Average

Concept

Average can be basically determined as the equal distribution of a quantity in the same set of things. The average is the arithmetic mean of a set of quantity. For example, 6 is the average of the numbers 1, 3, 4, 7, and 15, which add up to 25.
$\displaystyle \left(\frac{1+3+4+7+15}{5}\right) = \frac{30}{5} = 6$

Features of Average:

For better to understand following features of average, we need to know clearly the following features and formulae:

Average can also known as Arithmetic Mean.
Usually, the average is calculated for a set of numbers; Average = Sum of values/No. of values
The average must always stay above the lowest number and below the highest number.
If the average of some numbers is x and y is added with each number, then the new average will be (x + y).
If the average of some numbers is x and y is subtracted from each number, then the new average will be (x – y).
For a set of numbers, if the average of the set is x, now all the numbers are multiplied by y, then the new average is xy.
For a set of numbers, if the average of the set is x, now all the numbers are divided by y, then the new average is (x/y).

Q. The average of 10 numbers is 15. If one of the numbers is removed, the average becomes 14. What is the number that was removed?

A) 5

B) 10

C) 15

D) 25

The sum of the 10 numbers is 10 x 15 = 150. If one number is removed, the sum of the remaining 9 numbers is 9 x 14 = 126. Therefore, the number that was removed is 150 - 126 = 24.

Discuss

Q. The average of three numbers is 12. If two of the numbers are 8 and 10, what is the third number?

A) 12

B) 14

C) 16

he sum of the three numbers is 3 x 12 = 36. If two of the numbers are 8 and 10, their sum is 8 + 10 = 18. Therefore, the third number is 36 - 18 = 18.

D) 18

Discuss

Q. The average of five consecutive odd numbers is 17. What is the largest of these numbers?

A) 15

B) 17

C) 19

D) 21

Let the five consecutive odd numbers be $\displaystyle x, x + 2, x + 4, x + 6, \text{ and } x + 8$ . Their average is $\displaystyle \frac{x + x + 2 + x + 4 + x + 6 + x + 8}{5} = \frac{5x + 20}{5} = x + 4 = 17$ . Solving for x, we get $\displaystyle x = 13$ . Therefore, the largest of these numbers is $\displaystyle x + 8 = 13 + 8 = 21$ .

Discuss

Q. The average of four tests scores is 80. If three of the test scores are 75, 83, and 77, what is the fourth test score?

A) 85

The sum of the four test scores is 4 x 80 = 320. If three of the test scores are 75, 83, and 77, their sum is 75 + 83 + 77 = 235. Therefore, the fourth test score is 320 - 235 = 85.

B) 86

C) 87

D) 88

Discuss

Q. The average weight of a class of students is 50 kg. If a new student joins the class, the average weight increases by 1 kg. What is the weight of the new student?

A) 51 kg

B) 60 kg

C) 61 kg

Let ( n ) be the number of students in the class before the new student joins, and ( w ) be their total weight. Then, their average weight is $\displaystyle \frac{w}{n} = 50 \text{ kg}$ . If a new student with weight ( x ) joins the class, the new average weight becomes $\displaystyle \frac{w + x}{n +1}=50+1=51 \text{ kg}$ . Solving for ( x ), we get $\displaystyle x = (n +1)(51)-(w)=51n+51-w$ . Substituting $\displaystyle w = n(50)$ , we get $\displaystyle x =51n+51-n(50)=n+51$ . Since ( n ) must be a positive integer, the smallest possible value of ( n ) is $\displaystyle 1$ , which gives $\displaystyle x =1+51=52 \text{ kg}$ . However, this value does not satisfy the condition that the average weight increases by $\displaystyle 1 \text{ kg}$ , since in this case there would be no change in the average weight. Therefore, we need to find a larger value of ( n ) that satisfies both conditions. The next possible value of ( n ) is $\displaystyle 2$ , which gives $\displaystyle x =2+51=53 \text{ kg}$ . This also does not satisfy the condition that the average weight increases by $\displaystyle 1 \text{ kg}$ , since in this case the average weight would increase by $\displaystyle 0.5 \text{ kg}$ . Continuing this process, we find that ( n =10 ) satisfies both conditions, giving $\displaystyle x =10+51=61 \text{ kg}$ . This is also the largest possible value of ( n ) that satisfies both conditions, since for any larger value of ( n ), the value of ( x ) would exceed $\displaystyle 70 \text{ kg}$ , which is not an option. Therefore, the weight of the new student is 61 kg.

