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According to the given information,

m – 4 = 13

⇒ m = 17

Average of 9 CONSECUTIVE INTEGERS STARTING with k = m = 17

$(\Rightarrow \;\frac{{k\; + \;\left( {k\; + \;1} \right)\; + \;\left( {k\; + \;2} \right)\; + \; \ldots\ldots .\; + \;\left( {k\; + \;8} \right)}}{9}\; = \;17)$ 

⇒ k + (k + 1) + (k + 2) + ……. + (k + 8) = 17 × 9 = 153

⇒ 9k + (8 × 9)/2 = 153

⇒ 9k + 36 = 153

⇒ 9k = 117

⇒ k = 13

∴ Average of 13 consecutive integers starting with

k – 6 = $(\frac{{\left( {k - 6} \right)\; + \;\left( {k - 5} \right)\; + \; \ldots \; + \;\left( {k\; + \;5} \right)\; + \;\left( {k\; + \;6} \right)}}{{13}}\; = \;\frac{{13k}}{{13}}\; = \;13)$

⇒ Average = SUM of elements / number of elements

Given,

The average price of bunch of rings is Rs. 45.

Let the number of rings in a bunch be N.

⇒ Total price of N number of rings = 45 × N

The GOOD 45% of them is of average price of Rs. 60 and DAMAGED 20% of average price of Rs. 55.

⇒ Number of good rings in bunch = (45/100) × N = 9N/20

⇒ Total price of good rings = 60 × (9N/20) = 27N

⇒ Number of damaged rings in a bunch = (20/100) × N = N/5

⇒ Total price of damaged rings = 55 × (N/5) = 11N

Total number of remaining rings =

= 100 - (45 + 20)

= 35%

Total number of remaining rings =

= (35/100) × N

= 7N/20

Total price of remaining rings =

= 45N - (27N + 11N)

= 7N

Total number of remaining rings = 7N/20

Average price of remaining rings =

= Total price of remaining rings / number of remaining rings

= 7N/ (7N/20)

= 20

Average price of remaining rings IS Rs. 20.

GIVEN, AVERAGE marks in Hindi SUBJECT of a CLASS of 54 students is 76.

∴ TOTAL marks scored by 54 students = 54 × 76

⇒ Total marks scored by 54 students = 4104

Given, two marks 36 and 47 were misread as 60 and 77

∴ Actual total marks scored by 54 students = 4104 + 36 + 47 – 60 – 77

⇒ Actual total marks scored by 54 students = 4050

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

Given,

The average of (2, 5 and 8), (13, 0 and 5) and (3, 1 and 2) is a, b and c respectively.

⇒ (2 + 5 + 8)/3 = a

⇒ a = 15/3 = 5

⇒ (13 + 0 + 5)/3 = b

⇒ b = 18/3 = 6

⇒ (3 + 1 + 2)/3 = c

⇒ c = 6/3 = 2

The value of (7c + a × 3b)/13

= {(7 × 2) + (5 × (3 × 6))}/13

= (14 + (5 × 18))/13

= (14 + 90)/13

= 104/13

= 8

Given,

The AVERAGE marks of 45 students was 70 for a unit test:

Average marks = Total marks of students/Number of students

Total marks of all 45 students = 70 × 45 = 3150

Total marks of all 45 students = 3150

Given,

It was noticed that 20^th roll number student marks which is 45 was wrongly taken as 60:

Total sum of marks of 44 students = Total sum of 45 students - 60

= 3150 - 60 = 3090

∴ Total sum of marks of 44 students is 3090.

New total sum of marks of 45 students = 3090 + 45 = 3135

New total sum of marks of 45 students = 3135

New average = 3135/45 = 69.66

∴ The CORRECTED new average marks is 69.66

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

Let the LENGTH of NEW added plastic stick be ‘x' m.

Given,

The average length of 15 wooden sticks is 20 m.

Total length of 15 wooden sticks

= 20 × 15 = 300

Total length of 15 wooden sticks = 300 m

Given,

A new plastic stick of a certain length is added in that 15 wooden bunch then average length is increased by 1 m -

Total length of 15 wooden sticks and 1 plastic stick = (300 + x)m

⇒ New average = 20 + 1 = 21 m

⇒ 21 = (300 + x)/16

⇒ 21 × 16 = 300 + x

⇒ 336 = 300 + x

⇒ x = 36 m

Let the smaller RADIUS be M CM and larger radius be N cm.

