NCERT Solutions for Class 11 Maths Chapter 15 Statistics
NCERT Solutions for Class 11 Maths Chapter 15 Statistics: You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous class. In this chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In NCERT solutions for class 11 maths chapter 15 statistics, you will get questions related to all the above topics. You will also learn some methods for finding a representative value for the given data. This value is known as the measure of central tendency. This chapter is important for the CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT, etc. As statistics deal with the collected data for specific purposes and in the digital world, everything that we do comes in the form of data. This is a very useful technique to interpret, analyze the data, and make a conclusive decision based on the data. NCERT solutions for class 11 maths chapter 15 statistics will build your fundamentals of statistics which will be useful to learn advanced statistical techniques. There are 27 questions in 3 exercises in the NCERT textbook. All these questions are explained in NCERT solutions for class 11 maths chapter 15 statistics in a detailed manner. Check all NCERT solutions which will help you to understand the concepts in an easy way.
The complete NCERT Solutions for Class 11 Mathematics Chapter 15 is provided below:
NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.1
Question:1 . Find the mean deviation about the mean for the data.
Answer:
Mean ( ) of the given data:
The respective absolute values of the deviations from mean, are
6, 3, 2, 1, 0, 2, 3, 7
Hence, the mean deviation about the mean is 3.
Question:2. Find the mean deviation about the mean for the data.
Answer:
Mean ( ) of the given data:
The respective absolute values of the deviations from mean, are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Hence, the mean deviation about the mean is 8.4.
Question:3. Find the mean deviation about the median.
Answer:
Number of observations, n = 12, which is even.
Arranging the values in ascending order:
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
Now, Median (M)
The respective absolute values of the deviations from median, are
3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Hence, the mean deviation about the median is 2.33.
Question:4. Find the mean deviation about the median.
Answer:
Number of observations, n = 10, which is even.
Arranging the values in ascending order:
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Now, Median (M)
The respective absolute values of the deviations from median, are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Hence, the mean deviation about the median is 7.
Question:5 Find the mean deviation about the mean.
Answer:
5 | 7 | 35 | 9 | 63 |
10 | 4 | 40 | 4 | 16 |
15 | 6 | 90 | 1 | 6 |
20 | 3 | 60 | 6 | 18 |
25 | 5 | 125 | 11 | 55 |
= 25 | = 350 | =158 |
Now, we calculate the absolute values of the deviations from mean, and
= 158
Hence, the mean deviation about the mean is 6.32
Question:6. Find the mean deviation about the mean.
Answer:
10 | 4 | 40 | 40 | 160 |
30 | 24 | 720 | 20 | 480 |
50 | 28 | 1400 | 0 | 0 |
70 | 16 | 1120 | 20 | 320 |
90 | 8 | 720 | 40 | 320 |
= 80 | = 4000 | =1280 |
Now, we calculate the absolute values of the deviations from mean, and
= 1280
Hence, the mean deviation about the mean is 16
Question:7. Find the mean deviation about the median.
Answer:
5 | 8 | 8 | 2 | 16 |
7 | 6 | 14 | 0 | 0 |
9 | 2 | 16 | 2 | 4 |
10 | 2 | 18 | 3 | 6 |
12 | 2 | 20 | 5 | 10 |
15 | 6 | 26 | 8 | 48 |
Now, N = 26 which is even.
Median is the mean of and observations.
Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
Therefore, Median, M
Now, we calculate the absolute values of the deviations from median, and
= 84
Hence, the mean deviation about the median is 3.23
Question:8 Find the mean deviation about the median.
Answer:
15 | 3 | 3 | 13.5 | 40.5 |
21 | 5 | 8 | 7.5 | 37.5 |
27 | 6 | 14 | 1.5 | 9 |
30 | 7 | 21 | 1.5 | 10.5 |
35 | 8 | 29 | 6.5 | 52 |
Now, N = 30, which is even.
Median is the mean of and observations.
Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.
