Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 18 Mean, Median And Mode Of Ungrouped Data are provided here with simple step-by-step explanations. These solutions for Mean, Median And Mode Of Ungrouped Data are extremely popular among Class 9 students for Maths Mean, Median And Mode Of Ungrouped Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 9 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 669:

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:

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

#### Page No 669:

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

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

#### Page No 670:

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

#### Page No 670:

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

#### Page No 670:

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.

#### Page No 670:

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.

#### Page No 670:

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.

#### Page No 670:

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:

#### Page No 670:

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

Thus, we have:

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

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

#### Page No 670:

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

Thus, we have:

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

#### Page No 670:

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

Thus, we have:

After division, the numbers become:

#### Page No 670:

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:

#### Page No 670:

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

#### Page No 670:

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.

#### Page No 670:

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.

#### Page No 670:

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.

#### Page No 671:

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.

#### Page No 671:

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:

#### Page No 671:

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.

#### Page No 671:

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.

#### Page No 671:

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

#### Page No 671:

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.

#### Page No 671:

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.

#### Page No 671:

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.

#### Page No 671:

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

#### Page No 671:

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.

#### Page No 671:

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:

#### Page No 671:

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.

#### Page No 671:

Total salary of 75 workers = ₹ 5680 × 75
= ₹ 426000

Total salary of 25 workers = ₹ 5400 × 25
= ₹ 135000

Total salary of 30 workers = ₹ 5700 × 30
= ₹ 171000

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

Total salary of 20 workers = ₹ (426000 − 135000 − 171000)
= ₹ 120000

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

Hence, the mean weekly payment of the remaining workers is ₹ 6000.

#### Page No 671:

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.

#### Page No 672:

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.

#### Page No 676:

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 .

#### Page No 676:

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:

=

#### Page No 677:

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 .

#### Page No 677:

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:

#### Page No 677:

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:

#### Page No 677:

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

For the following data:

 Daily wages (in ₹) (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 .

#### Page No 677:

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.

#### Page No 677:

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:

#### Page No 677:

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:

#### Page No 677:

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:

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

#### Page No 678:

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.

#### Page No 678:

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

#### Page No 678:

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.

#### Page No 678:

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.

#### Page No 680:

(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,

#### Page No 680:

(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,

#### Page No 681:

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:

#### Page No 681:

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.

#### Page No 681:

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.

#### Page No 681:

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:

#### Page No 681:

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:

#### Page No 681:

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.

#### Page No 681:

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.

#### Page No 681:

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.

#### Page No 683:

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

#### Page No 683:

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

#### Page No 683:

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

#### Page No 683:

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

#### Page No 683:

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.

#### Page No 683:

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.

#### Page No 684:

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 .

#### Page No 684:

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.

#### Page No 684:

(c) 7

Mean of 5 observations = 11
We know:

#### Page No 684:

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

(b) 0

#### Page No 684:

(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.

#### Page No 684:

(c) 53 kg

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

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(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
$\therefore \mathrm{Mean}=\frac{1970}{50}=39.4\phantom{\rule{0ex}{0ex}}$

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(c) 64.91

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

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(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,

#### Page No 685:

(b)

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

#### Page No 685:

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

#### Page No 685:

(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

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

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(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:

#### Page No 685:

(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:

#### Page No 686:

(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:

#### Page No 686:

(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:

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(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.