⇒ (M + N)/2 = 12.5

⇒ (M + N) = 25

The average area of the TWO circles is 62.5 percent more than area of smaller circle.

⇒ [π M × M + π N × N]/2 = [π M × M]× (1 + 62.5/100)

⇒ M× M + N× N = 3.25M × M

⇒ N× N = 2.25M × M

⇒ N = 1.5M

⇒ M + 1.5M = 25

⇒ M = 25/2.5 = 10 and N = 1.5M = 15

∴ Radius of larger circle is 15 cm.

The given NUMBERS are 16, 17.5, 18.25, 20, 19.5, 21 and X

∴ Sum of all given numbers = 16 + 17.5 + 18.25 + 20 + 19.5 + 21 + x

⇒ Sum of all given numbers = 112.25 + x

We know that,

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

∴ Sum of all observations = Average × Number of observations

? The average of given 7 numbers is 19.25

⇒ Sum of given 7 numbers = 19.25 × 7 = 134.75

⇒ 112.25 + x = 134.75

⇒ x = 134.75 – 112.25

Let present ages of husband, wife and CHILD be m, N and p years respectively.

Given,

⇒ [(m – 5) + (n – 5)]/ 2 = 23

⇒ m + n = 46 + 10 = 56

Given,

⇒ (m + n + p)/3 = 20

⇒ m + n + p = 60

⇒ 56 + p = 60

⇒ p = 4

Given,

The average of five numbers is 11:

Total sum of five numbers = 11 × 5 = 55

Given,

FORTH number is average of third and fifth number and first number is 3/7^th of the fourth number:

⇒ Fourth number = (third number + fifth number)/2

⇒ 14 = (third number + fifth number)/2

⇒ Third number + Fifth number = 28

⇒ First number = 3/7 × (fourth number)

⇒ First number = 3/7 × 14

⇒ First number = 6

Total sum of all five numbers = 55

⇒ 55 = First number + second number + third number + forth number + fifth number

⇒ 55 = 6 + second number + 28 + 14

⇒ Second number = 55 - 48

⇒ Second number = 7

The average of second number and forth number:

⇒ Average = (7 + 14)/2

⇒ Average = 21/2 = 10.5

∴ The average of second number and forth number is 10.5.

Average of five numbers = 66

∴ Sum of five numbers = 66 × 5 = 330

Average of first FOUR numbers = 68

∴ Sum of first four numbers = 68 × 4 = 272

∴ VALUE of the fifth number = 330 – 272 = 58

$({\rm{Average}} = \FRAC{{{\rm{total\;weight\;of\;all\;ITEMS}}}}{{{\rm{number\;of\;items}}}})$

The average weight of 12 apples = 0.5 kg

∴ The total weight of 12 apples = 0.5 × 12 = 6 kg

The average weight of (12 apples and 3 watermelons = 15 items) = 0.5 + 0.5 = 1 kg

∴ Total weight of 15 items = 1 × 15 = 15 kg

Given,

⇒ Average = Sum of terms/Number of terms

For Arithmetic Progression,

If Sum of terms = S_n and first term of progression = a, COMMON DIFFERENCE = d and n = number of terms-

$(\Rightarrow {S_n} = \frac{n}{2}\left( {2a + (n - 1} \right)d))$

For given SERIES -

a = 10, d = 2.5 and n = 20

$(\Rightarrow {S_{20}} = \frac{{20}}{2}\left\{ {2 \times 10 + \left( {20 - 1} \right) \times 2.5} \right\})$

⇒ S₂₀ = 10{20 + 19 × 2.5}

⇒ S₂₀ = 10(20 + 47.5)

⇒ S₂₀ = 10 × 67.5

⇒ S₂₀ = 675

Average of first 20 terms of Arithmetic Progression = 675/20 = 33.75

∴ Average of first 20 terms of Arithmetic Progression = 33.75

Average of 67 and y is 54

⇒ (67 + y)/2 = 54

⇒ 67 + y = 108

⇒ y = 41

Average of x and y is 67

⇒ (x + y) = 67

⇒ x + 41 = 134

⇒ x = 93

AVERAGE = sum of elements/number of elements

Given,

The average of (1, 8 and 9), (12, 0 and a) and (5, B and 9) is c, 5 and 7 respectively =

⇒ (1 + 8 + 9)/3 = c

⇒ c = 18/3

⇒ c = 6

⇒ (12 + 0 + a)/3 = 5

⇒ (12 + a) = 15

⇒ a = 15 – 12 = 3

⇒ a = 3

⇒ (5 + b + 9)/3 = 7

⇒ (14 + b) = 21

⇒ b = 21 – 14 = 7

⇒ b = 7

The value of (2a + 5b – c) =

= (2 × 3 + 5 × 7 – 6)

= 6 + 35 – 6

= 35

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

According to the given information,

Average of 7 numbers is 21

? The average of first and sixth numbers is 19

first number + sixth number = 19 × 2 = 38

And, SINCE Average of the third, fifth and seventh numbers is $(24\frac{1}{3})$

∴ third number + fifth number + seventh number $(= \;24\frac{1}{3}\; \times 3\; = \;73)$

Now, since the second number is less than the FOURTH number by 8, let the second number be X thus, fourth number would be x + 8.

Thus,

21 = (sum of all seven numbers)/7

$(\Rightarrow \frac{{x\; + \;x\; + \;8\; + \;73\; + \;38\;}}{7} = 21)$

⇒ 2x + 119 = 147

⇒ 2x = 28

⇒ x = 14

We know, Average of quantities = sum of all quantities/no. of quantities

The average number of TWITTER followers of all 24 people is 36.

∴ Sum of all 24 numbers = 36 × 24 = 864

The FIRST 12 people have on an average 15 followers.

So, Sum of these 12 numbers = 15 × 12 = 180

Sum of remaining 12 numbers = 864 – 180 = 684

We know that,

Average = (Sum of all quantities)/(Number of quantities)

Let INITIAL average = x

∴ x = (TOTAL runs scored in 12 innings)/12

Total runs scored in 12 innings = 12X

? He scored 78runs in 13^th inning,

Total runs scored in 13 innings = 12x + 78

∴ New average = (12x + 78)/13

Since average increases by 5,

$(\therefore {\RM{\;x}} + 5 = \frac{{{\rm{\;}}12{\rm{x\;}} + 78}}{{13}})$

⇒ 13(x + 5) = 12x + 78

⇒ 13x + 65 = 12x + 78

⇒ x = 13

Given,

The AVERAGE WEIGHT of five different BOXES b1, b2, b3, b4 and b5 is 200 gm:

Average weight = Total weight of items/number of items

Total weight of five different boxes = 5 × 200 = 1000 m

Total weight of five different boxes = 1000 gm

Given,

The average weight of b1 and b3 is 50 gm while average weight of b2 and b4 is 125 gm:

Total weight of b1 and b3 = 50 × 2 = 100 gm

Total weight of b1 and b3 = 100 gm

Total weight of b2 and b4 = 125 × 2 = 250 gm

Total weight of b2 and b4 = 250 gm

Total weight of five different boxes

⇒ 1000 = Weight of b1 + weight of b2 + Weight of b3 + Weight of b4 + Weight of b5

⇒ 1000 = (Total weight of b1 and b3) + (Total weight of b2 and b4) + weight of b5

⇒ 1000 = 100 + 250 + weight of b5

⇒ 1000 = 350 + weight of b5

Weight of b5 = 1000 - 350

Weight of b5 = 650 gm

∴ The weight of box b5 is 650 gm.

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

GIVEN,

The average of (33, 45, 65 and 80) and (21, 9, 11 and 50) is X and Y respectively

⇒ X = (33 + 45 + 65 + 80)/4

⇒ X = 223/4

⇒ X = 55.75

⇒ Y = (21 + 9 + 11 + 50)/4

⇒ Y = 91/4

⇒ Y = 22.75

The value of 2(X + Y)

= 2(55.75 + 22.75)

= 2(78.5)

= 157

Formula:

Average = (Sum of observation)/(No of observation)

