FORMULAE USED:

$\Rightarrow \left\{A_E=\frac{S_E}{n_E}or{S}_E=A_E\times n_E\right\}$

Where,

S_E = sum of entities,

n_E = number of entities,

A_E = Average of entities

CALCULATION:

Let the average of prime numbers be P and average of composite numbers be C.

Again the number of prime numbers be x, then the number of composite numbers be 2x.

Then $\frac{Px+2Cx}{3x}=9\Rightarrow P+2C=27$                   …… (1)

And $\frac{2Px+Cx}{3x}=11$

⇒ 2P + C = 33                     ……………. (2)

On adding eq. (1) and (2)

We get,

P + C = 20

And on subtracting eq. (1) from (2)

We get,

P – C = 6

Therefore, P = 13 and C = 7

Thus, (C/P) = (7/13)

After investing 200 + 400 + 250 = 850,

The remaining sum = 1000 – 850 = 150

Let, he gets x% on 150.

Average interest = 3.85%, which shall be calculated on the total sum of 1000.

So, total interest = (3.85/100) × 1000 = 38.5

Interest on 200 = (3/100) × 200 = 6

Interest on 400 = (3.5/100) × 400 = 14

Interest on 250 = (5/100) × 250 = 12.5

Interest on 150 = (x/100) × 150 = 1.5x

∴ 38.5 = 6 + 14 + 12.5 + 1.5x

⇒ x = 4%

Alternatively,

We can use the weighted average concept as:

[(3 × 200) + (3.5 × 400) + (5 × 250) + (x × 150)]/1000 = 3.85

600 + 1400 + 1250 + 150x = 3850

We know that, formula for average-

$\Rightarrow \left\{A_E=\frac{S_E}{n_E}or{S}_E=A_E\times n_E\right\}$

Where,

S_E = sum of entities,

n_E = number of entities,

A_E = Average of entities.

Now, according to the question,

Average marks obtained by Amitabh, Vinod and Shashi = 80,

∴ Total marks obtained by Amitabh, Vinod and Shashi = 80 × 3 = 240

Similarly,

Average marks obtained by Praveen and Smita = 65.

∴ Total marks obtained by Praveen and Smita = 65 × 2 = 130.

Similarly,

Average marks obtained by Amitabh, Vinod, Shashi, Praveen, Smita, Hema, Rekha and jaya = 72.5,

∴ Total marks obtained by Amitabh, Vinod, Shashi, Praveen, Smita, Hema, Rekha and jaya = 72.5 × 8 = 580

Now,

Total marks obtained by Hema, Rekha and jaya

= Total marks obtained by all of them – total marks obtained by Amitabh, Vinod and Shashi – total marks obtained by Praveen and Smita

⇒ Total marks obtained by Hema, Rekha and jaya = 580 – 240 – 130 = 210

∴ Required average marks obtained by Hema, Rekha and Jaya = 210/3 = 70

Hence, the required average marks by Hema, Rekha and jaya are 70.

                                    Alternatively-

We know that, formula:

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:

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

∴ Average marks obtained by Hema, rekha and Jaya $=\frac{\left(72.5\times 8\right)-\left(80\times 3\right)-\left(65\times 2\right)}{3}$

$=\frac{580-240-130}{3}=70$

We know that, formula,

$\Rightarrow \left\{A_E=\frac{S_E}{n_E}or{S}_E=A_E\times n_E\right\}$

Where,

S_E = sum of entities,

n_E = number of entities,

A_E = Average of entities.

Let number of students in the sections A, B, C and D be a, b, c and d respectively.

Then, total weight of students of section

A =45a

Total weight of students of section B = 50b

Total weight of students of section C =72c

Total weight of students of section D = 80d

According to the question,

Average weight of students of sections A and B = 48 kg

$\therefore \frac{45a+50b}{a+b}=48$ 

⇒ 45a + 50b = 48a + 48b

⇒ 3a = 2b

⇒ 15a = 10b

And average weight of students of sections B and C = 60kg.

