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#### Question 1:

Find the mean of:
(i) the first eight natural numbers
(ii) the first ten odd numbers
(iii) the first seven multiples of 5
(iv) all the factors of 20
(v) all prime numbers between 50 and 80.

We know:

(i) The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7 and 8.
Mean of these numbers:

(ii) The first ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.
Mean of these numbers:

$\frac{1+3+5+7+9+11+13+15+17+19}{10}\phantom{\rule{0ex}{0ex}}=\frac{100}{10}\phantom{\rule{0ex}{0ex}}=10$

(iii) The first seven multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.
Mean of these numbers:

$\frac{5+10+15+20+25+30+35}{7}\phantom{\rule{0ex}{0ex}}=\frac{140}{7}\phantom{\rule{0ex}{0ex}}=20$

(iv) The factors of 20 are 1, 2, 4, 5, 10 and 20.
Mean of these numbers:

(v) The prime numbers between 50 and 80 are 53, 59, 61, 67, 71, 73 and 79.
Mean of these numbers:

$\frac{53+59+61+67+71+73+79}{7}\phantom{\rule{0ex}{0ex}}=\frac{463}{7}\phantom{\rule{0ex}{0ex}}=66.14$

#### Question 2:

The number of children in 10 families of a locality are
2, 4, 3, 4, 2, 0, 3, 5, 1, 6.
Find the mean number of children per family.

Numbers of children in 10 families = 2, 4, 3, 4, 2, 0, 3, 5, 1 and 6.
Thus, we have:

$=\frac{30}{10}\phantom{\rule{0ex}{0ex}}=3$

#### Question 3:

The following are the number of books issued in a school library during a week:
105, 216, 322, 167, 273, 405 and 346.
Find the average number of books issued per day.

Numbers of books issued in the school library: 105, 216, 322, 167, 273, 405 and 346
Thus, we have:

#### Question 4:

The daily minimum temperature recorded (in degree F) at a place during a week was as under:

 Monday Tuesday Wednesday Thursday Friday Saturday 35.5 30.8 27.3 32.1 23.8 29.9
Find the mean temperature.

Daily minimum temperatures = 35.5, 30.8, 27.3, 32.1, 23.8 and 29.9
Thus, we have:

#### Question 5:

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 13, find the value of x and hence find the mean of the last three observations.

We know that,

The first five observations are x, x + 2, x + 4, x + 6 and x + 8.

Mean of these numbers = $\frac{x+\left(x+2\right)+\left(x+4\right)+\left(x+6\right)+\left(x+8\right)}{5}$
$⇒13=\frac{5x+20}{5}\phantom{\rule{0ex}{0ex}}⇒13×5=5x+20\phantom{\rule{0ex}{0ex}}⇒65=5x+20\phantom{\rule{0ex}{0ex}}⇒5x=65-20\phantom{\rule{0ex}{0ex}}⇒5x=45\phantom{\rule{0ex}{0ex}}⇒x=\frac{45}{5}\phantom{\rule{0ex}{0ex}}⇒x=9\phantom{\rule{0ex}{0ex}}$

Hence, the value of x is 9.

Now, the last three observations are 13, 15 and 17.
Mean of these observations = $\frac{13+15+17}{3}$
= $\frac{45}{3}$
= 15

Hence, the mean of the last three observations is 15.

#### Question 6:

The mean weight of 6 boys in a group is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. Find the weight of the sixth boy.

The individual weights of five boys are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg.
Now,
Let the weight of the sixth boy be x kg.
We know:

Also,
Given mean = 48 kg
Thus, we have:

Therefore, the sixth boy weighs 53 kg.

#### Question 7:

The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.

Let the marks scored by 50 students be x1, x2,...x50.
Mean = 39
We know:

Thus, we have:

Also, a score of 43 was misread as 23.

#### Question 8:

The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?

Let the numbers be x1, x2,...x24.

We know:

Thus, we have:

After addition, the new numbers become (x1+3), (x2+3),...(x24+3).
New mean:

#### Question 9:

The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean?

Let the numbers be x1, x2,...x20.
We know:

Thus, we have:

Numbers after subtraction: (x1$-$6), (x2$-$6),...(x20$-$6)

∴ New Mean = $\frac{\left({x}_{1}-6\right)+\left({x}_{2}-6\right)+........+\left({x}_{20}-6\right)}{20}$

#### Question 10:

The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers?

Let the numbers be x1, x2,...x15
We know:

Thus, we have:

After multiplication, the numbers become 4x1, 4x2,...4x15
∴ New Mean = $\frac{4{x}_{1}+4{x}_{2}+......+4{x}_{15}}{15}$

#### Question 11:

The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?

