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⇒ Average age of family = 60

⇒ Sum of the age of the family member = 60 × 5 = 300

When the MAN leaves and new PERSON joins the family average age of the family = 45

Now,

⇒ Sum of the family = 45 × 5 = 225

⇒ Difference between the sum of the ages of person = 300 - 225 = 75

⇒ Age of new member = 15 years

Let the no. of bananas he bought be ‘X’

Price of bananas = (48/12) × x = 4X

Price of oranges = (66/12) × 16 = Rs. 88

Now,

TOTAL price paid = 4x + 88

⇒ 5(x + 16) = 4x + 88

⇒ 5x + 80 = 4x + 88

⇒ x = 8

∴ The man bought 8 bananas

Let AVERAGE runs of 11 INNINGS = X

∴ Total runs MADE by him in 11 innings = 11x

Score of 12th innings = 100

Average run of 12 innings = x + 5

Total runs of 12 innings = 12(x + 5) = 12x + 60

11x + 100 = 12x + 60

Let third number be 3x

So, the first number is 4X

And second number = 3x + 20

We know, average = SUM of all the observation/total number of observation

According to the question

Average = (4x + 3x + 20 + 3x)/3

⇒ (4x + 3x + 20 + 3x) = 40 × 3 = 120

⇒ 10X = 100

⇒ x = 10

We know that:

$(Average = \frac{{SUM\ of\ ELEMENTS}}{{number\ of\ elements}})$

⇒ Monday + Tuesday + Wednesday + Thursday = 58 × 4 = 232 ________ (1)

⇒ Tuesday + Wednesday + Thursday + Friday = 60 × 4 = 240 ________ (2)

⇒ Monday : Friday = 7 : 8

⇒ Monday = 7/8 × Friday_________(3)

Now, subtracting equation (1) from equation (2), we get

⇒ Friday - Monday = 8

⇒ Friday - 7/8 × Friday = 8(from equation 3)

⇒ Friday = 8 × 8 = 64

∴ Temperature on Friday was 64°C

Let his average before the 21^st match be X and average after the 21^st match will be x + 2

His total RUNS before the 21^st match can be given as 20x.

From the problem statement

⇒ x + 2 = (87 + 20x)/21

⇒ x = 45

First we need to FIND the total number of odd NUMBERS,

Last number = 159 and First number = 113

According to AP,

⇒ 159 = 113 + 2(n - 1)

⇒ n - 1 = 23

⇒ n = 24

∴ Sum of 24 odd numbers from 113 to 159 = 24/2 × [2 × 113 + 2(24 - 1)]

⇒ 12 × [226 + 46]

⇒ 3264

Let the 5 consecutive ODD numbers be x, x + 2, x + 4, x + 6 and x + 8

⇒ (x + x + 2 + x + 4 + x + 6 + x + 8)/5 = 81

⇒ 5x + 20 = 405

⇒ 5x = 405 – 20 = 385

⇒ x = 385/5 = 77

∴ The RATIO of second and third numbers = x + 2/x + 4 = 79/81

Let 7 odd CONSECUTIVE numbers are (n - 6), (n - 4), (n - 2), n, (n + 1), (n + 4), (n + 6)

GIVEN that, Average = 31

Average = MIDDLE term = n = 31

If we ADD previous and next odd number to these 7 odd numbers, again middle term will be

n = 31

∴ New average = 31

Weight of B1 = 65 kg.

Average weight of B1, B2 and B3 = 64 kg.

∴ Total weight of B1, B2 and B3 = 3 × 64 = 192 kg.

∴ Total weight of B2 and B3 = 186 – 65 = 127 kg.

Average weight of B1, B4 and B5 = 58 kg.

∴ Total weight of B1, B4 and B5 = 174 kg.

∴ Total weight of B1, B2, B3, B4 and B5 = 127 + 174 = 301 kg.

∴ Average weight of B1, B2, B3, B4 and B5 = 301/5 = 60.2 kg.

SUM of FIRST 25 MULTIPLES of 11 = 25/2[22 + 24 × 11] = 3575

Let total number of students be X.

Then, the sum of total weight of class = 40x

If a STUDENT with weight 60 kg joins the class, the new AVERAGE weight BECOMES 42 kg.

⇒ 40x + 60 = 42(x + 1)

⇒ 40x + 60 = 42X + 42

⇒ 42x - 40x = 60 - 42

⇒ 2x = 18

⇒ x = 9

Hence, the total number of students in class = 10.