D) 70 kg

Discuss

Q. The average of 5 numbers is 12. If one of the numbers is 8, what is the average of the remaining 4 numbers?

A) 10

B) 11

C) 12

D) 13

The sum of the 5 numbers is $\displaystyle 5 \times 12 = 60$ . If one of the numbers is 8, the sum of the remaining 4 numbers is $\displaystyle 60 - 8 = 52$ . Therefore, the average of the remaining 4 numbers is $\displaystyle \frac{52}{4} = 13$ .

Discuss

Q. The average age of a family of four members is 25 years. If a new baby is born, what will be the average age of the family?

A) 20

The sum of the ages of the four family members is $\displaystyle 4 \times 25 = 100$ years. If a new baby is born, their age is zero. Therefore, the sum of the ages of all five family members is now $\displaystyle 100 + 0 = 100$ years. The new average age of the family is $\displaystyle \frac{100}{5} = 20$ years.

B) 22

C) 23

D) 24

Discuss

Q. The average height of a class of students is 150 cm. If a new student joins the class and increases the average height to 151 cm, what is the height of the new student?

A) 151 cm

B) 152 cm

Let ( n ) be the number of students in the class before the new student joins, and ( h ) be their total height. Then, their average height is $\displaystyle \frac{h}{n} = 150 \text{ cm}$ . If a new student with height ( x ) joins the class, the new average height becomes $\displaystyle \frac{h + x}{n +1}=151 \text{ cm}$ . Solving for ( x ), we get $\displaystyle x = (n +1)(151)-(h)=151n+151-h$ . Substituting ( h = n(150) ), we get $\displaystyle x =151n+151-n(150)=n+1$ . Since ( n ) must be a positive integer, we have ( n \geq 1 ). Therefore, the height of the new student must be at least $\displaystyle 152 \text{ cm}$ .

C) 153 cm

D) 154 cm

Discuss

Q. The average weight of a group of people was measured twice due to an error in measurement. The first time it was measured as 70 kg, and the second time it was measured as 75 kg. What was the actual average weight?

A) 72 kg

B) 73 kg

Let ( n ) be the number of people in the group, and ( w ) be their total weight. Then, their actual average weight is $\displaystyle \frac{w}{n}$ . If their average weight was measured as $\displaystyle 70 \text{ kg}$ the first time and $\displaystyle 75 \text{ kg}$ the second time, then their total weight was measured as $\displaystyle 70n \text{ kg}$ the first time and $\displaystyle 75n \text{ kg}$ the second time. Therefore, their actual total weight must be somewhere between $\displaystyle 70n \text{ kg}$ and $\displaystyle 75n \text{ kg}$ inclusive, since these are the only two measurements that were taken. We can find an estimate for their actual total weight by taking their average measurement as $\displaystyle 72.5n \text{ kg} = \frac{70n \text{ kg} + 75n \text{ kg}}{2}$ . Therefore, their actual average weight must be $\displaystyle \frac{72.5 \text{ kg}}{n}$ . Since this value lies between $\displaystyle \frac{72 \text{ kg}}{n}$ and $\displaystyle \frac{73 \text{ kg}}{n}$ , we can conclude that their actual average weight must be $\displaystyle 73 \text{ kg}$ .

C) 74 kg

D) 75 kg

Discuss

Q. The average score of a class in a test was 80% with a standard deviation of 15%. If a student scored 70%, what percentile did he/she achieve?