Let the numbers be x1, x2,...x12.
We know:

Thus, we have:

After division, the numbers become:

#### Question 12:

The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of the new set of 20 numbers.

Let the numbers be x1, x2,...x20.
We know:

Thus, we have:

New numbers are:

(x1 + 3), (x2 + 3),...(x10 + 3), x11,...x20

New Mean:

#### Question 13:

The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.

Let the numbers be x1, x2,..., x6.
Mean = 23
We know:

Thus, we have:

${x}_{1}+{x}_{2}.......{x}_{6}=138$.....................(i)

If one number, say, x6, is excluded, then we have:

${x}_{1}+{x}_{2}......+{x}_{5}=100....................\left(\mathrm{ii}\right)$

Using (i) and (ii), we get:

Thus, the excluded number is 38

#### Question 14:

The average height of 30 boys was calculated to be 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.

We know that,

Mean of height of 30 boys = $\frac{\sum _{i=1}^{30}{x}_{i}}{30}$

It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean.

Hence, the correct mean is 151.

#### Question 15:

The mean weight of a class of 34 students is 46.5 kg. If the weight of the teacher is included, the mean is rises by 500 g. Find the weight of the teacher.

Mean weight of 34 students = 46.5 kg
Sum of the weights of 34 students = kg
Increase in the mean weight when the weight of the teacher is included = 500 g = 0.5 kg
∴ New mean weight = (46.5 + 0.5) kg = 47 kg
Now,
Let the weight of the teacher be x kg.
Thus, we have:

Therefore, the weight of the teacher is 64 kg.

#### Question 16:

The mean weight of a class of 36 students is 41 kg. If one of the students leaves the class then the mean is decreased by  200 g. Find the weight of the student who left.

Mean weight of 36 students = 41 kg
Sum of the weights of 36 students =
Decrease in the mean when one of the students left the class = 200 g = 0.2 kg
Mean weight of 35 students = (41 $-$ 0.2) kg = 40.8 kg
Now,
Let the weight of the student who left the class be x kg.

Hence, the weight of the student who left the class is 48 kg.

#### Question 17:

The average weight of a class of 39 students is 40 kg. When a new student is admitted to the class, the average decreases by 200 g. Find the weight of the new student.

Average weight of 39 students = 40 kg
Sum of the weights of 39 students =
Decrease in the average when new student is admitted in the class = 200 g = 0.2 kg
∴ New average weight = (40 $-$ 0.2) kg = 39.8 kg
Now,
Let the weight of the new student be x kg.
Thus, we have:

Therefore, the weight of the new student is 32 kg.

#### Question 18:

The average weight of 10 oarsmen in a boat is increased by 1.5 kg when one of the crew who weighs 58 kg is replaced by a new man. Find the weight of the new man.

Let the average weight of 10 oarsmen be x kg.
Sum of the weights of 10 oarsmen = 10x kg
∴ New average weight = (x + 1.5) kg
Now, we have:

#### Question 19:

The mean of 8 numbers is 35. If a number is excluded then the mean is reduced by 3. Find the excluded number.

Mean of 8 numbers = 35
Sum of 8 numbers = $35×8=280$
Let the excluded number be x.
Now,
New mean = 35 $-$ 3 = 32
Thus, we have:

Therefore, the excluded number is 56.

#### Question 20:

The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct men.

Mean of 150 items = 60
Sum of 150 items = $\left(150×60\right)=9000$
New sum = [9000 $-$ (52 + 8) + (152 + 88)] = 9180

Correct mean =

Therefore, the correct mean is 61.2.

#### Question 21:

The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result.

Mean of 31 results = 60
Sum of 31 results = $31×60=1860$
Mean of the first 16 results = 58
Sum of the first 16 results = $58×16=928$
Mean of the last 16 results = 62
Sum of the last 16 results = $62×16=992$
Value of the 16th result = (Sum of the first 16 results + Sum of the last 16 results) $-$ Sum of 31 results
= (928 + 992) $-$ 1860
= 1920 $-$ 1860
= 60

#### Question 22:

The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46, find the 6th number.

Mean of 11 numbers = 42
Sum of 11 numbers = 42$×$11 = 462
Mean of the first 6 numbers = 37
Sum of the first 6 numbers = 37$×$6 = 222
Mean of the last 6 numbers = 46
Sum of the last 6 numbers = 46$×$6 = 276
∴ 6th number = [(Sum of the first 6 numbers + Sum of the last 6 numbers) $-$ Sum of 11 numbers]
= [(222 + 276) $-$ 462]
= [498 $-$ 462]
= 36
Hence, the 6th number is 36.

#### Question 23:

The mean weight of 25 students of a class is 52 kg. If the mean weight of the first 13 students of the class is 48 kg and that of the last 13 students is 55 kg, find the weight of the 13th student.

Mean weight of 25 students = 52 kg
Sum of the weights of 25 students = (52$×$25) kg = 1300 kg
Mean weight of the first 13 students = 48 kg
Sum of the weights of the first 13 students = (48$×$13) kg = 624 kg
Mean weight of the last 13 students = 55 kg
Sum of the weights of the last 13 students = (55$×$13) kg =715 kg
Weight of the 13th student = (Sum of the weights of the first 13 students + Sum of the weights of the last 13 students) $-$ Sum of the weights of 25 students
= [(624+715)$-$1300] kg
= 39 kg
Therefore, the weight of the 13th student is 39 kg.

#### Question 24:

The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations.

Mean score of 25 observations = 80
Sum of the scores of 25 observations = 80$×$25 = 2000
Mean score of another 55 observations = 60
Sum of the scores of another 55 observations = 60$×$55 = 3300

$=\frac{2000+3300}{80}\phantom{\rule{0ex}{0ex}}=\frac{5300}{80}\phantom{\rule{0ex}{0ex}}=66.25$

Therefore, the mean score of the whole set of observations is 66.25.

#### Question 25:

Arun scored 36 marks in English, 44 marks in Hindi, 75 marks in mathematics and x marks in science. If he has secured an average of 50 marks, find the value of x.

Marks scored by Arun in English = 36
Marks scored by Arun in Hindi = 44
Marks scored by Arun in mathematics = 75
Marks scored by Arun in science = x
Average marks = 50
Thus, we have:

$⇒155+x=200\phantom{\rule{0ex}{0ex}}⇒x=200-155=45$

∴ Marks scored by Arun in science = 45

#### Question 26:

A ship sails out to an island at the rate of 15 km/h and sails back to the starting point at 10 km/h. Find the average sailing speed for the whole journey.

Let the distance from the starting point to the island be x km.
Speed of the ship sailing out to the island = 15 km/h
Speed of the ship sailing back to the starting point = 10 km/h
We know:

Therefore, the average speed of the ship in the whole journey was 12 km/h.

#### Question 27:

There are 50 students in a class, of which 40 are boys. The average weight of the class is 44 kg an that of the girls is 40 kg. Find the average weight of the boys.

Total students in the class = 50
Number of boys = 40
∴ Number of girls = (50 $-$ 40) = 10
Average weight of students in the class = 44 kg
Average weight of girls in the class = 40 kg
Sum of the weights of girls in the class = (40 $×$10) kg = 400 kg
Thus, we have:

#### Question 28:

The aggregate monthly expenditure of a family was â‚ą 18720 during the first 3 months, â‚ą 20340 during the next 4 months and â‚ą 21708 during the last 5 months of a year. If the total savings during the year be â‚ą 35340 find the average monthly income of the family.

The aggregate yearly expenditure of a family = Rs (18720 × 3) + Rs (20340 × 4) + Rs (21708 × 5)
= Rs (56160 + 81360 + 108540)
= Rs 246060

The total savings during the year = Rs 35340

The average yearly income = Rs 246060 + Rs 35340
= Rs 281400

∴ The average monthly income of the family =

Hence, The average monthly income of the family is Rs 23450.

#### Question 29:

The average weekly payment to 75 workers in a factory is â‚ą 5680. The mean weekly payment to 25 of them is â‚ą 5400 and that of 30 others is â‚ą 5700. Find the mean weekly payment of the remaining workers.

Total salary of 75 workers = â‚ą 5680 × 75
= â‚ą 426000

Total salary of 25 workers = â‚ą 5400 × 25
= â‚ą 135000

Total salary of 30 workers = â‚ą 5700 × 30
= â‚ą 171000

No. of remaining workers = 75 − (25 +30)
= 20

Total salary of 20 workers = â‚ą (426000 − 135000 − 171000)
= â‚ą 120000

∴ The mean weekly payment of the 20 workers = $\frac{120000}{20}$
= â‚ą 6000

Hence, the mean weekly payment of the remaining workers is â‚ą 6000.

#### Question 30:

The mean marks (out of 100) of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.

Let the number of girls be x and the number of boys be y.

Mean marks of boys =

Mean marks of girls =

Mean marks of all the students =

Hence, the ratio of the number of boys to the number of girls is 2:1.

#### Question 31:

The average monthly salary of 20 workers in an office is â‚ą 45900. If the manager's salary is added, the average salary becomes â‚ą 49200 per month. What's manager's monthly salary?

Average monthly salary of 20 workers = Rs 45900
Sum of the monthly salaries of 20 workers =
By adding the manager's monthly salary, we get:
Average salary = Rs 49200
Now,
Let the manager's monthly salary be Rs x.
Thus, we have:

Therefore, the manager's monthly salary is Rs 115200.

#### Question 1:

Obtain the mean of the following distribution:

 Variable (xi) 4 6 8 10 12 Frequency (fi) 4 8 14 11 3

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Variable (xi) 4 6 8 10 12 Frequency (fi) 4 8 14 11 3

Mean = $\frac{\left(4×4\right)+\left(6×8\right)+\left(8×14\right)+\left(10×11\right)+\left(12×3\right)}{4+8+14+11+3}$
= $\frac{16+48+112+110+36}{40}$
= $\frac{322}{40}$
= 8.05

Hence, the mean of the following distribution is 8.05 .

#### Question 2:

The following table shows the weights of 12 workers in a factory:

 Weight (in kg) 60 63 66 69 72 No. of workers 4 3 2 2 1
Find the mean weight of the workers.

We will make the following table:

 Weight (xi) No. of Workers (fi) (fi)(xi) 60 4 240 63 3 189 66 2 132 69 2 138 72 1 72 12 771

Thus, we have:

=

#### Question 3:

The measurements (in mm) of the diameters of the heads of 50 screws are given below:

 Diameter (in mm) (xi) 34 37 40 43 46 Number of screws (fi) 5 10 17 12 6
Calculate the mean diameter of the heads of the screws.

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Diameter (in mm) (xi) 34 37 40 43 46 Number of screws (fi) 5 10 17 12 6

Mean = $\frac{\left(34×5\right)+\left(37×10\right)+\left(40×17\right)+\left(43×12\right)+\left(46×6\right)}{5+10+17+12+6}$
= $\frac{170+370+680+516+276}{50}$
= $\frac{2012}{50}$
= 40.24

Hence, the mean diameter of the heads of the screws is 40.24 .

#### Question 4:

The following data give the number of boys of a particular age in a class of 40 students.

 Age (in years) 15 16 17 18 19 20 Frequency (fi) 3 8 9 11 6 3
Calculate the mean age of the students.

We will make the following table:

 Age (xi) Frequency (fi) (fi)(xi) 15 3 45 16 8 128 17 9 153 18 11 198 19 6 114 20 3 60 40 $\sum {f}_{i}{x}_{i}=$698

Thus, we have:

#### Question 5:

Find the mean of the following frequency distribution:

 Variable (xi) 10 30 50 70 89 Frequency (fi) 7 8 10 15 10

We will make the following table:

 Variable (xi) Frequency (fi) (fi)(xi) 10 7 70 30 8 240 50 10 500 70 15 1050 89 10 890 50 2750

Thus, we have:

#### Question 6:

Find the mean of daily wages of 40 workers in a factory as per data given below:

 Daily wages (in â‚ą) (xi) 250 300 350 400 450 Number of workers (fi) 8 11 6 10 5

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Daily wages (in â‚ą) (xi) 250 300 350 400 450 Number of workers (fi) 8 11 6 10 5

Mean = $\frac{\left(250×8\right)+\left(300×11\right)+\left(350×6\right)+\left(400×10\right)+\left(450×5\right)}{8+11+6+10+5}$
= $\frac{2000+3300+2100+4000+2250}{40}$
= $\frac{13650}{40}$
= 341.25

Hence, the mean of daily wages of 40 workers in a factory is 341.25 .

#### Question 7:

If the mean of the following data is 20.2, find the value of  p.

 Variable (xi) 10 15 20 25 30 Frequency (fi) 6 8 p 10 6

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Variable (xi) 10 15 20 25 30 Frequency (fi) 6 8 p 10 6

Mean = $\frac{\left(10×6\right)+\left(15×8\right)+\left(20×p\right)+\left(25×10\right)+\left(30×6\right)}{6+8+p+10+6}$
$⇒20.2=\frac{60+120+20p+250+180}{30+p}\phantom{\rule{0ex}{0ex}}⇒20.2\left(30+p\right)=610+20p\phantom{\rule{0ex}{0ex}}⇒606+20.2p=610+20p\phantom{\rule{0ex}{0ex}}⇒20.2p-20p=610-606\phantom{\rule{0ex}{0ex}}⇒0.2p=4\phantom{\rule{0ex}{0ex}}⇒p=\frac{4}{0.2}\phantom{\rule{0ex}{0ex}}⇒p=\frac{40}{2}\phantom{\rule{0ex}{0ex}}⇒p=20$

Hence, the value of  p is 20.

