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Given,

Average age of three BOYS = 22

Total Age of three boys = 66 YEARS

Let the ages of three boys is 6a, 9A and 7a respectively.

Then,

⇒ 6a + 9a + 7a = 66

⇒ 22a = 66

⇒ a = 3

We know,

Average of quantities =sum of all quantities/no. of quantities

The SCHOOL has 4 sections of Physics in Class XI having 45, 30, 50 and 52 students.

The mean marks obtained in Physics test are 60, 75, 55 and 80 RESPECTIVELY for the 4 sections.

The TOTAL marks of 45 students in Physics in 1^st section = 45 × 60 = 2700

The total marks of 30 students in Physics in 2^nd section = 30 × 75 = 2250

The total marks of 50 students in Physics in 3^rd section = 50 × 55 = 2750

The total marks of 52 students in Physics in 4^th section = 52 × 80 = 4160

Total number of students = 45 + 30 + 50 + 52 = 177

Total marks of 177 students in Physics = 2700 + 2250 + 2750 + 4160 = 11860

Total NUMBER of EMPLOYEES = 2 + 12 + 1 = 15

Bonus for Managers = 2 × (0.4 × 30L) = 24L

Bonus for EXECUTIVES = 12 × (0.2 × 8L) = 19.2L

Bonus for TRAINEE = 0.4L

Total bonus = 43.6L = 4360000

The average number of boys in 8 classes of a school is 42.

So, total number of boys = 8 × 42 = 336

If in three of the classes, 4 boys are replaced by GIRLS, the number of boys will decrease by 12, and number of girls will increase by 12.

Now, average number of girls BECOMES 47.5.

⇒ Total number of girls = 8 × 47.5 = 380

⇒ Total number of students = (336 – 12) + 380 = 704

Let the numbers be (2N – 1), (2n + 1) and (2n + 3).

AVERAGE of the numbers = ((2n – 1) + (2n + 1) + (2n + 3))/3 = 2n + 1

Given, this average is 14 more than 1/3^rd of 2n – 1

⇒ 2n + 1 = 14 + (2n – 1)/3

⇒ 6n + 3 = 42 + 2n – 1

⇒ 4n = 38

Let the seventh number is X.

Sum of NINE numbers = 18 × 9 = 162

X + (X + 5) + (X + 6) = 162 – 42 – (46/3 × 3) = 74

⇒ 3X + 11 = 74

⇒ X = 21

∴ 9th number = 21 + 6 = 27

The increase in total age = 8 × 3 = 24 YEARS

If two people whose ages are 21 and 23 years are removed, then they have to replace by 2 men whose ages have a total sum of 21 + 23 + 24 years.

Total ages of two NEW men = 68 years

$(Average = \frac{{Sum\;of\;all\;observations}}{{Number\;of\;observations}})$

∴ Average ages of two new men = 68/2 = 34 years

LET the numbers be x, x+2, x+4 and x+6 then:

Formula:

$(\eqalign{ & AVERAGE = {{SUM\ of\ elements} \over {number\ of\ elements}} \CR & Required\ average = {{x + x + 2 + x + 4 + x + 6} \over 4} \cr & \Rightarrow 29 = {{4x + 12} \over 4} \cr & \Rightarrow 29 = x + 3 \cr & \Rightarrow x = 26 \cr})$

Hence, the largest numbers = x+6 = 26 + 6 = 32

Let ages of P, Q and R be a, b and c years respectively.

According to the given INFORMATION:

a + b + 2c = 61 -------(i)

3a + b + c = 45 -------(ii)

a + 3B + 2c = 89------(iii)

Solving these three equations SIMULTANEOUSLY, we get:

AVERAGE HEIGHT of A, B, & C = 148

So, A + B + C = 148 × 3 = 444----(1)

Average height of A & B = 136

So, A + B = 136 × 2 = 272----(2)

Average height of B & C = 125

So, B + C = 125 × 2 = 250----(3)

Subtract equation (2) from equation (1)

A + B + C – A – B = 444 – 272

⇒ C = 172

Putting value of C in equation (3), we get

B + 172 = 250

The average age of A and B = 20 years

The SUM of age of A and B = 20 × 2 = 40 years

The average age of B and C =18 years

The sum of age of B and C = 18 × 2 = 36 years

The average age of A and C = 21 years

The sum of age of A and C = 21 × 2 = 42 years

The sum of age of A + B + C = (A + B + B + C + A + C) = 40 + 36 + 42 = 118

⇒ 2(A + B + C) = 118

⇒ (A + B + C) = 59

Average of A + B + C = 59/3 = 19.67

LET there be 100 students in the group.

