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LET the average weight of the boys be X KG.

In a class of 55 students there are 34 girls.

So, no. of boys = 55 – 34 = 21

The average weight of these girls is 51 kg and average weight of the full class is 55.2 kgs.

So, we can write now,

(51 × 34) + (21 × x) = 55 × 55.2

⇒ 1734 + 21x = 3036

⇒ 21x = 1302

⇒ x = 62

GIVEN, a mixture of water is obtained by mixing equal VOLUME of water at 4°C and 8°C.

Let the five consecutive EVEN numbers be X, X + 2, X + 4, X + 6, X + 8

Given the average of these numbers = 38

⇒ COMBINED Sum of the all the five consecutive even numbers = (Average) × (Total no. of numbers)

⇒ X + (X + 2) + (X + 4) + (X + 6) + (X + 8) = (38 × 5)

⇒ 5X + 20 = 190

⇒ 5X = 190 – 20 = 170

⇒ X = 170/5 = 34

So the LARGEST number = 34 + 8 = 42 and Smallest number = 34

Let the number of BOYS in the CLASS be 2A

Then, the number of girls will be 3a

Given: Average weight of boys in the class = 18 kg

Average weight of girls in the class = 21 kg

Total weight of boys = 18 × 2a = 36A

Total weight of girls = 21 × 3a = 63a

Total weight of the class = 36a + 63a = 99a

PRESENT age of Rajeev = 42 + 10 = 52

Present age of Mamta = 41 + 10 = 51

Present age of Rahul = 14 + 10 = 24

Present age of Rohan = 18 + 10 = 28

Sum of ages of all family members = 52 + 51 + 24 + 28 = 155

Average age of all family members = 155/4 = 38.75 ≈ 39

LET, the total distance travelled by train be ‘x’ km.

⇒ Time taken to travel one fourth part $(= \frac{{\frac{x}{4}}}{{50}} = \frac{x}{{200}})$

⇒ Time taken to travel one third part $( = \frac{{x/3}}{{60}} = \frac{x}{{180}})$

⇒ Time taken to travel rest part (5/12) $( = \frac{{\frac{{5x}}{{12}}}}{{100}} = \frac{x}{{240}})$

∴ Total time taken in the journey =$(\frac{{53\;x\;}}{{3600}})$

Average speed of the whole journey = total distance/total time

$(= \frac{x}{{\frac{{53x}}{{3600}}}}\; = \frac{{3600}}{{53}}\;\;{\rm{km}}/{\rm{h}})$

Total number of months = (4 + 3 + 5) months = 12 months

For FIRST 4 months,

Average MONTHLY EXPENDITURE = Rs. 2570

∴ Total expenditure in 4 months = 4 × Average expenditure = 4 × 2570 = Rs. 10280

For second 3 months,

Average monthly expenditure = Rs. 2490

∴ Total expenditure in 3 months = 3 × Average expenditure = 3 × 2490 = Rs. 7470

For last 5 months,

Average monthly expenditure = Rs. 3030

∴ Total expenditure in 5 months = 5 × Average expenditure = 5 × 3030 = Rs. 15150

∴ Total expenditure in 12 months = (10280 + 7470 + 15150) = Rs. 32900

Given, Savings during 12 months = Rs. 5320

? Income = Expenditure + Savings

∴ Income in 12 months = Rs. 32900 + Rs. 5320 = Rs. 38220

∴ Average income per month

= (Income in 12 months/Number of months)

Number of TERMS in the given series are EVEN,

∴ median = (sum of middle terms) ÷ 2

Given median = 2.5

(2 + x) ÷ 2 = 2.5

2 + x = 5

x = 5 - 2

Average of numbers = Sum of numbers/ TOTAL numbers

The numbers are 1, 2, 3, 4, 5…,50

Sum of n CONSECUTIVE TERMS = n(n + 1)/2

Sum of first 50 natural numbers = 50 × 51/2 = 1275

Average = 1275/50 = 25.5

The average age of a FATHER and a mother is 35 years.

The sum of their (father and mother) AGES = 35 × 2 = 70 = S (SAY)

The average age of the father, mother and their only son is 27 years.

