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 (d) Let the five consecutive numbers be x, x+1, x+2, x+3 and x+4 respectively.

 Then, according to the question, 

x + x + 1 + x + 2 + x + 3 + x + 4 = 5 × 48

 or 5x +10 = 5 × 48 

or 5 (x + 2) = 5 × 48 

or x + 2 = 48 

or x = 48 - 2 = 46

  A = x = 46

 and E = x + 4 = 46 + 4 = 50 

 A × E = 46 × 50 = 2300

(e) Let A = x
According to the question
x + x +1 + x + 2 + x + 3 + x +4
= 5 × 48

5x + 10 = 240
⇒ 5x = 230
⇒ x = 46
E = 46 + 4 = 50
 A × E = 46 × 50 = 2300

Given :

The average of three consecutive numbers is 27.

Calculation :

Let the first number be x

The second number = (x + 1)

The third number = (x + 2)

A.T.Q.

(x + x 1 + x + 2)/3 = 27

⇒ (3x + 3) = 27 × 3

⇒ 3x + 3 = 81

⇒ 3x = 78

⇒ x = 26

First number = 26

Second number = 27

Third number = 28

Now,

Each number is decreased by 3

First new number = 26 - 3 = 23

Second new number = 27 - 3 = 24

Third new number = 28 - 3 = 25

⇒ New average = (23 + 24 + 25)/3

⇒ New average = 72/3

⇒ New average = 24

∴ the required new average is 24

Short tricks :

If each number is decreased or increased by the given number the average will also decrease or increase by the given number.

Each number is decreased by 3 so, the average will also decrease by 3.

Given average = 27

Each number is decreased by 3 

⇒ New average = 27 - 3

∴ The required new average is 24.

Given:

Average marks of 40 students = 86

Calculations:

Total marks obtained by 40 students = 86 × 40

⇒ 3440

6 highest marks are removed

∴ the average marks of 34 students = 83

Total marks obtained by 34 students = 83 × 34

⇒ 2822

⇒ Required Average of top 6 students = (3440 – 2822)/6

⇒ 618/6

⇒ 103

Given:

The average age of 40 students = 16 years

On admission of 10 new students average = 15 years

Average of 5 new students = 11

Formula used:

\({\rm{Average}} = \frac{{{\rm{sum\;of\;all\;the\;values}}}}{{{\rm{number\;of\;values}}}}\)

Calculation:

Sum of all ages of 40 students = 40 × 16

⇒ 640 years

After admission of 10 new students, number of students now = 40 + 10

⇒ 50

Sum of all ages after the admission of 10 new students = 50 × 15

⇒ 750 years

Sum of ages of 5 new students = 5 × 11

⇒ 55 years

Now remaining ages of 5 new students = Sum of all ages after the admission of 10 new students - Sum of all ages of 40 students - Sum of ages of 5 new students

⇒ 750 – 640 – 55

⇒ 750 – 695

⇒ 55 years

Average of 5 new students = 55/5

⇒ 11

Hence, the Average of 5 new students is 11.

Given:

Average weight of 29 students = 28 kg

Reduced average after admission of a new student = 27.8 kg

Formula used:

Total weight = Average × Number of students

Calculation:

Total weight of 29 students = 29 × 28

⇒ 812 kg

Total weight of 30 students = 30 × 27.8

⇒ 834 kg

Weight of new students = (834 – 812) kg

⇒ 22 kg

∴ The weight of new student is 22 kg

Given:

2,4, 6,.........., 100

Formula used:

Average of first n even number = (n/2) + 1

where n is the last number

Calculation:

Average of first n even number = (n/2) + 1

⇒ n = 100

⇒ average = (100/2) + 1

⇒ average = 50 + 1 ⇒ 51

∴ The average of 2,4, 6,.........., 100 is 51.

Given:

weight of 10 persons = 50, 30, 40, 30, 60, 70, 80, 40, 50 and 20 kg

Concept:

Average weight = Total weight/Total persons

Solution:

Total weight = 50 + 30 + 40 + 30 + 60 + 70 + 80 + 40 + 50 + 20 ⇒ 470 kg

Total persons = 10

Average weight = 470/10

⇒ 47 kg

Given:

The average score of Subir in 15 tests = 29

The average score of Ruchira in 11 tests = 27

Concept used:

Total marks scored = average marks obtained in tests × number of tests

Calculation:

Total marks scored by Subir = 29 × 15 = 435

Total marks scored by Ruchira = 27 × 11 = 297

Marks to be scored by Ruchira to reach Subir’s score = 435 – 297 = 138

Average marks to be obtained by Ruchira in the next four tests = 138/4 = 34.5

∴ The average mark to be obtained by Ruchira on the next four tests is 34.5.

