Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3)
Data Handling – II (Central Values Exercise 23.3) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Find the median of the following data (1 – 8)
1. 83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
First we have to arrange given data into ascending order,
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (45 + 54)/2
= 49.5
Hence median for given data = 49.5
2. 133, 73, 89, 108, 94,104, 94, 85, 100, 120
Solution:
First we have to arrange given data into ascending order,
73, 85, 89, 94, 100, 104, 108, 120, 133
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (94 + 100)/2
= 97
Hence median for given data = 97
3. 31, 38, 27, 28, 36, 25, 35, 40
Solution:
First we have to arrange given data into ascending order
25, 27, 28, 31, 35, 36, 38, 40
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (31 + 35)/2
Page 2
Find the median of the following data (1 – 8)
1. 83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
First we have to arrange given data into ascending order,
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (45 + 54)/2
= 49.5
Hence median for given data = 49.5
2. 133, 73, 89, 108, 94,104, 94, 85, 100, 120
Solution:
First we have to arrange given data into ascending order,
73, 85, 89, 94, 100, 104, 108, 120, 133
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (94 + 100)/2
= 97
Hence median for given data = 97
3. 31, 38, 27, 28, 36, 25, 35, 40
Solution:
First we have to arrange given data into ascending order
25, 27, 28, 31, 35, 36, 38, 40
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (31 + 35)/2
= 33
Hence median for given data = 33
4. 15, 6, 16, 8, 22, 21, 9, 18, 25
Solution:
First we have to arrange given data into ascending order
6, 8, 9, 15, 16, 18, 21, 22, 25
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 16
5. 41, 43,127, 99, 71, 92, 71, 58, 57
Solution:
First we have to arrange given data into ascending order
41, 43, 57, 58, 71, 71, 92, 99, 127
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 71
6. 25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution:
First we have to arrange given data into ascending order,
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (26 + 29)/2
= 27.5
Hence median for given data = 27.5
7. 12, 17, 3, 14, 5, 8, 7, 15
Page 3
Find the median of the following data (1 – 8)
1. 83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
First we have to arrange given data into ascending order,
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (45 + 54)/2
= 49.5
Hence median for given data = 49.5
2. 133, 73, 89, 108, 94,104, 94, 85, 100, 120
Solution:
First we have to arrange given data into ascending order,
73, 85, 89, 94, 100, 104, 108, 120, 133
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (94 + 100)/2
= 97
Hence median for given data = 97
3. 31, 38, 27, 28, 36, 25, 35, 40
Solution:
First we have to arrange given data into ascending order
25, 27, 28, 31, 35, 36, 38, 40
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (31 + 35)/2
= 33
Hence median for given data = 33
4. 15, 6, 16, 8, 22, 21, 9, 18, 25
Solution:
First we have to arrange given data into ascending order
6, 8, 9, 15, 16, 18, 21, 22, 25
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 16
5. 41, 43,127, 99, 71, 92, 71, 58, 57
Solution:
First we have to arrange given data into ascending order
41, 43, 57, 58, 71, 71, 92, 99, 127
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 71
6. 25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution:
First we have to arrange given data into ascending order,
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (26 + 29)/2
= 27.5
Hence median for given data = 27.5
7. 12, 17, 3, 14, 5, 8, 7, 15
Solution:
First we have to arrange given data into ascending order,
3, 5, 7, 8, 12, 14, 15, 17
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n +1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (8 + 12)/2
= 10
Hence median for given data = 10
8. 92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Solution:
First we have to arrange given data into ascending order,
35, 42, 51, 56, 67, 69, 72, 81, 85, 92
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (67 + 69)/2
= 68
Hence median for given data = 68
9. Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and
their median is 25, find x.
Solution:
Here, the number of observations n is 9.
Since n is odd, the median is the n+12th observation, i.e., the 5
th
observation.
As the numbers are arranged in the descending order, we therefore observe from the
last.
Median = 5
th
observation.
