Average

\begin{array}{l} =\frac{2+3+5+7+11+13+17+19+23}{9} \\ =\frac{100}{9} \\ =11 \frac{1}{9} \end{array}

There are five prime numbers between 30 and 50.
This statement provides information about the quantity and range of prime numbers being considered.

They are 31, 37, 41, 43, and 47.
This list enumerates the prime numbers that fall within the specified range.

Therefore the required average=(31+37+41+43+47)5=1995 =39.8

To find the average of the first 40 natural numbers,
the sum of these numbers is calculated using the formula
"sum of first n natural numbers=n(n+1)/2",
which yields a sum of 820.
The average is then calculated by dividing this sum by the total number of numbers,
resulting in an average of 20.5

To find the average weight of the entire class,
given that section A has 36 students with an average weight of 40kg and section B has 44 students with an average weight of 35kg,
we can first calculate the total weight of all 80 students.
This can be done by adding the total weight of the students in each section, as follows:
Total weight = (36 * 40kg) + (44 * 35kg) = 2980kg
The average weight of the entire class can then be calculated by dividing the total weight by the number of students, as follows:
Average weight = 2980kg / 80 students = 37.25kg

To find the total money spent by the group of nine people at the hotel,
given that eight of them spent Rs.12 each on their meals and the ninth person spent Rs.8 more than the average expenditure of all nine people,
we can first determine the average expenditure of the group.
Let the average expenditure of all nine be represented by x.
Then, the total expenditure of the eight people who spent Rs.12 each can be represented as 12*8 = 96.
The ninth person's expenditure can be represented as x+8.
The total expenditure of the group can be represented as the sum of the expenditure of each person, or 9x.
We can set up the equation 9x = 96 + (x+8) to represent this situation.
Solving for x yields x = 13, so the average expenditure of the group is Rs.13.
The total money spent by the group is then equal to 9x = Rs.(9*13) = Rs.117.

To find the largest of the three numbers,
given that the second number is twice the first and is also thrice the third,
and that the average of the three numbers is 44,
we can assign variables to represent each of the numbers.
Let the third number be represented by x.
Then the second number can be represented as 3x, and the first number can be represented as 3x/2.
The average of the three numbers can be represented as (x+3x+(3x/2))/3 = 44.
Solving for x yields x = 24. The largest of the three numbers is then the second number,
which is 3x = 3 * 24 = 72

To find the 13th result, given that the average of 25 results is 18, 
the average of the first 12 results is 14, 
and the average of the last 12 results is 17, 
we can first determine the sum of the 25 results.

This can be done by multiplying the average of the results by the total number of results, as follows:

Sum of results = 18 * 25 = 450

We can then determine the sum of the first 12 results by multiplying the average of these results by the number of results, as follows:

Sum of first 12 results = 14 * 12 = 168

Similarly, 
we can determine the sum of the last 12 results by multiplying the average of these results by the number of results, as follows:

Sum of last 12 results = 17 * 12 = 204

The 13th result is then equal to the sum of the 25 results minus the sum of the first 12 results and the last 12 results, or:

13th result = 450 - (168 + 204) = 450 - 372 = 78

6th result = (58*6+63*6-60*11)=66

Let A, B, and C represent their individual weights.
Then:

A + B + C = 45 kg * 3 = 135 kg
A + B = 40 kg * 2 = 80 kg
B + C = 43 kg * 2 = 86 kg

B = (A + B) + (B + C) - (A + B + C) = (80 kg + 86 kg - 135 kg) = 31 kg

The total age of 39 students is 39 students * 15 years/student = 585 years.

The average age of 40 people (including the teacher) is 15 years 3 months = 61/4 years.

The total age of 40 people is 61/4 years/person * 40 people = 610 years.

The age of the teacher is the total age of 40 people - the total age of 39 students = 610 years - 585 years = 25 years

The total weight increase when the new man replaces one of the crew members is 
1.8 kg/person * 10 people = 18 kg.
The weight of the new man is
53 kg + 18 kg = 71 kg.

Let x be the average score of the batsman after the 17th inning.

Then the average score of the batsman after the 16th inning is (x - 3).

The total number of runs scored by the batsman in the first 16 innings is (x - 3) * 16 + 87 = 17x.