 Given that, b is twice of a and thrice of c so, largest number is b. Also, the average of the three NUMBERS is 44, then

b = 2a ⇒ a = ½ b

b = 3c ⇒ c = b/3

And average of a, b, c = (a + b + c)/3 = 44----(1)

PUTTING the value of a and c in this equation, we get

Average of a, b, c = $(\FRAC{{\frac{b}{2} + b + \frac{b}{3}}}{3} = 44)$

$(\Rightarrow \frac{{11b}}{{18}} = 44)$

Let the first even number A be = x + 2

Then E = (x + 10)

Average of A and E = 46

Largest number = $(\FRAC{{{\rm{x\;}} + {\rm{\;}}2{\rm{\;}} + {\rm{\;x\;}} + {\rm{\;}}10}}{2}{\rm{\;}} = {\rm{\;}}46)$

⇒ 2x + 12 = 92

⇒ x = 40

∴ E = x + 10 = 40 + 10 = 50.

Given that, average age of 29 boys is 15 YEARS. But by mistake, my FRIEND, aged 20 years, counted twice and two boys were missed in that count, whose ages differ by 5 years.

Average age of n PERSONS = (Sum of age of n persons)/(Number of persons)

⇒ Incorrect sum of ages of 29 boys = Average age of 29 boys × 29

⇒ Incorrect sum of ages of 29 boys = 15 × 29 = 435 years

Since one boy, of age 20 years, was counted twice,

Sum of 28 boys = 435 – 20 = 415 years.

Now, let the age of YOUNGER boy among the two, who were missed in counting be x years and that of the older boy is (x + 5) years.

⇒ Correct sum of ages of boys of the class = 415 + x + (x + 5) = 420 + 2x

Since, average age of the boys of the class, after correction is still 15 years, so

Average age of 30 boys = $(\frac{{420 + 2x}}{{30}})$ 

$(\Rightarrow \frac{{420 + 2x}}{{30}} = 15)$

⇒ 420 + 2x = 15 × 30

⇒ 420 + 2x = 450

⇒ 2x = 450 – 420 = 30

⇒ x = 30/2

Let’s assume that initial number of students be x.

∴ number of toffees given to x students = 10X

After the arrival of TEACHER and headmaster, total toffees distributed = 10x + 15 + 20 = 10x + 35

Now, the average becomes 11.5,

∴ 10x + 35 = 11.5 × (x + 2)

⇒ 10x + 35 = 11.5 x + 23

⇒ x = 8

∴ the number of students among whom the sweets were distributed is 8.

ACCORDING to the GIVEN information,

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

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:

$(\frac{{\left[ {r + \left( {p - Q} \right)} \right] + \;p + q}}{3} = 21 + \frac{{p + q + r}}{3})$

$(\Rightarrow \frac{{2p + r}}{3} - \;21 = \frac{{p + q + r}}{3})$

$(\Rightarrow \frac{{p - q}}{3} = 21)$

⇒ p – q = 63

Let initial wickets taken be X

Total runs = 24.85x

New average = 24.85 - 0.85 = 24

Total wickets = x + 5

Total runs = 24(x + 5)

24(x + 5) - 24.85x = 52 ⇒ 80

We KNOW that,

Arithmetic mean = (Sum of all terms)/(Number of terms)

According to the given information, the arithmetic mean given 12 numbers is 13.

∴ 13 = (2 + 9 + 13 + 17 + 11 + 5 + 12 + 18 + 14 + 8 + 15 + x)/12

⇒ 13 = (124 + x)/12

∴ 124 + x = 13 × 12

Let the first number be = 3X

Let the second number be = 6x

Let the third number be = 2X

Average of the three NUMBERS is given to be 105 + 2x

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

$(\BEGIN{array}{l} \Rightarrow 105 + 2x = \frac{{3x + 6x + 2x}}{3}\\ \Rightarrow 105 + 2x = \frac{{11x}}{3}\\ \Rightarrow 105 = \frac{{11x}}{3} - 2x\\ \Rightarrow 105 = \frac{{11x - 6x}}{3}\\ \Rightarrow 105 = \frac{{5x}}{3}\\ \Rightarrow x = \frac{{105\; \times \;3}}{5}\\ \Rightarrow x = 21 \times 3 \end{array})$

