# NCERT Solutions for Class 10 Maths Chapter 14 Statistics

NCERT solutions for class 10 maths chapter 14 Statistics - Today's world is highly dependent on data. Each and every sector has a group of data that represents various information. In the previous classes, you have studied the classification of the given data into grouped and ungrouped frequency distributions and have learnt to represent given data in the form of graphs such as histograms, bar graphs, etc and mean mode and median. Solutions of NCERT class 10 maths chapter 14 Statistics consists of the solutions for the questions based on the study of measures of central tendency, namely, mean, mode and median from the ungrouped data to that of grouped data. CBSE NCERT solutions for class 10 maths chapter 14 Statistics, consists detailed explanation of all 4 exercises comprising 25 questions. Statistics is the branch of applied mathematics that deals with the collection of data and represent it in a meaningful way. In NCERT class 10 maths chapter 14 Statistics you will also get the solutions of cumulative frequency, its distribution and how to draw cumulative frequency curves questions. Students face many real-life problems where the fundamentals of statistics are used to represent data in graphs or in tabular form. For classwise and subjectwise NCERT solutions, you can click on the given link.

## Types of questions asked from class 10 maths chapter 14 Statistics

• A direct method of finding mean
• Assumed mean method of finding mean
• Step deviation method of finding mean
• Question-related to median
• Questions based on the mode

## NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.1

Which method did you use for finding the mean, and why?

 Number of plants Number of houses     $f_i$ Classmark $x_i$ $f_ix_i$ 0-2 1 1 1 2-4 2 3 6 4-6 1 5 5 6-8 5 7 35 8-10 6 9 54 10-12 2 11 22 12-14 3 13 39 $\sum f_i$  =20 $\sum f_ix_i$ =162

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{162}{20} = 8.1$
We used the direct method in this as the values of $x_i\ and\ f_i$ are small.

Let the assumed mean be a = 550

 Daily  Wages Number of workers $f_i$ Classmark $x_i$ $d_i = x_i -a$ $f_id_i$ 500-520 12 510 -40 -480 520-540 14 530 -20 -280 540-560 8 550 0 0 560-580 6 570 20 120 580-600 10 590 40 400 $\sum f_i$  =50 $\sum f_ix_i$ =-240

Mean,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 550 + \frac{-240}{50} = 550-4.8 = 545.20$
Therefore, the mean daily wages of the workers of the factory is Rs. 545.20

## Statistics Excercise: 14.1

 Daily pocket allowance Number of children $f_i$ Classmark $x_i$ $f_ix_i$ 11-13 7 12 84 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 $\sum f_i$  =44+f $\sum f_ix_i$ =752+20f

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$\implies 18 = \frac{752+20f}{44+f}$

$\\ \implies 18(44+f) =( 752+20f) \\ \implies 2f = 40 \\ \implies f = 20$
Therefore the missing f = 20

### Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Let the assumed mean be a = 75.5

 No. of heartbeats per minute Number of women $f_i$ Classmark $x_i$ $d_i = x_i -a$ $f_id_i$ 65-68 2 66.5 -9 -18 68-71 4 69.5 -6 -24 71-74 3 72.5 -3 -9 74-77 8 75.5 0 0 77-80 7 78.5 3 21 80-83 4 81.5 6 24 83-86 2 84.5 9 18 $\sum f_i$ =30 $\sum f_ix_i$ =12

Mean,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$
Therefore, the mean heartbeats per minute of these women are 75.9

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Let the assumed mean be a = 57

 Number of  mangoes Number of boxes $f_i$ Classmark $x_i$ $d_i = x_i -a$ $f_id_i$ 50-52 15 51 -6 -90 53-55 110 54 -3 -330 56-58 135 57 0 0 59-61 115 60 3 345 62-64 25 63 6 150 $\sum f_i$  =400 $\sum f_ix_i$ =75