The sum of their ages = 27 × 3 = 81

⇒ S + age of son = 81

MEAN of the CLASS consisting of 12 STUDENTS = 67.4

Mean of the class consisting of 15 students = 72.3

The mode is the NUMBER that is repeated more often than any other, so 8 is the mode.

The Median ?-

If the TOTAL number of NUMBERS(n) is an odd number, then:

Median = (n + 1)/2^th term

If the total number of the numbers(n) is an even number, then the formula is given below?

Median = [(n/2)^th term + {(n/2) + 1}^th term]/2

The median is the middle value, so to rewrite the list in order?

0, 6, 6, 7, 7, 8, 8, 8

There are eight numbers in the list,

Quantity of rice REQUIRED by the HOSTEL MESS per day = 91/7 = 13 kg

Quantity of wheat required by the hostel mess per week = 42 kg

We know that,

Each month COMPRISES of 4 weeks.

⇒ Wheat is used 4 times in a month.

The months July and August have 31 days each.

⇒ TOTAL number of days in July and August = 31 + 31 = 62 days

Quantity of rice and wheat required for the months of July and August together = (62 × 13) + (42 × 8) = 1142 kg

∴ Quantity of rice and wheat required for the months of July and August together = 1142 kg

Due to the new PLAYER the average team weight has been REDUCED by 100 grams.

∴ Weight of new player would 100GRAMS × 11 = 1100 grams less than that of injured player’s weight.

GIVEN data,

Sachin scored average run of 45 in 9 matches.

TOTAL runs scored in 9 matches = average runs × 9

Total runs scored in 9 matches = 45 × 9 = 405

As per requirement,

Average runs scored in 10 matches should be 50

Total runs to be scored in 10 matches = average runs × 10

Total runs to be scored in 9 matches = 50 × 10 = 500

⇒ Runs to be scored in 10^th MATCH = Total runs to be scored in 10 matches - Runs scored in 9 matches

Runs to be scored in 10^th match = 500 – 405

⇒ Runs to be scored in 10^th match = 95

⇒ Sachin should score 95 runs in 10^th match so that his average runs scored to be 50.

Let assume current age of Suresh, Ramesh and Mahesh as x, y and z respectively,

For the sum of the age of Suresh and Ramesh,

[(x - 4) + (y - 4)]/2 = 36

⇒ x + y - 8 = 72

⇒ x + y = 80----(I)

From the data given in the question,

⇒ [x + y + z]/3 = 40

⇒ x + y + z = 120

Substitute the value of EQ(I) in above EQUATION,

⇒ 80 + z = 120

⇒ z = 120 - 80 = 40

∴ Age of Mahesh after 7 years = 40 + 7 = 47 years

Volume of water DISPLACED = (90 × 10 × 0.1) m³ = 90 m³

⇒ Weight of 1800 MEN = Volume of water displaced × DENSITY of water

⇒ 90 m³ × 1000kg/m³ = 90000kg

Let the initial population of town be 100.

Population after first YEAR = 100 + 25% of 100 = 100 + 25 = 125

Population after second year = 125 + 36% of 125 = 125 + 45 = 170

Total increment = (170 – 100)/100 = 7/10

∴ AVERAGE rate of INCREASE (in %) = (1/2) × (7/10) × 100 = 35%

LET the 7 consecutive number be (x - 3), (x - 2), (x - 1), (x), (x + 1), (x + 2), (x + 3)

The sum of above consecutive number = 7x

⇒ Average of numbers = The sum of above consecutive number/7 = A

⇒ x = A

After adding 2 numbers which are before these 7 numbers, we get

The numbers are (x - 5), (x - 4), (x - 3), (x - 2), (x - 1), (x), (x + 1), (x + 2), (x + 3)

⇒ Sum = 9X - 9

AVERAGE = Sum of quantities/Total number of quantities

Given, average temperature of the first three days is 27°C, and that of the next three days is 29° C.

Sum of temperatures for first SIX days = 27 × 3 + 29 × 3 = 168° C.

Let the temperature on the last day be ‘t’.

Thus average temperature of the WEEK $(= \frac{{168 + t}}{7})$

Given, average of the entire week is 28.5°C.

$(\therefore \frac{{168 + t}}{7} = 28.5)$

⇒ 168 + t = 199.5

LET the age of father be x YEARS and total age of 4 children is y.