Given Average of the runs of 133 PLAYERS of a team is 38

Sum of the runs of the team (Total Runs) = Average × NUMBER of players in the Team

Total Runs SCORED by the team = 38 × 133 = 5054

Let the number of male players be ‘M’ and number of female players be ‘F’

Number of runs scored by the male and female players together = 43M + 24F

⇒ 43M + 24F = 5054----(1)

⇒ M + F = 133----(2)

Substituting M = (133 - F) in Equation 1, we get

43 (133 - F) + 24F = 5054

⇒ 5719 - 43F + 24F = 5054

⇒ 43F - 24F = 5719 - 5054

⇒ 19F = 665

⇒ F = 665/19 = 35

M = 133 – 35 = 98

Ratio of the total runs of Male to Female players = (98 × 43)/(35 × 24) = 301/60

Sum of ages of 27 STUDENTS of a class = 27× 22 = 594

The AVERAGE age of 27 students of a class and a teacher = 22 + 1 = 23

Sum ofages of 27 students of a class and a teacher = 23 × 28 = 644

Age of the teacher = 644 – 594 = 50

Let the age of ARJUN is k years, so age of goal keeper is (k + 4)

Average age of team = (Total age of all the team players) / (No. of team players)

⇒ (22 + 2) = (22 × 9 + k + k + 4)/11

⇒ 264 = 202 + 2k

⇒ 2k = 62

⇒ k = 31 years(CAPTAIN’s age)

Prime NUMBERS before 19 = 17, 13, 11, 7, 5, 3, 2

We know, Average = SUM of all the number/total number

The batting AVERAGE of Batsman 1 = (42 + 51 + 9 + 78 + 63 + 20 + 12)/7 = 275/7 ≈ 39.3

The batting average of Batsman 2 = (30 + 22 + 91 + 76 + 84 + 11 + 7) = 321/7 ≈ 45.9

LET NUMBER of non-OFFICERS = a,

Then TOTAL salary of officers = 15 × 460 = 6900,

Total salary of non-officers = a × 110.

Total salary of all employees = 120 × (15 +a)

Then,

Salary of all employees = Salary of officers + Salary of non-officers

120 × (15 + a) = 15 × 460+110a

1800 + 120a = 6900 + 110a

120a – 110a = 6900 – 1800

10a = 5100

a = 510

Hence, number of non-officers = 510

Previous average of VIVEK = 65

Total number of matches = 13

Total runs SCORED by Vivek = 65 × 13 = 845

New total of Vivek = 845 – 60 + 99 = 884

∴ New average of Vivek = 884/13 = 68

Alternative method:

Difference between one MATCH in which he had scored 99 and the given average = 99 – 60 = 39

Change in average = 39/13 = 3

Here, difference value is LARGER than the previous value,

Let the average TILL 9^th MATCH = x

Total RUNS = 9x

BATSMAN score 111 runs in the 10^th match

So, (9x + 111)/10 = x + 5

⇒ 9x + 111 = 10x + 50

⇒ x = 111 - 50 = 61

Let 6 consecutive even NUMBERS be 2a, 2a + 2, 2a + 4, 2a + 6, 2a + 8, 2a + 10

Now,

Average = (2a + 2a + 2 + 2a + 4 + 2a + 6 + 2a + 8 + 2a + 10)/6

⇒ 221 = (12a + 30)/6

⇒ 1326 = 12a + 30

⇒ a = 108

∴ Highest even NUMBER = 2a + 10 = (2 × 108) + 10 = 226

We know the formula:

$({\rm{Average}} = {\rm{}}\FRAC{{{\rm{Sum\;of\;VALUES}}}}{{{\rm{NUMBER\;of\;values}}}})$

⇒ y = 4x

The given average

$(\Rightarrow {\rm{}}\frac{{5{\rm{x}} + 3{\rm{y}} + 7{\rm{x}}}}{3}{\rm{}} = {\rm{}}96)$

⇒ 5x + 3(4x) + 7x = 96 × 3

⇒ 24X = 96 × 3

⇒ x = 288/24 = 12

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

Now, we KNOW that,

⇒ Savings = Income – Expenditure

= Rs 25000 – Rs 23250

= Rs 1750

AVERAGE marks of 5 batches of 20, 25, 30, 15 & 35 students are 40, 35, 30, 40 & 35.

Therefore, the average Score for all the students is:

$(= \frac{{\left[ {\left( {20 \times 40} \RIGHT) + \left( {25 \times 35} \right) + \left( {30 \times 30} \right) + \left( {15 \times 40} \right) + \left( {35 \times 35} \right)} \right]}}{{\left( {20 + 25 + 30 + 15 + 35} \right)}} = \frac{{4400}}{{125}})$

= 35.20

Hence, the average score of all the students put TOGETHER is 35.20.