Therefore, Median, M
Now, we calculate the absolute values of the deviations from median, and
= 149.5
Hence, the mean deviation about the median is 5.1
Question:9. Find the mean deviation about the mean.
Answer:
Income per day | Number of Persons | Mid Points | |||
0 -100 | 4 | 50 | 200 | 308 | 1232 |
100 -200 | 8 | 150 | 1200 | 208 | 1664 |
200-300 | 9 | 250 | 2250 | 108 | 972 |
300-400 | 10 | 350 | 3500 | 8 | 80 |
400-500 | 7 | 450 | 3150 | 92 | 644 |
500-600 | 5 | 550 | 2750 | 192 | 960 |
600-700 | 4 | 650 | 2600 | 292 | 1168 |
700-800 | 3 | 750 | 2250 | 392 | 1176 |
=50 | =17900 | =7896 |
Now, we calculate the absolute values of the deviations from mean, and
= 7896
Hence, the mean deviation about the mean is 157.92
Question:10. Find the mean deviation about the mean.
Answer:
Height in cms | Number of Persons | Mid Points | |||
95 -105 | 9 | 100 | 900 | 25.3 | 227.7 |
105 -115 | 13 | 110 | 1430 | 15.3 | 198.9 |
115-125 | 26 | 120 | 3120 | 5.3 | 137.8 |
125-135 | 30 | 130 | 3900 | 4.7 | 141 |
135-145 | 12 | 140 | 1680 | 14.7 | 176.4 |
145-155 | 10 | 150 | 1500 | 24.7 | 247 |
=100 | =12530 | =1128.8 |
Now, we calculate the absolute values of the deviations from mean, and
= 1128.8
Hence, the mean deviation about the mean is 11.29
Question:11. Find the mean deviation about median for the following data :
Answer:
Marks | Number of Girls | Cumulative Frequency c.f. | Mid Points | ||
0-10 | 6 | 6 | 5 | 22.85 | 137.1 |
10-20 | 8 | 14 | 15 | 12.85 | 102.8 |
20-30 | 14 | 28 | 25 | 2.85 | 39.9 |
30-40 | 16 | 44 | 35 | 7.15 | 114.4 |
40-50 | 4 | 48 | 45 | 17.15 | 68.6 |
50-60 | 2 | 50 | 55 | 27.15 | 54.3 |
=517.1 |
Now, N = 50, which is even.
The class interval containing or item is 20-30. Therefore, 20-30 is the median class.
We know,
Median
Here, l = 20, C = 14, f = 14, h = 10 and N = 50
Therefore, Median
Now, we calculate the absolute values of the deviations from median, and
= 517.1
Hence, the mean deviation about the median is 10.34
Question:12 Calculate the mean deviation about median age for the age distribution of persons given below:
Answer:
Age (in years) | Number | Cumulative Frequency c.f. | Mid Points | ||
15.5-20.5 | 5 | 5 | 18 | 20 | 100 |
20.5-25.5 | 6 | 11 | 23 | 15 | 90 |
25.5-30.5 | 12 | 23 | 28 | 10 | 120 |
30.5-35.5 | 14 | 37 | 33 | 5 | 70 |
35.5-40.5 | 26 | 63 | 38 | 0 | 0 |
40.5-45.5 | 12 | 75 | 43 | 5 | 60 |
45.5-50.5 | 16 | 91 | 48 | 10 | 160 |
50.5-55.5 | 9 | 100 | 53 | 15 | 135 |
=735 |
Now, N = 100, which is even.
The class interval containing or item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.
We know,
Median
Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100
Therefore, Median
Now, we calculate the absolute values of the deviations from median, and
= 735
Hence, the mean deviation about the median is 7.35
NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.2
Question:1. Find the mean and variance for each of the data.
Answer:
Mean ( ) of the given data:
The respective values of the deviations from mean, are
-3, -2, 1 3 4 -5 -1 3
Hence, Mean = 9 and Variance = 9.25
Question:2. Find the mean and variance for each of the data.
