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There are five prime numbers between 30 and 50.

They are 31,37,41,43 and 47.

Therefore the required average=(31+37+41+43+47)/5 

= 199/5 

= 39.8.

Given:

15, 25, 35, 45 and 50

Formula used:

Average = (Sum of observation)/(Total number of observation)

Calculation:

Sum of numbers is

⇒15 + 25 + 35 + 45 + 50 = 170

Total number of observation is 5

So, average will be

⇒170/5 = 34

∴ The average of 15, 25, 35, 45 and 50 is 34.

Given:

The average number of cars with actors k, l and m is 27, which is 2 more than the number of cars with l.

The number of cars with actor k is 5 more than double the number of cars with actor m.

Formula:

Average of n numbers a₁, a₂ ... a_n = (a₁ + a₂ + ... a_n)/n

Calculation:

Total cars with k + l + m = 27 × 3 = 81

Number of cars with l = 27 – 2 = 25

K + m = 81 – 25

⇒ K + m = 56      ----(1)

Let number of cars with m be x 

Then with k = 2x + 5

⇒ x + 2x + 5 = 56

⇒ 3x = 51

⇒ x = 17

Number of cars with k = 2x + 5 = 39

Average cars with k and l = 39 + 25 = 64/2 = 32

∴ The average number of cars with k and l is 32.

Given:

The average speed of 3 bikes x, y and z is 80 km/hr. The average speed of bikes y and z is 90 km/hr.  The speed of bike y is 81 km/hr.

Formula:

Average of n numbers a₁, a₂ ... a_n = (a₁ + a₂ + ... a_n)/n

Calculation:

Given,

Average speed of 3 bikes = 80

Total speed of x + y + z = 240

Total speed of y + z = 180

Y = 81 (given)

Z = 180 – 81 = 99

x + y + z = 240

x + 180 = 240

x = 240 – 180 = 60

x + z = 60 + 99 = 159/2 = 79.5

So, the average speed of bikes x and z is 79.5

Given∶

Average height of 8 students = 80 cm

Average height of 12 students = 82 cm

Formula used∶

Average = Sum of all observation / Number of observation

Calculation∶

Sum of height of 8 students = 640 cm

Sum of height of 12 students = 984 cm

Sum of height of 4 students = 984 - 640

⇒ 344 cm

∴ The sum of the height of the new students is 344 cm.

Alternate method∶

If one or more quantities are added in a group, then the sum of new quantities added is calculated by given formula∶

Sum of new quantity = [Change in number of quantities × Original average] + [Change in average × Final number of quantities]

= [4 × 80 + 2 × 12]

⇒ 344

∴ The sum of the height of the new students is 344 cm.

Given∶

Average of first 5 terminal exam = 60

Average of first 2 terminal exam = 30

Calculation∶

Sum of marks in first 5 terminal exam = 60 × 5 = 300

Sum of marks in first 2 terminal exam = 30 × 2 = 60

⇒ Sum of marks in last 3 terminal exam = 300 - 60

⇒ Sum of marks in last 3 terminal exam = 240

Average marks of last 3 terminal exam = 240/3 = 80

∴ The average of last 3 terminal exam is 80.

Given:

Average of first 3 terminal exams = 50

Average of last 4 exams = 140

Formula used:

Average = Sum of all observation/Number of observation

Calculation:

Sum of the marks of first 3 terminal exam = 3 × 50 = 150

Sum of the marks of last 4 terminal exam = 4 × 85 = 340

Total marks in 7 exams = 490

Average marks = 490/7 = 70

GIven :

The average weight of 3 persons of group A is 60 kg 

The average weight of 2 persons of Group B is 70 kg 

Formula used :

Average = Total sum of all quantities/total number of quantities

Calculations :

Total weight of group A people = 60 × 3 = 180 kg 

Total weight of group B people = 70 × 2 = 140 kg 

⇒ Total weight = 140 + 180 = 320 kg 

⇒ Average weight of all persons = 320/5

⇒  64 kg 

∴ Average weight of all persons will be 64 kg  

Given:

The numbers of visitors every hour in a shop from 1 p.m. to 8 p.m. on Monday were recorded as 7, 10, 15, 22, 6, 4 and 13.