Mean,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 57+ \frac{75}{400} = 57+0.1875 = 57.1875 \approx 57.19$
Therefore, the mean number of mangoes kept in a packing box is approx 57.19

Let the assumed mean be a = 225 and h = 50

 Daily Expenditure Number of households $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 100-150 4 125 -100 -2 -8 150-200 5 175 -50 -1 -5 200-250 12 225 0 0 0 250-300 2 275 50 1 2 300-350 2 325 100 2 4 $\sum f_i$ =25 $\sum f_ix_i$ = -7

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 225 + \frac{-7}{25}\times50 = 225 -14 = 211$
Therefore, the mean daily expenditure on food is Rs. 211

Find the mean concentration of $SO_2$ in the air.

 Class  Interval Frequency  $f_i$ Classmark $x_i$ $f_ix_i$ 0.00-0.04 4 0.02 0.08 0.04-0.08 9 0.06 0.54 0.08-0.12 9 0.10 0.90 0.12-0.16 2 0.14 0.28 0.16-0.20 4 0.18 0.72 0.20-0.24 2 0.22 0.44 $\sum f_i$  =30 $\sum f_ix_i$ =2.96

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{2.96}{30} = 0.099$
Therefore, the mean concentration of $SO_2$ in the air is 0.099 ppm

 Number of days Number of Students $f_i$ Classmark $x_i$ $f_ix_i$ 0-6 11 3 33 6-10 10 8 80 10-14 7 12 84 14-20 4 17 68 20-28 4 24 96 28-38 3 33 99 38-40 1 39 39 $\sum f_i$  =40 $\sum f_ix_i$ =499

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{499}{40} = 12.475$$= \frac{499}{40} = 12.475\approx 12.48$
Therefore, the mean number of days a student was absent is 12.48 days.

Let the assumed mean be a = 75 and h = 10

 Literacy rates Number of cities $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 45-55 3 50 -20 -2 -6 55-65 10 60 -10 -1 -10 65-75 11 70 0 0 0 75-85 8 80 10 1 8 85-95 3 90 20 2 6 $\sum f_i$ = 35 $\sum f_ix_i$ = -2

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 70 + \frac{-2}{35}\times10 = 70 -0.57 = 69.43$
Therefore, the mean mean literacy rate is 69.43%

## Statistics Excercise: 14.2

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

The class having maximum frequency is the modal class.

The maximum frequency is 23 and hence the modal class = 35-45

Lower limit (l) of modal class = 35, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 23, frequency ( $f_0$ ) of class preceding the modal class = 21, frequency ( $f_2$ ) of class succeeding the modal class = 14.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 35 + \left(\frac{23-21}{2(23)-21-14} \right).10 \\ \\ = 35 + \frac{2}{11}.10$

$= 36.8$

Now,

 Age Number of patients $f_i$ Classmark $x_i$ $f_ix_i$ 5-15 6 10 60 15-25 11 20 220 25-35 21 30 630 35-45 23 40 920 45-55 14 50 700 55-65 5 60 300 $\sum f_i$  =80 $\sum f_ix_i$ =2830

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{2830}{80} = 35.37$
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.

Determine the modal lifetimes of the components.

The class having maximum frequency is the modal class.

The maximum frequency is 61 and hence the modal class = 60-80

Lower limit (l) of modal class = 60, class size (h) = 20

Frequency ( $f_1$ ) of the modal class = 61 frequency ( $f_0$ ) of class preceding the modal class = 52, frequency ( $f_2$ ) of class succeeding the modal class = 38.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 60 + \left(\frac{61-52}{2(61)-52-38} \right).20 \\ \\ = 60 + \frac{9}{32}.20$

$= 65.62$

Thus, the modal lifetime of 225 electrical components is 65.62 hours

The class having maximum frequency is the modal class.