First number to be DIVISIBLE by 7 in this range = 14

Last number to be divisible by 7 in this range = 70

This is an AP with DIFFERENCE = 7

Let the total NUMBERS in this AP be n

⇒ Average = Sum of AP/total numbers

⇒ Average = [n/2(last term + first term)]/n

⇒ Average = (last term + first term)/2

∴ Average = (14 + 70)/2 = 42

Let there be n people initially in the group

⇒ TOTAL earning of the group = n × 78

Let the highest earning be x + 38 and LOWEST earning be x

⇒ n × 78 = (n – 2) × 76 + (x + x + 38)

⇒ n × 78 = (n – 2) × 76 + (2x + 38)

⇒ 78n = 76n – 152 + 2x + 38

⇒ 2n = 2x – 114

Given,

⇒ 66 < x < 71

n is prime number

at x = 67

⇒ 2n = 2 × 67 – 114 = 20

⇒ n = 10 which is not a prime number

at x = 68

⇒ 2n = 2 × 68 – 114 = 22

⇒ n = 11 which is a prime number

? Average WEIGHT of P and Q is 32 kg

⇒ Sum of their weight = 2 × 32 = 64

⇒ P + Q = 64 kg (1)

Similarly,

Q + R = 86 kg (2)

P + R = 102 kg (3)

Adding EQUATION (1), (2) and (3)

2P + 2Q + 2R = 64 + 86 + 102 = 252

⇒ P + Q + R = 126 kg

Average of 12 NUMBERS = 6

⇒ Sum of 12 numbers = 6 × 12 = 72

A number 10 is REMOVED:

Then, the new sum = 72 - 10 = 62

∴ New Average = 62/11 = 5.63

LET the numbers be p, Q, R, s and t

According to question

⇒ (p + q + r)/ 3 = {(r + s + t)/3} – 20

⇒ p + q + r = r + s + t – 60

⇒ p + q = s + t – 60 ….. 1

Given,

⇒ r + s + t = 65

⇒ s + t = 65 – r …….2

Putting the value of s + t in 1 we get

⇒ p + q = 65 – r – 60

Let the age of grandfather and GRANDMOTHER is X and Y years and that of GRANDDAUGHTER is Z years.

According to question, the AVERAGE age of grandfather and grandmother is 65 years. i.e.

$(\Rightarrow \frac{{X + Y}}{2} = 65)$

⇒ X + Y = 130 years …..(1)

Also, Average age of G.father, G.mother and G.daughter is 50, thus

$(\Rightarrow \frac{{X + Y + Z}}{3} = 50)$

⇒ X +Y + Z = 150 …..(2)

Solving above two equations we get,

Let the three number be x, y and z.

Of the 3 numbers whose average is 112, the first is 3/13 TIMES the sum of other 2.

So, we can WRITE now,

x + y + z = 112 × 3 ---- (1)

x = (y + z) × (3/13)

⇒ y + z = 13x/3 ---- (2)

Substituting the VALUE of eq. (2) in eq. (1) we get,

x + (13x/3) = 112 × 3

⇒ 16x = 112 × 9

⇒ x = 63

∴ The first number is 63.

We know,

Average of quantities = (Sum of all quantities)/(No. of quantities)

The average temperature on SUNDAY, Monday and Tuesday was 45 °C.

∴ Total temperature of Sunday, Monday and Tuesday = 45° × 3 = 135 °C

The average temperature on Monday, Tuesday and Wednesday it was 42 °C.

∴ Total temperature of Monday and Tuesday and Wednesday = 42° × 3 = 126 °C

As on Wednesday, it was exactly 40 °C.

∴ Total temperature of Monday and Tuesday = (126° – 40°)C = 86 °C

∴ The temperature of Sunday = (Total temperature of Sunday, Monday and Tuesday) - (Total temperature of Monday and Tuesday) = (135° - 86°)C = 49 °C

FIRST 15 multiples of 12 are 12, 24, 36… 180

⇒ SUM of all multiples = 12 + 24 + 36 + ... + 180 = 12(1 + 2 + 3 + … + 15) = 12 × 120 = 1440

TOTAL MONTHLY INCOME of P and Q = 42000 × 2 = RS. 84,000

Total monthly income of P and Q when P quits his job and remains jobless = 12000 × 2 = Rs. 24,000

Let the four students be a, b, c, d

SUM of age of the LAST three students (b + c + d) = 20 × 3 = 60----(1)

Sum of age of the first three students (a + b + c) = 21 × 3 = 63----(2)

Subtraction Equation 2 with Equation 1, we GET (a - d) = 3

Given age of the first student ‘a’ = 26

Age of the last student ‘d’ = a – 3 = 26 – 3 = 23

Average age of 6 members of family = 25

⇒ Sum of AGES of all presently = 25 × 6 = 150

Youngest member of the family = 15 years

At the time of the birth of the youngest member, EVERYONE WOULD be 15 years YOUNGER and there will be five members only.