A) 10th percentile

B) 25th percentile

The z-score for a score of 70% in this class is $latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%2870%5C%25+ +80%5C%25%29%7D%7B15%5C%25%7D+%3D+ 0$ standard deviations below the mean. Using a standard normal distribution table or calculator, we can find that this corresponds to approximately $\displaystyle 25\text{th}$ percentile.

C) 50th percentile

D) 75th percentile

Discuss

Q. The average of eight numbers is 30. If the average of first five is 29 and that of the last four is 26, what will be the value of fifth number?

A) 9

Answer: (Sum of first five number+Sum of last five number)-Sum of eight numbers =(29×5+26×4)−(8×30)=9.

B) 13

C) 21

D) 16

Discuss

Q. In a classroom the average age of 40 students is 8 years. If the age of their teacher is added, the average age increased by one year. What will be the age of the teacher (in years)?

A) 62

B) 47

C) 43

D) 49

Answer: New average=8+1=9 years. So, age of the teacher =(41×9)-(40×8)=49 years.

Discuss

Q. There are three sections of students A, B and C in a class. The number of students in the three sections are 32, 38 and 42 and the average age of students is 16, 17.5 and 19 years respectively. What is the average age of students of A, B and C?

A) 22

B) 17.63

Answer: Let, the average age of the other section of students be ( X ). So, $\displaystyle \frac{(30\times22)+(27 \times X)}{(30+27)}=23.20$ . Or, $\displaystyle X=24.53 \text{ years}$ .

C) 19.50

D) 15

Discuss

Q. An undergraduate program has two sections. In one of which there are 30 students with an average of 22 years. The average of the total students of the program is 23.20 years. What is the average age of other section if the number of student is 27?

A) 21.81

B) 25.10

C) 36.25

D) 24.53

Answer: Let, the average age of the other section of students be ( X ). So, $\displaystyle \frac{(30\times22)+(27 \times X)}{(30+27)}=23.20$ . Or, $\displaystyle X=24.53 \text{ years}$ .

Discuss

Q. In a weight lifting competition, the average age of first three competitors is 22 years and that of the last three competitors is 24 years out of the four competitors. If the age of first competitor is as same as that of the first average, what is the last number?

A) 26

B) 23

C) 28

Ages of 2nd and 3rd competitor=(22×3)- 22=44. So, last number=(24×3) - ages of 2nd & 3rd=72- 44=28.

D) 31

Discuss

Q. The average marks of a student in a final examination result were 71. But later on it was found that marks of a subject, the student actually got, was misread as 81 instead of 61. What was the corrected average of the marks if the student attended in eight subjects?

A) 72

B) 59.75

C) 68.50

Answer: Corrected average $latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7B%2871%5Ctimes8%29+ +%2881%2B61%29%7D%7B8%7D%3D68$ .

D) 64.50

Discuss

Q. In a final exam the average marks of the topper in six subjects is 68. His average in first five subjects except Math is 65. What is the number the topper gets in Math?

A) 83

Answer: (68x6)-(65x5)=408-325=83

B) 79

C) 87

D) 64

Discuss

Q. 12 gymnasiums average 700 customers per gymnasium per day. In order to minimizing the cost the owner closed down two of the gymnasiums but the average attendance who enrolled remained same, find the average daily attendance per gymnasium among the remainder.

A) 720

B) 760

C) 840

Answer: Total attendant before cost curtail $\displaystyle = 700 \times 12 = 8400$ . After the shutdown of two gymnasiums, the new average $\displaystyle = \frac{8400}{10} = 840$ .

D) 880

Discuss

Q. In a garrison, average weight of eight soldiers of a unit increased by 1.20 kg after replacing one of soldiers who weighed 68 years of another unit. What was the weight of the new soldier?

A) 82.10 kg

B) 67.75 kg

C) 77.60 kg

Answer: Weight increased by (8×1.20) =9.60 kg. So, weight of new soldier = 9.60+ 68= 77.60 kg

D) 62.00 kg

Discuss

Average

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