How to find arithmetic mean - GRE Quantitative Reasoning
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Four groups of college students, consisting of 15, 20, 10, and 18 people respectively, discovered their average group weights to be 162, 148, 153, and 140, respectively. What is the average weight of all the students?
Four groups of college students, consisting of 15, 20, 10, and 18 people respectively, discovered their average group weights to be 162, 148, 153, and 140, respectively. What is the average weight of all the students?
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We know average = sum / number of students. Rearranging this formula gives sum = average * number of students. So to find the total average, we need to add up the four groups' sums and divide by the total number of students.
average = (15 * 162 + 20 * 148 + 10 * 153 + 18 * 140) / (15 + 20 + 10 + 18) = 150
We know average = sum / number of students. Rearranging this formula gives sum = average * number of students. So to find the total average, we need to add up the four groups' sums and divide by the total number of students.
average = (15 * 162 + 20 * 148 + 10 * 153 + 18 * 140) / (15 + 20 + 10 + 18) = 150
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Bill runs for 30 minutes at 8 mph and then runs for 15 minutes at 13mph. What was his average speed during his entire run?
Bill runs for 30 minutes at 8 mph and then runs for 15 minutes at 13mph. What was his average speed during his entire run?
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Rate = distance/time.
Find the distance for each individual segment of the run (4 miles and 3.25miles). Then add total distance and divide by total time to get the average rate, while making sure the units are compatible (miles per hour not miles per minute), which means the total 45 minute run time needs to be converted to 0.75 of an hour; therefore (4miles + 3.25 miles/0.75 hour) is the final answer.
Rate = distance/time.
Find the distance for each individual segment of the run (4 miles and 3.25miles). Then add total distance and divide by total time to get the average rate, while making sure the units are compatible (miles per hour not miles per minute), which means the total 45 minute run time needs to be converted to 0.75 of an hour; therefore (4miles + 3.25 miles/0.75 hour) is the final answer.
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Quantitative Comparison
The average weight of the 7 cats at the veterinarian's office is 8 pounds. The average weight of the 12 dogs at the vet is 16 pounds.
Quantity A: The average weight of all of the animals
Quantity B: The average weight of the cats plus the average weight of the dogs
Quantitative Comparison
The average weight of the 7 cats at the veterinarian's office is 8 pounds. The average weight of the 12 dogs at the vet is 16 pounds.
Quantity A: The average weight of all of the animals
Quantity B: The average weight of the cats plus the average weight of the dogs
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Quantity B has fewer calculations so let's look at that first. We just need to add up the two averages, so Quantity B = 8 + 16 = 24.
To calculate Quantity A, we need the formula for average = total sum / total number of animals = (7 * 8 + 12 * 16) / (7 + 12) = 248/19 = 13.05.
13.05 is less than 24, so Quantity B is greater.
Quantity B has fewer calculations so let's look at that first. We just need to add up the two averages, so Quantity B = 8 + 16 = 24.
To calculate Quantity A, we need the formula for average = total sum / total number of animals = (7 * 8 + 12 * 16) / (7 + 12) = 248/19 = 13.05.
13.05 is less than 24, so Quantity B is greater.
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Alice scored an 87, 85, 90, and 73 on her first four tests of the year. If she wants to have an 87% average in the class, what must she score on her 5th test, assuming the five tests are weighted equally?
Alice scored an 87, 85, 90, and 73 on her first four tests of the year. If she wants to have an 87% average in the class, what must she score on her 5th test, assuming the five tests are weighted equally?
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(87 + 85 + 90 + 73 + x) / 5 = 87
335 + x = 435
x = 100
(87 + 85 + 90 + 73 + x) / 5 = 87
335 + x = 435
x = 100
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Lucy averages 83% on her first 5 tests. What must she score on her sixth test to raise her class average to an 84, assuming all tests are weighted equally?
Lucy averages 83% on her first 5 tests. What must she score on her sixth test to raise her class average to an 84, assuming all tests are weighted equally?
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For the first 5 tests, Sum / 5 = 83, so Sum = 5 * 83 = 415.