#### Question 8:

If the mean of the following frequency distribution is 8, find the value of p.

 x 3 5 7 9 11 13 f 6 8 15 p 8 4

We will make the following table:

 (xi) (fi) (fi)(xi) 3 6 18 5 8 40 7 15 105 9 p 9p 11 8 88 13 4 52 41 + p 303 + 9p

We know:

Given:
Mean = 8
Thus, we have:

#### Question 9:

Find the missing frequency p for the following frequency distribution whose mean is 28.25.

 x 15 20 25 30 35 40 f 8 7 p 14 15 6

We will prepare the following table:

 (xi) (fi) (fi)(xi) 15 8 120 20 7 140 25 p 25p 30 14 420 35 15 525 40 6 240 50 + p 1445 + 25p

Thus, we have:

#### Question 10:

Find the value of p for the following frequency distribution whose mean is 16.6

 x 8 12 15 p 20 25 30 f 12 16 20 24 16 8 4

We will make the following table:

 (xi) (fi) (fi)(xi) 8 12 96 12 16 192 15 20 300 p 24 24p 20 16 320 25 8 200 30 4 120 100 1228 + 24p

Thus, we have:

$⇒24p=432\phantom{\rule{0ex}{0ex}}⇒p=18$

#### Question 11:

Find the missing frequencies in the following frequency distribution whose mean is 34.

 x 10 20 30 40 50 60 Total f 4 f1 8 f2 3 4 35

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 x 10 20 30 40 50 60 Total f 4 f1 8 f2 3 4 35

Mean = $\frac{\left(10×4\right)+\left(20×{f}_{1}\right)+\left(30×8\right)+\left(40×{f}_{2}\right)+\left(50×3\right)+\left(60×4\right)}{35}$

Also, 4 + f1 + 8 + f2 + 3 + 4 = 35
⇒ 19 + f1f2 = 35
f1f2 = 35 − 19
f1f2 = 16
⇒ 26 − 2f2f 2 = 16             (from (1))
⇒ 26 − f2 = 16
⇒ 26 − 16 =  f2
⇒  f2 = 10

Putting the value of f2 in (1), we get
f1 = 26 − 2(10) = 6

Hence, the value of f1 and  f2 is 6 and 10, respectively.

#### Question 12:

Find the missing frequencies in the following frequency distribution, whose mean is 50.

 x 10 30 50 70 90 Total f 17 f1 32 f2 19 120

We will prepare the following table:

 (xi) (fi) (fi)(xi) 10 17 170 30 f1 30f1 50 32 1600 70 f2 70f2 90 19 1710 120 3480 + 30f1 + 70f2

Thus, we have:

Also,
G
iven:
17
+ f1 + 32 + f2 + 19 = 120
68 + f1 + f2 = 120
f1 + f2 = 52
or, f2 = 52 $-$ f1
...(ii)
By putting the value of f2
in (i), we get:
2520
= 30f1 + 70(52 $-$ f1)
2520 = 30f1 + 3640 $-$ 70f1
40f1 = 1120
f1 = 28
Substituting the value in (ii), we get:
f2 = 52 $-$ f1 = 52 $-$ 28 = 24

#### Question 13:

Find the value of p, when the mean of the following distribution is 20.

 x 15 17 19 20 + p 23 f 2 3 4 5p 6

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 x 15 17 19 20 + p 23 f 2 3 4 5p 6

Mean = $\frac{\left(15×2\right)+\left(17×3\right)+\left(19×4\right)+\left(\left(20+p\right)×5p\right)+\left(23×6\right)}{2+3+4+5p+6}$
$⇒20=\frac{30+51+76+100p+5{p}^{2}+138}{15+5p}\phantom{\rule{0ex}{0ex}}⇒20\left(15+5p\right)=5{p}^{2}+100p+295\phantom{\rule{0ex}{0ex}}⇒300+100p=5{p}^{2}+100p+295\phantom{\rule{0ex}{0ex}}⇒5{p}^{2}=300-295\phantom{\rule{0ex}{0ex}}⇒5{p}^{2}=5\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=\frac{5}{5}\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=1\phantom{\rule{0ex}{0ex}}⇒p=±1$

Hence, the value of  p is ±1.

#### Question 14:

The mean of the following distribution is 50.

 x 10 30 50 70 90 f 17 5a + 3 32 7a – 11 19
Find the value of a and hence the frequencies of 30 and 70.

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 x 10 30 50 70 90 f 17 5a + 3 32 7a – 11 19

Mean = $\frac{\left(10×17\right)+\left(30×\left(5a+3\right)\right)+\left(50×32\right)+\left(70×\left(7a-11\right)\right)+\left(90×19\right)}{17+5a+3+32+7a-11+19}$
$⇒50=\frac{170+150a+90+1600+490a-770+1710}{60+12a}\phantom{\rule{0ex}{0ex}}⇒50\left(60+12a\right)=2800+640a\phantom{\rule{0ex}{0ex}}⇒3000+600a=2800+640a\phantom{\rule{0ex}{0ex}}⇒640a-600a=3000-2800\phantom{\rule{0ex}{0ex}}⇒40a=200\phantom{\rule{0ex}{0ex}}⇒a=\frac{200}{40}\phantom{\rule{0ex}{0ex}}⇒a=5$

Hence, the value of a is 5.

Also, the frequency of 30 is 28 and the frequency of 70 is 24.

#### Question 1:

Find the median of
(i) 2, 10, 9, 9, 5, 2, 3, 7, 11
(ii) 15, 6, 16, 8, 22, 21, 9, 18, 25
(iii) 20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22
(iv) 7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2

(i) Arranging the numbers in ascending order, we get:
2, 2, 3, 5, 7, 9, 9, 10, 11
Here, n is 9, which is an odd number.
If n is an odd number, we have:

Now,

(ii) Arranging the numbers in ascending order, we get:
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, n is 9, which is an odd number.
If n is an odd number, we have:

Now,

(iii) Arranging the numbers in ascending order, we get:
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, n is 11, which is an odd number.
If n is an odd number, we have:

Now,

(iv) Arranging the numbers in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, n is 13, which is an odd number.
If n is an odd number, we have:

Now,

#### Question 2:

Find the median of
(i) 17, 19, 32, 10, 22, 21, 9, 35
(ii) 72, 63, 29, 51, 35, 60, 55, 91, 85, 82
(iii) 10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27

(i) Arranging the numbers in ascending order, we get:
9, 10, 17, 19, 21, 22, 32, 35
Here, n is 8, which is an even number.
If n is an even number, we have:

Now,

(ii) Arranging the numbers in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here, n is 10, which is an even number.
If n is an even number, we have:

Now,

(iii) Arranging the numbers in ascending order, we get:
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here, n is 12, which is an even number.
If n is an even number, we have:

Now,

#### Question 3:

The marks of 15 students in an examination are:
25, 19, 17, 24, 23, 29, 31, 40, 19, 20, 22, 26, 17, 35, 21.
Find the median score.

Arranging the marks of 15 students in ascending order, we get:
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Here, n is 15, which is an odd number.
We know:

Thus, we have:

#### Question 4:

The heights (in cm) of 9 students of a class are 148, 144, 152, 155, 160, 147, 150, 149, 145.
Find the median height.

Arranging the given data in ascending order:
144, 145, 147, 148, 149, 150, 152, 155, 160

Number of terms = 9 (odd)

Hence, the median height is 149.

#### Question 5:

The weights (in kg) of 8 children are:
13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8.
Find the median weight.

Arranging the weights (in kg) in ascending order, we have:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here, n is 8, which is an even number.
Thus, we have:

Hence, the median weight is 13.85 kg.

#### Question 6:

The ages (in years) of 10 teachers in a school are:
32, 44, 53, 47, 37, 54, 34, 36, 40, 50.
Find the median age.

Arranging the ages (in years) in ascending order, we have:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, n is 10, which is an even number.
Thus, we have:

#### Question 7:

If 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41 are ten observation in an ascending order with median 24, find the value of x.

10, 13, 15, 18, x+1, x+3, 30, 32, 35 and 41 are arranged in ascending order.
Median = 24
We have to find the value of x.
Here, n is 10, which is an even number.
Thus, we have:

#### Question 8:

The following observations are arranged in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93.
If the median is 65, find the value of x.

Arranging the given data in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93

Number of terms = 10 (even)

Hence, the value of x is 64.

#### Question 9:

The numbers 50, 42, 35, (2x + 10), (2x – 8), 12, 11, 8 have been written in a descending order. If their median is 25, find the value of x.

Arranging the given data in ascending order:
8, 11, 12, (2x – 8), (2x + 10), 35, 42, 50

Number of terms = 8 (even)

Hence, the value of x is 12.

#### Question 10:

Find the median of the data
46, 41, 77, 58, 35, 64, 87, 92, 33, 55, 90.
In the above data, if 41 and 55 are replaced by 61 and 75 respectively, what will be the new median?

Arranging the given data in ascending order:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Number of terms = 11 (odd)

Hence, the median of the data is 58.

Now, In the above data, if 41 and 55 are replaced by 61 and 75 respectively.
Then, new data in ascending order is:
33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92

Number of terms = 11 (odd)

Hence, the new median of the data is 64.

