Real Numbers - GMAT Quantitative

Today, . In 3 years, .

is negative and is positive

Therefore, the solution cannot be negative.

Add 3 to both sides:

Divide both sides by 2:

To calculate the number of students taking at least 1 science course, add the number of students who are taking each course and subtract the number of students who are taking 2 (to ensure they're not counting twice).

To calculate the number of students NOT taking a class, subtract this number by the total number of students.

To find :

,

so

Since ,

and we choose the positive square root

This question does not require any calculation. Given that 32 (an even number) is a factor of , then 2 must be a prime factor. If is then multiplied by 2 (to get ) then has no additional unique prime factors (its only additional prime factor, 2, is NOT unique).

The lowest possible value of is the lowest possible value of divided by the highest possible value of :

The highest possible value of is the highest possible value of divided by the lowest possible value of :

so this is a composite number for all and .

is by definition a composite number.

the product of 2 numbers.

This leaves just . For a number to be prime, it must be odd (except for 2) so we need to have either or be odd (but not both). The first composite odd number is 9. . The smallest composite number is 4. .

is a prime number.

So the answer is

The definition of the set of real numbers is the set of all numbers that can fit into a/b where a and b are both integers and b does not equal 0.

So, since we see a fraction here, we know a non-real number occurs if the denominator is 0. Therefore we can find where the denominator is 0 by setting x-3 =0 and solving for x. In this case, x=3 would create a non-real number. Thus our answer is that x CANNOT be 3 for our expression to be a real number.

For the expression to be defined, the denominator needs to be different from 0. Therefore:

So the correct answer is 2.

To show the other answers aren't always true, we need to choose 2 numbers that satisfy the given inequalities but also contradict each answer one-by-one.

Let this choice shows that sometimes , which rules out the answers , is negative, and

Now Let , this will still satisfy the given inequalities, but now . This means that the answer " is positive" isn't always true either.

In the absence of grouping symbols, additions and subtractions are performed from left to right. Only the expression in (I) has an addition as its leftmost operation, so this is the only expression that is evaluated by adding first.

The expression is equal to , which is calculated as follows:

,

or seventeen twenty-eighths.

The last nonzero digit is located in the fourth place right of the decimal point - the ten-thousandths place. Put the number, without decimal point or leading zeroes, over 10,000. This number is , or four hundred seven ten-thousandths.

The one-thousandths and ten-thousandths places are the third places and the fourth places, respectively, to the right of the decimal point. Therefore:

Thirty-three one-thousandths =

Three hundred three ten-thousandths =

Add:

The last nonzero digit ends at the ten-thousandths place, so this is , or

six hundred thirty-three ten-thousandths.

The one-hundredths place is the second place to the right of the decimal point; therefore, three one-hundredths is equal to 0.03. One fourth can be converted to decimal form as follows:

,

which is twenty-five one-hundredths, or 0.25. Subtract:

,

or twenty-two one-hundredths.

. As the product of integers, must be an integer.

, making the sum of integers, and, consequently, an integer.

, making the sum of integers, and, consequently, an integer.

, making the difference of integers, and, consequently, an integer.

We demonstrate that need not be an integer through a counterexample. Let .

and , so the conditions of the problem are met. However,

, which is not an integer. This makes the correct response.

We can find each in terms of .

In ascending order, the numbers are:

The correct choice is .