Class 9 Exam > Class 9 Notes > Mathematics (Maths) Class 9 > RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1
RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 | Mathematics (Maths) Class 9 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
Page 2
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
Page 3
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
S o l u t i o n :
We will make the following table:
Age (x
i
) Frequency (f
i
) (f
i
)(x
i
)
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
? f
i
=
40
? f
i
x
i
=
698
Thus, we have:
Mean =
?f i x i
?
x i
=
698
40
= 17. 45 years
Q u e s t i o n : 3 6
Find the mean of the following frequency distribution:
Variable (x
i
) 10 30 50 70 89
Frequency (f
i
) 7 8 10 15 10
S o l u t i o n :
We will make the following table:
Variable (x
i
) Frequency (f
i
) (f
i
)(x
i
)
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
? f
i
=
50
? f
i
x
i
=
2750
Thus, we have:
Mean =
?f i x i
?
x i
=
2750
50
= 55
Q u e s t i o n : 3 7
Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
Mean =
(250×8)+(300×11)+(350×6)+(400×10)+(450×5)
8+11+6+10+5
=
2000+3300+2100+4000+2250
40
=
13650
40
= 341.25
Hence, the mean of daily wages of 40 workers in a factory is 341.25 .
Q u e s t i o n : 3 8
If the mean of the following data is 20.2, find the value of p.
Page 4
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
S o l u t i o n :
We will make the following table:
Age (x
i
) Frequency (f
i
) (f
i
)(x
i
)
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
? f
i
=
40
? f
i
x
i
=
698
Thus, we have:
Mean =
?f i x i
?
x i
=
698
40
= 17. 45 years
Q u e s t i o n : 3 6
Find the mean of the following frequency distribution:
Variable (x
i
) 10 30 50 70 89
Frequency (f
i
) 7 8 10 15 10
S o l u t i o n :
We will make the following table:
Variable (x
i
) Frequency (f
i
) (f
i
)(x
i
)
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
? f
i
=
50
? f
i
x
i
=
2750
Thus, we have:
Mean =
?f i x i
?
x i
=
2750
50
= 55
Q u e s t i o n : 3 7
Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
Mean =
(250×8)+(300×11)+(350×6)+(400×10)+(450×5)
8+11+6+10+5
=
2000+3300+2100+4000+2250
40
=
13650
40
= 341.25
Hence, the mean of daily wages of 40 workers in a factory is 341.25 .
Q u e s t i o n : 3 8
If the mean of the following data is 20.2, find the value of p.
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
Mean =
(10×6)+(15×8)+(20×p)+(25×10)+(30×6)
6+8+p+10+6
? 20. 2 =
60+120+20p+250+180
30+p
? 20. 2(30 +p) = 610 +20p ? 606 +20. 2p = 610 +20p ? 20. 2p -20p = 610 -606 ? 0. 2p = 4 ? p =
4
0.2
? p =
40
2
? p = 20
Hence, the value of p is 20.
Q u e s t i o n : 3 9
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
S o l u t i o n :
We will make the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
? f
i
=
41 + p
? f
i
x
i
=
303 + 9p
We know:
Mean =
?f i x i
?
x i
Given:
Mean = 8
Thus, we have:
8 =
303+9p
41+p
? 328 +8p = 303 +9p ? p = 25
Q u e s t i o n : 4 0
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
S o l u t i o n :
We will prepare the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
15 8 120
20 7 140
25 p 25p
30 14 420
35 15 525
40 6 240
? f
i
=
50 + p
? f
i
x
i
=
1445 + 25p
Thus, we have:
Mean =
?f i x i
?
x i
? 28. 25 =
1445+25p
50+p
? 28. 25(50 +p) = (1445 +25p) ? 1412. 5 + 28. 25p = 1445 +25p ? 3. 25 p = 32. 5 ? p = 10
Page 5
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
S o l u t i o n :
We will make the following table:
Age (x
i
) Frequency (f
i
) (f
i
)(x
i
)
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
? f
i
=
40
? f
i
x
i
=
698
Thus, we have:
Mean =
?f i x i
?
x i
=
698
40
= 17. 45 years
Q u e s t i o n : 3 6
Find the mean of the following frequency distribution:
Variable (x
i
) 10 30 50 70 89
Frequency (f
i
) 7 8 10 15 10
S o l u t i o n :
We will make the following table:
Variable (x
i
) Frequency (f
i
) (f
i
)(x
i
)
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
? f
i
=
50
? f
i
x
i
=
2750
Thus, we have:
Mean =
?f i x i
?
x i
=
2750
50
= 55
Q u e s t i o n : 3 7
Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
Mean =
(250×8)+(300×11)+(350×6)+(400×10)+(450×5)
8+11+6+10+5
=
2000+3300+2100+4000+2250
40
=
13650
40
= 341.25
Hence, the mean of daily wages of 40 workers in a factory is 341.25 .
Q u e s t i o n : 3 8
If the mean of the following data is 20.2, find the value of p.
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
Mean =
(10×6)+(15×8)+(20×p)+(25×10)+(30×6)
6+8+p+10+6
? 20. 2 =
60+120+20p+250+180
30+p
? 20. 2(30 +p) = 610 +20p ? 606 +20. 2p = 610 +20p ? 20. 2p -20p = 610 -606 ? 0. 2p = 4 ? p =
4
0.2
? p =
40
2
? p = 20
Hence, the value of p is 20.