The total score of the group = 5200

The total score of the brightest 20% students = 80 × 20 = 1600

The total score of the dullest 25% students = 31 × 25 = 775

The total score of the other students = 5200 – (1600 + 775) = 2825

From the given data

Average of the five subjects = total marks of the subjects/no of subjets = (65 + 55 + 80 + 75 + 85)/5 = 72

Given that TEN POINTS are to be deducted from this average marks due to LACK of attendance

⇒ student net average marks = 72 – 10 = 62

Half of STUDENTS, i.e., 60/2 = 30 have a height of 160 cm.

One third of students, i.e., 60/3 = 20 have a height of 162 cm.

Remaining students = 60 – 30 – 20 = 10

Let height of remaining students be T cm.

Average height = 161.2 cm

⇒ 161.2 = [(30 × 160) + (20 × 162) + 10T]/60

⇒ 9672 = 4800 + 3240 + 10T

⇒ T = 163.2

∴ Height of remaining students is 163.2 cm.

AVERAGE age of 5 children = 5 years

So, total sum of age of the 5 children = 25

Average age of 5 children including their father = 5 + 5 × 100/100 = 10 years

So, total sum of age of the 5 children including their father = 6 × 10 = 60 years

$(Average = \frac{{sum}}{{Number\;of\;terms}})$

⇒ Sum = Average × Number of terms

Let the sum of other six numbers be x, then as PER question

Sum of 9 numbers = Sum of 3 numbers + sum of other 6 numbers

⇒ 100 × 9 = 40 × 3 + x

⇒ x = 900 – 120 = 780

∴ Average of other six numbers = 780/6 = 130

Let the first number be x.

x = (3/10) × (sum of other three SIDES)

Therefore, sum of other three sides = (10/3) × x

Average = Sum of four sides/4

Average = (x + sum of other three sides)/4

Therefore, Average = (x + 10x/3)/4

Since average = 91

91 = (13x/3)/4

x = (91 × 12)/13

x = 84

We know that, AVERAGE = Sum of all quantities/number of quantities

Average of 7 NUMBERS = 28

∴ sum of all numbers = 28 × 7 = 196

Average of first 3 numbers is 23.

Sum of first 3 numbers = 3 × 23 = 69

Average of last 3 numbers os 42.

Sum of last 3 numbers = 3 × 42 = 126

We know that the first 5 odd numbers are 1, 3, 5, 7, 9

Sum of the squares of the five odd numbers = 1² + 3² +5² + 7² + 9² = 1 + 9 + 25 + 49 + 81 = 165

Average = Sum of all the numbers/Total numbers = 165/5 = 33

Let five positive numbers be a, b, c, d and e

The average of five positive numbers is 319.20

⇒ ( a + b + c + d + e )/5 = 319.2

⇒ a + b + c + d + e = 319.2 × 5 = 1596. . . . equation (1)

The average of first TWO numbers is 482.5

⇒ (a + b)/2 = 482.5

⇒ a + b = 482.5 × 2

⇒ a + b = 965. . . . equation (2)

The average of last two numbers is 258.5

⇒ (d + e)/2 = 258.5

⇒ d + e = 258.5 × 2 = 517. . . . equation (3)

Using equations (2) and (3) in equation (1), we get

⇒ 965 + c + 517 = 1596

⇒ c = 1596 – (965 + 517) = 1596 – 1482 = 114

We know that, average = SUM of all quantities/Number of quantities

Average age of family = Sum of ages of all family members/Number of members

∴ 16 = Sum of ages of 4 family members five years AGO/4

⇒ Sum of ages of 4 family members 5 years ago = 4 × 16 = 64

In five years, each of these 4 members’ ages have increased by 5.

∴ Sum of present ages of 4 family members = Sum of ages of 4 family members 5 years ago + (5 × 4)

∴ Sum of present ages of 4 family members = 64 + 20 = 84

Let the age of the BABY be X years.

∴ Sum of present ages of all family members including baby = 84 + x

According to the information given in the question, average present age of family members including baby, is 1 more than what it was 5 years ago excluding baby, i.e. 16 + 1 = 17

∴ Average = 17

⇒ 17 = (84 + x)/5

⇒ 84 + x = 17 × 5 = 85

⇒ x = 1

Let the first integer be 2X.

Average of 6 consecutive integers: (2x + (2x + 2) + ... + (2x + 10))/6= 12x + 2(1 + 2 + ... + 5) = 12x + 30 = 25 × 6 = 150

⇒ x = 10.

Average of NEW numbers: (2x + (2x + 2) + ... + (2x + 12) )/ 7 = (14x + 42)/7 = 2x + 6 = 20 + 6 = 26.

Let the FIFTH person expenditure be Rs x

According to the question

⇒ {(500 × 4) + x}/5 + 100 = x

⇒ x = 625

SUM of ages of 6 STUDENTS 1 YEAR ago = 16.5 × 6 = 99

Sum of CURRENT ages of 6 students = 99 + 6 × 1 = 105

Since the average remains same even after adding 1 more student:

∴ Sum of Current ages of 6 students = 16.5 × 7 = 115.5

∴ Age of NEW student = 115.5 - 105 = 10.5 years

Let the average after 13^th MATCH be ‘X’

Average before 13^th match = x - 4

Now, TOTAL runs after 13^th match = Total runs before 13^th match + runs SCORED in 13^th match

⇒ 13x = 12(x - 4) + 95

⇒ 13x - 12x = 95 - 48

⇒ x = 47

∴ Average after 13^th match = 47

The average of three NATURAL numbers, P, Q and R is 35.

⇒ (P + Q + R)/3 = 35

⇒ (P + Q + R) = 105

The average of P and Q is 42 while average of Q and R is 25.

⇒ (P + Q)/2 = 42 and (Q + R)/2 = 25

⇒ P + Q = 84 and Q + R = 50

Solving, we get Q = 84 + 50 – 105 = 29

⇒ P = 84 – 29 = 55 and R = 50 – 29 = 21

LET the number of STUDENTS be ‘x’.

⇒ Sum of AGES of all students/x = 8

⇒ Sum of ages of all students = 8x

Now, Sum of ages of teachers/12 = 45

⇒ Sum of ages of teachers = 540

Average of all;

⇒ (Sum of ages of students + Sum of ages of teachers)/(x + 12) = 9

⇒ 8x + 540 = 9x + 108

⇒ 9x – 8x = 540 – 108

⇒ x = 432

∴ Total number of students = x = 432

Let the INITIAL average be ‘y’

∴ The initial sum of the AGES of the 20 employees was 20Y

Let ‘x’ be the age of the person who retired

∴ After the person RETIRES, the sum will be 20y – x + 23

We know that, Sum/(Number of values) = Average

According to the given condition,

⇒ (20y – x + 23)/20 = y – 3

∴ 20y – x + 23 = 20y – 60

∴ x = 83

∴ The age of the person who retired was 83

Total COST of PRICE of three shirts = 1250 × 3 = RS. 3750

Cost of two more shirts = 1450 × 2 = Rs. 2900

The average price of all the 5 shirts he BUYS = (3750 + 2900)/5

We know that, Average of 20 students = (sum of N NUMBERS)/n

∴ x = (20 × 60)/20 = Rs60

Amount DEPOSITED by 21^st student = 60 + 15 = Rs. 75

After 5 years, the AGE of eldest person = 95 years

After 5 years, the age of youngest person = 15 years

After 5 years, average age of remaining 5 persons = (22 + 5) years = 27 years

After 5 years, SUM of AGES of remaining 5 persons = 27 × 5 = 135 years

When eldest member DIES, family will have six members.

Sum of the n^th term = n/2 × (2a + (n – 1) × d)

n = number of terms

a = first term

d = Common difference

Sum of the 30^th terms = 30/2 × (2 × 9 + (30 – 1) × 9)(n = 30, a = 9, d = 9)

= 15 × (18 + 29 × 9)

= 4185

Sum of the first 30 multiples of 9 = 4185

Actual weight of 45 kids

= 45 x 33 + 27 - 38

= 1485 - 11 = 1474 kg.

