# Average Practice Set 2

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# Average Practice Set 2

### Introduction

Average is a straight- forward concept and it can be solved easily by equal distribution method. Average is the sum of all the elements in a given data set divided by the total number of elements in the data set. The most commonly denoted term for average is Arithmetic Mean, simply termed as mean. The article Average Practice Set 2 very useful for for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

The Average (arithmetic mean) of a group or set of N numbers is defined as the sum of those numbers divided by N. Here N is the number of values or observations in a set

Average = $$\frac {Sum of Numbers / Values}{Numbers of Values / Observation}$$

A = $$\frac {Sum}{N}$$ Or Sum = N * Average

### Quiz

Q1. The average age of a family of 6 members is 22 years. If the age of the youngest member be 7 years, then what was the average age of the family at the birth of the youngest member?

A. 15
B. 18
C. 28
D. 24

Explanation:
Total age of all members = $$6 \times 22$$= 132 years

7 years ago, total sum of ages = 132 – $$(6 \times 7)$$ = 90 years

But at that time there were 5 members in the family

Average at that time = $$\frac {90} {5}$$ = 18 years

Q2. A batsman in his 17th innings makes a score of 85 and thereby increases his average by 3. What is his average after 17 innings?

A. 30
B. 37
C. 40
D. 45

Explanation:
Let the average after 16th innings be a

then total score after 17th innings $$\Rightarrow$$ 16a + 85 = 17 (a + 3)

a = 85 – 51 = 34

Average after 17 innings = a + 3 = 34 + 3 = 37

Q3. There were 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs 42 per day while the average expenditure per head diminishes by Rs 1. Find the original expenditure of the mess?

A. 450
B. 420
C. 430
D. 410

Explanation:
Suppose the average expenditure was Rs a.

Then total expenditure = 35 a

When 7 more students join the mess, total expenditure = 35a + 42

Now, the average expenditure= $$\frac {(35a + 42)}{(35 + 7)}$$

Now, we have $$\frac {(35a + 42)}{42}$$ = (a – 1)

or, 35a + 42 = 42a – 42

7a = 84

a = 12

Q4. The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of the remaining 55% is:

A. 51.4
B. 52.6
C. 56.1
D. 55.3

Explanation:
Let the required mean score be a

Then, $$20 \times 80 + 25 \times 31 + 55 \times a = 52 \times 100$$

1600 + 775 + 55a = 5200

55a = 2825

a = 51.4

Q5. The average of a non-zero number and its square is 5 times the number. The number is:

A. 0 , 7
B. 0 , 6
C. 5 , 7
D. 0 , 9

Explanation:
Let the number be x.

Then, $$\frac {(x + {x}^{2})}{2} = 5x$$

$${x}^{2} – 9x$$ = 0

$$x (x – 9)$$ = 0

$$x = 0 or x$$ = 9.

Q1. If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of $${a}^{2}, {b}^{2}, {c}^{2}$$is:

A. 3 M x M
B. 3 M
C. 9 M
D. 9 M x M

Explanation:
We have :

$$\frac {(a + b + c)}{3}$$ = M or (a + b + c) = 3 M.

Now, $${(a + b + c)}^{2}$$ = $${(3 M)}^{2} = {9 M}^{2}$$

$${a}^{2} + {b}^{2} + {c}^{2} + 2 (ab + bc + ca) = {9 M}^{2}$$

$${a}^{2} + {b}^{2} + {c}^{2} = {9 M}^{2}$$

Required mean = $$\frac {({a}^{2} + {b}^{2} + {c}^{2})} {3}$$ = $$\frac {{9 M}^{2}} {3} = {3 M}^ {2}$$

Q2. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one them weighing 65 kg. What might be the weight of the new person?

A. 65 kg
B. 70 kg
C. 85 kg
D. 92 kg

Explanation:
Total weight increased = $$(8 \times 2.5) kg = 20 kg$$

Weight of new person = $$(65 + 20) kg = 85 kg$$

Q3. The average age of the boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is:

A. 15
B. 16
C. 17

Explanation:
Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given

So, the data provided is inadequate

Q4. A cricketer has completed 10 innings and his average is 21.5 runs. How many runs must he make in his next innings so as to raise his average to 24?

A. 44
B. 45
C. 49
D. 48

Explanation:
Total of 10 innings = $$21.5 \times 10$$ = 215

Suppose he needs a score of a in $${11}^{th}$$ innings; then average in 11 innings = $$\frac {(215 + a)}{11}$$= 24

or, a = 264 – 215 = 49

Q5. Find the average of 13 + 26 + 39 + …….. + 260

A. 136.5
B. 136
C. 137
D. 135

Explanation:
$$\frac {13 (1 + 2 + 3 + ….. + 20)}{20 \times 2}$$

$$\frac {13 \times 20 \times 21}{40}$$

$$\frac {5460}{40}$$ = 136. 5

Q1. The average of six numbers is 3.95. The average of two of them is 3.4, while the average of the other two is 3.85. What is the average of the remaining two numbers?

A. 4.2
B. 4.6
C. 5.1
D. 5.6

Explanation:
Sum of the remaining two numbers = $$(3.95 \times 6) – [(3.4 \times 2) + (3.85 \times 2)]$$

= 23.70 – (6.8 + 7.7)

= 23.70 – 14.5 = 9.20

Average = $$\frac {9.2}{2}$$ = 4.6

Q2. The average salary of the entire staff in an office is Rs 120 per month. The average salary of officers is Rs 460 and that of non-officer is Rs 110. If the number of officers is 15, then find the number of non-officer in the office?

A. 450
B. 550
C. 510
D. 520

Explanation:
Let the required number of non-officers = a

Then, $$110a + 460 \times 15$$ = 120 (15 + a)

or, 120a – 110a = $$450 \times 15 – 120 \times 15$$ = 15 (460 – 120)

or, 10a = $$15 \times 340$$

a = $$15 \times 34$$ = 510

Q3. Find the average of first 20 multiple of 7?

A. 71.5
B. 73.5
C. 75.2
D. 76.6

Explanation:
Average = $$\frac { 7 (1 + 2 + 3 + ……. + 20)}{20} = 7 \times 20 \times \frac {21} {20 \times 2} = 73.5$$

Q4. The average of four consecutive ODD number is 28.Find the largest number.

A. 25
B. 31
C. 13
D. 27

Explanation:
$$\frac {x + (x + 2) + (x + 4) + (x + 6)}{4}$$ = 28

4x +12 = $$28 \times 4 = > 112$$

$$x = \frac {(112 – 12)}{4} = \frac {100}{4}$$ 24

Largest number $$x + 6 = 25 + 6 = 31$$

Q5. Find the average of first 60 natural numbers

A. 30.5
B. 31
C. 31.5
D. 32

Explanation:
Sum of First 60 Natural numbers = $$\frac { n (n + 1)}{2}$$

$$\frac {60 \times 61}{2}$$ = 1830

Average = $$\frac {1830}{60}$$ = 30. 5

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