Formula used:

Arithmetic mean = Average = Sum of values/Number of values

Calculation:

Sum of visitors = 7 + 10 + 15 + 22 + 6 + 4 + 13 = 77

Number of hours = 8 - 1 = 7 hours

Arithmetic mean of visitors = 77/7 = 11 visitors/hour

∴ The arithmetic mean of visitors is 11.

Given:

Run rate for first 10 overs =  3.2

Formula Used:

Run rate = (Total runs )/ Number of overs

Calcualtion:

for the 1st 10 overs the score will be 10*3.2=32
⇒  the remaining score is 282-32=250
⇒ remaining overs = 40

⇒  the run rate  required = 250/40=6.25

Conclusion:
∴ Run rate should be 6.25

Given:

Number of parcels = 15

Weight of each parcel = \(10 \dfrac{1}{2}\) kg

Formula Used:

Total weight = Number of parcels × Weight of 1 parcel

Calculations:

Number of parcels = 15

Weight of each parcel = 21/2 kg

Now, Total weight = Number of parcels × Weight of 1 parcel 

⇒ Total weight = 15 × 21/2 kg

⇒ Total weight = 157.5 kg

∴ The total weight of the parcels is 157.5 kg.

Required average speed = ((2xy)/(x+y)) km / hr 

= (2 x 84 x 56)/(84+56)km/hr 

= (2*84*56)/140 km/hr 

= 67.2 km/hr. 

Given:

Average marks in three subjects = 80

Average marks in math and Hindi = 90

Average marks in Hindi and English = 70

Formula used:

Total marks = Average marks × Number of subjects

Calculation:

Total marks in all three subjects = 80 × 3

⇒ 240

Total marks in math and Hindi = 90 × 2

⇒ 180

Total marks in Hindi and English = 70 × 2

⇒ 140

Total marks in Maths + Hindi and Hindi + English = 180 + 140

⇒ 320

Marks in Hindi = 320 – 240

⇒ 80 marks

∴ The marks in Hindi is 80

GIVEN:

Batsman score in 17th inning = 87 and his average increases by 3 

CALCULATION:

Batsman score in 17th inning = 87 and his average increases by 3 

⇒ Let the average after 17th inning = x

⇒ Average after 16th inning = (x - 3)

⇒ 16 × (x - 3) + 87 = 17x

⇒ x = 87 - 48 

⇒ x = 39

∴ His Average after the 17th inning will be 39

The average marks of a student in seven subjects is 41. 

(a) New average marks = (7x41-14+42) /7

=(287+28)/7 =315/7 =45

Given:

Runs scored by the batsman in 17^th innings = 87

Increased average = 3

Calculation:

Let the average after 17th inning = x

Then average after 16th inning = x - 3

⇒ 16(x - 3) + 87 = 17x 

⇒ x = 87 - 48 = 39

∴ Average after 17^th innings = 39 Runs.

Given:

Score in 17^th match = 87

In 17^th match increase in average is 3

Formula used:

Average = Sum of observations/Number of observations

Calculation:

Let the initial average be x

New average = x + 3

x + 3 = {(16 × x) + 87}/17

⇒ 17x + 51 = 16x + 87

⇒ 17x – 16x = 87 – 51

⇒ x = 36

⇒ New average = 36 + 3 = 39 

∴ His average after 17th match is 39.

Let the average after 17th inning = x. 

Then, average after 16th inning = (x - 3). 

∴ 16 (x - 3) + 87 = 17x or x = (87 - 48) = 39.

Total marks scored by Raju = 38 × 4 = 152

Total marks scored by Ghanshyam  = 41 × 4 = 164

Ratio of marks obtained by Raju and Ghanshyam in Maths = 5 ∶ 6

⇒ Let the marks obtained by Raju and Ghanshyam in Maths be 5x and 6x

Marks scored by Raju in Social = 40

Marks scored by Ghanshyam in Social = 45

Let the marks scored by them in English and Science = y

∴ The difference in marks scored by them 

⇒ (6x + y + y + 45) – (5x + y + y + 40) = 164 – 152

⇒ x = 7

∴ Marks scored by Raju in Maths = 7 × 5 = 35

∴ Marks obtained by Raju in English and Science 

⇒ y + y = 152 – 40 – 35 = 77

⇒ y = 77/2 = 38.5

∴ The marks scored by Ghanshyam and Raju in English and Science is 38.5

Given:

