by a number x. Here's an example of two data sets with the same standard deviation:

S = {0,3,6,9}, T = S + 2 = {2,5,8,11}

T is the same as S, except that every element has been "shifted" over by 2. T and S will have the same standard deviation.

Now let's look at the data sets we have:

S = The first 5 odd positive integers = {1,3,5,7,9}
T = The first 5 even positive integers = {2,4,6,8,10}

Just by looking at these sets, we can see that the elements of T are just the elements of S shifted over by 1! By the rule above, S and T have the same standard deviation, so the correct answer is C.

2) The GRE will never ask you to actually compute the standard deviation of a dataset, so we can use a little trick. For the purposes of the GRE, but not in real life, it is generally safe to assume that the standard deviation of a dataset is equal to its range. The range of the first 5 positive odd integers is 9-1 = 8, and the range of the first 5 postive even integers is 10-2 = 8. These two datasets have the same range, therefore we would guess that they have the same standard deviation. The answer is C.

Guide 5, Page 106 #5
Peggy will choose 5 of her 8 friends to join her for intramural volleyball. In how many ways can she do so?

This is a relatively simple combination question. We're looking at a team, so the order of the 5 people that she's chosen doesn't matter, and the answer will just be the number of combinations you can have when taking 5 elements from a set of 8, 8!/(3!5!) = 56.

I really like the anagram grid used to explain the problem on page 107, and I'll contribute this alternate.

When dealing with combinations and permutations, there are three groups we are interested in:
1) The entire population
2) The population that is chosen
3) The population that is not chosen.

We always care about the order of the entire population, and we never care about the order of the people that are not chosen. The difference between combinations and permutations is that for combinations, we do not care about the order of the population that are chosen, and for permutations we do care about the order. The formula for the number of combinations and permutations will be:

(The number of possible orderings of the whole population)/(The product of the number of possible orderings of the group that isn't chosen, and if we care about it, the number that are chosen).

So, for our problem, we have
1) The entire population = 8
2) The population that is chosen = 5
3) The population that is not chosen = 3

We are taking a combination so we care about both the order of the population that is chosen and the order of the population that is not chosen. The number of possible orderings of the whole population is 8!, the number of possible orderings of the population that is chosen is 5!, and the number of possible orderings of the population that isn't chosen is 3!. The result is 8!/(5!3!) = 56.

Guide 6, Page 107 #6
Three men (out of 7) and three women (out of 6) will be chosen to serve on a committee. In how many ways can this committee be formed?.

The total number of ways to choose the committee is equal to the number of ways to choose the men on the committee multiplied by the number of women. To see this, realize that the men and women are chosen independently, so they're like separate courses of a meal. If we have 5 options for an appetizer and 8 options for an entree, the total number of possible meals is 5*8. In this case, the choice of 3 men is our appetizer, and the choice of 3 women is our main course.

The number of ways to choose 3 men out of 7 is the number of combinations, 7!/(3!4!) = 35. The number of ways to choose 3 women out of 6 is 6!/(3!3!) = 20. Thus the total number of ways to choose the committee is 35*20 = 700.

[michael.k.bilow]

And now, the thrilling conclusion:

Guide 5, Page 125 #6
John invites 12 friends to a dinner party, half of which are men. Exactly one man and one woman are bringing desserts. If one person from this group is selected at random, what is the probability it is a woman, OR a man who is not bringing dessert.

This is complicated probability, because we're looking at both an "and" statement and an "or" statement. Also, the "and" is hidden in this little statement: "a man who is not bringing dessert."

A man who is not bringing dessert means that the person must be a man AND must be bringing dessert. Reading over the problem, half of the 12 people at the party are men, and one of them is bringing dessert. So there are 6 - 1 = 5 men who are not bringing dessert.

In general, when we see an "or" statement, we are going to add probabilities. However, the "or" statement in probability is also tricky, because we have to be careful of double-counting overlaps. For example, if we were looking at the set of numbers between 1 and 10 that are even OR are divisible by 3, and we just added up the number of even numbers (5) and the number of numbers divisible by 3 (3), we'd get 8/10, which is incorrect. The number 6 is an element of both sets, and just by adding the probabilities together, we'd count it twice. The correct answer is the number of even numbers plus the number of numbers divisible by 3 minus the size of the overlap set (1), which gives 7/10.

