GMAT Math : Arithmetic

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Example Question #11 : Understanding Arithmetic Sets

Sets

The above represents a Venn diagram. The universal set is the set of all positive integers.

Let be the set of all multiples of 5; let be the set of all perfect squares; let be the set of perfect cubes.

As you can see, the three sets divide the universal set into eight regions. Suppose each positive integer was placed in the correct region. Which of the following numbers would be in the same region as 1,225?

Possible Answers:

3,375

15,625

1,764

3,025

4,356

Correct answer:

3,025

Explanation:

1,225 is divisible by 5 (last digit is 5). It is a perfect square, since $1,225 = 35 ^{2}$ . It is not a perfect cube, however, since $10 ^{3} = 1,000 < 1,225 < 1,331 = 11 ^{3}$ .

Therefore, 1,225 is an element in and , but not . We are looking for an element in and , but not - that is, a multiple of 5 and a perfect square but not a perfect cube.

By looking at the last digits, we can immediately eliminate 1,764 and 4,356, since neither is a multiple of 5. We can eliminate 15,625, since it s a perfect cube - $15,625 = 25 ^{3}$ .

Of the two remaining numbers, 3,375 is not a perfect square, since

$58 ^{2} = 3,364 < 3,375 < 3,481 = 59^{2}$

3,025 is a perfect square: $35 ^{2} = 3,025$

3,025 is not a perfect cube: $14 ^{3} = 2,744 < 3,025 < 15^{3}= 3,375$

3,025 is a multiple of 5, as can be seen from the last digit.

This is the correct choice.

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Example Question #11 : Sets

Sets

The above represents a Venn diagram. The universal set is the set of all positive integers.

Let be the set of all multiples of 3; let be the set of all multiples of 5; let be the set of all multiples of 7.

Possible Answers:

510

476

595

777

736

Correct answer:

476

Explanation:

$728 \div 3 = 242\textrm{ R }2$

$728 \div 5 = 145\textrm{ R }3$

$728 \div 7 = 104$

728 is divisible by 7, but not 3 or 5; therefore, 728 is in but not or . To be in the same region, a number must also be in but not or - that is, divisible by 7 but not 3 or 5.

510 and 595 can be eliminated as multiples of 5 (from the last digit); 777 can be eliminated as a multiple of 3 (digit sum is 21).

Now let's look at the two reminaing choices.

736 is not divisible by 7, since $736 \div 7 = 105 \textrm{ R }1$ .

476 is not divisible by 3 or 5, but it is divisible by 7:

$476 \div 3 = 158\textrm{ R }2$

$476 \div 5 = 95\textrm{ R }1$

$476 \div 7 = 68$

476 is the correct choice.

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Example Question #13 : Understanding Arithmetic Sets

Let be the set $\{\Gamma ,\Delta ,\Lambda ,\Phi \}$ , and be the set $\{\Pi ,\Lambda ,\Delta ,\Theta ,\Gamma ,\Phi ,\Omega , C^2\}$ .

What are the elements in the set $A \cap B$ ?

Possible Answers:

$\{\Delta ,\Lambda ,\Phi , \Gamma \}$

$\{\Gamma, \Phi,\Lambda \}$

$\{C^2\}$

$\{\Gamma ,\Delta ,\Lambda ,\Phi ,\Pi ,\Theta ,\Omega , C^2\}$

$\phi$

Correct answer:

$\{\Delta ,\Lambda ,\Phi , \Gamma \}$

Explanation:

$A \cap B$ is the set of all elements that are in both and . So in this case the elements that are the same for both sets are $\Delta,\Lambda,\Gamma,\Phi$ , Note that the order that you present the elements in your set doesn't matter. What matters is that you don't exclude any necessary elements, or add any that don't belong.

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Example Question #14 : Understanding Arithmetic Sets

If is the set of the multiples of between and and is the set of multiples of between and ; what is $A \bigcap B$ ?

Possible Answers:

$\oslash$

[{6 , 12}]

[{2 , 3 , 4 , 6 , 8 , 9 , 10 , 12 , 15}]

[{2 , 3}]

[{6 , 9 , 12}]

Correct answer:

[{6 , 12}]

Explanation:

A = [2 , 4 , 6 , 8 , 10 , 12 , 14]

B = [3 , 6 , 9 , 12 , 15]

So the intersection of A and B is the numbers that are in both A and B.

Thus $A \bigcap B= [6 , 12]$

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Example Question #15 : Understanding Arithmetic 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: is negative.