⇒ 50b + 72c = 60(b + c)

⇒ 10b = 12c

Now, average weight of students of A, B, C and D = 60 kg

∴ 45a + 50b + 72c + 80d

= 60(a + b + c + d)

⇒ 15a + 10b – 12c -20d = 0

⇒ 15a = 20d

⇒ a : d = 4 : 3

We know that,

 $Averageage=\frac{Totalage}{numberofpeople}$

The average age of the grandparents = 67 years

The total age of grandparents = 67 × 2 = 134 years

The average age of parents = 35 years

The total age of parents = 35 × 2 = 70 years

The average age of three children = 6 years

The total age of three children = 6 × 3= 18 years

∴ Average age of the family = $\frac{Totalageofthefamily}{Totalno.offamilymemebers}$

⇒ (134 + 70 + 18)/7 = 222/7

$=31\frac{5}{7}\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}$

∴ The average age of the family is $31\frac{5}{7}\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}$.

Bowling average of a cricketer = 12.4 runs/wicket

We know that Bowling average = (Total runs conceded)/(Total wickets taken)

∴ Total runs conceded = 12.4 × total wickets taken ……………………(1)

Given that in his recent match he conceded 26 runs and takes 5 wickets which decreases his average by 0.4

 ∴ (Bowling average) – 0.4 = (Total runs conceded+26)/(Total wickets taken+5)

 ⇒ 12.4 – 0.4 = (Total runs conceded+26)/(Total wickets taken+5)

 ⇒ 12 = (Total runs conceded+26)/(Total wickets taken+5)

⇒ 12(total wickets taken) + 60 = Total runs conceded+26

⇒ 12(total wickets taken) + 60 = 12.4(total wickets taken) + 26 (∵ From (1))

⇒ 0.4(total wickets taken) = 60 – 26

⇒ 0.4(total wickets taken) = 34

∴ Total wickets taken = 34/0.4= 85

Hence the numbers of wickets taken by the bowler till the last match were 85

Average expenditure of days 1 to 10 = Rs 800

Total expenditure for days 1 to 10 = Rs 800 × 10 = 8000

Total expenditure for days 11 to 20 = Rs (800 – 25) × 10

= Rs 775 × 10

= Rs 7750

Total expenditure for days 21 to 30 = Rs (775 – 25) × 10

= Rs. 750 × 10

= Rs. 7500

Total expenditure= Rs (7500 + 7750 + 8000)

= Rs. 23250

Savings = Income – Expenditure

= Rs 25000 – Rs 23250

= Rs. 1750

Since the month has 30 days and it starts with Saturday hence there will be 5 Sundays in the month and rest 25 will be normal days.

So, total number of visitors who comes on Sunday = 250 × 5 = 1250

And total number of visitors on weekdays = 100 × 25 = 2500

Hence total number of visitors in that month = 2500 + 1250 = 3750

As we know that,

$\text{Average of given entities}=\frac{\text{Sum of the given entities}}{\text{Number of the given entities}}$

So the required average = 3750/30 = 125

Let’s assume that the total initial number of rows of the cinema are x, and each row has y seats.

∵ initially there’re 500 seats,

xy = 500          

⇒ x =500/y

Now, according to the given information:

After the renovation of the cinema, the total number of seats became 10% less

∴New number of seats = 500 – (10% of 500) = 500 – 50 = 450

The number of rows was reduced by 5

∴ New number of rows = x – 5

Each row contained 5 seats more than before

∴New number of seats = y +5

⇒ (x – 5)(y+5)= 450

⇒xy + 5x – 5y – 25 = 450

Substituting the value of xy,

⇒500 – 25 + 5x – 5y = 450

⇒ y – x = 5

Now, putting value from (2)

∴ y – 500/y = 5

⇒ y² – 500 = 5y

⇒ y² – 5y – 500 = 0

⇒ (y +20) (y – 25)

⇒ y = -20,25

∴ Seats of the cinema = 25     (∵ cannot be negative)

∴ 25x = 500

⇒ x= 500/25 = 20

Hence, initially there were 20 rows and 25 seats in the cinema.

First we will find Quantity A,

Quantity A:

Let the present age of Ram = X years

10 years ago, his age was = (X – 10) years

According to given condition,

7/5 × (X – 10) = X

⇒ 7 × (X – 10) = 5X

⇒ 7X – 70 = 5X

⇒ 2X = 70

⇒ X = 35

∴ Age of Ram at his marriage is 25 years.  