Using this strategy, let's break down the sets we have. The problem asks us for the number of women OR men who are not bringing dessert. Since half the 12 people at the party are women, there are 6 of them, and we computed above that the number of men who are not bringing dessert is 5. The question that remains is: What is the size of the overlap?

For the purposes of the GRE, no one can be both a man and a woman, so there can be no possible overlap between the two sets. Thus the number of people who are women or men that are not bringing desserts is 6 + 5 - 0 = 11. The probability of selecting one of these people is 11/12.


Guide 6, Page 134 #2
A salad dressing requires oil, vinegar, and water in the ratio 2:1:3. If Oliver has 1 cup of oil, 1/3 cup of vinegar, and 2 cups of water, what is the maximum number of cups of dressing that he can mix?

This is a problem that a lot of us had to do in high school chemistry class, that the GRE has lifted and rephrased in a way that's more than a little confusing.

What's going to help us do this problem is to set it up ratios between each of the ingredients and the total amount of dressing that could be made. Looking at the ratio, we can see that 2 cups of oil, 1 cup of vinegar, and 3 cups of water will make 2+1+3=6 cups of salad dressing. So we the ratio between oil and dressing is 2:6, the ratio between vinegar and dressing is 1:6, and the ratio between water and dressing is 3:6.

Using these ratios, we can compute the total amount of dressing we can make from each of our three ingredients.

1 cup oil/x cups dressing = 2 cups oil/6 cups dressing
1/3 cup vinegar/y cups dressing = 1 cup vinegar/6 cups dressing
2 cups water/z cups dressing = 3 cups water/6 cups dressing

x, y, and z represent the greatest amount of dressing we could make by using up the entire quantity of oil, vinegar, and water, respectively. The greatest total amount of dressing we could possibly make given these ingredients will be the smallest number among x, y, and z, because that will consume the limiting resource

Solving the equations above, x = 3 cups, y = 2 cups, and z = 4 cups. y is the smallest of our three possible yields, so the answer is 2 cups. To check this, see that in 2 cups of dressing, there is 1/3 cup of vinegar, 2/3 cup oil, and 1 cup water. We have exactly 1/3 cup of vinegar, so making exactly this much dressing will leave us with no more vinegar, which means it will be impossible for us to make any more dressing.

Guide 6, Page 164 #14
A biologist analyzes the number of paramecia visible under a microscope for a collection of protozoa samples. The average number of parmecia is 8.1 per sample, and the standard deviation is 2.4. The distribution of paramecia visible across the samples is approximately normal.
Quantity A: The number of paramecia visible at the 75th percentile in the distribution of the samples.
Quantity B: 10.5

We need 2 things to solve this problem: 1) The table of values for the normal distribution [see the previous reply] and 2) The meaning of the value 10.5.

10.5 seems like a random value, but it's actually very well-chosen. 10.5 = 8.4 + 2.1, so it is exactly one standard deviation above the mean. To find out what percentile this is, let's look at our table of values:

There will be 2.5% of the data more than 2 SDs below the mean.
There will be 13.5% of the data between 1 and 2SDs below the mean.
There will be 34% of the data between 1SD below the mean and the mean itself.
There will be 34% of the data between the mean and 1SD above the mean.

Adding up all these numbers, we get that 1 SD above the mean is roughly equivalent to the 84th percentile. So we would expect to see 10.5 paramecia on the slide at the 84th percentile, and therefore we would expect to see between 8.4 and 10.5 paramecia at the 75th percentile. The correct answer is B.


Guide 6, Page 170 #6
The frequency distribution of student scores is shown below: How many of the scores are above the class average?

64: 1
70: 2
72: 1
79: 4
83: 4
85: 2
90: 3
94: 2
95: 1

The first thing we absolutely must do on this problem is compute the average. The solution advocates the use of the M+/MR functions that are available on the calculator, but I would strongly advise against using them unless you've already practiced using them in school.

I don't mind spending a minute or more on this problem just being very careful with my computation, so I don't make an easy math error.