∴ REQUIRED actual average weight of those 45 kids

= 1474/45= 32.76 kg.

Quicker Approach

Decrease in weight = 38 - 27 = 11 kg

∴ Required actual average weight

$( = \left( {33\;-\;\frac{{11}}{{45}}} \right) = \;32.76\;kg)$

Let the three numbers be X, Y and Z

The average of the first and second number is (X + Y)/2

And the average of the second and THIRD numbers is (Y + Z)/2

We find the difference between these TWO, and it WOULD equal 5:

⇒ (X + Y)/2 - (Y + Z)/2 = 5

⇒ (X – Z)/2 = 5

TOTAL COST of 4 items = 4 × 1250 = RS. 5000

Total cost of 5 items = 5000 + 2000 = Rs. 7000

∴ New AVERAGE = 7000/5 = Rs. 1400

<P>LET the weight of A, B and C be p, q and r

According to the problem statement

⇒ (p + q + r)/3 = 49

⇒ p + q + r = 147----(1)

Now given that average weight of A and B is 35 kg

⇒ (p + q)/2 = 35

⇒ p + q = 70 ----(2)

Now, (1) – (2)

⇒ r = 147 – 70 = 77----(3)

Now given that average weight of B and C is 62 kg

⇒ (q + r)/2 = 62

⇒ q + r = 124 Using (3)

⇒ q = 124 – 77 = 47 YEARS

∴ the weight of B is 47 years

Total expenditure of 5 days = 130 × 5 = 650

Total expenditure of first 4 days = 100 + 125 + 85 + 160 = 470

∴ Expenditure of 5^th day = 650 – 470 = 180

Given data -

Average weight of Sunil and Ravi = 45 kg

Weight of Sunil and Ravi = 45 × 2

Weight of Sunil and Ravi = 90 kg- - - (1)

& Average weight of Ravi and Pranit = 41 kg

Weight of Ravi and Pranit = 41 × 2

Weight of Ravi and Pranit = 82 kg- - - - (2)

& Average weight of A & C = 42 kg

Weight of Sunil and Pranit = 42 × 2

Weight of Sunil and Pranit = 84 kg- - - - (3)

ADDING results (1), (2) & (3) -

Weight of Sunil and Ravi + Weight of Ravi and Pranit + Weight of Sunil and Pranit = 90 + 82 + 84

⇒ 2 × Weight of Sunil + 2 × Weight of Ravi + 2 × Weight of Pranit = 256 kg

⇒ 2(Weight of Sunil + Weight of Ravi + Weight of Pranit) = 256 kg

⇒ Weight of Sunil + Weight of Ravi + Weight of Pranit = 128 kg - - - - (4)

Subtracting result (2) from result (4) -

(Weight of Sunil + Weight of Ravi + Weight of Pranit) - (Weight of Ravi + Weight of Pranit) = 128 - 82

∴ Weight of Sunil = 46 kg

The AVERAGE age of a jury of 5 is 40.

∴The total age of the MEMBERS of the jury is (5 × 40) = 200.

A member aged 35 resigns.

Current total age of the jury of 4:

 = 200 – 35

 =165

A man aged 25 BECOMES a member of the jury.

New total age of the jury of 5:

 = 165 + 25

= 190

The new average age of the jury of 5 is:

= 190 ÷ 5

LET the numbers be ‘X’ and ‘y’

Sum of first 20 multiples of x = x(1 + 2 + … + 20) = (x × 20 × 21)/2 = 210x

Sum of first 20 multiples of y = y(1 + 2 + … + 20) = (y × 20 × 21)/2 = 210y

Their average = (210x + 210y)/40

⇒ 924 = 21(x + y)/4

⇒ x + y = 176

Since a, b, c, d, e, f, g be consecutive even numbers;

Suppose a = 2, b = 4, c = 6, d = 8, e = 10, f = 12 and g = 14.

Since j, K, l, m, n be consecutive odd numbers;

Suppose j = 1, k = 3, l = 5, m = 7 and n = 9.

