Problems on average play a major role in quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the topic problems on average. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve problems in this topic. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to learn few basic operations in this topic and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve problems related to average. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

**Here, we are going to have some problems on average . You can check your answer online and see step by step solution.**

1. Find the average of first 20 natural numbers which are divisible by 7.

Clearly, the first natural number which is divisible by 7 is 7. The next numbers which are divisible by 7 are 14, 21.....

Let us write the first twenty natural numbers which are divisible by 7. They are 7,14,21,28........ up to 20 terms.

Sum of all the above numbers

= 7+14+21+28.........up t0 20 term

Since all of the above numbers are divisible by 7, we can factor 7

= 7(1+2+3+4+.........+20)

Therefore, sum = 7(210)

Average = (sum of all 20 numbers)/20

Average = (7X210)/20 = 73.5

Hence average of first 20 natural numbers which are divisible by 7 is 73.5

Let us write the first twenty natural numbers which are divisible by 7. They are 7,14,21,28........ up to 20 terms.

Sum of all the above numbers

= 7+14+21+28.........up t0 20 term

Since all of the above numbers are divisible by 7, we can factor 7

= 7(1+2+3+4+.........+20)

Therefore, sum = 7(210)

Average = (sum of all 20 numbers)/20

Average = (7X210)/20 = 73.5

Hence average of first 20 natural numbers which are divisible by 7 is 73.5

2. The average of four consecutive even numbers is 27. Find the largest of these numbers.

If "x' be the first even number, then the four consecutive even numbers are x, x+2, x+4, x+6

Average of the four consecutive even numbers = 27

(x+x+2+x+4+x+6)/4 = 27

(4x+12) = 108

4x = 96 ===> x = 24

Hence the largest even number = x+6 = 30

Average of the four consecutive even numbers = 27

(x+x+2+x+4+x+6)/4 = 27

(4x+12) = 108

4x = 96 ===> x = 24

Hence the largest even number = x+6 = 30

3. There are two sections A and B of a class, consisting 36 and 44 students respectively. If the average weight of the section A is 40 kg and that of section B is 35 kg, find the average weight of the whole class.

For section A, average weight = 40 kg

That is, (sum of the weights of 36 students) /36 = 40

Sum of the weights of 36 students = 1440

For section B, average weight = 35 kg

That is, (sum of the weights of 44 students) /44 = 35

Sum of the weights of 44 students = 1540

Total weight of (for whole class = 44+36)80 students = 1440+1540 = 2980

Average weight of the whole class = 2980/80 = 37.25 kg

That is, (sum of the weights of 36 students) /36 = 40

Sum of the weights of 36 students = 1440

For section B, average weight = 35 kg

That is, (sum of the weights of 44 students) /44 = 35

Sum of the weights of 44 students = 1540

Total weight of (for whole class = 44+36)80 students = 1440+1540 = 2980

Average weight of the whole class = 2980/80 = 37.25 kg

4. In John's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doesn't agree with John and he thinks that John's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of John?

Let "x" be John's weight

According to John, we have 65 < x < 72

According to his brother, we have 60 < x < 70

According to his mother, we have x≤68

The values of "x" which satisfy all the above three conditions are 66, 67 and 68

Average of the above three values = (66+67+68)/3 = 201/3 = 67 kg

Hence average of different probable weights of John is 67 kg

According to John, we have 65 < x < 72

According to his brother, we have 60 < x < 70

According to his mother, we have x≤68

The values of "x" which satisfy all the above three conditions are 66, 67 and 68

Average of the above three values = (66+67+68)/3 = 201/3 = 67 kg

Hence average of different probable weights of John is 67 kg

5. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning.

Let "x" be the average after 17th inning

(Total run scored in 17 innings)/17 = x

Total run scored in 17 innings = 17x ----(1)

Average after 16th inning = x-3

(Total run scored in 16 innings)/16 = x-3

Total runs scored in 16 innings = 16(x-3)

Total runs scored in 17 innings = 16(x-3)+87 ----(2)

From (1) and (2), we get 16(x-3)+87 = 17x

Solving the above equation, we get x = 39

Hence the batsman's average after 17th inning is 39.

(Total run scored in 17 innings)/17 = x

Total run scored in 17 innings = 17x ----(1)

Average after 16th inning = x-3

(Total run scored in 16 innings)/16 = x-3

Total runs scored in 16 innings = 16(x-3)

Total runs scored in 17 innings = 16(x-3)+87 ----(2)

From (1) and (2), we get 16(x-3)+87 = 17x

Solving the above equation, we get x = 39

Hence the batsman's average after 17th inning is 39.

**6.The average salary of a group of unskilled workers is $10000 and that of a group of skilled workers is $15000. If the combined salary is $12000, then what is the percentage of skilled workers?**

Let x̄,x̄_{1} and x̄_{2} be the combined average, average salary of unskilled workers and skilled workers respectively.

Let n_{1} and n_{2} be the number of unskilled and skilled workers respectively.

Formula for combined mean is x̄ = [n_{1}x̄_{1}+n_{2}x̄_{2}]/[n_{1}+n_{2}]

Plugging the values of x̄,x̄_{1} and x̄_{2} in the above formula, we get

12000 = [10000n_{1}+15000n_{2}]/[n_{1}+n_{2}]

12000n_{1}+12000n_{2} = 10000n_{1}+15000n_{2}

2000_{1} = 3000n_{2} ---> n_{1}/n_{2} = 3/2 --->n_{1}:n_{2} = 3:2

No. of unskilled workers = 3x, no. of skilled workers = 2x and

total no. of workers = 3x+2x = 5x

Percentage of skilled workers = (2x/5x)X100 % = 40%

Let n

Formula for combined mean is x̄ = [n

Plugging the values of x̄,x̄

12000 = [10000n

12000n

2000

No. of unskilled workers = 3x, no. of skilled workers = 2x and

total no. of workers = 3x+2x = 5x

Percentage of skilled workers = (2x/5x)X100 % = 40%

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