Now to solve for the last test, (415 + x) / 6 = 84. Then 415 + x = 504, and x = 89.
For the first 5 tests, Sum / 5 = 83, so Sum = 5 * 83 = 415.
Now to solve for the last test, (415 + x) / 6 = 84. Then 415 + x = 504, and x = 89.
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There exists a function f(x) = 3_x_ + 2 for x = 2, 3, 4, 5, and 6. What is the average value of the function?
There exists a function f(x) = 3_x_ + 2 for x = 2, 3, 4, 5, and 6. What is the average value of the function?
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First we need to find the values of the function: f(2) = 3 * 2 + 2 = 8, f(3) = 11, f(4) = 14, f(5) = 17, and f(6) = 20. Then we can take the average of the five numbers:
average = (8 + 11 + 14 + 17 + 20) / 5 = 14
First we need to find the values of the function: f(2) = 3 * 2 + 2 = 8, f(3) = 11, f(4) = 14, f(5) = 17, and f(6) = 20. Then we can take the average of the five numbers:
average = (8 + 11 + 14 + 17 + 20) / 5 = 14
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What is the arithmetic mean (average) of the following set of numbers:
34, 26, 18, 12, 40
What is the arithmetic mean (average) of the following set of numbers:
34, 26, 18, 12, 40
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If in a set of numbers, the numbers are: 
, the average is automatically 
.
To find the average, add up the sum of all the numbers and divide by the number of items present.
If in a set of numbers, the numbers are:
To find the average, add up the sum of all the numbers and divide by the number of items present.
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What is the average (arithmetic mean) of all multiples of five from 5 to 45 inclusive?
What is the average (arithmetic mean) of all multiples of five from 5 to 45 inclusive?
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All multiples of 5 must first be added.
5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 = 225
Because we added 9 terms, the product must be divided by 9.
225 / 9 = 25.
25 is the average.
All multiples of 5 must first be added.
5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 = 225
Because we added 9 terms, the product must be divided by 9.
225 / 9 = 25.
25 is the average.
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Column A
The mean of the sample of numbers 2, 5, and 10.
Column B
The mean of the sample of numbers 1, 5, and 15.
Column A
The mean of the sample of numbers 2, 5, and 10.
Column B
The mean of the sample of numbers 1, 5, and 15.
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The arithmetic mean is the average of the sum of a set of numbers divided by the total number of numbers in the set. This is not to be confused with median or mode.
In Column A, the mean of 5.66 is obtained when the sum (17) is divided by the number of values in the set (3).
In Column B, the mean of 7 is obtained when 21 is divided by 3. Because 7 is greater than 5.66, Column B is greater. The answer is Column B.
The arithmetic mean is the average of the sum of a set of numbers divided by the total number of numbers in the set. This is not to be confused with median or mode.
In Column A, the mean of 5.66 is obtained when the sum (17) is divided by the number of values in the set (3).
In Column B, the mean of 7 is obtained when 21 is divided by 3. Because 7 is greater than 5.66, Column B is greater. The answer is Column B.
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Sample Set A has 25 data points with an arithmetic mean of 50.
Sample Set B has 75 data points with an arithmetic mean of 100.
Quantity A: The arithmetic mean of the 100 data points encompassing A and B
Quantity B: 80
Sample Set A has 25 data points with an arithmetic mean of 50.
Sample Set B has 75 data points with an arithmetic mean of 100.