=> 25 = 2x – 8
=> 2x = 25 + 8
=> 2x = 33
=> x = (33/2)
x = 16.5
Page 4
Find the median of the following data (1 – 8)
1. 83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
First we have to arrange given data into ascending order,
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (45 + 54)/2
= 49.5
Hence median for given data = 49.5
2. 133, 73, 89, 108, 94,104, 94, 85, 100, 120
Solution:
First we have to arrange given data into ascending order,
73, 85, 89, 94, 100, 104, 108, 120, 133
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (94 + 100)/2
= 97
Hence median for given data = 97
3. 31, 38, 27, 28, 36, 25, 35, 40
Solution:
First we have to arrange given data into ascending order
25, 27, 28, 31, 35, 36, 38, 40
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (31 + 35)/2
= 33
Hence median for given data = 33
4. 15, 6, 16, 8, 22, 21, 9, 18, 25
Solution:
First we have to arrange given data into ascending order
6, 8, 9, 15, 16, 18, 21, 22, 25
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 16
5. 41, 43,127, 99, 71, 92, 71, 58, 57
Solution:
First we have to arrange given data into ascending order
41, 43, 57, 58, 71, 71, 92, 99, 127
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 71
6. 25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution:
First we have to arrange given data into ascending order,
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (26 + 29)/2
= 27.5
Hence median for given data = 27.5
7. 12, 17, 3, 14, 5, 8, 7, 15
Solution:
First we have to arrange given data into ascending order,
3, 5, 7, 8, 12, 14, 15, 17
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n +1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (8 + 12)/2
= 10
Hence median for given data = 10
8. 92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Solution:
First we have to arrange given data into ascending order,
35, 42, 51, 56, 67, 69, 72, 81, 85, 92
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (67 + 69)/2
= 68
Hence median for given data = 68
9. Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and
their median is 25, find x.
Solution:
Here, the number of observations n is 9.
Since n is odd, the median is the n+12th observation, i.e., the 5
th
observation.
As the numbers are arranged in the descending order, we therefore observe from the
last.
Median = 5
th
observation.
=> 25 = 2x – 8
=> 2x = 25 + 8
=> 2x = 33
=> x = (33/2)
x = 16.5
10. Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92,
33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Solution:
Arranging the given data in ascending order, we have:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Here, the number of observations n is 11 (odd).
Since the number of observations is odd, therefore,
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 58.
Hence, median = 58.
If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending
order are:
33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99
New median = Value of the 6
th
observation = 58.
11. Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is
replaced by 85, what will be the new median?
Solution:
Arranging the given data in ascending order, we have:
41, 43, 57, 58, 61, 71, 92, 99,127
Here, the number of observations, n, is 9(odd).
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
Hence, the median = 61.
If 58 is replaced by 85, then the new observations arranged in ascending order are:
41, 43, 57, 61, 71, 85, 92, 99, 12
New median = Value of the 5
th
observation = 71.
12. The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44,
45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg,
find the new median.
Solution:
Arranging the given data in ascending order, we have:
Page 5
Find the median of the following data (1 – 8)
1. 83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
First we have to arrange given data into ascending order,
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (45 + 54)/2
= 49.5
Hence median for given data = 49.5
2. 133, 73, 89, 108, 94,104, 94, 85, 100, 120
Solution:
First we have to arrange given data into ascending order,
73, 85, 89, 94, 100, 104, 108, 120, 133
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (94 + 100)/2
= 97
Hence median for given data = 97
3. 31, 38, 27, 28, 36, 25, 35, 40
Solution:
First we have to arrange given data into ascending order
25, 27, 28, 31, 35, 36, 38, 40
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (31 + 35)/2
= 33
Hence median for given data = 33
4. 15, 6, 16, 8, 22, 21, 9, 18, 25
Solution:
First we have to arrange given data into ascending order
6, 8, 9, 15, 16, 18, 21, 22, 25
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 16
5. 41, 43,127, 99, 71, 92, 71, 58, 57
Solution:
First we have to arrange given data into ascending order
41, 43, 57, 58, 71, 71, 92, 99, 127
Given number of observations, n = 9 (odd)
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 71
6. 25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution:
First we have to arrange given data into ascending order,
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (26 + 29)/2
= 27.5
Hence median for given data = 27.5
7. 12, 17, 3, 14, 5, 8, 7, 15
Solution:
First we have to arrange given data into ascending order,
3, 5, 7, 8, 12, 14, 15, 17
Given number of observations, n = 8 (even)
Therefore median = (n/2)
th
term + ((n +1)/2)
th
term
Median = (value of 4
th
term + value of 5
th
term)/2
= (8 + 12)/2
= 10
Hence median for given data = 10
8. 92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Solution:
First we have to arrange given data into ascending order,
35, 42, 51, 56, 67, 69, 72, 81, 85, 92
Given number of observations, n = 10 (even)
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
= (67 + 69)/2
= 68
Hence median for given data = 68
9. Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and
their median is 25, find x.
