GMAT Math : DSQ: Understanding arithmetic sets

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Example Question #11 : Sets

Number_sets

Examine the above diagram, which shows a Venn diagram representing the sets of real numbers.

If real number were to be placed in its correct region in the diagram, which one would it be - I, II, III, IV, or V?

Statement 1: If $m = n \div 7$ , then would be placed in Region IV.

Statement 2: If $p = n \cdot 7$ , then would be placed in Region IV.

Possible Answers:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

Correct answer:

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Explanation:

Region IV comprises the rational numbers that are not integers. A number is rational if and only if it can be expressed as the quotient of integers.

From Statement 1 alone, it can be inferred that is rational, and that it is not an integer. Since $m = n \div 7$ , it follows that $n= 7 \cdot m$ . However, this is not sufficient to narrow it down completely.

For example:

If $m = \frac{1}{7}$ , then $n= 7 \cdot \frac{1}{7} = 1$ , a natural number, putting it in Region I.

If $m = \frac{1}{6}$ , then $n= 7 \cdot \frac{1}{6} = \frac{7}{6}$ , a rational number but not an integer, putting it in Region IV.

From Statement 2 alone, it can be inferred that is rational, and that it is not an integer. From $p= n \cdot 7$ , it follows that $n = p \div 7$ . The nonzero rational numbers are closed under division, so must be a rational number. However, since $p= n \cdot 7$ is not an integer, cannot be an integer, since the integers are closed under multiplication. Therefore, Statement 2 alone proves that belongs in Region IV.

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Example Question #12 : Sets

Number_sets

Examine the above diagram, which shows a Venn diagram representing the sets of real numbers.

If real number were to be placed in its correct region in the diagram, which one would it be - I, II, III, IV, or V?

Statement 1: If m = n+ 7 , then would be placed in Region I.

Statement 2: If p =n- 7 , then would be placed in Region I.

Possible Answers:

BOTH statements TOGETHER are insufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

Correct answer:

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Explanation:

Region I comprises the natural numbers - $\left \{ 1, 2, 3, 4,5,...\right \}$

From Statement 1 alone, m = n+ 7 is a natural number; since n = m - 7 , it follows that is the difference of a natural number and 7 - that is,

$n\in \left \{ -6, -5, -4,..., 0, 1, 2, 3,...\right \}$

could be in any of three regions - I, II, or III.

Conversely, from Statement 2 alone, is the sum of a natural number and 7 - that is,

$n \in \left \{8, 9, 10, 11, 12, 13,...\right \}$

must be a natural number and it must be in Region I.

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Example Question #13 : Sets

Number_sets

Examine the above diagram, which shows a Venn diagram representing the sets of real numbers.

If real number were to be placed in its correct region in the diagram, which one would it be - I, II, III, IV, or V?

Statement 1: If $k = n^{2}$ , then would be in Region I.

Statement 2: If $h = n^{3}$ , then would be in Region III.

Possible Answers:

EITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Correct answer:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Explanation:

Assume Statement 1 alone. It cannot be determined what region is in.

For example, suppose k = 4 , which is in Region I (the set of natural numbers, or positive integers). It is possible that n = 2 , putting it in Region I, or n =- 2 , putting it in Region III (the set of integers that are not whole numbers - that is, the set of negative integers).

Assume Statement 2 alone. It cannot be determined what region is in.

For example, suppose h = - 8 , which is in Region III; then n = -2 , which is also in Region III. But suppose h = - 4 ; then $n = - \sqrt[3]{4}$ , which, as an irrational number, is in Region V.

Now assume both statements. Then has an integer as a square and an integer as a cube. must either be an integer or an irrational number. But

$n = \frac{n^{3}}{n^{2}}$ , making it the quotient of integers, which is rational. Therefore, is an integer. Furthermore, its cube is negative, so is negative. The two statements together prove that is a negative integer, which belongs in Region III.

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Example Question #14 : Sets

How many elements are in set ?

Statement 1: has exactly subsets.

Statement 2: has exactly proper subsets.

Possible Answers:

EITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

Correct answer:

EITHER statement ALONE is sufficient to answer the question.

Explanation:

A set with elements has exactly $2^{N}$ subsets in all, and $2^{N} - 1$ proper subsets (every subset except one - the set itself).

From Statement 1, since has $64 = 2^{6}$ subsets, it follows that it has 6 elements. From Statement 2, since has 63 proper subsets, it has 64 subsets total, and, again, 6 elements. Either statement alone is sufficient.

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Example Question #15 : Sets

Which, if either, is the greater number: or ?

Statement 1: |A + 6| = |B|

Statement 2: -A = B

Possible Answers:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

Correct answer:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Explanation:

Statement 1 alone gives insufficient information. For example, if A = 0 , then:

|A + 6| = |B|

|B| = |0 + 6| = |6| = 6

B = 6 or B = -6

Since -6 < 0 = A < 6 , it is unclear which of and is greater, if either.

Statement 2 gives insufficient information; if is positive, is negative, and vice versa.

Assume both to be true. The two statements form a system of equations that can be solved using substitution:

|A + 6| = |B|

-A = B

|A + 6| = |-A |

Case 1:

A + 6 = -A

2A + 6 = 0

2A = -6

A = -3

B = -A = -(-3) = 3

Case 2:

A + 6 =-( -A)

A + 6 = A

6= 0

This equation has no solution.

Therefore, the only possible solution is A = -3, B = 3 . Therefore, it can be concluded that B > A .

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GMAT Math : DSQ: Understanding arithmetic sets

Study concepts, example questions & explanations for GMAT Math

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Example Question #11 : Sets

Example Question #12 : Sets

Example Question #13 : Sets

Example Question #14 : Sets

Example Question #15 : Sets

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