Arithmetic Mean - GRE Quantitative Reasoning

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.

We know the formula here is average = sum / number of values. Plugging in the values we have, 72 = sum / 5. Then the sum = 72 * 5 = 360, so Quantity A is greater.

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

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

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

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.

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.

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.

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.

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

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.

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.

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 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 expected value, we multiply the number by its corresponding probability.

expected value = 2(0.1) + 4(0.2) + 6(0.3) + 8(0.4) = 6

sum for all 7 tests = 85 * 7 = 595

sum for first 6 tests = 87 * 6 = 522

score on 7th test = 595 – 522 = 73

Let's assume that the three numbers that average 20 are x, y, and z. That means that the sum of x, y and z has to be 60. The average (in this case 20) is the sum of the numbers divided by the quantity of numbers (in this case 3). Thus the sum of the numbers must equal the average (in this case 20) times the number of numbers (in this case 3). Similarly the sum of x, y, z, and the fourth number have to equal 100. If x + y + z = 60 and x + y + z + 4th number = 100 then 4th number has to be 40 which is greater than Option B at 35.

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

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.