Q u e s t i o n : 3 9
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
S o l u t i o n :
We will make the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
? f
i
=
41 + p
? f
i
x
i
=
303 + 9p
We know:
Mean =
?f i x i
?
x i
Given:
Mean = 8
Thus, we have:
8 =
303+9p
41+p
? 328 +8p = 303 +9p ? p = 25
Q u e s t i o n : 4 0
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
S o l u t i o n :
We will prepare the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
15 8 120
20 7 140
25 p 25p
30 14 420
35 15 525
40 6 240
? f
i
=
50 + p
? f
i
x
i
=
1445 + 25p
Thus, we have:
Mean =
?f i x i
?
x i
? 28. 25 =
1445+25p
50+p
? 28. 25(50 +p) = (1445 +25p) ? 1412. 5 + 28. 25p = 1445 +25p ? 3. 25 p = 32. 5 ? p = 10
Q u e s t i o n : 4 1
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
S o l u t i o n :
We will make the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
8 12 96
12 16 192
15 20 300
p 24 24p
20 16 320
25 8 200
30 4 120
? f
i
=
100
? f
i
x
i
=
1228 + 24p
Thus, we have:
Mean =
?f i x i
?
x i
? 16. 6 =
(1228+24p)
100
? 16. 6 ×100 = (1228 +24p) ? 1660 = 1228 +24p
? 24p = 432 ? p = 18
Q u e s t i o n : 4 2
Find the missing frequencies in the following frequency distribution whose mean is 34.
x 10 20 30 40 50 60 Total
f 4 f
1 8 f
2 3 4 35
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
x 10 20 30 40 50 60 Total
f 4 f
1 8 f
2 3 4 35
Mean =
(10×4)+ 20×f 1 +(30×8)+ 40×f 2 +(50×3)+(60×4)
35
? 34 =
40+20f 1 +240+40f 2 +150+240
35
? 34(35) = 670 +20f
1
+40f
2
? 1190 -670 = 20f
1
+40f
2
? 20f
1
+40f
2
= 520 ? 20 f
1
+2f
2
= 520 ? f
1
+2f
2
=
520
20
? f
1
+2f
2
= 26 ? f
1
= 26 -2f
2
. . . (1)
Also, 4 + f
1
+ 8 + f
2
+ 3 + 4 = 35
? 19 + f
1
+ f
2
= 35
? f
1
+ f
2
= 35 - 19
? f
1
+ f
2
= 16
? 26 - 2f
2
+ f
2
= 16 from(1
)
? 26 - f
2
= 16
? 26 - 16 = f
2
? f
2
= 10
Putting the value of f
2
in 1
, we get
f
1
= 26 - 210
= 6
Hence, the value of f
1
and f
2
is 6 and 10, respectively.
Q u e s t i o n : 4 3
Find the missing frequencies in the following frequency distribution, whose mean is 50.
x 10 30 50 70 90 Total
f 17 f
1 32 f
2 19 120
S o l u t i o n :
We will prepare the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
10 17 170
30 f
1
30f
1
50 32 1600
( ) ( )
( )
Read More
|
44 videos|412 docs|54 tests
|
Related Exams
About this Document
|
4.73/5 Rating |
|
Dec 23, 2024 Last updated |
Document Description: RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 for Class 9 2024 is part of Mathematics (Maths) Class 9 preparation.
The notes and questions for RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 have been prepared according to the Class 9 exam syllabus. Information about RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 covers topics
like and RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 Example, for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1.
Introduction of RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 in English is available as part of our Mathematics (Maths) Class 9
for Class 9 & RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 in Hindi for Mathematics (Maths) Class 9 course.
Download more important topics related with notes, lectures and mock test series for Class 9
Exam by signing up for free. Class 9: RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 | Mathematics (Maths) Class 9
Description
Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 | Mathematics (Maths) Class 9 - Class 9 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 9 | Best notes, free PDF download
Information about RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1
In this doc you can find the meaning of RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 defined & explained in the simplest way possible. Besides explaining types of
RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 theory, EduRev gives you an ample number of questions to practice RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 tests, examples and also practice Class 9
tests
|
Explore Courses for Class 9 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
study material
,Semester Notes
,Median and Mode of Ungrouped Data- 1 | Mathematics (Maths) Class 9
,Viva Questions
,Median and Mode of Ungrouped Data- 1 | Mathematics (Maths) Class 9
,video lectures
,Important questions
,Extra Questions
,Median and Mode of Ungrouped Data- 1 | Mathematics (Maths) Class 9
,MCQs
,Sample Paper
,past year papers
,RS Aggarwal Solutions: Mean
,Summary
,RS Aggarwal Solutions: Mean
,Previous Year Questions with Solutions
,ppt
,shortcuts and tricks
,RS Aggarwal Solutions: Mean
,mock tests for examination
,Objective type Questions
,Free
,Exam
,practice quizzes
;
Additional Information about RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 for Class 9 Preparation
RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 Free PDF Download
The RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 is an invaluable resource that delves deep into the core of the Class 9 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 RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 now and kickstart your journey towards success in the Class 9 exam.
Importance of RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1
The importance of RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 cannot be overstated, especially for Class 9 aspirants.
This document holds the key to success in the Class 9 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.
RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 Notes
RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1.
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, RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 Notes on EduRev are your ultimate resource for success.
RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 Class 9 Questions
The "RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 Class 9 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 9 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 RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 on the App
Students of Class 9 can study RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1,
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 RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 is prepared as per the latest Class 9 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup