R S Aggarwal Solutions for Class 10 Maths Chapter 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive is one of the most important concepts in Class 10. Chapter 18 of class 10 Maths mainly deals with problems based on mean, median, mode and graphical representation of the data. Students must practice R S Aggarwal Solutions for Maths to score well in exams.
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Exercise 18A
Question 1: If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.
Solution:
We know that,
\(Mean = \frac{Sum\,of\, the\, given\, observations}{Total\, number\, of\, observations}\)Sum of the given observations = x + (x + 2) + (x + 4) + (x + 6) + (x + 8)
= x + x + 2 + x + 4 + x + 6 + x + 8
= 5x + 20
Total number of observations = 5
Therefore, Mean = (5x + 20)/5
Also, Mean = 11 (given)
⇨ 11 = (5x + 20)/5
⇨ 55 = 5x + 20
or x = 7
Question 2: If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?
Solution:
Mean of 25 observations = 27 (given)
Total observations = 25
\(Mean = \frac{Sum\,of\, the\, given\, observations}{Total\, number\, of\, observations}\)27 = (sum of 25 observations)/25
⇨ sum of 25 observations = 27 x 25 = 675
When each observation is decreased by 7, then
New Sum is 675 – 25 x 7 = 500
New Mean = 500/25= 20
Question 3: Compute the mean of following data:
Class | 1-3 | 3-5 | 5-7 | 7-9 |
Frequency | 12 | 22 | 27 | 19 |
Solution:
Now,
\(Mean = \frac{\sum fixi}{\sum fi}\)= 426/80
= 5.325
Question 4: Find the mean of the following data, using direct method:
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequency | 7 | 5 | 6 | 12 | 8 | 2 |
Solution:
\(Mean = \frac{\sum fixi}{\sum fi}\)= 1150/40
= 28.75
Question 5: Find the mean of the following data, using direct method:
Class | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
Frequency | 6 | 10 | 8 | 12 | 4 |
Solution:
\(Mean = \frac{\sum fixi}{\sum fi}\)= 1980/40
=49.5
Question 6: Compute the mean of the following data, using direct method:
Class | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
Frequency | 6 | 9 | 15 | 12 | 8 |
Solution:
\(Mean = \frac{\sum fixi}{\sum fi}\)= 13200/40
= 264
Question 7: Using an appropriate method, find the mean of following frequency distribution:
Class | 84−90 | 90−96 | 96−102 | 102−108 | 108−114 | 114−120 |
Frequency | 8 | 10 | 16 | 23 | 12 | 11 |
Which method did you use, and why?
Solution:
\(Mean = \frac{\sum fixi}{\sum fi}\)= 8244/80
= 103.05
Used direct method as it is easy to calculate.
Question 8: If the mean of the following frequency distribution is 24, find the value of p.
Class | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 |
Frequency | 3 | 4 | p | 3 | 2 |
Solution:
\(Mean = \frac{\sum fixi}{\sum fi}\)Mean = 24 ( given)
⇨ 24 = (270+25p)/(12+p)
⇨ 24(12+p) = 270 + 25p
or p = 18
Question 9: The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is Rs. 18, find the missing frequency f.
Daily pocket allowance (in Rs.) | 11−13 | 13−15 | 15−17 | 17−19 | 19−21 | 21−23 | 23−25 |
Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
Solution:
\(Mean = \frac{\sum fixi}{\sum fi}\)Mean = 18 (given)
18 = (752 + 20f)/(44+f)
792 + 18f = 752 + 20f
or f = 20
Question 10: If the mean of the following frequency distribution is 54, find the value of p.
Class | 0−20 | 20−40 | 40−60 | 60−80 | 80−100 |
Frequency | 7 | p | 10 | 9 | 13 |
Solution:
\(Mean = \frac{\sum fixi}{\sum fi}\)Mean = 54 (given)
⇨ 54 = (2370 + 30p)/(39+p)
⇨ 1053 + 27p = 1185 + 15p
or p = 11
Exercise 18B
Question 1: In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age (in years) | 0−15 | 15−30 | 30−45 | 45−60 | 60−75 |
Number of patients | 5 | 20 | 40 | 50 | 25 |
Solution:
Where cf = cumulative frequency
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Here:
N = 140
N/2 = 70
cf > 70 is 140
Median class = 45-60
So, l = 45, h = 15, f = 50 and
cf = cf of preceding class i.e. 65
Substitute all the value in the above formula, we get
Median = 45 + {15 x (70-65)/50}
= 45 + 1.5
= 46.5
Therefore, median age of diabetic patients is 46.5 years.