If you add up the total number of scores, you'll see there are 20 students in the class. Adding up the scores is a bit more complicated, and more prone to careless errors, so the first thing I would do is find out how many points were derived from each different score, by multiplying each score by the number of students who attained that score. That gives me this list:

64 x 1 = 64
70 x 2 = 140
72 x 1 = 72
79 x 4 = 316
83 x 4 = 332
85 x 2 = 170
90 x 3 = 270
94 x 2 = 188
95 x 1 = 95

I much prefer to write out answers on my scratch paper like this, and then add all the numbers up on the calculator, instead of messing around with calculator functions that I was never trained to use. Both techniques will yield 1,647 as the sum of the scores, I use my way because I know that I can do this process robotically without making mistakes. Writing out intermediate answers like I've done here also allows me to check my work as I'm going through the problem. The average score is then 82.35, and we have 4 + 2 + 3 + 2 + 1 = 12 scores that are above the average.


Guide 6, Page 171 #10

Asset Amount Invested Expected Return
Stock X $40 10%
Stock Y $40 8%
Stock Z $20 15%

Quantity A: The percent return an investor would expect for investing in the above stocks
Quantity B: 10.5%

The reason we use .4, .4, and .2, is because that corresponds to the fraction of the total investment in stock X ($40/$100), stock Y ($40/$100), and stock Z ($20/$100), respectively. These are the weights of our weighted average. 40% of our total return wll be the return of stock X, 40% will be the return of stock Y, and 20% will be the return of stock Z. That gives us .4 * 10% + .4 * 8% + .2 * 15% = 4% + 3.2% + 3% = 10.2%. This is less than 10.5%, so the answer is B.

Another way to do this problem is to compute the expected return for each stock, and then divide by the total investment. The expected return from stock X is $40 * 10% = $4, the expected return from stock Y is $40 * 8% = $3.2, and the expected return from stock Z is $20 * 15% = $3. Thus the total dollar value we expect to be returned is $10.2. Our investment was $40 + $40 + $20 = $100, so our expected return is $10.2/$100 = 10.2%.

Guide 5, Page 173 # 20
A pomegranate grower packages pomegranates in 10-pound and 20-pound boxes. If the grower fills more than twice as many 20-pound boxes as 10-pound boxes, which of the following could be the percentage of pomegranates, by weight, that are packaged in 10-pound boxes?

A) 15%
B) 20%
C) 25%
D) 40%
E) 60%

Before we do any math with this problem, let's try to reason are way to some good numbers to pick.

The problem tells us that there are more than 2 times as many 20-pound boxes as 10-pound boxes, so let's find out what happens when there are exactly 2 times as many 20-pound boxes as 10-pound boxes.

Let's say there are 2x 20 pound boxes and 1*x 10-pound box. Then we'd have 2x*20 + 1x*10 = 50x pounds of pomegranates in total, and the 10 pound boxes would make up 10x/50x * 100% = 20% of the total amount.

What does 20% represent? Well, it will be the maximum percent of the weight that we can have from 10-pound boxes. That eliminates all the answers but A and B. Can we have exactly 20%? No! The problem tells us that there are going to be more than two times as many 20-pound boxes as 10-pound boxes, so we're always going to have strictly less than 20% of our total weight in 10-pound boxes. We can eliminate B, leaving the correct answer, A.

There's also a dirty trick that solves this problem without doing any math at all; if you can make a shipment that has 25% of its weight coming from 10-pound boxes, then you can make any percentage that is less than 25%, by adding more 20-pound boxes. Therefore, the only possible way that there could be one answer for this problem is the smallest possible percentage.


Guide 6, Page 182 #5
A set of integers consists of 2,3,5,7,11,12,and x. If x increases by 1, the median of the set stays unchanged. However, if x decreases by 1, the median of the set also decreases by 1. What is the value of x?


This is a very hard problem, and it requires a lot of reasoning to figure out the value for x.

Let's ignore x for a second. We know that the median only cares about the middle value of the set, so let's see where the middle is without x. We have 6 numbers in our set, plus x, so the middle two values that we know are 5 and 7. Let's split the set up at 5 and 7, and check the different cases.

1) x < 5
2) 5 < x < 7
3) x > 7
4) x = 5
5) x = 7

1) x < 5:

If x < 5, then the median value will be 5, since 2, 3, and x will be smaller than 5 and 7, 11, and 12 will be larger. The problem says that if x decreases by 1, the median decreases by 1, but this will not happen in our case, because if x < 5 means that x -1 < 5. We can reject this range of values.

