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Average Tricks & Tips

Average Short-Cut Tricks & Tips : Average Short-Cut Tricks & Tips Question Pdf for Banking, SSC, RRB, FCI, Railway, UPSC, State PCS, Insurance & other Competitive exams. Average Short-Cut Tricks & Tips shortcut Tricks Pdf, Average Short-Cut Tricks & Tips MCQ, Average Short-Cut Tricks & Tips Objective Question & Answer Pdf. “Average Short-Cut Tricks & Tips Questions PDF” In this post we are providing you the Average Short-Cut Tricks & Tips pdf with detailed solution & Short Tricks. So that you can easily get the logic of question. This Average Short-Cut Tricks & Tips Pdf we are Providing is free to download. ” Most Important Average Short-Cut Tricks & Tips Question PDF with Answers“

Average Short-Cut Tricks & Tips Plays a vital role in Exam. In every exam you will get at least 5-10 questions from this topic. So candidates must focus on this topic and download this Average Short-Cut Tricks & Tips pdf to get important questions with best solution regarding Average Short-Cut Tricks & Tips. We have put all Previous Year Questions of Average Short-Cut Tricks & Tips that are Asked in various Govt & Private Exam.

What are Averages?

Averages can be defined as the central value in a set of data. Average can be calculated simply by dividing the sum of all values in a set by the total number of values. In other words, an average value represent the middle value of a data set. The data set can be of anything like age, monety, runs, etc.

Formula

We obtain the average of a number using the formula that is:
Average = (Sum of observations / Number of observations)

Examples:

Question 1

Find the average of numbers? 23, 25, 65, 12, 50
Solution:
According to the formula:
Average = (23+25+65+12+50)/5
= 175/5 = 35
Hence 35 is the average of these numbers
But this is the basic Average formula which helps us in our School days a lot to clear the exams but now we are in Competitive exams so we want to quick and precise. For these we need our basic and also we have to learn something new.

TYPE 1:

AVERAGE OF NUMBERS HAVING DIFF IN ARITHMETIC:
There are two cases: 

  • When a number of terms is odd then the average will be middle term. 
  • When a number of terms is even then the average will be the average of two middle terms. 

Examples:

Question 2
Find the average of  12, 13, 14, 15, 16

Solution:

  • Here the difference is in AP and number of terms is odd. So the average will be the middle number. 
  • There is another way to calculate the average for numbers having diff in Arithmetic 
  • Average can be calculated as ½ (first term + last term) 

Examples:

Question 3
Find the average of : 12, 13, 14, 15, 16, 17

Solution:

Here the number of terms is even and diff is in arithmetic So the average will be the average of middle two terms.
That is, 14 and 15
Average = (14 + 15)/2 = 14.5
ALTERNATIVE:
Average = ½ (first term + last term)
= ½ (14 + 15) = 14.5

TYPE 2:

If the average of some numbers is M and another number x is added, subtracted, multiplied or divided from each number then then average will be added, subtracted, multiplied or divided respectively.

  • If M is average and x is added to each number, then new average is “M + x” 
  • If M is average and x is multiplied to each number, then new average is “M * x” 
  • If M is average and x is subtracted to each number, then new average is “M – x” 
  • If M is average and x is divided to each number, then new average is “M / x” 

TYPE 3

If the average of n numbers is “M” and one number is included, then two cases arise:

If the included number ”X” is greater than average then average is subtracted from the included number and results will be divided by “n+1” 

Extra value = X – MDistributed value = (X – M) / (n+1)
New average = M + Distributed valueIf the included number ”X” is smaller than average then the included number is subtracted from the average and resultant is divided by the “n+1”

Extra value = M – XAdjusted value = (M – X) / (n+1)
New average = M – Adjusted value

TYPE 4

If the average of n numbers is “M” and one number is excluded, then two cases arise: 

If the excluded number ”X” is greater than average then average is subtracted from the excluded number and resultant will be divided by “n-1” 
Extra value = X – MAdjusted value = (X – M) / (n-1)
New average = M – Adjusted valueIf the excluded number ”X” is smaller than average then the excluded number is subtracted from the average and resultant is divided by the “n-1”

Extra value = M – XAdjusted value = (M – X) / (n-1)
New average = M + Adjusted value

TYPE 5:

If the average of n numbers is “M” and one number is excluded and another number is included then two cases arise: 


If the excluded number ”X” is greater than included number “Y” then included number is subtracted from the excluded number and resultant will be divided by “n”
Extra value = X – YAdjusted value = (X – Y) / (n)
New average = M – Adjusted valueIf the excluded number ”X” is smaller than included number then the excluded number is subtracted from the included number and resultant is divided by the “n” 

Extra value = Y – XAdjusted value = (Y – X) / (n)
New average = M + Adjusted value

EXAMPLES:

Question 4
The average age of 15 students in class is 42 and the age of teacher which is 58 is also included with them. Find the new average age of teacher and students. 