Question 2: Compute the median from the following data:
Marks | 0-7 | 7-14 | 14-21 | 21-28 | 28-35 | 35-42 | 42-49 |
Number of students | 3 | 4 | 7 | 11 | 0 | 16 | 9 |
Solution:
Where cf = cumulative frequency
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Here:
N = 50
N/2 = 25
Here cumulative frequency is 25
Median class = 21-28
So, l = 21, h =7, f = 11 and
cf = cf of preceding class i.e. 14
Substitute all the value in the above formula, we get
Median =21 + {7x (25-14)/11}
= 28
Question 3: The following table shows the daily wages of workers in a factory:
Daily wages (in Rs) | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
Number of workers | 40 | 32 | 48 | 22 | 8 |
Find the median daily wage income of the workers.
Solution:
Where cf = cumulative frequency
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Here:
N = 150
N/2 = 75
cf just greater than 75 is 120
Median class = 200-300
So, l = 200, h =100, f = 48 and
cf = cf of preceding class i.e. 72
Substitute all the value in the above formula, we get
Median =200+ {100 x (75-72)/48}
= 200 + 6.25
= 206.25
Median of daily wages is Rs. 206.25.
Question 4: Calculate the median from the following frequency distribution:
Solution:
Where cf = cumulative frequency
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Here:
N = 49
N/2 = 24.5
cf just greater than 24.5 is 26
Median class = 15-20
So, l = 15, h = 5, f = 15 and
cf = cf of preceding class i.e. 11
Substitute all the value in the above formula, we get
Median = 15+ {5 x (24.5-11)/15}
= 15 + 4.5
= 19.5
Thus, median of frequency distribution is 19.5.
Question 5: Given below is the number of units of electricity consumed in a week in a certain locality:
Calculate the median.
Solution:
Where cf = cumulative frequency
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Here:
N = 67
N/2 = 33.5
cf just greater than 33.5 is 42
Median class = 125-145
So, l = 125, h = 20, f = 20 and
cf = cf of preceding class i.e. 22
Substitute all the value in the above formula, we get
Median = 125+ {20 x (33.5-22)/20}
= 125 + 11.5
= 136.5
Thus, median of electricity consumed is 136.5.
Question 6: Calculate the median from the following data:
Solution:
Where cf = cumulative frequency
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Here:
N = 100
N/2 = 50
cf just greater than 50 is 56
Median class = 150-155
So, l = 150, h = 5, f = 22 and
cf = cf of preceding class i.e. 34
Substitute all the value in the above formula, we get
Median = 150+ {5 x (50-34)/22}
= 150 + 3.64
= 153.64
Question 7: Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Solution:
Let the missing frequency be x.
Median = 24
N/2 = (55+x)/2 = 27.5 + x/2
Median class = 20-30
l = 20, h = 10, f = x and cf = cf of preceding class i.e. 30
Using below formula:
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)⇨ 24 = 20 + {10 x (27.5+ x/2 – 30)/x}
⇨ 24x = 20x + 5x – 25
or x = 25
Missing frequency is 25.
Question 8: The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.
Class | 0−5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 |
Frequency | 12 | a | 12 | 15 | b | 6 | 6 | 4 |
Solution:
Here:
N = 70
N/2 = 35
Median = 16, Median class = 15-20
So, l = 15, h = 5, f = 15 and
cf = cf of preceding class i.e. 24+a
We know,
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Substitute all the value in the above formula, we get
16 = 15 + {5x (35-24-a)/15}
⇨ 16 = 15 + (11-a)/3
or a = 8
Now, 70 = 55 + a + b
70 = 55 + 8 + b
⇨ b = 7
So, missing freqiuncies are 8 and 7.
Question 9: In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.
