Arithmetic - GMAT Quantitative

Order and repetition do NOT change a set. Therefore, the set we want to describe contains the numbers 1, 3, and 4. The only set that doesn't contain all 3 of these numbers is , so it is the set that does not equal the rest of the sets.

Add each element of to each element of .

= {1 + 3, 1 + 4, 1 + 6, 4 + 3, 4 + 4, 4 + 6} = {4, 5, 7, 8, 10}

We are looking for the union of the sets. That means we want the elements of A OR B.

So = {2, 3, 4, 5, 7}.

We are looking for the union of the sets. That means we want the elements of A OR B.

So = {2, 3, 4, 5, 7}.

We are looking for the union of the sets. That means we want the elements of A OR B.

So = {2, 3, 4, 5, 7}.

We are looking for the union of the sets. That means we want the elements of A OR B.

So = {2, 3, 4, 5, 7}.

We are looking for the union of the sets. That means we want the elements of A OR B.

So = {2, 3, 4, 5, 7}.

We need to add 3 to every element in .

Then:

We are looking for the union of the sets. That means we want the elements of A OR B.

So = {2, 3, 4, 5, 7}.

comprises the set of elements common to all three sets. However, since is the set of all even integers and comprises the set of all odd integers, no element can be common to all three sets. The correct response is 0.

comprises the set of all odd integers except 1; comprises the set of all integers of the form , a natural number. Therefore, any number that is not in the union of these two sets must be in neither one.

, so is even or 1 (although 1 is not a choice). We can eliminate odd choices 147, 149, and 151 immediately.

, so we determine which number cannot be expressed as , a natural number.

148 is elminated, since it is two less than a multiple of 3. 150 is the correct choice.

,

which is the correct choice.

The arithmetic mean of a set is the sum of its elements divided by the number of elements, which here is . This makes

the correct choice.

There are nine elements, so the median is the fifth-highest element. For this fifth-highest element to be 70, first of all, 70 must be in the set; since none of the known elements are equal to 70, then one of the two unknowns must be 70.

Assume without loss of generality that . Then four of the elements are already known to be less than 70. Since four elements must be greater than or equal to 70, must be one of them.

Therefore, the correct choice is that one must be equal to 70 and the other must be greater than or equal to 70.

The median of an odd number of data values is the middle valie when the scores are arranged in descending order. Since the scores are arranged, the middle score, and the median, is .

Assume both statements to be true, and examine two cases.

Case 1:

The arithmetic mean of and is

The conditions of both statements are satisfied.

The mean of the five numbers is their sum divided by 5:

Case 2:

The arithmetic mean of and is

The conditions of both statements are satisfied.

But the mean of the five numbers is

Therefore, the mean may or may not be equal to .

comprises the set of elements common to all three sets. However, since is the set of all even integers and comprises the set of all odd integers, no element can be common to all three sets. The correct response is 0.

The midrange of a data set is the arithmetic mean of its greatest element and least element. Here, those elements are and , so we can find the midrange as follows:

Add the numbers and divide by 6: