 # Average Practice Set 1 5 Steps - 3 Clicks

# Average Practice Set 1

### 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 1 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. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?

A. 28$$\frac {4}{7}$$
B. 31$$\frac {5}{7}$$
C. 32$$\frac {1}{7}$$
D. None of these

Explanation:
Required Average = $$\frac {67 \times 2 + 35 \times 2 + 6 \times 3}{2 + 2 + 3}$$

$$\frac {134 + 170 + 18}{7}$$

$$\frac {222}{7}$$

31$$\frac {5}{7}$$ years

Q2. A grocer has a sales of Euro 6435, Euro 6927, Euro 6855, Euro 7230 and Euro 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Euro 6500?

A. Euro 4991
B. Euro 5991
C. Euro 6001
D. Euro 6991

Explanation:
Total sale for 5 months = Euro (6435 + 6927 + 6855 + 7230 + 6562) = Euro 34009.

Required sale = Euro [ $$(6500 \times 6) – 34009$$]

= Euro (39000 – 34009)

= Euro 4991.

Q3. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

A. 0
B. 1
C. 10
D. 19

Explanation:
Average of 20 numbers = 0.

Sum of 20 numbers $$(0 \times 20)$$= 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a then $${20}^{th}$$ number is (-a).

Q4. The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?

A. 76 kg
B. 76.5 kg
C. 85 kg

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

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

Q5. The captain of a cricket team of 11 members is 26 years old and the wicket-keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

A. 23 years
B. 24 years
C. 25 years
D. None of these

Explanation:
Let the average age of the whole team by x years.

$$11 x – (26 + 29) = 9(x -1)$$

$$11 x – 9 x$$= 46

$$2 x$$ = 46

$$x$$ = 23.

So, the average age of the team is 23 years.

Q1. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

A. 35 years
B. 40 years
C. 50 years
D. None of these

Explanation:
Sum of the present ages of husband, wife and child = $$(27 \times 3 + 3 \times 3)$$ years = 90 years.

Sum of the present ages of wife and child = $$(20 \times 2 + 5 \times 2)$$ years = 50 years.

Husband’s present age = (90 – 50) years = 40 years.

Q2. A car owner buys petrol at Euro 7.50, Euro 8 and Euro 8.50 per liter for three successive years. What approximately is the average cost per liter of petrol if he spends Euro 4000 each year?

A. Euro 7.98
B. Euro 8
C. Euro 8.50
D. Euro 9

Explanation:
Total quantity of petrol = $$(\frac {4000}{7.50} + \frac {4000}{8} + \frac {4000}{8.50})$$litres

= 4000 $$(\frac {2}{15} + \frac {1}{8} + \frac {2}{17})$$ litres

= $$(\frac {76700}{51})$$ litres

Total amount spent = Euro $$(3 \times 4000)$$= Euro 12000. Average cost = Euro $$\frac {(12000 \times 51)}{76700}$$= Euro 6120/767 = Euro 7.98

Q3. In Jessica’s opinion, her weight is greater than 65 kg but less than 72 kg. Her brother does not agree with Jessica and he thinks that Jessica’s weight is greater than 60 kg but less than 70 kg. Her mother’s view is that her weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Jessica?

A. 67 kg.
B. 68 kg.
C. 69 kg.

Explanation:
Let Jessica’s weight by X kg.

According to Jessica, 65 < X < 72

According to Jessica’s brother, 60 < X < 70.

According to Jessica’s mother, X <= 68

The values satisfying all the above conditions are 66, 67 and 68.

therefor, Required Average = $$(\frac {66 + 67 + 68}{3}) = (\frac {201}{3})$$ = 67 kg

Q4. 3 years ago, the average of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is:

A. 5 years
B. 2 years
C. 1 years
D. 4 years

Explanation:
Total age of 5 members, 3 years ago = $$(17 \times 5)$$ years = 85 years

Total age of 5 members now = $$(85 + 3 \times 5)$$ years = 100 years

Total age of 6 members now = $$(17 \times 6)$$ years = 102 years

Age of the baby = (102 – 100) years = 2 years

Q5. Of the four numbers, the first is twice the second, the second is one-third of the third and the third is 5 times the fourth. The average of the numbers is 24.75. The largest of these numbers is:

A. 45
B. 25
C. 30
D. 45

Explanation:
Let the fourth number be a

then hen, third number = 5 a, second number= $$\frac {5a} {3} and first= \frac {10a} {3}$$

a + 5a + $$\frac {5a}{3} + \frac {10}{3} = 24.75 \times 4$$

so,a = 9

So, the numbers are 9, 45, 15 and 30

Q1. The average age of students in a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is:

A. 1: 4
B. 2 : 3
C. 3 : 4
D. 4 : 2

Explanation:
Let the ratio be k:1

Then, $$k \times 16.4 + 1 \times 15.4 = (k + 1) \times 15.8$$

(16.4 – 15.8) k = (15. 8 – 15. 4)

k = $$\frac {2}{3}$$

Q2. The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60 % more than the price of the other, what is the price of each of these two books?

A. Rs 10 and Rs 16
B. Rs 12 and Rs 24
C. Rs 24 and Rs 18
D. Rs 28 and Rs 12

Explanation:
Total price of the two books = $$Rs. [(12 \times 10) – (11.75 \times 8)]$$ = Rs. (120 – 94) = Rs. 26

Let the price of one book be Rs.x

Then, the price of other book = $$Rs. (x + 60 Percent of x )= x + (\frac {3}{5}) x = (\frac {8}{5}) x$$

so, $$x + (\frac {8}{5}) \times$$= 26 , x = 10

The prices of the two books are Rs. 10 and Rs. 16

Q3. The average age of 30 boys in a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher?

A. 50
B. 44
C. 45
D. 42

Explanation:
Total ages of 30 boys = $$14 \times 30$$ = 420 years

Total age when class teacher is included = $$15 \times 31$$= 465 years

Age of class teacher = 465 – 420 = 45 years

Q4. The average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates are 39 and that of failed candidates is 15, what is the number of candidates who passed the examination?

A. 90
B. 100
C. 108
D. 115

Explanation:
Let the number of passed candidates be a

Then total marks $$\Rightarrow 120 \times 35 = 39 a + (120 – a) \times 15$$

4200 = 39 a + 1800 – 15 a

a = 100

Q5. The average of 11 results is 50. If the average of the first 6 results is 49 and that of last 6is 52, find the sixth result?

A. 55
B. 56
C. 65
D. 62

Explanation:
The total of 11 results = $$11 \times 50$$ = 550

The total of first 6 results = $$6 \times 49$$ = 294

The total of last 6 results = $$6 \times 52$$ = 312

The sixth result is common to both:

Sixth result = 294 + 312 – 550 = 56

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