Solution:

So the included age of teacher is greater than the average age of students.
Extra value = 58 – 42 = 16
Adjusted value = 16/16 = 1
New average = 42 + 1 = 43
Hence required new average is 43.

Question 5

An average marks o a student in 7 subjects is 94 and the mark of English which is 70 is excluded. Find the new average marks of a student. 

Solutions:

So the excluded mark is smaller than the average marks of a student.
Extra value = 94 – 70 = 24
Adjusted value = 24/6 = 4
New average = 94 – 4 = 90
Hence the new average marks of 6 subjects of student are 90.

Question 6

The average weight of 10 persons is 65 kg. But one person weighing 75 kg left and another person weighing 65 kg joined the group. Find the new average weight. 

Solutions:

So the weight of leaving person is greater than the weight of joining person. So the average will also decrease.

Extra value = 75 – 65 = 10
Adjusted value = 10/10 = 1
New average = 65 – 1 = 64
Hence the new required average weight of 10 persons is 64.

Question 7

The average age of 10 students is 25. But the age of teacher is also included and the new average age is increased to 29. Find the age of teacher. 

Solutions:

We can also find any missing value through this process.
Acc. to the ques.
New average age is greater than the previous average age. This implies that the age of teacher must be greater than the average. We have to find how much.
Extra value = 29 – 25 = 4
Adjusted value = 4 * 11 = 44
Age of teacher = 25 + 44 = 69
Hence the required age of teacher is 69.

Some More Examples

Example 1

The average age of the family of 6 members is 20 years. If the age of the youngest member is 6 years. Find the average age of the family at the birth of youngest member?

Solution:

During the birth of the youngest child, there will be only 5 members of the family. According to Type 2 if a particular number is subtracted from every number then the number is subtracted from the average.
So the average of 6 members = 20 – 6 = 14
But there will be only 5 members during birth.
So, Required average = Total age/ no. of members
=14×65=845 = 16.8
Hence the average age is 16.8 years

Example 2. 

The average score of batsmen in 10 innings is 21.5 runs.
How many runs a batsmen must score in his next innings So that his average score will jump to 25 runs?
Solution:
This  ques. Is based on TYPE 3 where one innings is to be included and with that average will increase.
Increased average= 25 – 21.5 = 3.5
This increase will be due to more runs scored then the average score.
Hence the 11th inning score = 21.5 + 3.5×11 = 60

Try out some Examples

  1. A batsman in his 16th inning scores 80 runs and thereby increases the average by 4 runs. Find the average after the 16th inning. (HINT: Firstly find average of 16 innings)
  2. The average age of 30 boys in the class is equal to 15 years. If the age of teacher is also included the average becomes 16 years. Find the age of teacher.
  3. The average of 45 observations is 35. Later it was found that one observation which is 56 is wrongly taken as 24. Find the actual average of 45 observations.
  • Till now we have considered only one person either included or excluded or both. But there can be a possibility of more than one person.

Example 3.

The average age of 8 persons is increased by 3 years when 2 men whose age is 40 years and 50 years are substituted by two other men. Find the average age of two new men.
Solution:
In this case, there are two persons including and excluding. So we have to consider both simultaneously.
Total age of excluding men= 40 + 50 = 90 years
Increased age = 3 × 8 = 24 years
This increased age is because the of including men is more than the exclexcluded.
Average age of new men =(90+24)2 = 57
In this case we cannot calculate individual value of each included men.