Quantity A: The arithmetic mean of the 100 data points encompassing A and B
Quantity B: 80
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Note that:
The arithmetic mean of the 100 data points encompassing A and B =
(total data of Sample Set A + total data of Sample Set B)/100
We have Mean of Sample Set A = 50, or:
(total of Sample Set A) / 25 = 50
And we have Mean of Sample Set B = 100, or:
(total of Sample Set B) / 75 = 100
We get denominators of 100 by dividing both of the equations:
Divide \[(total of Sample Set A) / 25 = 50\] by 4:
(total of Sample Set A) / 100 = 50/4 = 25/2
Multiply \[(total of Sample Set B)/75 = 100\] by 3/4:
(total of Sample Set B)/100 = 75
Now add the two equations together:
(total data of Sample Set A + total data of Sample Set B)/100
= 75 + 25/2 > 80
Note that:
The arithmetic mean of the 100 data points encompassing A and B =
(total data of Sample Set A + total data of Sample Set B)/100
We have Mean of Sample Set A = 50, or:
(total of Sample Set A) / 25 = 50
And we have Mean of Sample Set B = 100, or:
(total of Sample Set B) / 75 = 100
We get denominators of 100 by dividing both of the equations:
Divide \[(total of Sample Set A) / 25 = 50\] by 4:
(total of Sample Set A) / 100 = 50/4 = 25/2
Multiply \[(total of Sample Set B)/75 = 100\] by 3/4:
(total of Sample Set B)/100 = 75
Now add the two equations together:
(total data of Sample Set A + total data of Sample Set B)/100
= 75 + 25/2 > 80
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The average (arithmetic mean) of x, y and z is 15. If w is 10, then what is the average of w, x, y and z?
The average (arithmetic mean) of x, y and z is 15. If w is 10, then what is the average of w, x, y and z?
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We can calculate the arithmetic mean by adding up the numbers in a set, and dividing that total by the count of numbers in the set.
Thus, we know that (x + y + z) / 3 = 15. (Multiply both sides by 3.)
x + y + z = 45
We add w = 10 to that, and divide by the new count, 4.
55 / 4 = 13.75
We can calculate the arithmetic mean by adding up the numbers in a set, and dividing that total by the count of numbers in the set.
Thus, we know that (x + y + z) / 3 = 15. (Multiply both sides by 3.)
x + y + z = 45
We add w = 10 to that, and divide by the new count, 4.
55 / 4 = 13.75
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The combined height of John and Sandy is 130 inches. Sandy, John, and Allen together have a combined height of 215 inches. Sandy and Allen have combined height of 137 inches. How tall is John?
The combined height of John and Sandy is 130 inches. Sandy, John, and Allen together have a combined height of 215 inches. Sandy and Allen have combined height of 137 inches. How tall is John?
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Translate the question into a series of equations:
J + S = 130; J + S + A = 215; S + A = 137
Although there are several ways of approaching this, let us choose the path that is most direct. Given that J, S, A are all involved in the second equation, we can isolate J if we eliminate S and A - which can be done by using the data we have in the third equation. Since S + A = 137, we can rewrite J + S + A = 215 as:
J + 137 = 215.
Now, we only need to solve for J:
J = 215 - 137
J = 78 inches.
Translate the question into a series of equations:
J + S = 130; J + S + A = 215; S + A = 137
Although there are several ways of approaching this, let us choose the path that is most direct. Given that J, S, A are all involved in the second equation, we can isolate J if we eliminate S and A - which can be done by using the data we have in the third equation. Since S + A = 137, we can rewrite J + S + A = 215 as:
J + 137 = 215.
Now, we only need to solve for J:
J = 215 - 137
J = 78 inches.
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On a given exam, four students have an average score of 81 points. If another student takes the exam, what must he or she score in order to increase the overall average to at least 83 points?
On a given exam, four students have an average score of 81 points. If another student takes the exam, what must he or she score in order to increase the overall average to at least 83 points?
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Begin by translating the current state of affairs into an equation. We know that for four students, the average is ascertained by taking the sum of the scores (s) and dividing that by four:
s / 4 = 81; s = 324
Now, when we add the additional student, that person's score (x) will be added to the value for s. The average of this new group will be divided among five students. We must solve for the case in which the average is 83 points (thus giving us the case for the minimum score x necessary.) This yields the following equation:
(324 + x) / 5 = 83
Solve for x:
324 + x = 415; x = 91
The minimum score necessary is 91 points.