Given:

Total inning = 12

Total score = 972

Concept:

Average score = Sum of scores / Total number of innings

Solution:

Average score = 972 / 12⇒ 81

Calculation:

Let the average weight of the 20 contestants be x kg

Total weight = 20x kg

Let the weight of the contestant who left the show be y kg

∴ According to the question:

⇒ 20x – y = (x + 5.75) × 19

⇒ x = 109.25 + y      ----(i)

Again, when one of the new contestants joined the show then let the weight of the new contestant who joined the class be z kg

⇒ 20x + z = (x – 2.75) × 21

⇒ z = x – 57.75      ----(ii)

Putting the value of x from equation (i) in equation (ii),

⇒ z = 109.25 + y – 57.75

∴ the required difference = z – y

⇒ 51.5 kg

The difference between the age of the new contestant who joined the show and the contestant who left the show is 51.5 kg.

Given:

First 10 multiples of 7

Formula used:

Average = a × (n + 1)/2

Where, ‘a’ denotes the first number and ‘n’ denotes number of terms

Calculation:

Average = 7 × (10 + 1)/2

⇒ 7 × 11/2 = 38.5

∴ The average of the first 10 multiples of  is 38.5

Alternate method:

Among 10 terms, the middle term will be the  average of 5^th term and 6^th term and it will be the average of the whole series.

The 5^th term is 7 × 5 = 35

The 6^th term is 7 × 6 = 42

Average of 5^th and 6^th term = (35 + 42)/2 = 38.5

∴ The average of the first 10 multiples of 7 is 38.5 

Given

Number of students = 15

Average marks scored by 15 students = 46

Average marks scored by 7 students = 42

Concept

Average = Sum of all observation/ Number of observation

Calculation

Total marks scored by 15 students = 15 × 46

= 690

Total marks scored by 7 students = 7 × 42

= 294

Total marks scores by 8 students = 690 - 294

= 396

Average marks scored by 8 students = 396/8

= 49.6

Given

First 6 multiples of 4

Concept

Average = Sum of all observation/ Number of observation

Calculation

First 6 multiples of 4 are 4, 8, 12, 16, 20 and 24

Average = (4 + 8 + 12 + 16 + 20 + 24)/6

= 14

Given: The average pocket money of 3 friends A, B and C is 80 in a particular month. B spends double and C spends triple of what A spends during that month and the average of their unspent pocket money is 60.

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

Calculation:

Total pocket money of (A + B + C) = 80 × 3 = 240

Total unspent pocket money = 60 × 3 = 180

Total spent money = 240 – 180 = 60

Let A spends Rs. x then B = 2x, and C = 3x

According to question

6x = 60

⇒ x = 10

∴ A spends Rs. 10.

Given:

Average age of boys = 16 years

Average age of girls = 15 years

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:

As per the above formula, number of girls and boys are necessary to find the average

As number of girls and number of boys is not given in the question

∴ Average age of all can’t be computed 

Given:

Mahesh’s opinion = weight is greater than 55 kg but less than 62 kg

His brother’s opinion = weight is greater than 50 kg but less than 60 kg

His mother’s opinion = weight is less than 58 kg

Calculations:

Let Mahesh’s weight be x kg.

According to Mahesh his age is 55 < x < 62

According to Mahesh’s brother 50 < x < 60

According to Mahesh’s mother x < 58

⇒ From above three statements 55 < x < 58

⇒ Probable weights are either 56 kg or 57 kg

Required Average = (56 + 57)/2

⇒ 113/2

⇒ 56.5 kg

The average of different probable weights of Mahesh is 56.5 kg.

Given :-

Average income of A and B = Rs. 24,000

Average income of B and C = Rs. 36000

C income is twice the A income.

Concept :-

Average = Sum of all numbers value/Total number

Calculation :-

⇒ (A + B)/2 = 24000

⇒ A + B = 2 × 24000

⇒ A + B = 48000    ....(1)

Now 

⇒ (B + C)/2 = 36000

⇒ B + C = 2 × 36000

⇒ B + C = 72000    ....(2)

Now,

Let A's income be x 

⇒ C's income = 2x

Subtract equation (1) from equation (2)

⇒ B + C - (A + B) = 72000 - 48000

⇒ B + C - A - B = 24000

⇒ C - A = 24000

From question 

⇒ 2x - x = 24000 

⇒ x = 24000

Now 

⇒ A's income = Rs. 24,000

⇒ C's income = Rs. 48,000 

Put the value of A in equation (1)

⇒  B + 24000 = 48000

⇒ B = 48000 - 24000

⇒ B's income = 24,000

Now 

⇒ Average of all A , B and C income = (24000 + 24000 + 48000)/3

⇒ Average of all A , B and C income = 96000/3

⇒ Average of all A , B and C income = Rs. 32,000

∴ Average of all A , B and C income is Rs. 32,000

Given:

The average of 11 results = 60

The average of the first six results = 52

The average of the last six results = 64

Formula:

6^th result = total no of first six results + total no. of last six results – total no. of 11 results

Calculations:

6^th result = Total no of first six results + total no. of last six results – total no. of 11 results

⇒ (6 × 52) + (6 × 64) – (11 × 60)

⇒ 312 + 384 - 660

⇒ 696 - 660

⇒ 36

∴ The 6^th number is 36

Given:

A’s marks = 40

B’s marks = 60

Concept:

Average marks = Total marks / Total persons

Solution:

A’s marks = 40

B’s marks = 60

Total marks = 40 + 60 = 100

No. of persons = 2

Average = 100 / 2 = 50

∴ Average is 50.

Given:

Total number of girls = 20

Average marks of all girls = 50

Total number of boys = 30

Average marks of all boys = 60

Concept:

Average marks = Total marks / Total persons

Solution:

Total marks obtained by all girls = 20 × 50 = 1,000

Total marks obtained by all boys = 30 × 60 = 1,800

Average marks of whole class = (1,000 + 1,800)/(20 + 30)

⇒ 2,800/50 = 56

Given:

Marks of 10 persons = 40, 60, 50, 20, 30, 40, 50, 50, 10, 20

Concept:

Average marks = Total marks / Total persons

Solution:

Total marks = 40 + 60 + 50 + 20 + 30 + 40 + 50 + 50 + 10 + 20 = 370

Total number of persons = 10

Given:

Ram’s Hindi marks = 20

Ram’s English marks = 2 × Hindi

Ram’s Science marks = 20 + English

Ram’s Math marks = 50

Concept:

Average marks = Total marks / Total persons

Solution:

Ram’s Hindi marks = 20

Ram’s English marks = 2 × Hindi

Ram’s English marks = 2 × 20 ⇒ 40

Ram’s Science marks = 20 + English

Ram’s Science marks = 20 + 40 ⇒ 60

Ram’s Math marks = 50

Ram’s total marks = 20 + 40 + 60 + 50 ⇒ 170

Total subject = 4

Given:

Average of 17 number is 15.

Average of first 9 numbers is 14.

Average of last 9 numbers is 16.

Formula Used:

Average = Sum of all observations/Total number of observations

Calculation:

For all observations,

15 = Sum of all 17 observations/17

⇒ Sum of all 17 observations = 15 × 17

⇒ Sum of all 17 observations = 255

For first 9 numbers,

14 = Sum of first 9 numbers/9

⇒ Sum of first 9 numbers = 14 × 9

⇒ Sum of first 9 numbers = 126

For last 9 numbers,

16 = Sum of last 9 numbers/9

⇒ Sum of last 9 numbers = 16 × 9

⇒ Sum of last 9 numbers = 144

Sum of first 9 number + Sum of last 9 number,

⇒ Sum of total 18 numbers

I.e. Middle number is added twice

For middle number,

⇒ Sum of 18 numbers – Sum of 17 numbers

⇒ (126 + 144) – 255

⇒ 270 – 255

⇒ 15

∴ The middle number is 15.

Given:

Total persons = 8

Total height = 2,072 cm

Concept:

Average height = Total height / Total persons

Solution:

Given:

Jyoti’s weight = 60 kg

Neel’s weight = 50 kg

Concept:

Average weight = Total weight / Total persons

Solution:

Total weight = 60 + 50 ⇒ 110 kg

Total persons = 2

Given:

A’s height = 140 cm

B’s height = 130 cm

Concept:

Average height = Total height / Total persons

Solution:

Total height = 140 + 130 ⇒ 270

Total persons = 2

Given:

B’s weight = 45 kg

A’s weight = 2 × B’s weight

C’s weight = 10 + average weight of A and B

Concept:

Average weight = Total weight / Total persons

Solution:

B’s weight = 45 kg

A’s weight = 2 × B’s weight

A’s weight = 2 × 45 ⇒ 90 kg

Average weight of A and B = (45 + 90) / 2 ⇒ 135 / 2 ⇒ 67.5

C’s weight = 10 + average weight of A and B

= 10 + 67.5

= 77.5 kg

Given:

Average of four numbers = 12

Two other number with average 15 is added to the collection

Concept:

Total sum = Total number × average

Calculation:

Total sum of 4 numbers = 4 × 12 = 48

Total sum of other 2 numbers = 2 × 15 = 30

Total sum of 6 numbers = 48 + 30 = 78

New average of resulting six numbers = 78/6 = 13

Given:

The average of twelve numbers = 39.

The average of the last five numbers = 35,

The average of the first four numbers = 40.

The fifth number is 6 less than the sixth number and 5 more than the seventh number. 

Concept Used:

Sum = Average × total number

Solution:

The Sum of the twelve numbers = 12 × 39 = 468

The sum of the last five number = 5 × 35 = 175

The sum of the first four number = 4 × 40 = 160

The sum of the fifth, sixth and seventh number = (468 – 175 – 160) = 133

Let, the fifth number is x

Then, the sixth number is (x + 6) and the seventh number is (x – 5)

x + (x + 6) + (x – 5) = 133

⇒ 3x + 1 = 133

⇒ 3x = 132

⇒ x = 44

The fifth number is 44 and the sixth number is (44 + 6) = 50

Average of fifth and sixth number is (44 + 50)/2 = 47

∴ The average of fifth and sixth number is 47

Given:

The average age of a teacher and three students is 20 years.

Formula used:

Average = Total sum of all numbers/Number of values

Calculation:

Let the age of teachers be y.

And the age of Student be x.

Difference y = (x + 20)

According to the question:

Average age = (x + x + x + y)/4 = 20 years

⇒ (3x + y) = 20 × 4

⇒ 3x + y = 80

⇒ 3x + x + 20 = 80

⇒ 4x = 60

⇒ x = 15

⇒ y =  (x + 20)

⇒ y = 15 + 20 = 35

∴ The age of the teacher is 35 years.

Given:

Total number of boys and girls in the class = 50

The average age of all the boys = 20 years

The average age of all the girls = 15 years

Total number of boys = Total number of girls + 6

Formula Used:

Average age of a group of people  = Sum of age of all the people in the group/Total number of people in the group

Calculation:

Let the total number of girls in the class be x

So, the total number of boys in the class is (x + 6)

Hence, we get: 

x + (x + 6) = 50

⇒ x = 22

So, the total number of girls in the class = 22, and 

The total number of boys in the class = 22 + 6 = 28

So, we obtain the total age of all the boys and girls in the class = (28 × 20) + (22 × 15) = 890

The average age of all the 50 students in the class = 890/50 = 17.8 years

∴ The average age of the whole class is 17.8 years

Given:

There are 4 adults and the consumption of rice per month = 10 kg per head.

The average consumption of rice per head per month is 8.5 kg.

The average consumption for adults is 10 kg per head and for minor it is 5.5 kg per head.

Solution:

Let the number of minors be x.

Number of adults = 4

Total consumption of adults = 4 × 10 kg = 40 kg

Total consumption of minors = 5.5 × x = 5.5x kg

According to the question:

(40 + 5.5x)/(4 + x) = 8.5

⇒ 40 + 5.5x = 34 + 8.5x

⇒ 40 – 34 = 8.5x – 5.5x

⇒ 3x = 6

⇒ x = 2

 ∴ The number of minors in the family is 2.

Given:

Average of x numbers is y2.

Average of y numbers is x2.

Concepts used:

Average = Sum of all observations/Number of observations

Calculation:

Sum of x numbers = xy²

Sum of y numbers = x²y

Total sum of x and y observations = xy² + x²y = xy × (x + y)

Average of x + y observations = xy × (x + y)/(x + y) = xy

∴ The average of x + y observations is xy.

Given:

M is 4 years younger than twice the average of his current age and his age 3 years ago.

Concepts used:

Average = Sum of all observations/Number of observations

Calculation:

Let the current age of M be x years.

M's age 3 years ago = x – 3 years

Average = Sum of all observations/Number of observations

⇒ Average of M's current age and his age 3 years ago = (x + x – 3)/2 = (2x – 3)/2

Twice the average of M's current age and his age 3 years ago = 2 × [(2x – 3)/2] = 2x – 3

M today is 4 years younger than twice the average of his current age and his age 3 years ago.