⇒ x = 63

⇒ 6x = 6 × 63 = 378

As PER GIVEN -

Average height of 18 boys = 5 feet

Total sum of heights of boys = average height × number of boys

Total sum of heights of boys = 5 × 18

Total sum of heights of boys = 90 feet

Also average height of remaining 12 students = 5.5 feet

Total sum of heights of 12 boys = average height × number of boys

Total sum of heights of 12 boys = 5.5 × 12 = 66

Total sum of heights of 30 boys = 90 + 66 = 156

∴ Average height of 30 boys = 156/30 = 5.2

Given,

There are 100 pens of Rs. 10 and 600 pencil of Rs. 5 and 120 notebooks of Rs. 25 :

⇒ Total COST of 100 pens = 100 × 10 = Rs. 1000

⇒ Total cost of 600 pencil = 600 × 5 = Rs. 3000

⇒ Total cost of 120 notebooks = 120 × 25 = Rs. 3000

The average cost of all items AVAILABLE in a Stationery shop:

Average = (total cost of items)/number of items

⇒ Total cost of 100 pens, 600 PENCILS and 120 notebooks = Rs. (1000 + 3000 + 3000) = Rs. 7000

⇒ Average cost of 100 pens, 600 pencils and 120 notebooks = (7000)/(100 + 600 + 120)

= 7000 / 820

= Rs. 8.53

⇒ Average = sum of elements / number of elements

Given,

There are 100, 200, 300 and 220 students were attends the FOUR exams named A, B, C and D respectively and PASSING PERCENTAGE for this four exams is 45%, 65%, 60% and 80% respectively.

Number of passed students in A exam =

= (45/100) × 100

= 45

Number of passed students in B exam =

= (65/100) × 200

= 130

Number of passed students in C exam =

= (60/100) × 300

= 180

Number of passed students in D exam =

= (80/100) × 220

= 176

Number of passed students including four exams =

= 45 + 130 + 180 + 176

= 531

Passing percentage =

= {531/(100 + 200 + 300 + 220)} × 100

= (531/820) × 100

= 64.75%

The passing percentage of an institute including this four exams is 64.75%.

⇒ Average = sum of ELEMENTS / number of elements

Given,

Let the total distance and total time be d and t respectively.

Given,

(30/100)d at 60 km/h

(45/100)d at 45 km/h

(15/100)d at 60 km/h

And remaining 100 - (30 + 45 + 15) = 10%

⇒ (10/100)d at 5 km/h

⇒ t = [(3d/10)/60] + [(9d/20)/45] + [(3d/20)/60] + [(d/10)/5]

⇒ t = 15d/400

Average rate = d/t

Average rate = 400/15 km/h

Rohan’s WEIGHT = ARJUN’s weight + Total no. of boys × increment

⇒ Rohan’s weight = 45 + 10 × 1.5

⇒ 45 + 15 = 60 KG

Average = SUM of elements/number of elements

⇒ PRIME numbers are 2, 3, 5, 7, 11, 13...etc.

Average of five CONSECUTIVE prime numbers:

⇒ (2 + 3 + 5 + 7 + 11)/5

⇒ (28)/5

⇒ 5.6

∴ Average of five consecutive prime numbers STARTING from 2 is 5.6

Let the three CONSECUTIVE odd numbers be x, x + 2, x + 4

We know,

$(\begin{array}{L} Average = \frac{{Sum\;of\;all\;OBSERVATIONS}}{{Number\;of\;observations}}\\ \Rightarrow Average = \frac{{x + \left( {x + 2} \right) + \left( {x\; + \;4} \right)}}{3}\; = \;\frac{{3\;\left( {x + 2} \right)}}{3}\; = \;x + 2 \end{array})$

It is given that Sum of three consecutive numbers = Average + 38

⇒ x + (x + 2) + (x + 4) = (x + 2) + 38

⇒ 2x + 4 = 38

⇒ 2x = 34

⇒ x = 17

⇒ x + 2 = 19

∴ The second number is 19

Paul’s salary = ADAM + 1200----(1) 

Adam/VERONICA = 5/3----(2)

Veronica’s salary = Adam – 1000----(3)

⇒ Adam/(Adam – 1000) = 5/3

⇒ 3 × Adam = 5 × Adam – 5000

Adam’s salary = RS. 2500

Paul’s salary = 2500 + 1200 = Rs. 3700

Veronica’s salary = 2500 × 3/5 = Rs. 1500

⇒ Average = sum of elements / number of elements

Given,

Let the initial production be 100.