Solving for x, we get x = (87 - 48) / 17 = 39

Average = (12 + 24 + 36 + 48 + 60)/5
= 180/5
= 36

The new average = Initial average ∗ 2
= 20 ∗ 2 = 40

If the average of three consecutive even numbers is 34, 
and the first number is X, 
then the next two even numbers would be X+2 and X+4. 
Using this information, 
we can set up the following two equations:
Apti Average 17 = 34
Apti Average 18 = 34
Solving these equations for X, we find that X = 32. 
Therefore, the largest of the three numbers is X+4 = 32+4 = 36.

The total marks obtained by Sohan in six subjects is 444,
and the total marks obtained by him in five subjects excluding science is 350.
Therefore, the marks obtained by Sohan in science are 444 - 350 = 94.

The age of the new boy is 13 years.
To find this, we can use the following steps: Let the age of the new boy be X years.
The total age of the 30 boys is 480 years.
The average age of the group becomes 15.8 years when the boy of age 19 leaves and the new boy joins.
Therefore, 480 - 19 + X = 474.
Solving for X, we find that X = -6 + 19 = 13 years.

The weight of the new man is 40 kg.
To find this, we can use the following steps:
The average decrease in weight per person is 2 kg.
The total number of men is 10, so the total decrease in weight when one of them is replaced by a new man is 10 * 2 = 20 kg.
The weight of the man who is replaced is 60 kg.
Therefore, the weight of the new man is 60 - 20 = 40 kg.

The sum of the ages of the two teachers is 62 years.
To find this, we can use the following steps:
The total age of the 30 boys is 450 years.
The total age of the 30 boys and the two teachers is 512 years.
Therefore, the sum of the ages of the two teachers is 512 - 450 = 62 years.

The total age of A and B four years ago: 40 years
The present total age of A and B: 48 years
The present total age of A, B, and C: 75 years
The present age of C: 75 - 48 = 27 years
The age of C after 7 years: 27 + 7 = 34 years.

The weight of the new woman is 85 kg.
To find this, we can use the following steps:
The total weight increased is 20 kg (8 x 2.5 = 20).
Therefore, the weight of the new woman is 65 + 20 = 85 kg.

The average age of the team is 23 years.
To find this, we can use the following steps:
Let the average age of the whole team be x years.
The difference between the total ages of the captain and wicketkeeper and the average age of the whole team is 9(x - 1).
The difference between the total ages of the captain and wicketkeeper is 55 years.
Therefore, 11x - 55 = 9(x - 1).
Solving for x, we find that x = 23 years.

Rakesh's monthly income is Rs. 4000. To find this, we can use the following steps:
The total income of Rakesh and Suresh is 10100.
The total income of Suresh and Ramesh is 12500.
The total income of Rakesh and Ramesh is 10400.
Adding these three equations gives us 2(P + Q + R) = 33000 or P + Q + R = 16500.
Subtracting the equation for Suresh and Ramesh from this equation gives us P = 4000.
Therefore, Rakesh's monthly income is Rs. 4000.

The average of the different possible weights of Varun is 67 kg.
To find this, we can use the following steps:
Let Varun's weight be X kg.
According to Varun, 65 < According to Varun's father, 60 < X < 70.
According to Varun's sister, X <= 68.
The different possible weights of Varun, or the values that satisfy all of these conditions, are 66, 67, and 68.
The average of these weights is (66 + 67 + 68) / 3 = 201 / 3 = 67 kg.

The weight of Q is 31 kg.
To find this, we can use the following steps:
Let P, Q, and R represent their respective weights.
The sum of the weights of P, Q, and R is 135.
The sum of the weights of P and Q is 80.
The sum of the weights of Q and R is 86.
Adding these two equations gives us P + 2Q + R = 166.
Subtracting the equation for P, Q, and R from this equation gives us Q = 31.

The average weight of all the students in the class is 48.55 kg.
To find this, we can use the following steps:
The average weight of 16 students in the class is 50.25 kg.
The average weight of the remaining 8 students is 45.15 kg.
The total weight of all the students in the class is 16 * 50.25 + 8 * 45.15 = 1165.20 kg.
The total number of students in the class is 24.
Therefore, the average weight of all the students in the class is 1165.20 / 24 = 48.55 kg.

The average number of visitors per day in a month of 30 days beginning with a Sunday is 285.
To find this, we can use the following steps:
There are 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday.
The average number of visitors on Sunday is 510, and the average number of visitors on other days is 240.
Therefore, the total number of visitors in the month is 5 * 510 + 25 * 240 = 5250 + 6000 = 11250.
The total number of days in the month is 30.
Therefore, the average number of visitors per day in the month is 11250 / 30 = 285.

The average marks of all the students is 54.68.
To find this, we can use the following steps:
The average marks of the three classes of 45, 60, and 55 students are 60, 55, and 50, respectively.
The total number of students in the three classes is 45 + 60 + 55 = 160.
The total marks of the three classes are 45 * 60 + 60 * 55 + 55 * 50 = 2700 + 3300 + 2750 = 8750.
Therefore, the average marks of all the students is 8750 / 160 = 54.68.

The new average will be 10.5 years.
To find this, we can use the following steps:
The sum of the ages of the 15 boys is 165.
The sum of the ages of the 5 boys is 45.
The total age of the 20 boys is 165 + 45 = 210.
The total number of boys is 20.
Therefore, the average of the ages of the 20 boys is 210 / 20 = 10.5 years.

The combined average of the two groups is 20.44.
To find this, we can use the following steps:
The number of quantities in group A is 10, and the number of quantities in group B is 8.
The individual average of group A is 24, and the individual average of group B is 16.
The combined total of the two groups is 10 * 24 + 8 * 16 = 240 + 128 = 368.
The total number of quantities in the two groups is 10 + 8 = 18.
Therefore, the combined average of the two groups is 368 / 18 = 20.44.

The average of the squares of the first 6 consecutive even numbers is 60.66.
To find this, we can use the following steps:
The first 6 consecutive even numbers are 2, 4, 6, 8, 10, and 12.
The average of the squares of these numbers is 2 * (6 + 1) * (2 * 6 + 1) / 3 = 2 * 7 * 13 / 3 = 182 / 3 = 60.66.

The average of the first 4 consecutive even numbers is 5.
To find this, we can use the following steps:
The first 4 consecutive even numbers are 2, 4, 6, and 8.
The average of these numbers is (2 + 4 + 6 + 8) / 4 = 20 / 4 = 5.

The average of the first 5 consecutive even numbers starting from 2 to 10,
where the last even number is 10, is 6. To find this, we can use the following steps:
The first 5 consecutive even numbers starting from 2 to 10,
where the last even number is 10, are 2, 4, 6, 8, and 10.
The average of these numbers is (2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6

The average of the squares of the first 6 consecutive even numbers starting from 2 to 12,
where the last even number is 12, is 60.66. To find this, we can use the following steps:
The first 6 consecutive even numbers starting from 2 to 12, where the last even number is 12, are 2, 4, 6, 8, 10, and 12.
The average of the squares of these numbers is (2 * 2 + 4 * 4 + 6 * 6 + 8 * 8 + 10 * 10 + 12 * 12) / 6 =
(4 + 16 + 36 + 64 + 100 + 144) / 6
= 13 * 14 / 3
= 182 / 3
= 60.66

The sixth number is 30 if the average of 11 numbers is 30,
the average of the first six numbers is 17.5,
and the average of the last six numbers is 42.5.
To find this, we can use the following steps:
Calculate the total of the 11 numbers by multiplying the average value of 30 by 11, which gives us 330.
Calculate the total of the first six numbers by multiplying the average value of 17.5 by 6, which gives us 105.
Calculate the total of the last six numbers by multiplying the average value of 42.5 by 6, which gives us 255.
Find the sixth number by adding the value of the first six and last six numbers and subtracting it from the total value of the 11 numbers,
which gives us (105 + 255) - 330 = 30.

The middle number is 11 if the average of 15 numbers is 15,
the average of the first five numbers is 14, and the average of the other 9 numbers is 16.
To find this, we can use the following steps:
Calculate the total of the 15 numbers by multiplying the average value of 15 by 15, which gives us 225.
Calculate the total of the first five numbers by multiplying the average value of 14 by 5, which gives us 70.
Calculate the total of the other 9 numbers by multiplying the average value of 16 by 9, which gives us 144.
Find the middle number by subtracting the total of the first five and other nine numbers from the total value of the 15 numbers,
which gives us (225) - [(70) + (144)] = 11.