Average marks of a student in 7 subjects = 75

Average marks in 6 subjects excluding Science = 73

Formula used:

Average = Sum of values/number of values

Calculation:

The average marks of a student in seven subjects = 75

The sum of marks of a student in seven subjects = 75 × 7

⇒ 525

The average marks in six subjects, excluding Science = 73

The sum of marks of a student in seven subjects = 73 × 6

⇒ 438

Science scored in Science = The sum of marks in seven subjects - the sum of marks without Science

Marks scored in Science = 525 - 438

⇒ 87

∴ Marks scored in Science is 87.

Formula used:

Average = Sum of all observations/Total number of all observations

Calculation:

Sum of marks of 6 students = 50 × 6

⇒ 300

Sum of marks of 8 students = 70 × 8

⇒ 560

Sum of marks of 14 students = 300 + 560

⇒ 860

Average marks of 14 students = 860/14

⇒ 61.4

∴ The average marks of 14 students is 61.4

Given: The average age of the class of 30 students increases by one year when 15 new students join.

Formula: Average of n numbers a1, a2, ..., an = (a1 + a2 + ... + an)/n

Calculation:

Let the average age of the class be x.

Total age of 30 students = 30x

New average age of the class after joining 15 students = x + 1

Total age of 45 students = 45 × (x + 1)

Let the average age of 15 new students be y.

Total age of 15 students = 15y

According to question

45(x + 1) = 30x + 15y

⇒ 45x + 45 = 30x + 15y

⇒ 15x + 45 = 15y

⇒ y – x = 3

The difference in the average age of the new students and the new average of the whole class =

y – (x + 1) = (x + 3) – (x + 1) = 2

The difference in the average age of the new students and the new average of the class is 2 years.

These are two sets of series :

Series 1: It contains four consecutive odd numbers, whose average is 28.

Numbers - 25, 27, 29 ,31

The sum of the second and third numbers of series 1 is equal to the largest number of series 2.

⇒ 27 + 29 = 56

Series 2: It contains three consecutive even numbers.

⇒ 52, 54, 56

The Sum of numbers of series 2.

⇒ 52 + 54 + 56

⇒ 162

∴ The sum of all the numbers of series 2 is 162.

Given:

Average age of the class = 12 years

Total number of students in the class = 20

Age of the student leaving the class = 15 years

Age of the student entering the class = 13 years

Formula Used:

Total of ages of all the students = Average age of the class × Total number of students in the class

Calculation:

We obtain the new total age of the class as:

(12 × 20) - 15 + 13 = 238 years

So, the new average age of the class = 238/20 = 11.9 years

∴ The new average age of the class is 11.9 years

Formula used:

Average of n numbers = Sum of all numbers/n

Calculation:

Odd numbers from 1 to 13 are 1, 3, 5, 7, 9, 11, 13

Sum of squares of 1, 3, 5, 7, 9, 11, 13 is 1 + 9 + 25 + 49 + 81 + 121 + 169 = 455

∴ Average of these numbers = 455/7 = 65

Given:

Average of four consecutive odd natural numbers is eight less than the average of three consecutive even natural numbers,

The sum of these three even numbers is equal to the sum of above four odd numbers

Formula used:

\(Average=(Sum of all numbers)/(total numbers)\)

Calculation:

Four consecutive odd natural number is

⇒ X₁ – 2, X₁, X₁ + 2 and X₁ + 4

Average = (X₁ – 2 + X₁ + X₁ + 2 + X₁ + 4)/4

⇒ (4X₁ + 4)/4

⇒ X₁ + 1

Three consecutive even natural numbers is

⇒ X₂ – 2, X₂, X₂ + 2

Average = (X₂ – 2 + X₂ + X₂ + 2)/3

⇒ 3X₂/3

⇒ X₂

According to question

⇒ X₁ + 1 = X₂ – 8

⇒ X₁ – X₂ = -9      ….(i)

And the sum of these three even numbers is equal to the sum of above four odd numbers