We know that, AVERAGE age = sum of ages of all persons/ number of persons

∴ 14 = y/4

⇒ y = 56 years

Now as given in the question, (x + y)/5 = 20

⇒ (x + 56)/5 = 20

⇒ x = 100 – 56

Total TOFFEES PURCHASE on first day = Rs. spend/price of 1 toffee = 20/1 = 20

Total toffees purchased on next day = Rs. Spend/price of 1 toffee = 40/(1 × 50/100) = 80

Total toffees purchased = 20 + 80 = 100

Total money SPENT = 20 + 40 = 60

∴ Average price PER toffee = Rs. spent/No. of toffees purchased = 60/100 = Rs. 0.60 = 60 PAISE

We KNOW that, Average weight = (SUM of weight of all students)/(Total no. of students)

Initially, total NUMBER of students = 20, and Average weight = 25 Kg

∴ Sum of weight of all students initially = 20 × 25

= 500 Kg

Let the weight of new STUDENT be X Kg. Then,

⇒ New sum of weight of all students = (500 + X) Kg

Given: Average weight after entry of new student = 24.8 Kg

$(\therefore 24.8 = \frac{{500 + X}}{{20 + 1}})$

Average of a QUANTITY = SUM of quantities/Total NUMBER of quantities

The average temperature of first THREE days in a week is 25°C

Sum of temperatures of first three days = 25° × 3 = 75°C

The average temperature of the next three days is 29°C

Sum of temperature of the next three days = 29° × 3 = 87°C

The average temperature of the whole week is 28.5°C

Sum of temperature of the whole week is = 28.5° × 7 =199.5°C

The temperature of the last day of the week

= Sum of temperatures of the whole week – Sum of temperatures of first three days – Sum of temperatures of the next three days

= 199.5°C – 75°C – 87°C

Given: AVERAGE RUNS scored by 7 players = 53

⇒ Total runs scored by players = 53 × 7 = 371

⇒ Sum of runs scored by 6 players = 121 + 40 + 26 + 56 + 37 + 48 = 328

⇒ Runs scored by 7^th player = 371 – 328 = 43

∴ the correct OPTION is 3)

LET the numbers be x, y and z

As per the given data,

y = 3x

x = y/3

⇒ y = 2z

⇒ z = y/2

Given that average of the three numbers is 66

⇒ (x + y + z)/3 = 66

⇒ (y/3 + y + y/2)/3 = 66

⇒ (2y + 6 y + 3y)/(3 × 6) = 66

⇒ 11y = 66 × 18

⇒ y = 108

∴ First number x = y/3 = 108/3 = 36

Let the TOTAL money spent by them be RS. y

⇒ Total money spent by them can be given as = 14 × 7 + y/8 + 21 this is equal to y

⇒ 119 + y/8 = y

⇒ y = 119 × 8/7

⇒ y = 136

∴ the total money spent by them is Rs. 136

Given, (X + Y)/2 = 200

⇒ X + Y = 400 …(1)

ALSO, (Y + Z)/2 = 238.5

⇒ Y + Z = 477 …(2)

Also, (X + Z)/2 = 152.5

⇒ X + Z = 305 …(3)

Adding all the EQUATIONS,

⇒ 2(X + Y + Z) = 400 + 477 + 305

⇒ X + Y + Z = 1182/2 = 591

⇒ (X + Y + Z)/3 = 591/3 = 197

Given that average marks of 40 students is 25

⇒ SUM of marks of 40 students/40 = 25

⇒ sum of marks of students = 40 × 25 = 1000

According to the PROBLEM statement, correct average can be given as

⇒ (1000 – 73 + 37)/40 = 964/40 = 24.1

∴ the correct average marks are 24.1

Average of NUMBERS = SUM of numbers/ Total numbers

(18 + 20 + 32 + 8 + X + 9 + 13)/7 = 17

100 + x = 119 ⇒ x = 19

Given, AVERAGE of 6 NUMBERS = 20

∴ Sum of 6 numbers = Average × 6 = 20 × 6 = 120

Let, a number ‘X’ is removed.

