Real Numbers - GMAT Quantitative

Looking at statements (1) and (2) separately, there is no information about the distance between and . Looking at statements (1) and (2) together, there are two possibilities: 1) and are on different sides of , and the distance between them is 19; 2) and are on the same side of , and the distance between them is 5. Therefore, we cannot get the only possible answer to this question based on the two statements.

Statement 1 precisely tells how many children are attending the party while Statement 2 only provides a range.

The price of 6 sandwiches can be expressed as . Regrouping is needed since this is not in the answer choices. If $3.00 is used instead of $2.99, each of the five additional sandwiches is .01 too high and the total is .05 too high. Therefore, this amount needs to be subtracted from the $4.95.

If it is known that , it can be determined that .

Then

Since , , and the question is answered.

But if we only know that , it is not sufficient.

Case 1:

yields

Case 2:

yields

The answer is that Statement 1 is sufficient, but not Statement 2.

A number is rational if and only if its decimal representation is terminating or repeating.

and are terminating decimals, which are rational (note: the latter number is not equal to ). Both can be eliminated.

has a repeating digit and has a repeating group of five digits, and are therefore repeating decimals, which are also rational. Both can be eliminated.

has an infinite pattern - the number of zeroes that precede each one is increasing by one with each group - but since the pattern is not repeating, it is neither a terminating decimal nor a repeating decimal. It is irrational.

To find the real solutions, we need to know what c is. An easy way to do that would be if we had a point on the function.

I) gives us such a point. Therefore, we are able to substitute in our values and solve for c using algebraic operations.

Thus our equation becomes:

.

From here we can try to factor to find the real solutions or you can use the quadratic formula.

Plugging in our values we get the solution .

II) Is not really relevant, we are dealing with a parabola, so it shouldn't be undefined anyway.

We can find each in terms of .

In ascending order, the four values are:

The correct choice is .

From Statement 1 alone, since 1 raised to the power of any real number is equal to 1, we know that

.

However, we cannot determine the value of knowing only that . If is a nonzero number, then . However, if , then

,

which is an undefined expression.

Assume Statement 1 alone. The multiplicative inverse of a number is the number which, when multiplied by that number, yields a product of 1. If the multiplicative inverse of is not an integer, it is possible for to be an integer or to not be one, as is shown in these examples:

If , which is an integer, then, since , the multiplicative inverse of is , which is not an integer, If , which is not an integer, then, since , the multiplicative inverse of is , which is not an integer.

Assume Statement 2 alone. The additive inverse of a number is the number which, when added to that number, yields a sum of 0. If is the additive inverse of the number , then

, or

by Statement 2, is an integer; ., the product of integers, is itself an integer.

Assume Statement 1 alone.

, since 0 multiplied by any number yields a product of 0.

Assume Statement 2 alone.

, since 0 added to any number yields the sum 0. However, without knowing and , or their product, it is impossible to determine the result.

From Statement 1 alone, since a side of a square with area has as its sidelength

,

.

This is the quotient of two integers; by definition, this is rational.

From Statement 2 alone, if we let , then . is rational, and so is 6, so their product, , is also rational.

Statement 1 provides insufficient information to determine whether is rational or not.

If , which, being an integer, is rational, then , which, being an integer, is rational.

If . which is irrational, then , which, being an integer, is rational.

Statement 2 is, however, sufficient. If for some rational number , then . 1, being an integer, is rational, and as the difference of rational numbers, is rational.

Assume Statement 1 alone. , the quotient of integers, is rational. The sum of any two rational numbers is rational, so if is rational, then is a rational number. However, is irrational, so, contrapositively, is irrational.

Assume Statement 2 alone. , the quotient of integers, is rational. The product of any two rational numbers is rational, so if is rational, then is a rational number. However, is irrational, so, contrapositively, is irrational.

Assume Statement 1 alone. The polynomial expression in

can be factored as follows:

Either , in which case , which, as an integer is also rational, or

, in which case:

or either or , both irrational.

Therefore, Statement 1 provides insufficient information.

Assume Statement 2 alone. If , then , making a fourth root of 4. A rational root of an integer must itself be an integer, and there is no integer which, when taken to the fourth power, is equal to 4. Therefore, must be irrational.

The two statements together provide insufficient information to answer the question. For example, 36 and 72 are both multiples of both 9 and 12. However, the positive square root of 36 is 6, making an integer and, consequently, rational. However, the positive square root of 72 is not an integer, since 72 falls between consecutive perfect squares 64 and 81 (the squares of 8 and 9, respectively), and any square root of an integer that is not itself an integer must be irrational.

Assume Statement 1 alone. The polynomial expression in

can be factored as follows:

Either , in which case , or

- which is impossible for any real value of , since

, and

Therefore, is an integer.

Assume Statement 2 alone. Again, we factor:

Similarly to the polynomial in Statement 1, , , or . All three possible values of are integers.

Assume Statement 1 alone. By the commutative property of addition,

.

By substitution,

.

Assume Statement 2 alone. By the commutative property of multiplication,

and

By two substitutions,

By the distributive property of multiplication over addition,

Assume Statement 1 alone. , so, by substitution,

.

But, without knowing , , or their product, it is impossible to determine the value of the expression. Statement 2 provides insufficient information, for similar reasons.

Now assume both statements to be true. Then

Since and , by substitution,

.

Assume Statmemt 1 alone. By the commutative property of multiplication:

By the addition property of equality,

By substitution,

Statement 2 alone, however, does not answer the question. If , then by substitution,

;

however, without further information, we cannot evaluate this expression.