⇒ X₂ – 2 + X₂ + X₂ + 2 = X₁ – 2 + X₁ + X₁ + 2 + X₁ + 4

⇒ 3X₂ = 4X₁ + 4

⇒ 3X₂ - 4X₁ = 4      ….(ii)

Solve the equation (i) and (ii),

  ⇒ X₁ = 23 and X₂ = 32

Average of odd number = X₁ + 1

⇒ 23 + 1

⇒ 24

∴ Average of odd number is 24

Given:

Three years ago the age of the family = 27 years

Formula use:

Average of the ages = (Total age/number of members)

Calculations:

⇒ Three years ago the total age of all the family members = 27 × 5 years

⇒ Three years ago the total age of all the family members = 135 years

⇒ After three years total age of the family members = 135 + 3 × 5 years

⇒ After three years total age of the family members = 135 + 15 = 150 years

⇒ The total present age of all the family members = 6 × 27 years

⇒ The total present age of all the family members = 162 years

⇒ The present age of the child = 162 - 150 years

⇒ The present age of the child = 12 years

∴ The present age of the child is 12 years.

Given: 

The all odd number up to 100

Concept:   

The sum of consecutive odd number up to N terms

1 + 3 + 5 + 7 + . . . . . + N terms = N² 

Average = (Sum of all observation/number of observation)

Calculation: 

The all odd number up to 100 

1, 3, 5, 7, . . . . 99

The number o observation(N) is

⇒ (99 + 1)/2 = 100/2 = 50

The sum of all odd number up to 100 is 

⇒ 1 + 3 + 5 + 7 + . . . + 99 = (50)² 

⇒ 2500

The average of all odd number up to 100 is 

⇒ 2500/50

⇒ 50

∴ The required average is 50.

Given: 

The average price of gold during last year was Rs. 1,120 per gram. In the first five months average was Rs. 1,105 and in the next three months, the average was Rs. 1,125

Concept: 

Average = (Sum of all observations/number of observations)

Calculation:

Let the average price of gold for the remaining four months is 'Y' rupees

According to the question

The average price of gold during last year is 

⇒ [(5 × 1105) + (3 × 1125) + (Y × 4)]/12 = 1120

⇒ (5525 + 3375 + 4Y) = 1120 × 12

⇒ 8900 + 4Y = 13440

⇒ 4Y = 13440 - 8900

⇒ Y = 4540/4

⇒ 1135 rupees

∴ The average price of gold in the remaining month of the year is 1135 rupees.

GIVEN:

There are five prime numbers between 30 and 50

FORMULA USED:

Average = (Sum of the observation)/Number of observation

CALCULATION:

Prime numbers between 30 and 50 are 31, 37, 41, 43, and 47

⇒ Average = (Sum of the observation)/Number of observation

⇒ Average = (31 + 37 + 41 + 43 + 47)/5

⇒ 199/5

⇒ 39.8

∴ Average of all prime numbers b/w 30 and 50 is 39.8

Given:

First 10 prime numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Formula used:

Average = (sum of total observations)/ (numbers of observations)

Calculation:

Sum of first 10 prime numbers = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29

⇒ Sum of first 10 prime numbers = 129

The average of first 10 prime number = (Sum of first 10 prime numbers)/10

⇒ The average of first 10 prime number = 129/10

⇒ 12.9

∴  The average of first 10 prime number is 12.9.

Given:

Average marks of 18 students = 45

Marks of 3 students were misread as 48, 31 and 95 instead of 98, 71 and 59

Formula used:

Total marks = Average × Number of students

Calculation:

Total marks = 18 × 45

⇒ 810

Correct total = 810 – [(48 + 31 + 95) + (98 + 71 + 59)]

⇒ 810 – 174 + 228

⇒ 864

Correct average = 864/18

⇒ 48

∴ The correct average is 48.

Given:

Average marks = 71

Total students = 14

Formula used:

Average = Sum of total terms/Total number of the terms

Calculation:

The total marks of the 14 students = 71 × 14 = 994

Total miscalculated marks of the students = 56 – 42 + 32 – 74 = -28

Correct total marks of the 14 students = 994 – 28 = 966

Correct average = 966/14 = 69

∴ The correct average marks is 69

Given:

One student marks were entered 88 instead of 68

Average of the students increased by 0.5

Formula used:

Average = Sum of marks of all the students/Total number of students

Calculation:

Let the total number of students in the class be n

Let the sum of the marks of the students except the wrong entered be x

The Original sum of marks of the students = x + 68

The sum of the marks of the student after wrong entered = x + 88

Now, A/Q

(x + 88)/n – (x + 68)/n = 0.5 

⇒ (88 – 68)/n = 0.5

⇒ 20/n = 0.5

⇒ n = 40

∴ The total number of students in the class is 40.