Solution:
Here, the number of observations n is 9.
Since n is odd, the median is the n+12th observation, i.e., the 5
th
observation.
As the numbers are arranged in the descending order, we therefore observe from the
last.
Median = 5
th
observation.
=> 25 = 2x – 8
=> 2x = 25 + 8
=> 2x = 33
=> x = (33/2)
x = 16.5
10. Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92,
33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Solution:
Arranging the given data in ascending order, we have:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Here, the number of observations n is 11 (odd).
Since the number of observations is odd, therefore,
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
= 58.
Hence, median = 58.
If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending
order are:
33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99
New median = Value of the 6
th
observation = 58.
11. Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is
replaced by 85, what will be the new median?
Solution:
Arranging the given data in ascending order, we have:
41, 43, 57, 58, 61, 71, 92, 99,127
Here, the number of observations, n, is 9(odd).
Therefore median = ((n+1)/2)
th
term
Median = value of 5
th
term
Hence, the median = 61.
If 58 is replaced by 85, then the new observations arranged in ascending order are:
41, 43, 57, 61, 71, 85, 92, 99, 12
New median = Value of the 5
th
observation = 71.
12. The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44,
45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg,
find the new median.
Solution:
Arranging the given data in ascending order, we have:
27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45
Here, the number of observations n is 15(odd).
Since the number of observations is odd, therefore,
Therefore median = ((n+1)/2)
th
term
Median = value of 8
th
term
Hence, median = 35 kg.
If 44 kg is replaced by 46 kg and 27 kg by 25 kg, then the new observations arranged in
ascending order are:
25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46
? New median = Value of the 8
th
observation = 35 kg.
13. The following observations have been arranged in ascending order. If the median
of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Here, the number of observations n is 10. Since n is even,
Therefore median = (n/2)
th
term + ((n + 1)/2)
th
term
Median = (value of 5
th
term + value of 6
th
term)/2
63 = x + (x + 2)/2
63 = (2x + 2)/2
63 = 2 (x + 1)/2
63 = x + 1
x = 63 – 1
x = 62
Read More
|
76 videos|345 docs|39 tests
|
Related Exams
About this Document
|
4.83/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) covers topics
like and RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3).
Introduction of RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: Data Handling – II (Central Values Exercise 23.3) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Data Handling – II (Central Values Exercise 23.3) RD Sharma Solutions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3)
In this doc you can find the meaning of RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Important questions
,video lectures
,Data Handling – II (Central Values Exercise 23.3) RD Sharma Solutions | Mathematics (Maths) Class 7
,Extra Questions
,practice quizzes
,Objective type Questions
,Exam
,Semester Notes
,MCQs
,Data Handling – II (Central Values Exercise 23.3) RD Sharma Solutions | Mathematics (Maths) Class 7
,study material
,past year papers
,Data Handling – II (Central Values Exercise 23.3) RD Sharma Solutions | Mathematics (Maths) Class 7
,Sample Paper
,Free
,ppt
,Summary
,Viva Questions
,mock tests for examination
,shortcuts and tricks
,Previous Year Questions with Solutions
;
Additional Information about RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) for Class 7 Preparation
RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) Free PDF Download
The RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3)
The importance of RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) Notes
RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) Class 7 Questions
The "RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) on the App
Students of Class 7 can study RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Data Handling – II (Central Values Exercise 23.3) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup