Mode - GMAT Quantitative

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

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Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

Tap to reveal answer

Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

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Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

Tap to reveal answer

Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

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Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

Tap to reveal answer

Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

Tap to reveal answer

Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

Tap to reveal answer

Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

Tap to reveal answer

Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

Tap to reveal answer

Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

Tap to reveal answer

Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

Tap to reveal answer

Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

True or false: is the arithmetic mean of the set $S = \left \{ A, B, C, D, E \right \}$ .

Statement 1:

Statement 2: is the arithmetic mean of and .

Tap to reveal answer

Answer

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

Case 1:

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

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

$\mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C$

Case 2:

The arithmetic mean of and is

$\frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C$

The conditions of both statements are satisfied.

But the mean of the five numbers is

$\mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C$

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

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

Tap to reveal answer

Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

Tap to reveal answer

Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

Consider the data set $\left \{ 2, 4,5, 6, 6, 6, 7, 8, 8, 10, N, N\right \}$ . It is known that . How many modes does this data set have, and what are they?

Tap to reveal answer

Answer

Of the known elements, 6 occurs the most frequently - three times. Since the unknown occurs only twice, and it cannot be equal to any of the other elements, its value does not affect the status of 6 as the most frequent element. Therefore, regardless of , 6 is the only mode.

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Question

For which of the following values of would the median and the mode of the data set be equal?

$\left \{ 13, 11, 19, 17, 13, 15, 13, 17, N \right \}$

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Answer

If the known values are ordered from least to greatest, the set looks like this:

$\left \{11, 13, 13, 13, 15, 17, 17, 19 \right \}$

Below are each of the choices, followed by the set that results if it is added to the above set, followed by the median - the middle element - and the mode - the most frequently occurring element.

$11: \left \{11, 11, 13, 13, 13, 15, 17, 17, 19 \right \}; median ; 13, mode; 13$

$15: \left \{11, 13, 13, 13, 15, 15, 17, 17, 19 \right \}; median ; 15, mode; 13$

$17: \left \{11, 13, 13, 13, 15, 17, 17, 17, 19 \right \}; median ; 15, modes; 13, 17; (bimodal)$

$19: \left \{11, 13, 13, 13, 15, 17, 17, 19, 19 \right \}; median ; 15, mode; 13$

Only the addition of 11 yields a set with median and mode equal to each other.

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Question

Give the mode of the set $\left \{ A, B, C, D, E \right \}$ .

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Answer

The mode of a set, if it exists, is the value that occurs most frequently. The inequality

means that the set

$\left \{ A, B, C, D, E \right \}$

can be rewritten as

$\left \{ A,A, C, D, E \right \}$ .

The most frequently occurring value is , making this the mode.

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Question

.

Give the mode of the set $\left \{ A, B, C, D, E,F\right \}$ .

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Answer

The mode of a set, if it exists, is the value that occurs most frequently. The inequality

means that the set

$\left \{ A, B, C, D, E,F\right \}$

can be rewritten as

$\left \{ A, A, C, D,D,F\right \}$

and occur as values twice each; the other values, and , are unique. Therefore, the set has two modes, and .

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Question

Which of these values is not a mode of the set $\left\{ A, B, C, D \right \}$ ?

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Answer

The mode of a set is the value that occurs most frequently in that set. Since

, it follows that

$\left\{ A, B, C, D \right \}$

can be rewritten as

$\left\{ A, A, C, C \right \}$ .

This makes and both modes, since both occur twice. Equivalently, since and , and are modes.

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