Answer:
Mean ( ) of first n natural numbers:
We know, Variance
We know that
Hence, Mean = and Variance =
Question:3. Find the mean and variance for each of the data
Answer:
First 10 multiples of 3 are:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Mean ( ) of the above values:
The respective values of the deviations from mean, are
-13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5
Hence, Mean = 16.5 and Variance = 74.25
Question:4. Find the mean and variance for each of the data.
Answer:
6 | 2 | 12 | -13 | 169 | 338 |
10 | 4 | 40 | -9 | 81 | 324 |
14 | 7 | 98 | -5 | 25 | 175 |
18 | 12 | 216 | -1 | 1 | 12 |
24 | 8 | 192 | 5 | 25 | 200 |
28 | 4 | 112 | 9 | 81 | 324 |
30 | 3 | 90 | 13 | 169 | 363 |
= 40 | = 760 | =1736 |
We know, Variance,
Hence, Mean = 19 and Variance = 43.4
Question:5. Find the mean and variance for each of the data.
Answer:
92 | 3 | 276 | -8 | 64 | 192 |
93 | 2 | 186 | -7 | 49 | 98 |
97 | 3 | 291 | -3 | 9 | 27 |
98 | 2 | 196 | -2 | 4 | 8 |
102 | 6 | 612 | 2 | 4 | 24 |
104 | 3 | 312 | 4 | 16 | 48 |
109 | 3 | 327 | 9 | 81 | 243 |
= 22 | = 2200 | =640 |
We know, Variance,
Hence, Mean = 100 and Variance = 29.09
Question:6 Find the mean and standard deviation using short-cut method.
Answer:
Let the assumed mean, A = 64 and h = 1
60 | 2 | -4 | 16 | -8 | 32 |
61 | 1 | -3 | 9 | -3 | 9 |
62 | 12 | -2 | 4 | -24 | 48 |
63 | 29 | -1 | 1 | -29 | 29 |
64 | 25 | 0 | 0 | 0 | 0 |
65 | 12 | 1 | 1 | 12 | 12 |
66 | 10 | 2 | 4 | 20 | 40 |
67 | 4 | 3 | 9 | 12 | 36 |
68 | 5 | 4 | 16 | 20 | 80 |
=100 | = 0 | =286 |
Mean,
We know, Variance,
We know, Standard Deviation =
Hence, Mean = 64 and Standard Deviation = 1.691
Question:âââââââ7. Find the mean and variance for the following frequency distributions.
Answer:
Classes | Frequency | Mid point | ||||
0-30 | 2 | 15 | 30 | -92 | 8464 | 16928 |
30-60 | 3 | 45 | 135 | -62 | 3844 | 11532 |
60-90 | 5 | 75 | 375 | -32 | 1024 | 5120 |
90-120 | 10 | 105 | 1050 | 2 | 4 | 40 |
120-150 | 3 | 135 | 405 | 28 | 784 | 2352 |
150-180 | 5 | 165 | 825 | 58 | 3364 | 16820 |
180-210 | 2 | 195 | 390 | 88 | 7744 | 15488 |
= N = 30 | = 3210 | =68280 |
We know, Variance,
Hence, Mean = 107 and Variance = 2276
Question:âââââââ8. Find the mean and variance for the following frequency distributions.
Answer:
Classes | Frequency | Mid-point | ||||
0-10 | 5 | 5 | 25 | -22 | 484 | 2420 |
10-20 | 8 | 15 | 120 | -12 | 144 | 1152 |
20-30 | 15 | 25 | 375 | -2 | 4 | 60 |
30-40 | 16 | 35 | 560 | 8 | 64 | 1024 |
40-50 | 6 | 45 | 270 | 18 | 324 | 1944 |
= N = 50 | = 1350 | =6600 |
We know, Variance,
Hence, Mean = 27 and Variance = 132
Question:9. Find the mean, variance and standard deviation using short-cut method.
