Let us simplify :

We can get alternate forms of the same fraction by multiplying the denominator and the numerator by the same number:

Now let's look at :

, but .

Therefore, is the correct answer, as it is not equivalent to .

If the largest of the three consecutive odd integers is d, then the three numbers are (in descending order):

d, d – 2, d – 4

This is true because consecutive odd integers always differ by two. Adding the three expressions together, we see that the sum is 3d – 6.

Since  and must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where .  This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.

Consecutive odd integers differ by 2. If the smallest integer is x, then

x + (x + 2) + (x + 4) = 93

3x + 6 = 93

3x = 87

x = 29

The three numbers are 29, 31, and 33, the largest of which is 33.

Add the ones digits:

Since there is no tens digit to carry over, proceed to add the tens digits:

The answer is .

If there are  students taking Greek, then there are  or  students taking Latin. However, the question asks how many total students there are in the school, so you must add these two values together to get:

 or  total students.

Add the ones digit.

Since the there is a tens digit, use that as the carryover to the next term.

Add the tens digit including the carryover.

The hundreds digit is 7.

Combine the ones digit of each calculation in order.  

The answer is:  

Add the ones digit.

Carry over the one from the tens digit to the next number.

Add the tens digit with the carry over.

Carry over the one from the tens digit to the hundreds digit.

Add the hundreds digit with the carry over.

The thousands digit has no carry over.  The second number has no thousands digit.  This means that the thousands is one.  Combine all the ones digits from each of the previous calculations.

The correct answer is: 

When one multiplies two negative numbers (or any even multiple) the result is a positive number. However, when one multiplies three negative numbers (or any odd multiple) the product is negative. If the result of multiplying 8 negatives is odd, the largest number of negative integers will be the largest odd number, in this case 7.

Let us analyze I, II, and III one at a time.

Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.

Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.

We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.

Only choice I and II will always produce odd numbers.

The answer is I and II only.