⇒ 2x – 3 – 4 = x

⇒ 2x – x = 7

⇒ x = 7 years

∴ Current age of M is 7 years.

Given:

The average of four consecutive even numbers a, b, c  and d = 33

Formula used:

Average = the sum of observations/the number of observations.

Calculation:

Let the four even consecutive numbers with respect to a, b, c and d be x - 2, x, x + 2  and x + 4

Average of four consecutive even number = [(x - 2) + x + (x + 2) + (x + 4)]/4

⇒ 33 = (4x + 4)/4

⇒ 4(x + 1)/4 = 33

⇒ x + 1 = 33

⇒ x = 33 - 1

⇒ x = 32

Then b = x

⇒ b = 32

And d = x + 4

⇒ d = 32 + 4

⇒ d = 36

Product of b and d = 32 × 36

⇒ 1152

∴ The product of b and d is 1152

Given:

Average = 27

Number of values = 8

Formula used:

Average = sum of value/number of values

Calculation:

according to question

⇒ (24 + 8 + 35 + 40 + x + 26 + 32 + 52)/8 = 27

⇒ (217 + x)/8 = 27

⇒ 217 + x = 216

⇒ x = 216 - 217

⇒ x = -1

∴ Value of x is -1.

Given:

Average age of all the employees in a company = 32 years.

The ratio of the average age of all female to average age of all male employees = 6 : 7.

 total employees in the company = 50

Formula Used:

Average = Sum of the value of the numbers/number of numbers

Calculation:

Let the average age of female and male be 6x and 7x respectively.

Total no. of employees = 50

No. of females = 50 × 60/100 = 30

Then, No. of males = 50 - 30 = 20

According to the question;

⇒ 32 × 50 = 6x × 30 + 7x × 20 

⇒ 1600 = 180x + 140x

⇒ 320x = 1600

⇒ x = 5

The average age of male = 7 × 5 = 35 years

∴ The average age of the male employees is 35 years.

(d) Let A = x,
According to the question,
 A + B + C + D + E
= x + (x + 2) + (x + 4) + (x + 6) + (x 8)
⇒5x + 20 = 34 × 5 = 170
⇒B × D = 32 × 36 = 1152

(b) Age of the fourth friend = 31 × 4 - 32 × 3
= 124 - 96 = 28 years

Given:

Average height of five boys = 153 cm

The height of last boy is one-fourth of the sum of four boys.

Formula used:

Average = ( sum of the observations)/(number of the observations)

Calculation:

Let the height of the boys be a, b, c, d, and e.

According to the question.

e = (a + b + c + d)/4

⇒ a + b + c + d = 4e

Average = ( sum of the observations)/(number of the observations)

⇒ sum of the observations = average × (number of the observations)

⇒ a + b + c + d + e = 153 × 5

⇒ 4e + e = 765

⇒ 5e = 765

⇒ e = 153

∴ The height of the last boy is 153 cm.

Given:

The average height of three boys is thrice height of the fourth boy.

The average height of all boys = 165 cm.

Formula used:

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

Calculation:

Let the height of boys be w, x, y and z.

According to the question,

(w + x + y)/3 = 3z

⇒ w + x + y = 9z      ----(I)

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

⇒ Sum of total observations = average × (numbers of observations)

⇒ w + x + y + z = 165 × 4

⇒ 9z + z = 660      (using equation (I))

⇒ 10z = 660

⇒ z = 66 cm

∴ The height of fourth boy is 66 cm.

Given:

Pooja’s total subject = 6

Pooja’s total marks = 480

Concept:

Average marks = Total marks / Total persons

Solution:

Required average =7(1+2+3+…….+20)/20 

(7*20*21)/(20*2) 

(147/2)

= 73.5.

Let there be n numbers in the original list and let their mean be y.
Then, sum of n numbers = ny

According to Question

 \(⇒ (ny+40)/(n+1)=y+4 ……(1 ) \)

\(⇒(ny+40+30)/(n+2 )=y+5 ……(2 )\)

On solving Eq 1

ny + 4n +y+4 = ny + 40

4n + y = 36× 5              ……(3 )

Similarly eq 2

5n + 2y = 60× 4           ……(4 )

On solving eq 3 and 4 by elimination

y =  20 , n = 4

Sum of first n natural numbers=n(n+1)/2; 

So,sum of 40 natural numbers=(40*41)/2 = 820. 

Therefore the required average=(820/40) = 20.5.