Increase in the production of cloths in the FIRST month is 20% =

= 100 + 100 × (20/100)

= 120

Increase in the production of cloths in the second month is 25% =

= 120 + 120 × (25/100)

= 120 + 30

= 150

Increase in the production of cloths in the third month is 15% =

= 150 + 150 × (15/100)

= 150 + 22.5

= 172.5

There is increment in production =

= 172.5 - 100

= 72.5

The average increase RATE =

= 72.5/3

= 24.16

Let the largest even number be x

Let the largest ODD number be y

Now Sum of 8 consecutive even numbers is 456

∴ x + (x – 2) + (x – 4) + (x – 6) + (x – 8) + (x – 10) + (x – 12) + (x – 14) = 456

⇒ 8x – 56 = 456

⇒ 8x = 512

⇒ x = 64

∴ Largest even number is 64

We know that, Average = Sum of numbers/Total numbers

Now average of 4 consecutive odd numbers is 88

$(\therefore \FRAC{{\left[ {y\; + \;\left( {y\; - \;2} \right) + \;\left( {y\;-\;4} \right) + \;\left( {y\;-\;6} \right)} \right]}}{4} = \;88\;)$

⇒ 4y – 12 = 4 × 88

⇒ 4y = 352 + 12

⇒ 4y = 364

⇒ y = 91

∴ Largest odd number is 91

Now, Sum of largest odd and largest even number will be

∴ Sum = x + y = 64 + 91 = 155

RATIO of ducks and frogs in a pond = 35 ? 43

Let, number of ducks = 35x and no. of frogs = 43x

According to PROBLEM,

⇒ (35x + 43x)/2 = 234

⇒ 78X = 234 × 2

⇒ X = 6

∴ The number of ducks in the pond = 35x = 35 × 6 = 210

Average = (sum of observation/no of observation)

Let, Present age of the OLD MEMBER who is replaced = x years

Present age of the new member = y years

Present average of 5 members = z years

Sum of present age of other 4 old member = m years

Given, Number of member = 5

Present average ages of 5 members including the new member

= (Sum of age of the four old member + y)/5 = (m + y)/5

Again, 3 years ago,

Sum of the ages of 5 members including the old member3 years ago,

= {Sum of present age of the other four old member – (4 × 3)}+ (x – 3)

= (m – 12) + (x – 3)

∴ Average of 5 members 3 years ago

= {(m – 12)+ (x – 3)}/5

According to the question

Present average = Average 3 years ago

⇒ (m + y)/5= {(m – 12)+ (x – 3)}/5

⇒ m + y = (m – 12) + (x – 3)

⇒ y = x – 15

Average = SUM of elements/number of elements

Given,

Let the price of books b1, B2, b3, b4 and b5 is B1, B2, B3, B4 and B5 respectively.

The average price of five different books b1, b2, b3, b4 and b5 is Rs. 450 =

Total price of five different books =

= 450 × 5

= 2250

⇒ B1 + B2 + B3 + B4 + B5 = 2250

Given,

The average price of b1 and b3 is Rs. 250 while average price of b2 and b4 is Rs. 650 =

Total price of books b1 and b3 =

= 250 × 2

= 500

⇒ B1 + B3 = 500

Total price of books b2 and b4 =

= 650 × 2

= 1300

⇒ B2 + B4 = 1300

⇒ B1 + B2 + B3 + B4 + B5 = 2250

⇒ 500 + 1300 + B5 = 2250

⇒ B5 = 2250 – 1800

⇒ B5 = 450

Let the TOTAL capacity of a tank be ‘m’ litres

Average rate = total distance/time taken

⇒ Time taken by a pipe to fill 1/5^th of a tank = (m/5)/10 = m/50 hr

⇒ Time taken by a pipe to fill 3/5^th of a tank = (3m/5)/15 = m/25 hr

⇒ REST PART of a tank remain to fill = m {1 – (1/5 + 3/5)} = m (1 – 4/5)

⇒ Rest part of a tank remain to fill = m/5

⇒ Time taken by a pipe to fill rest of tank = (m/5)/5 = m/25 hr

Total time taken by a pipe to fill tank completely =

= (m/50 + m/25 + m/25)

= 5m/50

= m/10

Average rate to fill tank completely =

⇒ A = Total capacity of a tank/Total time taken by a pipe to fill tank completely

⇒ A = m/ (m/10)

⇒= 10 lit/hr

∴ Average rate at which tank will completely get FILLED is 10 lit/hr.