The maximum frequency is 40 and hence the modal class = 1500-2000

Lower limit (l) of modal class = 1500, class size (h) = 500

Frequency ( $f_1$ ) of the modal class = 40 frequency ( $f_0$ ) of class preceding the modal class = 24, frequency ( $f_2$ ) of class succeeding the modal class = 33.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 1500 + \left(\frac{40-24}{2(40)-24-33} \right).500 \\ \\ = 1500 + \frac{16}{23}.500$

$= 1847.82$

Thus, the Mode of the data is Rs. 1847.82

Now,

Let the assumed mean be a = 2750 and h = 500

 Expenditure Number of families $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 1000-1500 24 1250 -1500 -3 -72 1500-2000 40 1750 -1000 -2 -80 2000-2500 33 2250 -500 -1 -33 2500-3000 28 2750 0 0 0 3000-3500 30 3250 500 1 30 3500-4000 22 3750 1000 2 44 4000-4500 16 4250 1500 3 48 4500-5000 7 4750 2000 4 28 $\sum f_i$ =200 $\sum f_ix_i$ = -35

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 2750 + \frac{-35}{200}\times500 = 2750 -87.5 = 2662.50$

Thus, the Mean monthly expenditure is Rs. 2662.50

The class having maximum frequency is the modal class.

The maximum frequency is 10 and hence the modal class = 30-35

Lower limit (l) of modal class = 30, class size (h) = 5

Frequency ( $f_1$ ) of the modal class = 10 frequency ( $f_0$ ) of class preceding the modal class = 9, frequency ( $f_2$ ) of class succeeding the modal class = 3

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 30 + \left(\frac{10-9}{2(10)-9-3} \right).5 \\ \\ = 30 + \frac{1}{8}.5$

$= 30.625$

Thus, Mode of the data is 30.625

Now,

Let the assumed mean be a = 32.5 and h = 5

 Class Number of states $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 15-20 3 17.5 -15 -3 -9 20-25 8 22.5 -10 -2 -16 25-30 9 27.5 -5 -1 -9 30-35 10 32.5 0 0 0 35-40 3 37.5 5 1 3 40-45 0 42.5 10 2 0 45-50 0 47.5 15 3 0 50-55 2 52.5 20 4 8 $\sum f_i$ =35 $\sum f_ix_i$ = -23

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 32.5 + \frac{-23}{35}\times5= 29.22$

Thus, the Mean of the data is 29.22

Find the mode of the data.

The class having maximum frequency is the modal class.

The maximum frequency is 18 and hence the modal class = 4000-5000

Lower limit (l) of modal class = 4000, class size (h) = 1000

Frequency ( $f_1$ ) of the modal class = 18 frequency ( $f_0$ ) of class preceding the modal class = 4, frequency ( $f_2$ ) of class succeeding the modal class = 9

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ =4000 + \left(\frac{18-4}{2(18)-4-9} \right).1000 \\ \\ = 4000 + \frac{14}{23}.1000$

$= 4608.70$

Thus, Mode of the data is 4608.70

The class having maximum frequency is the modal class.

The maximum frequency is 20 and hence the modal class = 40-50

Lower limit (l) of modal class = 40, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 20 frequency ( $f_0$ ) of class preceding the modal class = 12, frequency ( $f_2$ ) of class succeeding the modal class = 11

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ =40+ \left(\frac{20-12}{2(20)-12-11} \right).10 \\ \\ = 40 + \frac{8}{17}.10$

$= 44.70$

Thus, Mode of the data is 44.70

## CBSE NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.3

Let the assumed mean be a = 130 and h = 20

 Class Number of consumers $f_i$ Cumulative Frequency Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 65-85 4 4 70 -60 -3 -12 85-105 5 9 90 -40 -2 -10 105-125 13 22 110 -20 -1 -13 125-145 20 42 130 0 0 0 145-165 14 56 150 20 1 14 165-185 8 64 170 40 2 16 185-205 4 68 190 60 3 12 $\sum f_i = N$ = 68 $\sum f_ix_i$ = 7

