Calculating arithmetic mean - GMAT Quantitative

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Question

Some balls are placed in a large box, which include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1".

A carnival wants to set up a game by which a player can pay to draw a ball and win an amount of money equal to the number marked on the ball drawn. What should the carnival charge per play in order to make the expected value $1 per game in the carnival's favor?

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Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$

The expected value of the payment that the church will have to make per game can be calculated by multiplying the frequency of each outcome by the respective payment:

$1: \frac{10}{55} \cdot 1 = \frac{10}{55}$

$2: \frac{9}{55} \cdot 2 = \frac{18}{55}$

$3: \frac{8}{55} \cdot 3 = \frac{24}{55}$

$4: \frac{7}{55} \cdot 4 = \frac{28}{55}$

$5: \frac{6}{55} \cdot 5 = \frac{30}{55}$

$6: \frac{5}{55} \cdot 6 = \frac{30}{55}$

$7: \frac{4}{55} \cdot 7= \frac{28}{55}$

$8: \frac{3}{55} \cdot 8 = \frac{24}{55}$

$9: \frac{2}{55} \cdot 9 = \frac{18}{55}$

$1: \frac{1}{55} \cdot 10 = \frac{10}{55}$

...and then adding these products.

$\frac{10}{55}+ \frac{18}{55}+ \frac{24}{55}+ \frac{28}{55}+ \frac{30}{55}+ \frac{30}{55}+ \frac{28}{55}+ \frac{24}{55}+ \frac{18}{55}+ \frac{10}{55}$

$= \frac{220}{55}$

The carnival should expect to pay a prize of $4 per draw, so to expect an average profit of $1 per game, it should charge $5 per play.

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Question

The arithmetic mean of the set $\left \{ a, b, c, d \right \}$ is 250.

Which of the following is the arithmetic mean of the set $\left \{ a, b, c, d, e\right \}$ ?

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Answer

$\frac{a+b+c+d}{4} = 250$

$\frac{a+b+c+d +e }{5}= \frac{1,000 + e}{5} = 200 + \frac{1}{5}e$ ,

which is the correct choice.

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Question

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

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Answer

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

$\frac{A +B+ C+D+E }{5}$

the correct choice.

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Question

What is the mean of this data set?

$\left \{ 1, 0.1, 0.01, 0.001, 0.0001, 0.00001 \right \}$

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Answer

Add the numbers and divide by 6:

$\left (1 + 0.1 + 0.01 + 0.001+ 0.0001 + 0.00001 \right )\div 6$

$= 1.11111\div 6 = 0.185185$

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Question

Consider the data set $\left \{ 1, 3, 4, 6, 6, 7, 7, 7, 9, 10\right \}$ .

Which of the following is true about the median and the mean of the data set?

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Answer

The mean of the data set is the sum of the elements divided by the size of the set, which is ten:

$\frac{1+3+ 4+ 6+ 6+ 7+ 7+ 7+ 9+ 10}{10} = \frac{60}{10} = 6$

The median of a data set with an even number of elements is the arithmetic mean of the two elements that fall in the middle when the elements are arranged in ascending order. These two elements are 6 and 7, so the median is $\frac{6+7}{2} = 6.5$ .

The median is greater than the mean by .

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Question

Consider the data set

$\left \{ 65, 32, 89, 44, 90, 33, M, N\right \}$

For the set to have mean greater than or equal to 50, what must hold true about and ?

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Answer

Divide the sum of the eight elements by 8 to get the mean, which must be greater than or equal to 50:

$\frac{65+32+89+44+90+33+M+N }{8} \geq 50$

$\frac{353+M+N }{8} \geq 50$

$\frac{353+M+N }{8} \cdot 8\geq 50 \cdot 8$

$353+M+N \geq 400$

$353+M+N -353 \geq 400-353$

$M+N \geq 47$

This is all that is needed for the set to have mean 50.

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Question

What is the mean of this data set?

$\left \{ 1, 0.1, 0.01, 0.001, 0.0001, 0.00001 \right \}$

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Answer

Add the numbers and divide by 6:

$\left (1 + 0.1 + 0.01 + 0.001+ 0.0001 + 0.00001 \right )\div 6$

$= 1.11111\div 6 = 0.185185$

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Question

Consider the data set $\left \{ 1, 3, 4, 6, 6, 7, 7, 7, 9, 10\right \}$ .

Which of the following is true about the median and the mean of the data set?

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Answer

The mean of the data set is the sum of the elements divided by the size of the set, which is ten:

$\frac{1+3+ 4+ 6+ 6+ 7+ 7+ 7+ 9+ 10}{10} = \frac{60}{10} = 6$

The median of a data set with an even number of elements is the arithmetic mean of the two elements that fall in the middle when the elements are arranged in ascending order. These two elements are 6 and 7, so the median is $\frac{6+7}{2} = 6.5$ .

The median is greater than the mean by .

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Question

Consider the data set

$\left \{ 65, 32, 89, 44, 90, 33, M, N\right \}$

For the set to have mean greater than or equal to 50, what must hold true about and ?

Tap to reveal answer

Answer

Divide the sum of the eight elements by 8 to get the mean, which must be greater than or equal to 50:

$\frac{65+32+89+44+90+33+M+N }{8} \geq 50$

$\frac{353+M+N }{8} \geq 50$

$\frac{353+M+N }{8} \cdot 8\geq 50 \cdot 8$

$353+M+N \geq 400$

$353+M+N -353 \geq 400-353$

$M+N \geq 47$

This is all that is needed for the set to have mean 50.

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Question

Consider the data set $\left \{ 1, 3, 4, 6, 6, 7, 7, 7, 9, 10\right \}$ .

Which of the following is true about the median and the mean of the data set?

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Answer

The mean of the data set is the sum of the elements divided by the size of the set, which is ten:

$\frac{1+3+ 4+ 6+ 6+ 7+ 7+ 7+ 9+ 10}{10} = \frac{60}{10} = 6$

The median of a data set with an even number of elements is the arithmetic mean of the two elements that fall in the middle when the elements are arranged in ascending order. These two elements are 6 and 7, so the median is $\frac{6+7}{2} = 6.5$ .

The median is greater than the mean by .

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Question

Consider the data set

$\left \{ 65, 32, 89, 44, 90, 33, M, N\right \}$

For the set to have mean greater than or equal to 50, what must hold true about and ?

Tap to reveal answer

Answer

Divide the sum of the eight elements by 8 to get the mean, which must be greater than or equal to 50:

$\frac{65+32+89+44+90+33+M+N }{8} \geq 50$

$\frac{353+M+N }{8} \geq 50$

$\frac{353+M+N }{8} \cdot 8\geq 50 \cdot 8$

$353+M+N \geq 400$

$353+M+N -353 \geq 400-353$

$M+N \geq 47$

This is all that is needed for the set to have mean 50.

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Question

The arithmetic mean of the set $\left \{ a, b, c, d \right \}$ is 250.

Which of the following is the arithmetic mean of the set $\left \{ a, b, c, d, e\right \}$ ?

Tap to reveal answer

Answer

$\frac{a+b+c+d}{4} = 250$

$\frac{a+b+c+d +e }{5}= \frac{1,000 + e}{5} = 200 + \frac{1}{5}e$ ,

which is the correct choice.

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Question

The arithmetic mean of the set $\left \{ a, b, c, d \right \}$ is 250.

Which of the following is the arithmetic mean of the set $\left \{ a, b, c, d, e\right \}$ ?

Tap to reveal answer

Answer

$\frac{a+b+c+d}{4} = 250$

$\frac{a+b+c+d +e }{5}= \frac{1,000 + e}{5} = 200 + \frac{1}{5}e$ ,

which is the correct choice.

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Question

What is the mean of the following data set in terms of and ?

$\left \{a -x, a, a + x, a + 2x, 2a-x, 2a, 2a+x, 2a+2x \right \}$

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Answer

Add the expressions and divide by the number of terms, 8.

The sum of the expressions is:

$\left (a -x \right ) + a + \left ( a + x \right ) +\left ( a +2x \right ) + \left ( 2a-x \right ) + 2a + \left ( 2a+x \right ) +\left ( 2a+2x \right )$

Divide this by 8:

$\left ( 12a + 4x \right )\div 8 =\frac{3}{2} a +\frac{1}{2} x$

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Question

What is the mean of this data set?

$\left \{ 1, 0.1, 0.01, 0.001, 0.0001, 0.00001 \right \}$

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Answer

Add the numbers and divide by 6:

$\left (1 + 0.1 + 0.01 + 0.001+ 0.0001 + 0.00001 \right )\div 6$

$= 1.11111\div 6 = 0.185185$

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Question

What is the mean of the following data set in terms of and ?

$\left \{a -x, a, a + x, a + 2x, 2a-x, 2a, 2a+x, 2a+2x \right \}$

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Answer

Add the expressions and divide by the number of terms, 8.

The sum of the expressions is:

$\left (a -x \right ) + a + \left ( a + x \right ) +\left ( a +2x \right ) + \left ( 2a-x \right ) + 2a + \left ( 2a+x \right ) +\left ( 2a+2x \right )$

Divide this by 8:

$\left ( 12a + 4x \right )\div 8 =\frac{3}{2} a +\frac{1}{2} x$

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Question

Consider the data set $\left \{ 1, -2, 3, -4, 5, -6, 7, -8, 9, -10 \right \}$

What is its mean?

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Answer

Add the ten elements and divide the sum by 10. The mean is

$\left [ 1+ \left ( -2 \right )+3 +\left ( -4 \right )+5+ \left ( -6 \right )+7+ \left ( -8 \right )+9+\left ( -10 \right ) \right ] \div 10$

$= -5 \div 10 = -0.5$

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Question

The arithmetic mean of the set $\left \{ a, b, c, d \right \}$ is 250.

Which of the following is the arithmetic mean of the set $\left \{ a, b, c, d, e\right \}$ ?

Tap to reveal answer

Answer

$\frac{a+b+c+d}{4} = 250$

$\frac{a+b+c+d +e }{5}= \frac{1,000 + e}{5} = 200 + \frac{1}{5}e$ ,

which is the correct choice.

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Question

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

Tap to reveal answer

Answer

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

$\frac{A +B+ C+D+E }{5}$

the correct choice.

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Question

The arithmetic mean of the set $\left \{ a, b, c, d \right \}$ is 250.

Which of the following is the arithmetic mean of the set $\left \{ a, b, c, d, e\right \}$ ?

Tap to reveal answer

Answer

$\frac{a+b+c+d}{4} = 250$

$\frac{a+b+c+d +e }{5}= \frac{1,000 + e}{5} = 200 + \frac{1}{5}e$ ,

which is the correct choice.

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