Height in cms | 70-75 | 75-80 | 80-85 | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |
No. of students | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
Answer:
Let the assumed mean, A = 92.5 and h = 5
Height in cms | Frequency | Midpoint | ||||
70-75 | 3 | 72.5 | -4 | 16 | -12 | 48 |
75-80 | 4 | 77.5 | -3 | 9 | -12 | 36 |
80-85 | 7 | 82.5 | -2 | 4 | -14 | 28 |
85-90 | 7 | 87.5 | -1 | 1 | -7 | 7 |
90-95 | 15 | 92.5 | 0 | 0 | 0 | 0 |
95-100 | 9 | 97.5 | 1 | 1 | 9 | 9 |
100-105 | 6 | 102.5 | 2 | 4 | 12 | 24 |
105-110 | 6 | 107.5 | 3 | 9 | 18 | 54 |
110-115 | 3 | 112.5 | 4 | 16 | 12 | 48 |
=N = 60 | = 6 | =254 |
Mean,
We know, Variance,
We know, Standard Deviation =
Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275
Question:10. The diameters of circles (in mm) drawn in a design are given below:
Answer:
Let the assumed mean, A = 92.5 and h = 5
Diameters | No. of circles | Midpoint | ||||
32.5-36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |
36.5-40.5 | 17 | 38.5 | -1 | 1 | -17 | 17 |
40.5-44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |
44.5-48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |
48.5-52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |
=N = 100 | = 25 | =199 |
Mean,
We know, Variance,
We know, Standard Deviation =
Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553
NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.3
Question:1. From the data given below state which group is more variable, A or B?
Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |
Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |
Answer:
The group having a higher coefficient of variation will be more variable.
Let the assumed mean, A = 45 and h = 10
For Group A
Marks | Group A | Midpoint | | |||
10-20 | 9 | 15 | -3 | 9 | -27 | 81 |
20-30 | 17 | 25 | -2 | 4 | -34 | 68 |
30-40 | 32 | 35 | -1 | 1 | -32 | 32 |
40-50 | 33 | 45 | 0 | 0 | 0 | 0 |
50-60 | 40 | 55 | 1 | 1 | 40 | 40 |
60-70 | 10 | 65 | 2 | 4 | 20 | 40 |
70-80 | 9 | 75 | 3 | 9 | 27 | 81 |
=N = 150 | = -6 | =342 |
Mean,
We know, Variance,
We know, Standard Deviation =
Coefficient of variation =
C.V.(A) =
Similarly,
For Group B
Marks | Group A | Midpoint | | |||
10-20 | 10 | 15 | -3 | 9 | -30 | 90 |
20-30 | 20 | 25 | -2 | 4 | -40 | 80 |
30-40 | 30 | 35 | -1 | 1 | -30 | 30 |
40-50 | 25 | 45 | 0 | 0 | 0 | 0 |
50-60 | 43 | 55 | 1 | 1 | 43 | 43 |
60-70 | 15 | 65 | 2 | 4 | 30 | 60 |
70-80 | 7 | 75 | 3 | 9 | 21 | 72 |
=N = 150 | = -6 | =375 |
Mean,
We know, Variance,
We know, Standard Deviation =
Coefficient of variation =
C.V.(B) =
Since C.V.(B) > C.V.(A)
Therefore, Group B is more variable.
Question:2 From the prices of shares X and Y below, find out which is more stable in value:
Answer:
X( ) | Y( ) | ||
35 | 108 | 1225 | 11664 |
54 | 107 | 2916 | 11449 |
52 | 105 | 2704 | 11025 |
53 | 105 | 2809 | 11025 |
56 | 106 | 8136 | 11236 |
58 | 107 | 3364 | 11449 |
52 | 104 | 2704 | 10816 |
50 | 103 | 2500 | 10609 |
51 | 104 | 2601 | 10816 |
49 | 101 | 2401 | 10201 |
=510 | = 1050 | =26360 | =110290 |
For X,
Mean ,
Variance,
We know, Standard Deviation =
C.V.(X) =
Similarly, For Y,
Mean ,
Variance,
We know, Standard Deviation =
C.V.(Y) =
Since C.V.(Y) < C.V.(X)
Hence Y is more stable.