MEDIAN:
$N= 68 \implies \frac{N}{2} = 34$
$\therefore$ Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125;

c.f. = 22; f = 20; h = 20
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 125 + \left (\frac{34-22}{20} \right ).20 \\ \\ = 125 + 12$

$= 137$

Thus, the median of the data is 137

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125-145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ($f_1$) of the modal class = 20; frequency ($f_0$) of class preceding the modal class = 13, frequency ($f_2$) of class succeeding the modal class = 14.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 125 + \left(\frac{20-13}{2(20)-13-14} \right).20 \\ \\ = 125 + \frac{7}{13}.20$

$= 135.76$

Thus, Mode of the data is 135.76

MEAN:

Mean,
$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 130 + \frac{7}{68}\times20 = 137.05$

Thus, the Mean of the data is 137.05

 Class Number of consumers $f_i$ Cumulative Frequency 0-10 5 5 10-20 x 5+x 20-30 20 25+x 30-40 15 40+x 40-50 y 40+x+y 50-60 5 45+x+y $\sum f_i = N$ = 60

$\dpi{100} N= 60 \implies \frac{N}{2} = 30$

Now,
Given median = 28.5 which lies in the class 20-30

Therefore, Median class = 20-30
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ \implies28.5= 20 + \left (\frac{30-5-x}{20} \right ).10 \\ \\ \implies8.5=\frac{25-x}{2} \\ \implies 25-x = 8.5(2) \\ \implies x = 25 - 17 = 8$

Also,

$\\ 60 = 45 + x+y \\ \implies x+y = 60-45 = 15 \\ \implies y = 15-x = 15-8 \ \ \ (\because x =8) \\ \implies y = 7$

Therefore, the required values are: x=8 and y=7

 Class Frequency  $f_i$ Cumulative Frequency 15-20 2 2 20-25 4 6 25-30 18 24 30-35 21 45 35-40 33 78 40-45 11 89 45-50 3 92 50-55 6 98 55-60 2 100

$\dpi{100} N= 100 \implies \frac{N}{2} = 50$
Therefore, Median class = 35-45
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 35 + \left (\frac{50-45}{33} \right ).5 \\ \\$

$= 35.75$

Thus, the median age is 35.75 years.

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.

 Class Frequency  $f_i$ Cumulative Frequency 117.5-126.5 3 3 126.5-135.5 5 8 135.5-144.5 9 17 144.5-153.5 12 29 153.5-162.5 5 34 162.5-171.5 4 38 171.5-180.5 2 40

$\dpi{100} N= 40 \implies \frac{N}{2} = 20$
Therefore, Median class = 144.5-153.5

Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 144.5 + \left (\frac{20-17}{12} \right ).9 \\ \\$

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

 Class Frequency  $f_i$ Cumulative Frequency 1500-2000 14 14 2000-2500 56 70 2500-3000 60 130 3000-3500 86 216 3500-4000 74 290 4000-4500 62 352 4500-5000 48 400

$\dpi{100} N= 400 \implies \frac{N}{2} = 200$
Therefore, Median class = 3000-3500

Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 3000 + \left (\frac{200-130}{86} \right ).500 \\ \\ = 3000+406.97$

$= 3406.97$

Thus, the median lifetime of a lamp is 3406.97 hours

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

Determine the median number of letters in the surnames. Find the mean number of
letters in the surnames? Also, find the modal size of the surnames.