Given:

The average marks obtained by 120 students are 35

The average marks obtained by the students who have passed is 39

The average marks obtained by the students who have failed is 19

Concept Used:

Sum = Average × Number of data

Calculation:

The average marks obtained by 120 students are 35

⇒ Total marks obtained by all 120 students = 120 × 35

⇒ 4200

Let, the number of students passed in the examination be x

⇒ Total number of students failed in the examination is (120 – x)

Total marks obtained by the passed student = 39x

Total marks obtained by the students who have failed is 19 × (120 – x)

⇒ 19 × 120 – 19x

⇒ 2280 – 19x

Accordingly,

39x + 2280 – 19x = 4200

⇒ 20x = 4200 – 2280

⇒ 20x = 1920

⇒ x = 96

∴ The number of students who have passed is 96.

Given:

Average of 50 students = 60

Marks of one student had been wrongly written as 36 instead of 63

Formula used:

Sum of observations = Average × Number of observations

Calculation:

Total marks of all students = 50 × 60 = 3000

But one student’s marked wrongly entered 36 instead of 63

So exact total marks of all students = 3000 – 36 + 63 = 3027

Correct average of all the students = 3027/50 = 60.54

∴ The correct average is 60.54

Given:

The average marks obtained by 56 students = 15

Mean of passed students = 18

Mean of failed students = 11

Concept used:

Total marks = Average × number of students

Calculations:

The total marks obtained by 56 students = 56 × 15

⇒ 840

Let the number of students failed be 'x'

840 = 11 × x + 18 × (56 - x)

⇒ 840 = 11x - 18x + 1008

⇒ 7x = 1008 - 840

⇒ 7x = 168

⇒ x = 168/7

⇒ x = 24

∴ The number of students failed is 24

Given:

Marks of A = 215

Marks of B = 105

Marks of C = 202

Calculation:

⇒ (215 + 105 + 202)/3

⇒ 522/3

⇒ 174

∴ The required result will be 174.

Given:

Maximum marks in the exam = 100

Minimum cutoff marks to clear the exam = 60

Marks scored by Rahane = Marks scored by Virat + 13

Marks scored by Rahane = Marks scored by Pujara – 10

Formula Used:

Average marks scored by all the candidates = Total of marks scored by all the candidates together/Total number of candidates

Calculation:

Let the marks scored by Rahane = x

So, the marks scored by Virat = x – 13, and

The marks scored by Pujara = x + 10

Total of the marks scored by them = 55 × 3 = 165

So, we get:

x + x – 13 + x + 10 = 165

⇒ 3x = 168

⇒ x = 56

Hence, we get:

Marks obtained by Rahane = 56

Marks obtained by Virat = 56 – 13 = 43

Marks obtained by Pujara = 56 + 10 = 66

∵ The minimum cutoff marks is 60, only Pujara, who has scored 66 marks, cleared the exam.

∴ The total number of candidates, among these three who cleared the SBI PO Prelims exam is 1

Given:

The age of person 2 years from now will be double the average of his current age and his age 10 years ago.

Concepts used:

Average = Sum of all observations/Number of observations

Calculation:

Let the current age of the person be x years.

His age 2 years from now = x + 2 years

His age 10 years ago = (x – 10) years

Average of his current age and his age 10 years ago = (x + x – 10)/2 = (2x – 10)/2 = x – 5

According to the question,

⇒ x + 2 = 2 × (x – 5)

⇒ x + 2 = 2x – 10

⇒ 2x – x = 10 + 2

⇒ x = 12 years

His age 4 years ago = 12 – 4 years = 8 years

∴ His age 4 years ago was 8 years.