∴ Average of all the numbers = (a + b + c + d + e + f + g + j + k + l + m + n)/12

Putting the values of every term as SUPPOSED above;

∴ Average of all the numbers = (56 + 25)/12 = 81/12 = 27/4

On checking with the options, any OPTION will not be satisfied with this data.

Average age after adding the age of servant will be = 20 × (1 + 25/100)

⇒ average age after adding SERVANTS age = 25 years

Let servants age be x, from the QUESTIONS statement

⇒ sum of age of all family MEMBERS except servant/6 = 20

⇒ sum of age of all family members except servant = 120….(i)

Now,

Given that (sum of age of all family members except servant + x)/7 = 25

⇒ sum of age of all family members except servant + x = 7 × 25

From (i) we get

⇒ x = 7 × 25 – 120

⇒ x = 55

∴ servant’s age is 55 years

We know that, formula for average:

Weighted average of K groups with averages (A_1, A_2, A_{3, ………}A_k)and having elements (N_1, n_2, n_3……….n_k) respectively would be given by:

$({A_k} = \frac{{\LEFT( {{{\bf{n}}_1}{{\bf{A}}_1},\; + {{\bf{n}}_2}{{\bf{A}}_2} + {{\bf{n}}_3}{{\bf{A}}_3} +\ldots\ldots\ldots {{\bf{n}}_k}{{\bf{A}}_k}} \right)}}{{\left( {{n_1} + {n_2} + {n_3} +\ldots\ldots .{n_k}} \right)}})$

∴ Required average marks = (45 × 75 + 40 × 78)/(45 + 40)

= (3375 + 3120)/85

= 6495/85 = 76.41

Given,

3 years AGO, the average age of 3 siblings was 13 years

Average = (Sum of ages of PERSONS)/Number of persons

Sum of ages of 3 siblings before 3 years = 13 × 3 = 39 years

Sum of ages of 3 siblings at PRESENT = 39 + 9 = 48 years.

Given,

At present the sum of age of father and MOTHER is 82 years.

At present Sum of ages of whole family

= Sum of ages of 3 siblings at present + Sum of ages of Father and mother at present

⇒ 48 + 82 = 130 years

Average = (Sum of ages of whole family after 4 years)/5

Average = 130/5 = 26 years.

Sum of 12 numbers = 12 × 35 = 420

Remaining average = [420 - (20 + 27 + 15 + 29)]/8 = 41.125

Average = (sum of elements)/(number of elements)

Let the number of ITEMS sold by Ram in a day be x and total amount earned after selling all items EXCEPT one WRONGLY calculated price item be Rs. m.

Average price of all items sold in a day = OLD average = (total amount earned after selling all items)/number of items

= (m + 150)/x

Given,

The cost of one sold item is wrongly calculated as 150 INSTEAD of 250, so average price is decreased by Rs. 10

Average of all items with new price 250 = new average = (total amount earned after selling all items)/(x)

= {(m + 250)/x}

Old average + 10 = new average

⇒ {(m + 150)/x} + 10 = (m + 250)/x

⇒ (m + 150) + 10x = (m + 250)

⇒ 150 + 10x = 250

⇒ 10x = 100

⇒ x = 10

The total weight of all the employees = NUMBER of employees × average weight

Total weight = 45 × 64 = 2880 kg

Given that, fewer employees have boarded the BUS,

Let the number of employees on that DAY be ‘x’

As the total weight MUST REMAIN the same,

New increased average weight = 72/kg

⇒ x × 72 = 2880

⇒ x = 40

Total expenditure for 7 shirts = 7 × 300 = Rs. 2100

Total expenditure for 2 trousers = 2 × 450 = Rs. 900

Total expenditure for 1 pair of SHOES = Rs. 500

Total amount spent = 2100 + 900 + 500 = Rs. 3500

Total article = 7 + 2 + 1 = 10

Now,

The average of 23, 34, X, 21, 19 and 18 = 21

$(\RIGHTARROW \;\frac{{23 + 34 + x + 21 + 19 + 18}}{6} = 21)$

⇒ 23 + 34 + x +21 + 19 + 18 = 21 × 6

⇒115 + x = 126