 Class Number of surnames $f_i$ Cumulative Frequency Classmark $x_i$ $f_ix_i$ 1-4 6 6 2.5 15 4-7 30 36 5.5 165 7-10 40 76 8.5 340 10-13 16 92 11.5 184 13-16 4 96 14.5 51 16-19 4 100 17.5 70 $\sum f_i = N$ = 100 $\sum f_ix_i$ = 825

MEDIAN:
$N= 100 \implies \frac{N}{2} = 50$
$\therefore$ Median class = 7-10; Lower limit, l = 7;

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 7+ \left (\frac{50-36}{40} \right ).3 \\ \\$

$= 8.05$

Thus, the median of the data is 8.05

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency ($f_1$) of the modal class = 40; frequency ($f_0$) of class preceding the modal class = 30, frequency ($f_2$) of class succeeding the modal class = 16

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 7 + \left(\frac{40-30}{2(40)-30-16} \right).3 \\ \\ = 125 + \frac{10}{34}.3$

$= 7.88$

Thus, Mode of the data is 7.88

MEAN:

Mean,
$\overline x =\frac{\sum f_ix_i}{\sum f_i}$
$= \frac{825}{100} = 8.25$

Thus, the Mean of the data is 8.25

 Class Number of students $f_i$ Cumulative Frequency 40-45 2 2 45-50 3 5 50-55 8 13 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30

MEDIAN:
$N= 30 \implies \frac{N}{2} = 15$
$\therefore$ Median class = 55-60; Lower limit, l = 55;

Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 55+ \left (\frac{15-13}{6} \right ).5 \\ \\ = 55+\frac{2}{6}.5$

$= 56.67$

Thus, the median weight of the student is 56.67 kg

## NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.4

 Daily Income (Upper-Class Limit) Cumulative Frequency Less than 120 12 Less than 140 26 Less than 160 34 Less than 180 40 Less than 200 50

Now,

Taking upper-class interval on the x-axis and their respective frequencies on the y-axis,

Taking upper-class interval on the x-axis and their respective frequencies on the y-axis,

$N= 35 \implies \frac{N}{2} = 17.5$

Marking a point on the curve whose ordinate is 17.5 gives an x-ordinate= 46.5.

Hence, the Median of the data is 46.5

Now,

 Weight (Class) Frequency Cumulative Frequency >38 0 0 38-40 3 3 40-42 2 5 42-44 4 9 44-46 5 14 46-48 14 28 48-50 4 32 50-52 3 35

$N= 35 \implies \frac{N}{2} = 17.5$
$\therefore$ Median class = 46-48; Lower limit, l = 46;

Cumulative frequency of preceding class, c.f. = 14; f = 14; h = 2
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 46+ \left (\frac{17.5-14}{14} \right ).2 \\ \\ = 46+\frac{7}{14}$

$= 46.5$

Thus, the median using formula is 46.5 which verifies the result.

 Production yield (Upper-Class Limit) Cumulative Frequency More than or equal to 50 100 More than or equal to 55 98 More than or equal to 60 90 More than or equal to 65 78 More than or equal to 70 54 More than or equal to 75 16

Now,

Taking lower class limit on the x-axis and their respective frequencies on the y-axis,

## NCERT solutions for class 10 maths chapter wise

 Chapter No. Chapter Name Chapter 1 CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers Chapter 2 NCERT solutions for class 10 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables Chapter 4 CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations Chapter 5 NCERT solutions for class 10 chapter 5 Arithmetic Progressions Chapter 6 Solutions of NCERT class 10 maths chapter 6 Triangles Chapter 7 CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry Chapter 8 NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Chapter 9 Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry Chapter 10 CBSE NCERT solutions class 10 maths chapter 10 Circles Chapter 11 NCERT solutions  for class 10 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles Chapter 13 CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes Chapter 14 NCERT solutions for class 10 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 10 maths chapter 15 Probability

Solutions of NCERT class 10 - subject wise

## How to use NCERT solutions for class 10 maths chapter 14 Statistics?

• First of all, go through NCERT Textbook to have an understanding of data.

• Learn the techniques to calculate the mean, median, mode.

• Once you learn the techniques and formulae, then move to the practice exercises available in the chapter.

• While practicing, if you get stuck anywhere then you can take the help of NCERT solutions for class 10 maths chapter 14 Statistics.

Keep working hard & happy learning!