Given:

Average height of 25 girls = 1.2 m

Formula Used:

Average = (Sum of heights)/Number of girls

Calculation:

Sum of the height of the 25 girls = 25 × 1.2

⇒ 30 m

Sum of the height of the 20 girls after 5 left = 20 × 1.36

⇒ 27.2 m

Sum of the heights of the 5 girls = 30 – 27.2

⇒ 2.8 m

Required average = 2.8/5

⇒ 0.56 m

Given:

Total 10 person’s height = 2,100 cm

Excluded person’s height = 300 cm

Concept:

Average height = Total height/Total persons

Solution:

Total 10 person’s height = 2,100 cm

Excluded person’s height = 300 cm

Remaining person’s total height = 2,100 - 300 ⇒ 1,800 cm

Remaining persons = 9

Given:

Average marks of batch with 55 students = 50

Average marks of batch with 60 students = 55

Average marks of batch with 45 students = 60

Formula used :

If the values a, b, c, and d are the individual averages of each of the groups and n₁, n₂, n₃ and n₄ are the numbers of observations then

Weighted average = (an₁ + bn₂ + cn₃ + dn₄)/( n₁ + n₂ + n₃ + n₄)

Calculation:

Average = [(50 × 55) + (55 × 60) + (60 × 45)]/(55 + 60 + 45)

⇒ Average = (2750 + 3300 + 2700)/(160)

Average = 8750/160 = 54.68

∴ The average marks of all student is 54.68

Given:

The mean of 5 numbers is 28

One number is excluded so mean decreased by 5 i.e. 23

Formula Used:

Sum of numbers = Mean × No. of numbers

Calculation:

⇒ Sum of the 5 numbers = 5 × 28 = 140

Mean of 4 numbers is 23

⇒ Sum of the remaining four numbers = 4 × 23 = 92

∴ Excluded number = 140 – 92 = 48  

Given:

Average age of mother and 6 children = 12 years

When mother is excluded, average is reduced by 5 years

Formula used:

Average =(Sum of values)/(Total number of values)

Calculation:

Average when mother is excluded = 12 – 5 = 7 years

Sum of ages of mother and her 6 children = 12 × 7 = 84 years  

Sum of ages of 6 children = 7 × 6 = 42 years

Age of mother = 84 – 42 = 42 years

∴ The age of mother is 42 years. 

Given: 

The average age of 14 girls and their teacher's is 16 years.

If the teacher's age is excluded average reduces by 2

Concept: 

Average = (Sum of all observation/number of observation)

Calculation: 

The sum of age of 14 girls and their teacher's is 

⇒ 15 × 16

⇒ 240 years

The sum of age of 14 girls is 

⇒ 14 × (16 - 2)

⇒ 14 × 14

⇒ 196 years 

Now, 

The age of teacher's is 

⇒ 240 - 196

⇒ 44 years 

∴ The required age of the teacher's is 44 years.

Given:

Average age of 3 children of a family = 11 years old.

Average age of children together with their parents = 23 years.

If the father is older than the mother by 2 years.

Formula used:

Average = (Sum of total observations)/(Total number of observation)

Calculation:

Let the age of the mother is 'x' years old.

Then the age of father = (x + 2) years old

Average = (Sum of total observations)/(Total number of observation)

⇒ (Sum of total observations) = Average × (Total number of observation)

⇒ Sum of age of three children = 11 × 3 = 33 years

Sum of age of whole family = 5 × 23 = 115 years

So, sum of the age of father and mother = 115 – 33 = 82 years

According to the question,

x + 2 + x = 82

⇒ 2x = 80

⇒ x = 40

∴ The age of mother is 40 years old.

Given:

Average weight of 12 people and the trainer = 66 kg

Calculation:

The weight of 12 students and the trainer = 13 × 66

⇒ 858 kg

Total weight excluding the trainer = 68 × 12

⇒ 816

The weight of the trainer

⇒ 858 – 816

⇒ 42 kg

∴ The weight of the trainer is 42 kg. 

GIVEN:

The average of X, Y and the Z = 22 more than the average of Y, Z and W.

FORMULA USED:

Average = Sum of numbers/Total numbers

CALCULATION:

According to the question:

(X + Y + Z)/3 = (Y + Z + W)/3 + 22

⇒ X + Y + Z = Y + Z + W + 66

⇒ X = W + 66

⇒ X - W = 66