Some of the problems of average can directly be solved through formulas without using any trick

  • Average of first n natural numbers = (n+1)2
  • Average of first n even numbers = (n+1)
  • Average of first n odd numbers = n
  • Average of consecutive numbers = (FIRST+LAST NUMBER)2
  • Average of the square of first n natural no. = [(n+1)(2n+1)]6
  • Average of a cube of first n natural no. = [n(n+1) 2]4
  • If the average of m numbers is a and the average of n numbers (m>n) is b then the average of remaining numbers will be: (ma-nb)(m-n)

Try out some Examples:

  • Find the average of first 100 natural numbers.
  • Find the average of first 100 even numbers.
  • Find the average of first 100 odd numbers
  • Find the average of all natural numbers between 101 and 200.
  • Find the average of square of first 20 natural numbers
  • Find the average of cube of first 20 natural numbers 
  • The average of 25 numbers is 45 and the average of first 15 numbers is 35. Find the average of remaining numbers.

Some More Tricky Questions:

Example 1.

The average score of 45 exams is 470. Out of which average score of first 23 exams is 490 and the average score of last 23 exams is 445. Find the score of the 23rd exam?
Solution:
Here 23rd exam is included in both first and last averages. So firstly we will find the change in average of both from overall average and add both the differences
Then we will multiply the added difference with 23 as we have to find the 23rd exam score.
And finally, we will add the result to the overall average. That will provide us with the required value.
Exam Average
Exam Average Total 45470 First23490+20 Last23445−25
-5 ×23 = 115
23rd exam score = 470-115 = 355

Example 2. 

In a class average age of some students is 40 years. After some time 12 students with an average age of 32 years joins the class due to which average age of class will decrease by 4 years. Then find the number of students in the starting.
Solution:
Students average
X40X+1236
Average decreases due to the inclusion of these 12 students. And this is because there average age is less then the average age of class.
Adjusted value = 4×2 = x3
New average = average of included students + adjusted value
Hence, 36 = 32 + x3
x3 = 4
x = 12
Hence the students in the starting is 12.

Q. 1) The average weight of 4 men is increased by 3 kg when one of them who weighs 120 kg is replaced by another man. What is the weight of the new man?

Weight of new person = Weight of removed person + No. of persons * Increase in Average = 120+3*4
Weight of new person = 132 kg

Q. 2) The average marks obtained by 120 candidates in a certain examination is 35. If the average marks of passing candidates is 39 and that of the failed candidates is 15, what is the number of candidates who passed the exam?

Such questions are best dealt with allegation formula:
Mean value = 35, Dearer value = 39 and Cheaper = 15
Failed students : Passed students = (Dearer – Mean) : (Mean – Cheaper) = 39 – 35 : 35 – 15 = 4 : 20
No. of students who passed = (5/6) * 120 = 100

Q. 3)
Correct Average = Incorrect average + (Correct value – Incorrect value)/Total observations
Correct Average = 35 + (85 – 45)/20 = 37
Answer: (D)
 
Q. 4)
Take x = 1
Then A = (1 + 1)/2 = 1
Average of x^3 + 1/x^3 = (1 + 1)/2 = 1
Now put A = 1 in all the four options and check which one is giving ‘1’ as the output
Answer: (D)
 
Q. 5)
 
 

For consecutive numbers:
If the number of terms is odd (e.g. A, B, C), then their average is the middle term(i.e. B).
If the number of terms is even(e.g. A, B, C, D) then their average is (B+C)/2
So here, the average of eight consecutive even numbers = (5th term + 4th term)/2
Given, (5th term + 4th term)/2 = 93
5th term + 4th term = 186
We know, 5th term – 4th term = 2
Adding the equations, we get 5th term = 94
Hence 8th term = 5th term + 2 + 2 + 2 = 100
Answer: (C)

Q. 6)


Average = (8*3 + 20*2 + 26*m + 29*1)/(3 + 2 + m + 1)
17 = (93 + 26m)/(6 + m)
102 + 17m = 93 + 26m
m = 1
Answer: (A)

Q. 7)
Given, (3a + 4b)/2 > 50
3a + 4b > 100
5a > 100  [Since a = 2b]
a > 20
Hence the smallest value of a is 21
Answer: (D)
 
Q. 8)
Average of six numbers a,b,c,d,e,f = (Average of a,b + Average of c,d + Average of e,f)/3
3.95*3 = 3.4 + 3.85 + x
x = 4.6
Answer: (C)
 

Q. 9) The average of 11 numbers is 50. The average of first six numbers is 49, and that of last six is 52. Find the sixth result.
This is a common CGL question and a short-cut to solve it is:
Sixth result = 50 + 6{(52 – 50) + (49 – 50)} = 50 + 6(2 – 1) = 56
Answer: 56


Q. 10)

Answer: (B)


Q.11)

Shortcut techniques in Averages:

Questions based on averages can be easily solved using shortcuts. By using shortcuts, any question can be solved quickly and efficiently which can save a lot of time. So, some shortcuts to solve average questions are explained below along with illustrations.