Begin by translating the current state of affairs into an equation. We know that for four students, the average is ascertained by taking the sum of the scores (s) and dividing that by four:
s / 4 = 81; s = 324
Now, when we add the additional student, that person's score (x) will be added to the value for s. The average of this new group will be divided among five students. We must solve for the case in which the average is 83 points (thus giving us the case for the minimum score x necessary.) This yields the following equation:
(324 + x) / 5 = 83
Solve for x:
324 + x = 415; x = 91
The minimum score necessary is 91 points.
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Find the arithmetic mean of the following series of numbers:
1, 2, 2, 3, 4, 5, 11, 12
Find the arithmetic mean of the following series of numbers:
1, 2, 2, 3, 4, 5, 11, 12
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To find the arithmetic mean of a series of numbers, add up all of the numbers and divide by the number of numbers in the series. Adding all the numbers gives us 40 and there are 8 numbers in the series. 40/8 = 5
To find the arithmetic mean of a series of numbers, add up all of the numbers and divide by the number of numbers in the series. Adding all the numbers gives us 40 and there are 8 numbers in the series. 40/8 = 5
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Consider the given set of numbers 70, 81, 91, 83, 88, and 55.
A new number equal to a 12% increase in the median of this set is added to the list. What percentage larger is the mean of the new list in comparison with the original?
Consider the given set of numbers 70, 81, 91, 83, 88, and 55.
A new number equal to a 12% increase in the median of this set is added to the list. What percentage larger is the mean of the new list in comparison with the original?
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Reorder the set: 55, 70, 81, 83, 88, 91
Since the set is even, we find the median by taking the average of the middle two values: (81 + 83)/2 = 82
The average of the first set = (55 + 70 + 81 + 83 + 88 + 91)/6 = 468/6 = 78
The new value to be added is 82 * 1.12 = 91.84
This means the new mean is 559.84/7 = 79.9771
The increase in mean value is 79.9771 – 78 = 1.9771
This is (1.9771/78) 100, or 2.535% larger than the original mean.
Reorder the set: 55, 70, 81, 83, 88, 91
Since the set is even, we find the median by taking the average of the middle two values: (81 + 83)/2 = 82
The average of the first set = (55 + 70 + 81 + 83 + 88 + 91)/6 = 468/6 = 78
The new value to be added is 82 * 1.12 = 91.84
This means the new mean is 559.84/7 = 79.9771
The increase in mean value is 79.9771 – 78 = 1.9771
This is (1.9771/78) 100, or 2.535% larger than the original mean.
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A group of ten candy bars has an average cost of $0.89 per candy bar. How many bars must be bought at the cost of $0.72 a piece to bring the average down to $0.80?
A group of ten candy bars has an average cost of $0.89 per candy bar. How many bars must be bought at the cost of $0.72 a piece to bring the average down to $0.80?
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We know to begin with that for x candy bards:
x / 10 = 0.89; x = 8.90
We are trying to ascertain how many bars (y) must be added to this to amount to have an average of 0.80. This is represented:
(8.9 + 0.72y) / (10 + y) = 0.8
8.9 + 0.72y = 0.8 (10 + y)
8.9 + 0.72y = 8 + 0.8y
8.9 – 8 = 0.8y - 0.72y
0.9 = 0.08y
y = 0.9 / 0.08 = 11.25
However, since we cannot have a partial amount of candy bars, we will have to buy 12.
We know to begin with that for x candy bards:
x / 10 = 0.89; x = 8.90
We are trying to ascertain how many bars (y) must be added to this to amount to have an average of 0.80. This is represented:
(8.9 + 0.72y) / (10 + y) = 0.8
8.9 + 0.72y = 0.8 (10 + y)
8.9 + 0.72y = 8 + 0.8y
8.9 – 8 = 0.8y - 0.72y
0.9 = 0.08y
y = 0.9 / 0.08 = 11.25
However, since we cannot have a partial amount of candy bars, we will have to buy 12.
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In a given translation project, three translators each took sections of a book to translate. The first translator took 15000 words, which he translated at a rate of 500 words per 20 minutes. The second translator took 200000 words, which he translated at a rate of 1250 words per half hour. The third took 10000 words, which he translated at a rate of 250 words per 15 minutes. In terms of words per hour, what was the overall average translation rate for this project?