2) 5 < x < 7:

If 5 < x < 7, then the median value of the set is x, because 2,3,5 < x and x < 7,11,12. We know that if x increases by 1, the value of the median is unchanged, but there's no way that can be true because x < 7. If x = 6.5, for example, the new value of x (7.5) will no longer be the median, but the median will still increase to 7. We can reject this range of values.

3) x > 7
If x > 7, then the median value of the set is 7, because 2,3,5 < 7 and 7 < 11, 12, x. If x decreases by 1, the median may decrease--for example, if x = 7.2, the new median will be 6.2, but then the median has only decreased by 7 - (6.2) = .8. If we pick a value for x > 8, then x-1 > 7, and the median will remain unchanged at 7 when we subtract 1.

4) If x = 5, then the median is 5, but if x increases by 1, then the new median of the set will be 6. This contradicts the problem, which says that the median is unchanged when we add 1 to x.

5) If you're extra perceptive, you might have noticed that when we picked a number close to 7 in part (3), we got nearly all the behavior we wanted. Let's try 7 itself. If x = 7, the set is {2,3,5,7,7,11,12}, which has median value 7. If x decreases by 1, the set becomes {2,3,5,6,7,11,12}, and the median value will be 6 = 7 - 1, as the problem dictates. If x increases by 1, the set becomes {2,3,5,7,8,11,12} which has median 7, which is unchanged as dictated by the problem. x = 7 is the correct answer.


Guide 5, Page 183 #8
John is to select a committee of 5 individuals from among a group of 5 candidates. The committee will have a president, a Vice President, and two treasurers. How many different committees can John select from the 5 candidates?

This is a really, really challenging combinatorics problem, and the solution on page 189 is worth reading a few times.

Let me give you two alternate ways of doing this problem.

For the first way, let's select the people in order.
If we start by selecting the president, we have will have 5 choices for a person.
Once we've chosen a president, there will be 4 choices of vice president.
Now we have 3 people left and need to choose 2 treasurers. This is just the number of 2-person combinations we can take out of 3 people, which will be equal to 3!/(2!1!) = 3.

Multiplying our number of options together, we have 5 * 4 * 3 = 60 total different committees.

Another way to do this problem is to set up an anagram table. Out of 5 people, we want 1 President (P), 1 Vice President (V), 2 Treasurers (T), and 1 person who has no job (N). Asking how many committees we can make is the same question as asking how many distinct words we can make out of the letters PVTTN. Using the formula for the number of distinct words given on pages 105-106, we get 5!/(2!1!1!1!) = 60.

Okay, that's it for right now. I'll be back with more solutions later.

Keep studying!

[michael.k.bilow]

And now, the thrilling conclusion!

Guide 6, Page 185 # 16
The 500 students in a class took an examination. Scores were given on an integer scale of 0-100.Joe's score was 2 standard deviations above the mean score on the examination, and Charlie's score was exactly the 5th percentile. The distribution of exam scores was approximately normal. Which of the following statements must be true?

Indicate all such statements:

A: Joe scored closer to the mean than Charlie.
B: More than 400 students achieved scores less than or equal to Joe's score and greater than or equal to Charlie's score.
C: Fewer than 450 students achieved scores less than or equal to Joe's score and greater than or equal to Charlie's score.
D: At least one other person received the same score as Charlie.

A lot of ground to cover here:
See the solutions above for information on the percentiles of the normal distribution. You can use the table of values given to compute that 2 SDs above the mean is 98th percentile.

Thus, Joe (98th percentile) is farther from the mean than Charlie (5th percentile), so A is false.

For B and C, the total percentage of the data between Joe and Charlie's scores is 98 - 5 = 93%. 93% of 500 students is 465 scores. Therefore, B is true and C is false.

D is true, but I had no idea how to solve it at first glance. I was surprised to learn that there is a difference between a percentage and a percentile, and if you want to find out what that is, I encourage you to check out the solution on page 192.



Guide 5, Page 185 #17
The probability of Tom rolling a strike while bowling is 40% on any given frame. If Tom rolls 4 frames in a row, which of the following statements is true?


A: The probability of Tom rolling a strike on all 4 frames is greater than 3%
B: The probability of Tom rolling no strikes is less than 10%.
C: Tom is equally likely to roll exactly 1 strike as to roll exactly 2 strikes in those 4 frames.
D: Tom will roll 2 or more strikes less than half the time.