⇒ Required average = $(\frac{{Sum\;of\;ages\;of\;5\;members}}{5})$

Now, sum of 5 members = Sum of 6 members - (6 × 15) = Sum of 6 members - 90

⇒ Sum of 5 members = 150 - 90 = 60

Average marks in 6 exams = 62

Total marks scored in 6 exams = average marks × number of exams

Total marks scored in 6 exams = 62 × 6 = 372

The targeted average is 66 marks.

Marks to be obtained in total 7 exams to get average of 68 = 66 × 7 = 462

∴ ANIL should score 462 - 372 = 90 marks in 7^th exam so that his average marks should be 66.

As we know AVERAGE = SUM of all the observation/total number of observation

Average of (A, B) – Average of (B, C) = 45

⇒ (A + B)/2 - (B + C)/2 = 45

⇒ A + B - B - C = 45 × 2

⇒ A - C = 90

∴ DIFFERENCE between A and C = 90

LET first number = 2x

Second number = x

Third number = 2x/3

Average of three number is 198

(x + 2x + 2x/3)/3 = 198

11x = 1782

x = 162

First number = 2 × 162 = 324

Third number = 324/3 = 108

When WEIGHT of STUDENT is wrongly recorded as 84 INSTEAD of 48 the average of weight of class increased by 0.5 kg.

Let the strength of class be X.

Total increased in weight = 0.5 × X = 0.5X----(1)

This increased in weight statistics is equal to DIFFERENCE between actual and misreported weight of student.

Increase in weight = 84 - 48 = 36 kg----(2)

Equating results (1) & (2) -

0.5X = 36

⇒ X = 72

According to the given INFORMATION,

Number of male MEMBERS = 4

Let the number of FEMALE members be y.

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

The average monthly income of male members is Rs.24000 per head

∴ 24000 = TOTAL monthly income of males/4

⇒ Total monthly income of males = 24000 × 4 = 96000

The average monthly income of female members is Rs.18750 per head

∴ 18750 = Total monthly income of females/y

⇒ Total monthly income of females = 18750y

Now, total average monthly income = (Income of males + Income of females)/(males + females)

$(\therefore 21750 = \;\frac{{96000 + 18750y}}{{4 + y}})$

⇒ 87000 + 21750y = 96000 + 18750y

⇒ 3000y = 9000

⇒ y = 3

Average = Sum of observations/Number of observations

Given,

A man worked 15 hours a day for the FIRST 4 days

∴ Time the man worked in first 4 days = (15 × 4) hours = 60 hours

A man worked 14 hours a day for the next 3 days

∴ Time the man worked in next 3 days = (14 × 3) hours = 42 hours

And, he did not work on the 8^th days

∴ Total AMOUNT of time the man worked in 8 days

= 60 hours + 42 hours + 0 hours = 102 hours

∴ Average working time

= 102/8 hours

= 12.75 hours

= 12 hours + 0.75 hours

∴ Average = (36 + 43 + 51 + 68 + 77)/5 = 275/5 = 55

Total weight of all STUDENTS = 56 × 62.875 = 3521 kg

Total weight of girls = 21 × 56 = 1176 kg

Total weight of boys = 3521 – 1176 = 2345 kg

Average of N NUMBERS = 21

The sum of N numbers = 21N

Now,

⇒ 21N - 57 = 17(N - 1)

⇒ 21N - 17N = 57 - 17

⇒ 4N = 40

⇒ N = 10

Hence, the value of N = 10

SINCE given that AVERAGE of 41 consecutive odd numbers is 49,

Which implies that middle number is 49 and there are 20 numbers greater than that,

⇒ The largest number = 49 + 20 × 2 = 89

Given,

4 years ago, average age of 6 members = 27

∴ Sum of age of 6 members 4 years ago

= Average age × Number of member

= 27 × 6 = 162

∴ Sum of present age of 6 members = 162 + 4 × 6 = 162 + 24 = 186

Let, Present age of the baby = x years

Given,

Average age of the family today = 27

⇒ Sum of age of 7 members today /7 = 27

⇒ (186 + x)/7 = 27

⇒ 186 + x = 189

Given, AVERAGE YIELD per tree is 10,

⇒ 10 = $(\FRAC{{Total\;yield}}{{Total\;no.\;of\;trees}} = \frac{{60\; + \;120\; + \;180}}{{\left( {x - 2} \right)\; + \;x\; + \;\left( {x + 2} \right)}} = \frac{{360}}{{3x}})$

⇒ 10 = 120/x

Total weight of 26 girls = 42 × 26 = 1092 KG

Total weight of 39 students = 48 × 39 = 1872 kg

Total weight of (39 – 26) = 13 boys = 1872 – 1092 = 780