  • Shortcut to find average or change in average from a set of values

Example 1:

The average of a batsman in 16 innings is 36. In the next innings, he is scoring 70 runs. What will be his new average?

  1. a) 44
  2. b) 38
  3. c) 40
  4. d) 48

Solution:

Conventionally solving:

New average = (old sum+ new score)/(total number of innings) = ((16 ×36)+70)/((16+1)) = 38

Shortcut technique:

Step 1) Take the difference between the new score and the old average = 70 – 36= 34

Step 2) This is 34 extra runs which is spread over 17 innings. So, the innings average will increase by 34/17 = 2

Step 3) Hence, the average increases by => 36+2 = 38.

Here are a few more average questions and their solutions using the same technique.

Example 2:

The average marks of 19 children in a particular school is 50. When a new student with marks 75 joins the class, what will be the new average of the class?

Solution:

Step 1) Take the difference between the old average and the new marks = 75-50=25

Step 2) This score of 25 is distributed over 20 students => 25/20 = 1.25

Step 3) Hence, the average increases by 1.25=> 50+1.25 = 51.25.

Here is another question where the average dips.

Example 3:

The average age of Mr. Mark’s 3 children is 8 years. A new baby is born. Find the average age of all his children?

Solution:

The new age will be 0 years. The difference between the old average and the new age = 0-8= -8

This age of 8 years is spread over 4 children => (-8/4= -2) Hence, the average reduces to 8-2= 6 years.

  • Shortcut to find new value when average is given

Now here is a technique which will help to compute the new value when the average is given. Take this question for example:

Example 1:

The average age of 29 students is 18. If the age of the teacher is also included the average age of the class becomes 18.2. Find the age of the teacher?

  1. a) 28
  2. b) 32
  3. c) 22
  4. d) 24

Solution:

Conventionally solving,

Let the average age of the teacher = x

(29 × 18 + x × 1)/30

Solving for x, we get x = 24.

Shortcut Technique

Using the shortcut, based on the same method used previously:

Step 1: Calculate the change in average = 18.2 – 18 = 0.2.

This change in 0.2 is reflected over a sample size of 30.

Shortcut Techniques In Averages

The new age is increased by 30 × 0.2 = 6 years above the average i.e. 18 + 6 = 24; which is the age of the teacher.

Thinkquest- The average age of 26 students in an MBA school is 30. One student among these quits the school in between. Can you find the age of that student if the new average is 29.8?

Illustrations On Averages

Question 1:

Average goals scored by 15 selected players in EPL is 16.Maximum goals scored by a player is 20 and minimum is 12.Goals scored by players is between 12 and 20. What can be maximum number of players who scored at least 18 goals ?

  1. a) 10
  2. b) 5
  3. c) 9
  4. d) 6
  5. e) None of these

Solution: Option (c)

To maximize the number of players who scored 18 and above number of goals, one should assume that only one person has scored 20. To counter him, there will be one person who will score 12 goals.

i.e. 15 – 2 = 13 players left.

Now to maximize the 18 and above goals for every two players who are scoring 18, there will be one player scoring 12. This is done, to arrive at the average of 16. We will have 8 players with a score of 18 and 4 players with a score of 12.The last player will have a score of 16 Thus, the maximum number of people with 18 and more goals = 9.

Question 2:

The average weight of a group of 8 girls is 50 kg. If 2 girls R and S replace P and Q, the new average weight becomes 48 kg.The weight of P= Weight of Q and the weight of R= Weight of S.Another girl T, is included in the group and the new average weight becomes 48 kg. Weight of T= Weight of R. Find the weight of P?

  1. a) 48 kgs
  2. b) 52 kgs
  3. c) 46 kgs
  4. d) 56 kgs

Solution: Option (d)

8 x 50 +R+S-P-Q= 48×8 R+S-P-Q=-16

P+Q-R-S= 16 R=S and P=Q

P-R=8

One more person is included and the weight = 48 kg

Let the weight be a = (48 × 8 + a)9/9 = 48

A = 48 kg= weight of R

=> Weight of P= 48+8= 56 kg.

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