Which firm A or B pays larger amount as monthly wages?
Answer:
Given, Mean of monthly wages of firm A = 5253
Number of wage earners = 586
Total amount paid = 586 x 5253 = 30,78,258
Again, Mean of monthly wages of firm B = 5253
Number of wage earners = 648
Total amount paid = 648 x 5253 = 34,03,944
Hence firm B pays larger amount as monthly wages.
Question:3(ii) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
Which firm, A or B, shows greater variability in individual wages?
Answer:
Given, Variance of firm A = 100
Standard Deviation =
Again, Variance of firm B = 121
Standard Deviation =
Since , firm B has greater variability in individual wages.
Question:4 The following is the record of goals scored by team A in a football session:
Answer:
No. of goals scored | Frequency | |||
0 | 1 | 0 | 0 | 0 |
1 | 9 | 1 | 9 | 9 |
2 | 7 | 4 | 14 | 28 |
3 | 5 | 9 | 15 | 45 |
4 | 3 | 16 | 12 | 48 |
=N = 25 | = 50 | =130 |
For Team A,
Mean,
We know, Variance,
We know, Standard Deviation =
C.V.(A) =
For Team B,
Mean = 2
Standard deviation, = 1.25
C.V.(B) =
Since C.V. of firm B is more than C.V. of A.
Therefore, Team A is more consistent.
Which is more varying, the length or weight?
Answer:
For lenght x,
Mean,
We know, Variance,
We know, Standard Deviation =
C.V.(x) =
For weight y,
Mean,
Mean,
We know, Variance,
We know, Standard Deviation =
C.V.(y) =
Since C.V.(y) > C.V.(x)
Therefore, weight is more varying.
NCERT solutions for class 11 maths chapter 15 statistics-Miscellaneous Exercise
Answer:
Given,
The mean and variance of 8 observations are 9 and 9.25, respectively
Let the remaining two observations be x and y,
Observations: 6, 7, 10, 12, 12, 13, x, y.
∴ Mean,
60 + x + y = 72
x + y = 12 -(i)
Now, Variance
(Using (i))
-(ii)
Squaring (i), we get
(iii)
(iii) - (ii) :
2xy = 64 (iv)
Now, (ii) - (iv):
-(v)
Hence, From (i) and (v):
x – y = 4 x = 8 and y = 4
x – y = -4 x = 4 and y = 8
Therefore, The remaining observations are 4 and 8. (in no order)
Answer:
Given,
The mean and variance of 7 observations are 8 and 16, respectively
Let the remaining two observations be x and y,
Observations: 2, 4, 10, 12, 14, x, y
∴ Mean,
42 + x + y = 56
x + y = 14 -(i)
Now, Variance
(Using (i))
-(ii)
Squaring (i), we get
(iii)
(iii) - (ii) :
2xy = 96 (iv)
Now, (ii) - (iv):
-(v)
Hence, From (i) and (v):
x – y = 2 x = 8 and y = 6
x – y = -2 x = 6 and y = 8
Therefore, The remaining observations are 6 and 8. (in no order)
Answer:
Given,
Mean = 8 and Standard deviation = 4
Let the observations be
Mean,
Now, Let _{ } be the the resulting observations if each observation is multiplied by 3:
New mean,
= 24
We know that,
Standard Deviation =
-(i)
Now, Substituting the values of in (i):
Hence, the variance of the new observations =
Therefore, Standard Deviation = = = 12
Answer:
Given, Mean = and variance =
Now, Let _{ } be the the resulting observations if each observation is multiplied by a:
Hence the mean of the new observations is
We know,
Now, Substituting the values of :
Hence the variance of the new observations is
Hence proved.
