Average - Atreya Roy


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    Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.

    Averages is a basic chapter that is covered in the Arithmetic portion of the CAT QA syllabus. This is an easy topic from which a limited variety of questions regarding costs, age and numbers come. Average means the central number which generally has a number of elements lesser and greater than it. So it can be said as the mid point of the data set given.

    Average =  The sum value of a number of terms / quantities in a sample.

    The basic formula for average is :

    Average = Sum of all elements present in the sample / Number of total samples.

    Average = c1+c2+c3+c4…..cn/ n

    Question : What is the average of 10,20,30,40

    Answer : Average = Sum/Number of elements = 100/4 = 25

    If the elements are in an arithmetic progression, then the average of the series = First term + Last Term /2

    Question: What is the average of the following elements : 1,2,3,4,5,6,7,8,9,10 ?

    Answer : Since it’s in an AP, Average = First term + Last Term /2 = (1+10) /2 = 5.5

    Question:

    The average of 10 peaches is 0.3 Kg. If the weight of the lightest and heaviest peaches are removed, the weight will be 0.27. If the weight of the lightest peach is 0.12 kg, what is the weight of the heaviest peach ?

    Solution:

    Total Weight of peaches= .3*10 = 3kg

    Weight after 2 peaches removed = .27*8 = 2.16

    Weight of Lightest + heaviest = 3-2.16 = .84 kg

    Weight of lightest = .12 kg

    Weight f heaviest = .84-.12 = .72 Kg

    Concept of Weighted Average :

    As per the concept of weighted average, we are given a data where the quantity/amount of a second variable is given to us, and we need to find the average of that data .

    Suppose let us take an example that In class V of ABC school, we have two sections. There was a maths test conducted in the class where both the sections participated. Suppose Section A has 20 students and the average marks obtained by the students is 30, and in section B, which has 30 students , the average marks is 40. So what is the total average of the marks obtained in the test in the whole class.

    For solving this type of sums we will, first multiply the quantities of the same section, then add up the quantities we got and finally divide the whole thing by the number of students who took the test. As per the basic formula, average = Total marks / Total students . Hence for this sum we will do : 20*30 + 30+40 / 20+30 = (600+1200) / 50 = 1800/50 = 36

    This can also be done using the allegation method, which we will discuss later on.

    Few points to remember :

    1. If all terms of the data are multiplied or divided by a certain number n, then the average of the series will get multiplied or divided by the number n.
    2. If all the terms of the data series are increased/decreased by a certain number p, then we have 2 cases:
    1. In case of increase  : New Average= Old Average + p
    2. In case of decrease : New Average = Old Average – p
    1. If the series contain 0 as the middle term and there are equal number of terms left and right to zero, in an arithmetic series : Then the average of the terms = 0
    2. If There are even terms in the data series, and Half of them are multiplied by a definite number and the other half are divided by the same number then the average of the series will remain the same.

    Average - Practice questions

    If a person with age 45 joins a group of 5 persons with an average age of 39 years. What will be the new average age of the group?

    Solution :

    Total age will be 45 + 5×39 = 240. And there will be 6 persons now.

     So the average will be 240/6 = 40.

    Two students with marks 50 and 54 leave class VIII A and move to class VIII B. As a result the average marks of the class VIII A fall from 48 to 46. How many students were there initially in the class VIII A?

    Solution :

    The average of all the students of class VIII A is 46, excluding these two students.  They have 4 and 8 marks more than 46. So with the addition of these two students, 12 marks are adding more, and hence the average is increasing 2. There should be 6 students in that class including these two. This is the initial number of students.

    The average of 10 consecutive numbers starting from 21 is ?

    Solution:

    The average is simply the middle number, which is the average of 5th & 6th no. i.e, 25 & 26 i.e. 25.5.

    There are 30 consecutive numbers. What is the difference between the averages of first and last 10 numbers?

    Solution :

    The average of first 10 numbers is the average of 5th & 6th no. Whereas the average of last 10 numbers is the average of 25th & 26th no. Since all are consecutive numbers, 25th number is 20 more than the 5th number. We can say that the average of last 10 numbers is 20 more than the average of first 10 nos. So, the required answer is 20.

    If one student from group A is shifted to group B, which of the following is necessarily true? 

    1) The average weight of both groups increases

    2) The average weight of both groups decreases.

    3) The average weight of class remains the same.

    4) The average weight of group A decreases and that of group B increases.

    5) None of these

    Solution:

    Options (1) & (2) are not possible. Average of both cannot increase or decrease. Option (4) can be eliminated because we are not sure, whether the average of A increases & B decreases or A decreases & B increases or both remains unchanged. It depends on the weight of the student, who shifted from A to B.Option (3) is always true because even the student shifts from one group to other. The average weight of the whole class does not change.

    Mean, Median and Mode

    Mean, Median and Mode are part of Statistics which is not included in the syllabus for CAT, but the ideas revolving around these help give you an extra edge to solve problems in the platform. We will learn to solve typical problems that revolve around these concepts.

    Mean : It simply means the average of the set of elements we are considering. So if we look at it as a formula, Mean = Total Sum of Elements / Total Elements

    Median: If the elements we have are arranged in ascending order (Smallest to Largest), then the term appearing in the middle of the set is called the median. If we have “n” elements in arranged order as mentioned, we have the median as :

    1. (n/2) th + ((n/2)+1 )th elements : If “n” is even
    2. (n+1)/2 th element : If “n” is odd

    So if we have 6 elements : Medians are 3rd and 4th elements

    If we have 5 elements : Median will be the 3rd element

    Mode : Mode simply means the element/term that is present in the highest number (in measure of occurrence) in our series.

     If we have a set : 1,1,2,3,4,4,4 : we can see 1 appears twice, while 4 appears thrice. Hence Mode = 4 (the element appearing the most)

    These concepts are very useful to do a variety of sums.

    Example :

    In a set of 5 numbers, the mean is calculated to be 3.6. If Mode of the set is 4 and we have 4 different natural number in the set, what can be our highest number in the set ?

    Solution :

    Mean = 3.6

    Number of elements = 5. So sum of the elements = 18

    We have a mode, 4 And it is also said there are 4 different natural numbers in the series. Hence 4 can appear at most twice, and hence will appear twice to satisfy both the conditions. 2*4 =8

    Sum-8 = 18-8=10

    Sum of rest 3 numbers =10. Since natural numbers are asked, Min number =1, second min number =2

    Sum of these= 1+2=3. So, to get the highest number in the set, we will subtract the sum from the total remaining sum. Hence Our required number = 10-3 = 7

    Hence Answer = 7

    The median of 6 positive integers is 20. The only mode is less than 20. What is the largest number in the series?

    Solution :

    Given the average of the series is 20. Let the series be  a, b, c, d, e, f.

    Since, the mode is less than 20 and it is the only mode, c+d 40 but c=d=20 is not possible because only mode is less than 20

    So, c = 19 d = 21

    For maximum f; a=1   b = 1 (mode)

    E = 22 [ next value to median ]

    So the series is 1, 1, 19, 21, 22, x.

    Sum = 120

    Therefore, x = 120- (1+1+19+21+22)

    x = 120-(64)

    x= 56

    Hence 56 is the answer.

    Note : You might find sums where there may be 2 or more Modes. So, we will have 2 or more elements occurring the same maximum number of times amongst others.

     

     


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