Now,

Quantity B:

We know that, formula for average:

⇒ $\left\{A_E=\frac{S_E}{n_E}or{S}_E=A_E\times n_E\right\}$

Where,

S_E = sum of entities,

n_E = number of entities,

A_E = Average of entities.

Now, according to the question,

Total of A, B and C = 25 × 3 = 75

⇒ A + B + C = 75       ---- (1)

Total of A and B = 20 × 2 = 40

⇒ A + B = 40       ---- (2)

From equation (1) – (2) we get,

(A + B + C) – (A + B) = 75 – 40

⇒ C = 35

Clearly, Quantity B > Quantity A

Let the share of X, Y and Z be Rs. ‘x’, Rs. ‘y’ and Rs. ‘z’ respectively

⇒ x + y + z = 8000      ----(1)

Average share of X & Y = 200 + Average share of Y & Z

⇒ (x + y)/2 = 200 + (y+ z)/2

⇒ x + y = 400 + y + z

⇒ x = 400 + z      ----(2)

Also, average share of X & Z = Average share of X & Y – 125

⇒ (x + z)/2 = (x + y)/2 – 125

⇒ x + z = x + y – 250

⇒ y = z + 250      ----(3)

Substituting (2) and (3) in (1),

⇒ 400 + z + z + 250 + z = 8000

⇒ z = 7350/3 = Rs. 2450

⇒ x = 2450 + 400 = Rs. 2850

⇒ y = 2450 + 250 = Rs. 2700

Solving for Quantity A:

⇒ Quantity A = (x + y)/2 = (2850 + 2700)/2 = Rs. 2775

Solving for Quantity B:

⇒ Quantity B = (x + z)/2 = (2850 + 2450)/2 = Rs. 2650

Solving for Quantity C:

⇒ Quantity C = (y + z)/2 = (2700 + 2450)/2 = Rs. 2575

Let the no. of bricks used in building the pillars X, Y & Z be ‘x’, ‘y’ and ‘z’ respectively

Given, x = 45

Average no. of bricks per pillar = 49

⇒ x + y + z = 49 × 3 = 147

⇒ y + z = 102     ––– (1)

Also,

Average no. of bricks used in pillars Y & Z – Average no. of bricks used in pillars X & Y = 12

⇒ (y + z)/2 – (x + y)/2 = 12

⇒ z – x = 24

⇒ z = 24 + 45 = 69

Substituting in (1),

⇒ y = 102 – 69 = 33

⇒ Difference between no. of bricks used in Z & Y = 69 – 33 = 36

According to the given information,

Let the Number of male members = 9

Let the number of female members be y.

We know that, Average = Sum of all observations/Number of observations

The average expenses on shopping by male members is Rs.1500 per head

∴ 1500 = Expense by males/9

⇒ Expense by males = 1500 × 9 = 13500

The average expenses on shopping by female members is Rs.3200per head

∴ 3200 = Expense by females/y

⇒ Expense by females = 3200y

Now, overall average expense = (Expense by males + Expense by females)/(males + females)

$\therefore 2300=\frac{13500+3200y}{9+y}$

⇒ 20700 + 2300y = 13500 + 3200y

⇒ 900y = 7200

⇒ y = 8

Let the original average of the marks of all the candidates = x

And let the highest and least marks be a and b respectively.

According to the question,

98% of x = (76x – a)/75

98% of 75x + a = 76x      ----(1) 

And 103% of x = (76x – b)/75

103% of 75x + b = 76x    ----(2)

From equations (1) and (2)

98% of 75x + a = 103% of 75x + b

⇒ 5% of 75x = a – b

According to the question,

(a – b) = 33 

∴ 5% of 75x = 33

⇒ 5 × 75x/100 = 33 

By solving, x = 8.8

Total employees in the company = 40 

Average age of employees = 35 

Total age of employees = (40 × 35) = 1400 

In next 2 years, 

Total remaining employees = 40 – 10 = 30 

Retirement age = 60 years 

Total age of 30 employees after two years = 1400 + (40 × 2) – (60 × 10) 

= 1400 + 80 – 600 = 880 

∴ Average age after two years = 880/30 = 88/3 years