There's a lot of work to be done here, so let's start by computing the probability of each individual outcome. Remember that to compute the probability of a particular outcome, we need to compute the probability of an instance of the outcome, and multiply by the number of ways that the outcome of interest can occur.

For example, with the probabilities as they are, the probability of rolling 2 strikes and 2 not-strikes in one particular way is (.4)^2*(.6)^2. To compute the number of ways that we can roll 2 strikes out of 4 throws, we need to find the number of 2-element combinations of 4 items, which is 4!/(2!2!) = 6. Thus the total probability of throwing 2 strikes is 6*(.4)^2*(.6)^2 = .3456.

Pr(0 strikes) = .6^4 = .1296
Pr(1 strike) = 4*(.4)^1*(.6)^3 = .3456
Pr(2 strikes) = 6*(.4)^2 *(.6)^2 = .3456
Pr(3 strikes) = 4*(.4)^3*(.6)^1 = .1536
Pr(4 strikes) = (.4)^4 = .0256

Now, let's look at the answer choices.

A is false. The probability of rolling 4 strikes is 2.56%, which is less than 3%.
B is true. The probability of rolling no strikes is 12.96%, which is greater than 10%.
C is true. The probability of rolling exactly two strikes is equal to the probability of rolling exactly one strike, 34.56%.
D is true. The probability that Joe rolls 2 or more strikes is greater than 50%.


Guide 5, Page 185 #19
1500 individuals attended a marathon held in Town A. Of those, only y participated in the marathon. If x of the 1500 individuals were from Town A, and z of the individuals participated in the marathon but were not from Town A, which of the following represents the number of individuals who did not participate in the marathon and were not from Town A?
A) 1500 - x + 2y
B) 1500 - x + 2z
C) 1500 - x - y + z
D) 1500 - x + y - z
E) 1500 - x - z

Another super challenging problem here. Take a second to read it over and think of the correct strategy. Got it?

If you said pick numbers, you are correct! There are variables in the answer choices, and you can mess around with Venn Diagrams (which are impossible for me to draw in this forum anyway), but picking numbers is the easy and obvious choice.

You have to be careful picking values here, because they all depend on each other. Here are my choices: 1500 people at the marathon, 1200 from Town A, 800 marathon participants, 200 people not from Town A in the marathon. With these choices, there are 100 people not from Town A who are not participating in the marathon. These choices are basically random, though I tried to make sure that the difference between any pair of values is different.

To reiterate:
y = marathon participants = 800
x = Town A population = 1200
z = Not Town A marathon participants = 200.
Answer = Not Town A Not marathon participants = 100.

All we have to do is run through the choices.

A) 1500 - x + 2y = 1500 - 1200 + 2*800 = 1900 X
B) 1500 - x + 2z = 1500 - 1200 + 2*200 = 700 X
C) 1500 - x - y + z = 1500 - 1200 - 800 + 200 = -300 X
D) 1500 - x + y - z = 1500 - 1200 + 800 - 200 = 900 X
E) 1500 - x - z = 1500 - 1200 - 200 = 100 :-)

E is the correct answer. To see this from a set theory perspective, the people we are looking for are not from Town A and they are not marathon runners. That population is the same as the entire population minus everyone from Town A, minus everyone not from town A who is in the marathon. Translating this into the quantities we know, this is 1500 - x - z. This method is a little faster, but I prefer picking numbers because it is more mechanical and less prone to errors.


Final question!
Guide 5, Page 186 #20
How many 3-digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero?

Let's start by forgetting about the numbers where there are more than two repeated digits, and just finding out how many 3 digit integers have none of their digits equal to zero. In this case, there are 9 choices for each digit (i.e. the numbers 1-9, inclusive). The total number of these numbers is 9^3 = 729.

Now, let's examine the other condition: "none of the digits appear more than twice." What does that mean. Take a second to rephrase that statement in your head. Which 3-digit numbers have digits that appear more than twice?

The 3-digit numbers that have digits that appear more than twice are the 3-digit numbers that have all 3 digits the same. There are 10 of these numbers (000, 111, 222,...,999), but note that we've already excluded 000. We need to get rid of the other 9. Thus the total number of 3 digit numbers that satisfy the conditions stipulated by the problem is 729 - 9 = 720.

Whew! Great questions!