∴ Sum of remaining 5 numbers = 120 – x

Average of remaining 5 numbers

= Sum of remaining 5 numbers/5

= (120 – x)/5

According to the question,

Average of remaining 5 numbers = 15

∴ (120 – x)/5 = 15

⇒ 120 – x = 75

Average of 6 BOYS = Sum of AGE of 6 boys/6

⇒ 14 = sum of AGES of 6 boys/6

⇒ Sum of age of 6 boys = 84

Now,

Average of 11 GIRLS = sum of age of 11 girls/11

⇒ 12 = sum of age of 11 girls/11

⇒ Sum of age of 11 girls = 132

Now,

Sum of age of all girls and boys together = 84 + 132 = 216

Average of numbers = SUM of numbers/ TOTAL numbers

The numbers are 2, 4, 6, 8, 10…

Average = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)/10 = 110/10 = 11

LET the original strength of the CLASS be x

 Average age of x students = 42

∴ Sum of age of x students = 42x years

When, 12 students join the class, then number of students in the class = x + 12

Average of 12 new student’s age = 30yrs

∴ Sum of age of 12 students = 30 × 12 = 360 years

∴ Total sum of age of (x + 12) students = 42x + 360 years

∴ Average of (x+ 12) student’s age = $(\frac{{42{\RM{x\;}} + {\rm{\;}}360{\rm{\;}}}}{{{\rm{x\;}} + {\rm{\;}}12}})$

Given,

After the joining of 12 new students in the class, average age decrease by 8 years

$(\Rightarrow \;\frac{{42{\rm{x\;}} + {\rm{\;}}360{\rm{\;}}}}{{{\rm{x\;}} + {\rm{\;}}12}}\; = \;42 - 8)$

⇒ 42x + 360 = 34x + 408

⇒ 8x = 48

⇒ x = 6

∴ The original strength of the class was 6 students

Let the TOTAL capacity of a cistern be ‘m’ lit.

Average RATE = total distance / time taken

⇒ Time taken by a PIPE to fill 1/3^rd of a cistern = (m/3)/20 = m/60 HR

⇒ Time taken by a pipe to fill 1/5^th of a cistern = (m/5)/24 = m/120 hr

⇒ REST part of a cistern remain to fill = m {1 - (1/3 + 1/5)} = m (1 - 8/15)

⇒ Rest part of a cistern remain to fill = 7m/15

⇒ Time taken by a pipe to fill rest of cistern = (7m/15)/14 = m/30 hr

Total time taken by a pipe to fill cistern completely = (m/60 + m/120 + m/30) hr = 7m/120hr

Average rate to fill cistern completely:

⇒ Total capacity of a cistern / Total time taken by a pipe to fill cistern completely

= m/(7m/120)

= 120/7

= 17.14 lit/hr

∴ Average rate at which cistern will completely get filled is 17.14 lit/hr.

We know that,

Given, $({{p + q} \over 2} = 5.8)$ ⇒ p + q = 11.6 ……… (i)

Also, $({{q + R} \over 2} = 1.4)$ ⇒ q + r = 2.8……… (ii)

And, $({{r + s} \over 2} = 0.7 \Rightarrow)$ r + s = 1.4 …………….. (III)

From operation, [(i) – (ii) – (iii)] we GET,

⇒ p + q - q - r - r - s = 11.6 - 2.8 - 1.4

⇒p – 2r - s = 7.4

Let the THREE NUMBERS be a, b, C

Given the AVERAGE of the ‘a and ‘b’ is 37 more than the average of ‘b’ and ‘c’

⇒ [(a + b)/2] = 37 + [(b + c)/2]

⇒ [(a + b)/2] = [74 + b + c]/2

⇒ (a + b) = 74 + b + c

⇒ a - c = 74

⇒ Sum of age of Sita and Geeta = average × 2 = 30 × 2

⇒ S + G = 60----(1)

⇒ Sum of age of Geeta and Rita = average × 2 = 28 × 2

⇒ G + R = 56----(2)

⇒ Sum of age of Sita and rita = average × 2 = 32 × 2

⇒ S + R = 64----(3)

 Now, subtracting (2) from (1) we get

⇒ S – R = 4----(4)

Now, ADDING (3) and (4) we get,

⇒ 2S = 68

⇒ S = 34

Now,

⇒ R = 64 – S = 64 – 34 = 30

⇒ G = 56 – R = 56 – 30 = 26

? Average of three numbers is 112

This means sum of 3 numbers is = 112 × 3 = 336

Let the first number be X

Then X = (1/6) × (Sum of other TWO numbers) = (1/6) × (336 - X)

⇒ X + X/6 = 336/6

⇒ X = 336/7 = 48

∴ First number = 48

SINCE the average has increased, the new MAN will have weight more than 68.

