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Averages can be defined as the central value in a given set of data. They can be calculated by division of the sum of all values which are given in a set by the total number of values. It can also be said that the average value is equivalent to the middle value of a data set. The data can be of anything like runs, money, age, numbers etc.
The formula for Average = [Sum of observations or Data in a set / Number of observations or Data present in that set]
The list of exams in which questions from average are asked includes RRB NTPC, SSC, IBPS PO, CET, SBI PO, IBPS Clerk, SBI Clerk, RBI Grade B, SEBI Grade A etc.
Average Questions PDF:
Average PDF Set 1
Average Level 1
An average defines the sum of the observations of data given divided by the total number of observations. The concept of averages can be used to solve questions related to data interpretation, problems related to ages, problems related to replacement, problems related to time & speed, numbers as well as miscellaneous problems. The questions based on averages are 4-5 in number and appear in different competitive examinations.
Properties of Averages:
Some of the important properties of averages include the following:
1. The Average of a group always is present between the minimum and the maximum value of a group.
2. If there are odd numbers in an Arithmetic progression then the average will always be the central value. For example, let’s say there are 9 terms in an A.P then the average will be the 5th term of the progression)
3. If all numbers of a group increase or decrease by some value than the average will also increase or decrease by the same.
4. If all the numbers of the group are multiplied or divided by some constant number then the average will also be multiplied or divided by the same number.
1. The Average of first n natural number = (n + 1) / 2
2. The Average of first n even number = (n + 1)
3. The Average of first n odd number = n
4. The Average of consecutive number = (1st number + Last number) / 2
5. The Average of 1 to n odd numbers = (Last odd number + 1) / 2
6. The Average of 1 to n even numbers = (Last even number + 2) / 2
7. The Average of squares of first n natural numbers = [ (n + 1) (2n + 1) ] / 6
8. The Average of the cubes of first n natural number = [n (n + 1) ^2] / 4
9. The Average of n multiples of any number = [Number X (n + 1) ] / 2
Problems based on Averages:
Vibhor obtained 66, 74, 55, 92, 79 (out of 100) in English, Maths, Physics, Chemistry and Biology. what's his average marks?
So, let's make equation from the given information :
|Average =||(||66 + 74 + 55 + 92 + 79||)||=||366|
Average = 73.2%
Hence, option C is correct.
In Ashish’s opinion, his weight is greater than 55 kg but less than 62 kg. His father does not agree with Ashish and he thinks that Ashish’s weight is greater than 50 kg but less than 60 kg. His sister’s view is that his weight can't be greater than 58 kg. If all of them are correct in their estimation, what is the average of different probable weights of Ashish ?
A. 56.5 kg
B. 68 kg
C. 69 kg
D. Data inadequate
Let Ashish weight X kg.
According to Ashish , 55 < X < 62
According to Ashish's father, 50 < X < 60
According to Ashish's sister, X < 58
The values satisfying all the above conditions are 56 & 57.
|Therefore, Required average =||(||56 + 57||)||=||(||113||)|
So, the Ashish's weight is 56.5 kg.
The average height of the first six students is 170 cm, the average height of the last eight students is 175 cm. The average height of the total 16 students is 180 cm. Find the average height of the rest two students.
A. 210 cm
B. 250 cm
C. 240 cm
D. 230 cm
E. 260 cm
Sum of the heights of the first six students = 170 × 6 = 1020 cm
Sum of the heights of the last eight students = 175 × 8 = 1400 cm
Sum of the heights of the total 16 students = 180 × 16 = 2880
Sum of the height of the left 2 students = 2880 – 1020 – 1400 = 460
|Average height of the left 2 students =||460||= 230 cm|
Hence, option D is correct.
The average weight of five friends P, Q, R, S, and T is (x + 6) kg while the average weight of R and T is (x – 6) kg. If the weight of another person U is also added, then average weight of all of them is reduced by 5 kg. Find the value of ‘x’ if average weight of P, Q, S and U is 94.5 kg.
Total weight of friends P, Q, R, S and T = (x + 6) x 5 = 5 (x + 6) kg
So, total weight of P, Q and S = 5 (x + 6) – 2 (x – 6) = (3x + 42) = 3 (x + 14)
Weight of U = (x + 6 – 5) × 6 – 5 (x + 6) = 6 (x + 1) – 5 (x + 6) = (x – 24) kg
According to the question,
[3 (x + 14) + (x – 24) ] = 94.5 × 4
4x + 18 = 378
4x = 360; x = 90
Hence, option D is correct.
In a class with a certain number of students if one new student weighing 50 kg is added, then average weight of class is increased by 1 kg. If one more student weighing 50 kg is added, then the average weight of the class increases by 1.5 kg over the original average. What is the original average weight (in kg) of the class?
E. None of these
Let x be the number of students in the class and y be the average weight of the classNow according to question,
|xy + 50||= y + 1|
|x + 1|
x + y = 49 .......... (i) Again,
|xy + 50 + 50||= y + 1.5|
|x + 2|
1.5x + 2y = 97 .......... (ii) From equation (i) and (ii), we get y = 47
Hence, option (D) is correct.