#### Question 1:

Find the mode of the following items.
0, 6, 5, 1, 6, 4, 3, 0, 2, 6, 5, 6

On arranging the items in ascending order, we get:
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Clearly, 6 occurs maximum number of times.
∴ Mode = 6

#### Question 2:

Determine the mode of the following values of a variable.
23, 15, 25, 40, 27, 25, 22, 25, 20

On arranging the values in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
Clearly, 25 occurs maximum number of times.
∴ Mode = 25

#### Question 3:

Calculate the mode of the following sizes of shoes sold by a shop on a particular day.
5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9

On arranging the shoe sizes in ascending order, we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Clearly, 9 occurs maximum number of times.
∴ Mode = 9

#### Question 4:

A cricket player scored the following runs in 12 one-day matches:
50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35.
Find his modal score.

On arranging the runs in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Clearly, 50 occurs maximum number of times.
∴ Modal score = 50

#### Question 5:

If the mean of the data 3, 21, 25, 17, (x + 3), 19, (x – 4) is 18, find the value of x. Using this value of x, find the mode of the data.

We know that,

The given data is 3, 21, 25, 17, (x + 3), 19, (x – 4).

Mean of the given data = $\frac{3+21+25+17+\left(x+3\right)+19+\left(x-4\right)}{7}$
$⇒18\left(7\right)=84+2x\phantom{\rule{0ex}{0ex}}⇒126-84=2x\phantom{\rule{0ex}{0ex}}⇒2x=42\phantom{\rule{0ex}{0ex}}⇒x=21$

Hence, the value of x is 21.

Now, the given data is 3, 21, 25, 17, 24, 19, 17
Arranging this data in ascending order:
3, 17, 17, 19, 21, 24, 25

Here, 17 occurs maximum number of times.
∴ Mode = 17

Hence, the mode of the data is 17.

#### Question 6:

The numbers 52, 53, 54, 54, (2x + 1), 55, 55, 56, 57 have been arranged in an ascending order and their median is 55. Find the value of x and hence find the mode of the given data.

Arranging the given data in ascending order:
52, 53, 54, 54, (2x + 1), 55, 55, 56, 57

Number of terms = 9 (odd)

Hence, the value of x is 27.

Arranging the given data in ascending order:
52, 53, 54, 54, 55, 55, 55, 56, 57

Here, 55 occurs maximum number of times.
∴ Mode = 55

Hence, the mode of the data is 55.

#### Question 7:

For what value of x is the mode of the data 24, 15, 40, 23, 27, 26, 22, 25, 20, x + 3 found 25? Using this value of x, find the median.

Given: Mode = 25
∴ 25 occurs maximum number of times.

Arranging the given data in ascending order:
15, 20, 22, 23, 24, x + 3, 25, 26, 27, 40

x + 3 = 25
x = 25 − 3
x = 22

Hence, the value of x is 22.

Arranging the given data in ascending order:
15, 20, 22, 23, 24, 25, 25, 26, 27, 40

Number of terms = 10 (even)

Hence, the median is 24.5 .

#### Question 8:

The numbers 42, 43, 44, 44, (2x + 3), 45, 45, 46, 47 have been arranged in an ascending order and their median is 45. Find the value of x. Hence, find the mode of the above data.

Arranging the given data in ascending order:
42, 43, 44, 44, (2x + 3), 45, 45, 46, 47

Number of terms = 9 (odd)

Hence, the value of x is 21.

Arranging the given data in ascending order:
42, 43, 44, 44, 45, 45, 45, 46, 47

Here, 45 occurs maximum number of times.
∴ Mode = 45

Hence, the mode of the data is 45.

#### Question 1:

If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
(a) 5
(b) 6
(c) 7
(d) 8

(c) 7

Mean of 5 observations = 11
We know:

#### Question 2:

If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
(a) $10\frac{1}{3}$
(b) $10\frac{2}{3}$
(c) $11\frac{1}{3}$
(d) $11\frac{2}{3}$

(c) 11$\frac{1}{3}$
Mean of 5 observations = 9
We know:

#### Question 3:

If $\overline{)x}$ is the mean of then
(a) −1
(b) 0
(c) 1
(d) n − 1

(b) 0

#### Question 4:

If each observation of the data is increased by 8, then their mean
(a) remains the same
(b) is decreased by 8
(c) is increased by 5
(d) becomes 8 times the original mean

(b) is decreased by 8

1, x2,...xn.

Now the new numbers after decreasing every number by 8 : (x1−8) , (x2−8)...,(xn−8)

Hence, mean is decreased by 8.

#### Question 5:

The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
(a) 52 kg
(b) 52.8 kg
(c) 53 kg
(d) 47 kg

(c) 53 kg

Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.