Answer:
Given,
Number of observations, n = 20
Also, Incorrect mean = 10
And, Incorrect standard deviation = 2
Thus, incorrect sum = 200
Hence, correct sum of observations = 200 – 8 = 192
Therefore, Correct Mean = (Correct Sum)/19
= 10.1
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old sum - (8x8)
= 2080 – 64
= 2016
Hence, Correct Standard Deviation =
=2.02
Answer:
Given,
Number of observations, n = 20
Also, Incorrect mean = 10
And, Incorrect standard deviation = 2
Thus, incorrect sum = 200
Hence, correct sum of observations = 200 – 8 + 12 = 204
Therefore, Correct Mean = (Correct Sum)/20
= 10.2
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old sum - (8x8) + (12x12)
= 2080 – 64 + 144
= 2160
Hence, Correct Standard Deviation =
= 1.98
Which of the three subjects shows highest variability in marks and which shows the lowest?
Answer:
Given,
Standard deviation of physics = 15
Standard deviation of chemistry = 20
Standard deviation of mathematics = 12
We know ,
C.V.
The subject with greater C.V will be more variable than others.
Hence, Mathematics has lowest variability and Chemistry has highest variability.
Answer:
Given,
Initial Number of observations, n = 100
Thus, incorrect sum = 2000
Hence, New sum of observations = 2000 - 21-21-18 = 1940
New number of observation, n' = 100-3 =97
Therefore, New Mean = New Sum)/100
= 20
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)
= 40900 - 441 - 441 - 324
= 39694
Hence, Correct Standard Deviation =
= 3.036
NCERT solutions for class 11 mathematics
NCERT solutions for class 11- Subject wise
NCERT solutions for class 11 biology |
NCERT solutions for class 11 maths |
NCERT solutions for class 11 chemistry |
NCERT solutions for Class 11 physics |
e important terms and formulas to be remembered form NCERT solutions for class 11 maths chapter 15 statistics -
- Measures of dispersion- Range, mean deviation, Quartile deviation, standard deviation, variance are measures of dispersion.
- Range- The difference between the maximum and the minimum values of each series is called the ‘Range’ of the data. Range = Maximum Value – Minimum Value.
- Mean- The arithmetic average of a range of values or quantities, found by adding all values and dividing by the number of values. For example, the mean of 4, 10, and 1 is (4+10+1)/3=15/3=5.
X represents values or scores,
N represents the number of values or scores.
- Median- If the total number of values(n) is an odd number, then the formula is-
- Median- If the total number of values(n) is an even number, then the formula is-
- Mode- It is the most frequently occurring value or number.
- Mean deviation for ungrouped data-
- Mean deviation for grouped data-
This chapter seems very easy but it very useful in the field of science, research reliable predictions, or forecasts for future use, decision making, etc. There are 7 questions are given in the miscellaneous exercise. Try to solve them also to get command in this chapter. In NCERT solutions for class 11 maths chapter 15 statistics, you will get solutions to miscellaneous exercise too.
Happy Reading !!!
Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 15 Statistics
Question: What are important topics of the chapter Statistics ?
Answer:
Measures of Dispersion, Range, Mean Deviation, Variance, and Standard Deviation are the important topics of this chapter.
Question: What is the weightage of the chapter Statistics in Jee main ?
Answer:
This chapter has about 4% weightage in the Jee main exam.
Question: How does the NCERT solutions are helpful ?
Answer:
NCERT solutions are helpful for the students if they stuck while the NCERT problems. As these solutions are provided in a very simple language, it can be understood by an average student also.
Question: Does CBSE provides the solutions of NCERT class 11 maths ?
Answer:
No, CBSE doesn’t provide NCERT solutions for any class or subject.
Question: Where can I find the complete solutions of NCERT for class 11 maths ?
Answer:
Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.
Question: Which is the official website of NCERT ?
Answer:
http://ncert.nic.in/ is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.
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