Statement 2: |n| < 1

Possible Answers:

BOTH statements TOGETHER are insufficient 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

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 insufficient to answer the question.

Explanation:

If both statements are assumed, then -1 < n < 0 - that is, $n \in (-1, 0)$ . could fall in Region IV or V, as is shown in these two examples:

Example 1: $n= -\frac{1}{2}$ .

-1 < n < 0 , and is rational, so would be in Region IV.

Example 2: $n= -\frac{\pi}{4}$

-1 < n < 0 , and is irrational, so would be in Region V.

The two statements together are insufficient.

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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.

and are both numbers in Region II, and they may or may not be equal. In how many of the five regions could the number possibly fall?

Possible Answers:

One

Five

Two

Three

Four

Correct answer:

One

Explanation:

Region II comprises the whole numbers that are not natural numbers; however, there is only one such number, which is 0. Since and are both numbers in Region II, m = n = 0 . mn = 0 , forcing to be in Region II.

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Example Question #17 : Understanding Arithmetic Sets

Number_sets

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

is a number in Region I, and is a number in Region V. In how many of the five regions could the number m+n possibly fall?

Possible Answers:

Three

Four

One

Two

Five

Correct answer:

One

Explanation:

The key to answering this question is to know that the difference of any two rational numbers is also a rational number.

Suppose m+n is rational. Since , being a natural number, is also rational, the difference

(m+n)-m= n

must be rational.

But it is given that is in Region V, making it irrational. This produces a contradiction. Therefore, m+n must be irrational, and it can only be in one region, Region V.

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Example Question #13 : Sets

What is the fewest number of elements that a set can have in order to have more than 100 subsets?

Possible Answers:

Correct answer:

Explanation:

A set with elements has exactly $2^{N}$ subsets.

$64= 2^{6 }$ , so a set with 6 elements has 64 subsets.

$128 = 2^{7}$ , so a set with 7 elements has 128 subsets.

The correct response is 7.

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Example Question #1571 : Problem Solving Questions

Number_sets

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

and are both numbers in Region I; also, $m \ne n$ . In how many of the five regions could the number m - n possibly fall?

Possible Answers:

Three

Five

Two

Four

One

Correct answer:

Two

Explanation:

The numbers in Region I are exactly the natural numbers $\left \{ 1, 2, 3, 4, 5,...\right \}$ . All natural numbers are integers, and the integers are closed under subtraction, so m - n cannot fall in Region IV or Region V. Also, since it is given in the problem that $m \ne n$ , it follows that $m - n \ne 0$ , so the difference cannot be in Region II (the only whole number that is not a natural number is ).

It is possible for m - n to be in Region I. Example:

$m = 2, n = 1 \Rightarrow m - n = 1$

It is possible for m - n to be in Region III (the integers that are not whole numbers, or the negative integers). Example:

$m = 1, n = 2 \Rightarrow m - n= -1$

m - n can fall in either of two different regions.

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Example Question #1572 : Problem Solving Questions

Number_sets

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

and are both numbers in Region I; also, $m \ne n$ . In how many of the five regions could the number $m \div n$ possibly fall?

Possible Answers:

One

Two

Four

Five

Three

Correct answer:

Two

Explanation:

As natural numbers, and are also rational numbers; since the set of rational numbers is closed under division, and neither nor is equal to zero (zero not being a natural number), $m \div n$ is rational and cannot fall in Region V. Regions III (negative integers) and II (zero only) can be eliminated, since both and are positive. This leaves Regions I and IV.

Examples can be produced that would place $m \div n$ in Region I:

$m = 2, n = 1 \Rightarrow m \div n = 2$

Examples can be produced that would place $m \div n$ in Region IV (the rational numbers that are not integers):

$m = 1, n = 2 \Rightarrow m \div n = \frac{1}{2}$

$m \div n$ can fall in either of two different regions.

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GMAT Math : Arithmetic

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #11 : Understanding Arithmetic Sets

Example Question #11 : Sets

Example Question #13 : Understanding Arithmetic Sets

Example Question #14 : Understanding Arithmetic Sets

Example Question #15 : Understanding Arithmetic Sets

Example Question #12 : Sets

Example Question #17 : Understanding Arithmetic Sets

Example Question #13 : Sets

Example Question #1571 : Problem Solving Questions

Example Question #1572 : Problem Solving Questions

All GMAT Math Resources

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