Let, the original average be a, and let the weight of new man be x.

Average = Total weight / No. of people

∴a = (Total weight of 9 people + 68)/10

⇒ 10a = 68 + Total weight of 9 people

(Weight of remaining 9 people remains unchanged)

New average will be (1.5 + a) = (Total weight of 9 people + x)/10

⇒1.5 + a = [(10a – 68) + x]/10

⇒ 15 + 10a = 10a – 68 + x

∴ AVERAGE = 1 + 2 + ……..+ 99 = (99 + 1)/2 = 50

Let the AVERAGE marks scored by boys be a.

According to the PROBLEM statement

⇒ Average of class = 42 = (22 × 35 + 28 × a)/50

⇒ 2100 – 770 = 28a

⇒ a = 1330/28 = 47.5

Since the average WEIGHT of X, Y and Z is 74 kg,

⇒ X + Y + Z = 74 × 3 = 222---- (1)

The average weight of X and Y be 68 kg and that of Y and Z be 78 kg,

⇒ X + Y = 68 × 2 = 136---- (2)

⇒ Y + Z = 78 × 2 = 156---- (3)

Adding equation 2 and 3,

⇒ X + Y + Y + Z = 136 + 156 = 292

Subtracting equation 1 from it,

⇒ Y = 292 - 222 = 70

According to the PROBLEM statement

⇒ New AVERAGE = (50 × 65 – 83 + 38)/50

⇒ New average = 3205/50 = 64.1

Given: Number of members EARLIER = 120

⇒ TOTAL age of all the members earlier = 120 X 60.7 = 7284…..(i)

⇒ Total age of all the members presently = 150 x 56.3 = 8445…..(II)

⇒ Total age of new members = (ii) – (I)

⇒ 8445 – 7284 = 1161

∴ AVERAGE age of newly joined members = 1161/30 = 38.7

Let the average price of costliest metal be RS. T per gram.

⇒ SUM of price of all 5 metals = 5 × Rs. 432 = Rs. 2160

And, Sum of price of all 4 metals(EXCLUDING costliest metal) = 4 × Rs. (432-132) = Rs. 1200

⇒ Price of a gram of costliest metal = Rs. 2160 – Rs. 1200 = Rs. 960

⇒ Price of 20 grams of costliest metal = 20 × Rs. 960 = Rs. 19200

We know, AVERAGE of quantities = SUM of all quantities/no. of quantities

The average age of 7 boys is 14 YEARS. Then total age of 7 boys = 7 × 14 = 98 years.

A boy AGED 14 years died, So now the total age of remaining 6 boys = 98 – 14 = 84 years.

∴ The average age of remaining 6 boys = 84/6 = 14 years.

Let the TOTAL number of students appeared be n

And let the sum of MARKS of total STUDENT be S

∴ average marks = S/n = 50

⇒ S = 50n...........Equation. (1)

Now, after deducting 30 marks as the score is REDUCED from 90 to 60 of 100 students, the new sum of marks = S - 3000

∴ new average = (S - 3000)/n

But new average = 45

So, (S - 3000)/n = 45

Now, from equation. (1), we know that S = 50n

⇒ 50n - 3000 = 45n

⇒ n = 3000/5 = 600

∴ total number of students who appeared for an exam = 600

Average for a batsman $(= \FRAC{{Total\;runs}}{{number\;of\;innings}})$

Let his NEW average be x.

Given, a batsman’s average INCREASES by 3 on scoring 87 runs in his 17^th innings.

∴ x – 3 = (Total runs – 87)/16

⇒ Total runs = 16x + 39 ----- eq1

and x = Total runs/17

⇒ Total runs = 17x -------eq2

Solving eq1 and eq2 we get,

17x = 16x + 39

⇒ x = 39