The largest of four consecutive even numbers is 30 if their average is 27.
To find this, we can use the following steps:
Let the consecutive even numbers be represented by x, (x + 2), (x + 4), and (x + 6).
Calculate the average of the four numbers using the formula: Average = Sum of Quantities / Number of Quantities.
Solve the equation to find the value of x.
The largest number is (x + 6) and the smallest number is x. In this case, the largest number is 30 and the smallest number is 24.

The present age of the baby can be found by using the following steps:
Calculate the total age of the family of 5 members 2 years ago: 16 years * 5 members = 80 years
Calculate the present age of the family of 5 members: 80 years + (2 years * 5 members) = 90 years
Calculate the total age of the family of 6 members, including the baby: 16 years * 6 members = 96 years
Calculate the present age of the baby: 96 years - 90 years = 6 years

In this question, we are given that John's marks were wrongly entered as 83 instead of 63 and that the average marks for the whole class increased by half as a result.
We are asked to find the number of students in the class.
To solve this problem,
we can first assume the number of students in the class to be x.
We know that the total increase in marks for x students is x/2,
since the average increase is 1/2.
We can then use this information to find the value of x by setting up an equation using the given information.
Specifically, we can set up the equation x/2 = 83 - 63,
which gives us x = 40 students.
This means that there are 40 students in the class.

The formula for average is: average = sum of quantities / number of quantities.
In this case, the mean of 40 observations was given as 46. It was later found that one of the observations, 38, was wrongly recorded as 33.
To find the corrected mean value, we first need to calculate the sum of the 40 observations. Using the formula for average, we know that the sum of the 40 observations is 46 x 40 = 1840.
Next, we need to find the correct sum by adding the difference between the correct observation value of 38 and the wrongly recorded value of 33. The correct sum is 1840 + (38 - 33) = 1845.
Finally, we can use the formula for average to find the corrected mean value by dividing the correct sum of 1845 by the number of observations, which is 40.
This gives us a corrected mean value of 1845 / 40 = 46.125.

Output:

Let the total number of workers be x.then

6000x=(12000×150)+3750(x-150)

2250x=12375000

x=550

Let the number of boys in the class be x and the number of girls be y,

Then, 71x+73y=71.8(x+y)

Percentage of boys=[x/(x+y) ×100]%

=[3/2y/(3/2)y+y×100]%

=( 3y/2×2/5y×100) %

Percentage of girls =(100%-60%)

Output:

Required average=(1+2+3+...+100)/100

=1/100×100×101/2

=50.5

Output:

Sum of odd numbers =1+3+5+7+....+99

=25[2+(50-1)×2]

=2500

Required average=2500/50=50

½n[2a+(n-1)d]

Output:

Third number=(sum of 5 numbers)-(sum of 4 numbers)

Output:

• total weight of 50 students=(45×50)kg=2250kg

Average weight of 49 students=[45kg-100g]

=44.9kg

Total weight of 49 students=(44.9×49)kg=2200.1kg.

Weight of the student who left the class=(2250-2200.1)kg=49.9kg.

Output:

Let the average expenditure of all six be x

Then ,32×5+(x+80)=6x

240+x=6x

5x=240

Total money spent=6x=(6×48)=rs.288

Output:

Total cost of 5 tables=1227×5=6135

Total cost of 13 chairs=8280-6135=2145

Output:

Let the ages of the man and his son be 11x and 5x years respectively.

Given

Let the third number be x.Then second number be 3x

solution

Given:

Let the average after 17 th innings =x.

Then the average after 16 th innings=(x-3)

Given:

Sum of two numbers=2xy.

Solution

Total age increased=(2×11)months= 22 months=1 year 10 months

Solution

Marks in the seventh paper=(42×7×+35×7-40×13)

(294+245-520)=539-520=19

Solution

Total age 4 members,10 years ago=(24×4)=96

Total age of 4 members now=(98+10×4)=136

Total age of 6 members now=(24×6)=144

Sum of the ages of 2 children=(144-136)=8 years

Given:

There are five prime numbers between 30 and 50

Required average=[31+37+41+43+47]/5

Solution

Average body weight =(54+64+75+67+45+91)/6