Consider each condition:

The first if statement has:  a > 12 && b < -14 && c < d

While it is true that 14 > 12 and -15 < -14, it is not true that 11 < 22.  Therefore, the conjunction of all of these cannot be true.  For &&, all of the expressions must be true.

Next, consider: d - 20 < b

This is really:  11 - 20 < -15 or -9 < -15.  This is not true either.

Next, consider: a + 12 >= c

This is really: 12 + 12 >= 22 or 24 >= 22.  This is true.  Therefore, your output is "This is vexing......"!

First of all, note that your loop will run for values from 0 to 9, inclusive. 

Now, the first condition is relatively easy:

i % 2 == 0

This is true when the number i is even. When it is divisible by 2, the remainder will be 0. (Recall that the modulus gives you the remainder.)

Now, the next two else conditions will execute only for odd numbers. The first one can be rewritten:

i > 6 && i % 3 == 1

This means that we need all of the numbers from 7 to 9 that have a remainder of 1 when divided by 3. This only applies to i = 7. This will be the eighth line output. Therefore, you know that you will have alternating lines saying, "This is my favorite case!" All of the other ones will be "Well, at least it is something...", except for the eighth line, which will be "There once was a pool in the back yard!"

The easiest way to begin is by looking at the general form of the first if statement. Notice that it is made up of three expressions that have a boolean AND between them. This means that if any of those members is false, the whole thing will be false. Well, consider the very first term:

s2.charAt(4) == ' '

This is asking about the fifth character in the string s2. (Remember, the index starts at 0.) Well, this is true for s but not for s2, which has 'e' for its fifth member.  Thus, we know that the first if must be false.

Now, to the second:

s2.charAt(5) == 's'

This also is not true, for the sixth character in s2 is ' '.

The third if is also false. j + 12 is equal to 0!

However, it is indeed the case that (45 - 20 < 100) is true. Therefore, the fourth if statement will execute, giving you the output:

Where will the group be having lunch tomorrow?

Let us consider each conditional in turn. When we reach one that is true, we know our output.

First Conditional: i < 15 && j >= -4

Recall that you must have both of these true in order for the whole expression to be true. Well, while it is the case that , it is not true that .  Thus, this conditional is false.

Second Conditional:  !(i < 100 && k > 4)

For this, you must first attempt the expression in parentheses. Both of these are true. However, !(true) is false. (It makes sense in English: "That is not true!" This means, "That is false!")

Third Conditional:  !(i == 3 && k == 4)

Well, here, we know that both 12 == 3 and 103 == 4 are both false. Therefore, the expression is !(false). Therefore, it is true! This is your output.

Let's consider the cases in order until we find our matching true expression.

Conditional 1: i > 34 || c < 'B'

All we need is for one of these to be true for the whole expression to be true (given the nature of the logical OR represented by ||. However, neither of these are true.  

Conditional 2: s.charAt(9) == c

If you carefully count the letters in the String s, you will see that index 9 (i.e. the tenth character) is M. Therefore, this is equal to c. Thus, you know that your program will output I read William of Ockham yesterday!

The "!" symbol means not. Therefore, 7 does not equal 14 is true.

The expression !(y < 10 || y > 10) evaluates to (y >= 10 && y <= 10).

Since y = 10, the expression evaluates to true.

a and b are both primitive types of integer. a is equal to 2 and b is equal to 3. 2 does not equal 3, so the answer is false.

Since both objects are not primitive types, double equals, ==, would return whether the pointers of the type objects are equivalent. The string object requires the string equals method which is String.eq("string")

3 is the correct answer because a = 5 and b = 10 so 5 + 10 = 15 which is d.