#### Question 6:

The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
(a) 38.6
(b) 39.4
(c) 39.8
(d) 39.2

(b) 39.4

Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = $\left(39×50\right)=1950$
Correct sum = (1950 + 43 $-$ 23) = 1970
$\therefore \mathrm{Mean}=\frac{1970}{50}=39.4\phantom{\rule{0ex}{0ex}}$

#### Question 7:

The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is
(a) 64.86
(b) 65.31
(c) 64.91
(d) 64.61

(c) 64.91

Mean of 100 items = 64
Sum of 100 items = $64×100=6400$
Correct sum = (6400 + 36 + 90 $-$ 26 $-$ 9) = 6491

#### Question 8:

The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be
(a) 50.5
(b) 51
(c) 51.5
(d) 52

(b) 51

Mean of 100 observations = 50
Sum of 100 observations = $100×50=5000$
It is given that one of the observations, 50, is replaced by 150.
∴ New sum = (5000 $-$ 50 + 150) = 5100
And,

#### Question 9:

Let $\overline{)x}$ be the mean of and $\overline{)y}$ be the mean of .
If $\overline{)z}$ is the mean of then
(a)
(b)
(c)
(d)

(b)

$\overline{z}=\frac{\left({x}_{1}+{x}_{2}+...+{x}_{n}\right)+\left({y}_{1}+{y}_{2}+...+{y}_{n}\right)}{2n}$

#### Question 10:

If $\overline{)x}$ is the mean of then for , the mean of is
(a) $\left(a+\frac{1}{a}\right)\overline{)x}$

(b) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{2}$

(c) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{n}$

(d) $\frac{\left(a+\frac{1}{a}\right)\overline{)x}}{2n}$

(b) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{2}$

#### Question 11:

If are the means of n groups with number of observations respectively, then the mean $\overline{)x}$ of all the groups taken together is
(a) $\sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}$

(b) $\sum _{\frac{i=1}{{n}^{2}}}^{n}{n}_{i}{\overline{)x}}_{i}$

(c) $\frac{\sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}}{\sum _{i=1}^{n}{n}_{i}}$

(d) $\frac{\sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}}{2n}$

#### Question 12:

The mean of the following data is 8.

 x 3 5 7 9 11 13 y 6 8 15 p 8 4
The value of p is
(a) 23
(b) 24
(c) 25
(d) 21

(c) 25

 x y x$×$y 3 6 18 5 8 40 7 15 105 9 p 9p 11 8 88 13 4 52 Total 41 + p 303 + 9p

$\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Mean}=\frac{303+9p}{41+p}\phantom{\rule{0ex}{0ex}}\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mean}=8\phantom{\rule{0ex}{0ex}}\therefore \frac{303+9p}{41+p}=8\phantom{\rule{0ex}{0ex}}⇒303+9p=328+8p\phantom{\rule{0ex}{0ex}}⇒p=25$

#### Question 13:

The runs scored by 11 members of a cricket team are
15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0
The median score is
(a) 27
(b) 29
(c) 31
(d) 20

(b) 29

Arranging the weight of 10 students in ascending order, we have:
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56
Here, n is 11, which is an odd number.
Thus, we have:

#### Question 14:

The weight of 10 students (in kgs) are
55, 40, 35, 52, 60, 38, 36, 45, 31, 44
The median weight is
(a) 40 kg
(b) 41 kg
(c) 42 kg
(d) 44 kg

(c) 42 kg

Arranging the numbers in ascending order, we have:
31, 35, 36, 38, 40, 44, 45, 52, 55, 60
Here, n is 10, which is an even number.
Thus, we have:

#### Question 15:

The median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is
(a) 4
(b) 5
(c) 6
(d) 7

(c) 6

We will arrange the given data in ascending order as:
3, 4, 4, 5, 6, 7, 7, 7, 12
Here, n is 9, which is an odd number.
Thus, we have:

#### Question 16:

The median of the numbers 84, 78, 54, 56, 68, 22, 34, 45, 39, 54 is
(a) 45
(b) 49.5
(c) 54
(d) 56

(c) 54

We will arrange the data in ascending order as:
22, 34, 39, 45, 54, 54, 56, 68, 78, 84
Here, n is 10, which is an even number.
Thus, we have:

#### Question 17:

Mode of the data 15, 17, 15, 19, 14, 18, 15, 14, 16, 15, 14, 20, 19, 14, 15 is
(a) 14
(b) 15
(c) 16
(d) 17

(b) 15

Here, 14 occurs 4 times, 15 occurs 5 times, 16 occurs 1 time, 18 occurs 1 time, 19 occurs 1 time and 20 occurs 1 time. Therefore, the mode, which is the most occurring item, is 15.

#### Question 18:

The median of the data arranged in ascending order 8, 9, 12, 18, (x + 2), (x + 4), 30, 31, 34, 39 is 24. The value of x is
(a) 22
(b) 21
(c) 20
(d) 24