In a given translation project, three translators each took sections of a book to translate. The first translator took 15000 words, which he translated at a rate of 500 words per 20 minutes. The second translator took 200000 words, which he translated at a rate of 1250 words per half hour. The third took 10000 words, which he translated at a rate of 250 words per 15 minutes. In terms of words per hour, what was the overall average translation rate for this project?
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To find the answer, we need to know the total words and the total number of hours involved.
The first is easy: 15000 + 200000 + 10000 = 225000 words
To ascertain the number of hours, we have to look at each translator separately. Although there are several ways to do this, let's consider it this way:
Translator 1 can translate at 500 words per 20 minutes OR 1500 words per hour.
Translator 2 can translate at 1250 words per half hour OR 2500 words per hour.
Translator 3 can translate at 250 words per 15 minutes OR 1000 words per hour.
Therefore, we know each translator took the following amount of time:
Translator 1: 15000 / 1500 = 10 hours
Translator 2: 200000 / 2500 = 80 hours
Translator 3: 10000 / 1000 = 10 hours
The total number of hours was therefore 10 + 80 + 10 = 100 hours.
The average rate was 225000 words/100 hours, or 2250 words per hour.
To find the answer, we need to know the total words and the total number of hours involved.
The first is easy: 15000 + 200000 + 10000 = 225000 words
To ascertain the number of hours, we have to look at each translator separately. Although there are several ways to do this, let's consider it this way:
Translator 1 can translate at 500 words per 20 minutes OR 1500 words per hour.
Translator 2 can translate at 1250 words per half hour OR 2500 words per hour.
Translator 3 can translate at 250 words per 15 minutes OR 1000 words per hour.
Therefore, we know each translator took the following amount of time:
Translator 1: 15000 / 1500 = 10 hours
Translator 2: 200000 / 2500 = 80 hours
Translator 3: 10000 / 1000 = 10 hours
The total number of hours was therefore 10 + 80 + 10 = 100 hours.
The average rate was 225000 words/100 hours, or 2250 words per hour.
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A certain function takes the value 1 with probability 1/2, 2 with probability 1/3, and 3 with probability 1/6. What is the mean value of the function?
A certain function takes the value 1 with probability 1/2, 2 with probability 1/3, and 3 with probability 1/6. What is the mean value of the function?
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To find the mean or expected value, we multiply each value by its corresponding probability and add them up. So the mean = 1(1/2) + 2(1/3) + 3(1/6) = 5/3.
To find the mean or expected value, we multiply each value by its corresponding probability and add them up. So the mean = 1(1/2) + 2(1/3) + 3(1/6) = 5/3.
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In the number set {12, 7, 2, 14, 12, 8, 9, 6, 11} f equals the mean, g equals the median, h equals the mode, and j equals the range.
Which statement is true?
In the number set {12, 7, 2, 14, 12, 8, 9, 6, 11} f equals the mean, g equals the median, h equals the mode, and j equals the range.
Which statement is true?
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The answer is f = g < j = h.
First rearrange the number set in a convenient form:
{2, 6, 7, 8, 9, 11, 12, 12, 14}
f = 9
g = 9
h = 12
j = 12
The answer is f = g < j = h.
First rearrange the number set in a convenient form:
{2, 6, 7, 8, 9, 11, 12, 12, 14}
f = 9
g = 9
h = 12
j = 12
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In a regular 52-card deck of cards, what is the expected number of aces in a 5-card hand?
In a regular 52-card deck of cards, what is the expected number of aces in a 5-card hand?
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There are 4 aces in the 52-card deck so the probability of dealing an ace is 4/52 = 1/13. In a 5-card hand, each card is equally likely to be an ace with probability 1/13. So together, the expected number of aces in a 5-card hand is 5 * 1/13 = 5/13.
There are 4 aces in the 52-card deck so the probability of dealing an ace is 4/52 = 1/13. In a 5-card hand, each card is equally likely to be an ace with probability 1/13. So together, the expected number of aces in a 5-card hand is 5 * 1/13 = 5/13.
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