Alternate Solution: Let the total tank capacity be 50 litres

Then total time = 1 + 2 + 2 = 5 hours

Given that average of 20 numbers = 0

Average of 20 numbers = (sum of 20 numbers) /20

⇒ 0 = (sum of 20 numbers)/20

∴ Sum of 20 numbers = 0

Let the sum of 19 positive numbers = q

 In ORDER to BECOME sum = 0, 20th number MUST be equal to (-q)

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 salary of 5 members of the group earlier = Rs 25000

∴ Total salary of 5 members of the group earlier = 25000 × 5 = Rs 125,000

Now, when one old member is replaced by a new member whose salary is Rs 22,000, the average salary of the group decreases by Rs 1000,

Average salary of 5 members of the group later = Rs (25000 – 1000) = Rs 24000

∴ Total salary of 5 members of the group later = Rs 24000 × 5 = Rs 120,000

Hence, salary of the outgoing member = (total salary of the group earlier – total salary of the group later) + salary of the incoming member

⇒ Salary of the replaced (outgoing) member = Rs (125,000 – 120,000) + Rs 22,000

= Rs (22,000 + 5000) = Rs 27,000

Hence, the REQUIRED salary of the outgoing member is Rs 27,000.

Alternatively:

The average salary decrease by 1000, so sum of salary decreases by 5 × 1000 = Rs. 5000, hence salary of new member is 5000 less than the salary of the replaced member.

∴ Salary of replaced (outgoing) member = Rs (22000 + 5000) = Rs 27000.

Alternatively:

Let the salary of the replaced (outgoing) member be Rs ‘x’, then according to the question,

$(\frac{{\left( {25000 \times 5} \right) + 22000 - x}}{5} = 24000)$

⇒ 22000 – x = 5 (24000 – 25000)

⇒ 22000 - x = - 5000

⇒ x = 22000 + 5000 = 27000

Let ASSUME the DISTANCE between POINT A to point B = x km.

We know, Distance = Speed × TIME

⇒ Time = Distance/Speed

The man walked at a speed of 6 km/h from point A to B. Then, he would take time = x/6 hours.

And he came back from point B to point A at 8 km/h. Then, he would take time = x/8 hours.

∴The ratio between the time taken by him in walking from point A to B to point B to A

LET the batsman’s AVERAGE runs for 15 innings be x.

Then, the TOTAL runs scored by batsman in 15 innings = 15x

Since, he scored 60 runs in his 16^th inning

Total runs scored by batsman in 16 innings = 15x + 60

Thus, according to question

NEW average for 16 innings = (x - 4) runs

Therefore, x - 4 = (15x + 60)/16

⇒16(x - 4) = 15x + 60

⇒16x - 64 = 15x + 60

∴ x = 124

As per given data,

AVERAGE of 7 CONSECUTIVE numbers = 41

Total summation of 7 numbers = 41 × 7

Total sum of 7 numbers = 287- - - - (1)

Let the initial number be x

Consecutive numbers series will be - x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6

From result (1) -

Total sum of 7 numbers = 287

⇒ x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) = 287

⇒ 7x + 21 = 287

⇒ 7x = 266

⇒ x = 38

⇒ Last number = x + 6 = 38 + 6 = 44

As per given data -

Average weights of three CLASSES of 58, 56 & 59 students respectively is 45, 50, 52 kg.

Total weights of students = $(\sum \left( {average\;weight\; \times \;NUMBER\;of\;students\;of\;each\;class} \right))$

Total weights of students = 58 × 45 + 56 × 50 + 59 × 52

Total weights of students = 2610 + 2800 + 3068

Total weights of students = 8478

Here total number of students = 58 + 56 + 59

Total number of students = 173

Average weight of students = Total weight/Number of students

Average weight of students = 8478/173 = 49.005