Runs scored | 2500−3500 | 3500−4500 | 4500−5500 | 5500−6500 | 6500−7500 | 7500−8500 |
Number of batsmen | 5 | x | y | 12 | 6 | 2 |
Solution:
Here:
N = 60
N/2 = 30
Median = 5000, Median class = 4500-5500
So, l = 4500, h = 1000, f = y and
cf = cf of preceding class i.e. 5+x
We know,
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Substitute all the value in the above formula, we get
5000 = 4500 + {1000 x (30-5-x)/y}
⇨ 500 = 1000 x (25-x)/y
⇨ 0.5 = (25-x)/y
⇨ x + 05.y = 25 …(1)
Again, from table
25 + x + y = 60
⇨ x + y = 35 …(2)
Solving (1) and (2), we have
x = 15 and y = 20
Missing frequencies are 15 and 20.
Question 10: If the median of the following frequency distribution is 32.5, find the values of f1 and f2.
Class
interval |
0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
Frequency | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
Solution:
Sum of all the frequencies = f1+ 5 + 9 +12 + f2 + 3 + 2 = 40 (given)
⇨ f1+ f2 = 9 …(1)
Median is 32.5 lies in 30 – 40, so the median class is 30 – 40
N = 40
N/2 = 20
l = 30, h = 10, f = 12
cf = f1 + 5+9 = f1+14
We know,
\(Median = I +\left \{ h\times\frac{\left ( \frac{N}{2} – cf\right )}{f} \right \}\)Substitute all the value in the above formula, we get
32.5 = 30 + {10 x (20-f1-14)/12 }
⇨ 32.5 = 30 + (30-5f1)/6
⇨ 2.5 = (30-5f1)/6
or f1 = 3
From (1)
⇨ 3 + f2 = 9
⇨ f2 = 6
Exercise 18C
Question 1: Find the mode of the following frequency distribution:
Marks | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
Frequency | 12 | 35 | 45 | 25 | 13 |
Solution:
Here:
Maximum class frequency is 45.
The class corresponding to Maximum class frequency is the modal class.
So, Modal class = 30 – 40
Lower limit = l = 30
Modal class size =h = 10
Frequency of class preceding the modal class =f0 = 35
Frequency of the modal class= f1 = 45
Frequency of class succeeding the modal =f2= 25
Mode formula is given by,
\(Mode = I +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h\) \(Mode = 30 +\left ( \frac{45-35}{2\times 45-35-25} \right )\times 10\)= 30 + 100/30
= 33.33
Question 2: Compute the mode of the following data:
Class | 0−20 | 20−40 | 40−60 | 60−80 | 80−100 |
Frequency | 25 | 16 | 28 | 20 | 5 |
Solution:
Here:
Maximum class frequency = 28
The class corresponding to Maximum class frequency is the modal class.
So, Modal class = 40-60
Lower limit = l = 40
Modal class size =h = 20
Frequency of class preceding the modal class =f0 = 16
Frequency of the modal class= f1 = 28
Frequency of class succeeding the modal =f2= 20
Mode formula is given by,
\(Mode = I +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h\) \(Mode = 40 +\left ( \frac{28-16}{2\times 28-16-20} \right )\times 20\)= 40 + 12/20 x 12
= 52
Question 3: Heights of students of Class X are given in the following frequency distribution:
Height (in cm) | 150−155 | 155−160 | 160−165 | 165−170 | 170−175 |
Number of students | 15 | 8 | 20 | 5 | 13 |
Find the modal height. Also, find the mean height. Compare and interpret the two measures of central tendency.
Solution:
Here:
Maximum class frequency = 20
The class corresponding to Maximum class frequency is the modal class.
So, Modal class = 160-165
Lower limit = l = 160
Modal class size =h = 5
Frequency of class preceding the modal class =f0 = 8
Frequency of the modal class= f1 = 20
Frequency of class succeeding the modal =f2= 12
Mode formula is given by,
\(Mode = I +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h\) \(Mode = 160 +\left ( \frac{20-8}{2\times 20-8-12} \right )\times 5\)= 160 + 12/20 x 5
= 163
The modal height is 163 cm, which implies maximum number of students have model height 163 cm
Use Direct Method to find the mean:
= 9670/60
= 161.17
Mean is 161.17
Thus, 161.7 cm is the average height of all the students.
Question 4: Find the mode of the following distribution:
Class interval | 10-14 | 14-18 | 18-22 | 22-26 | 26-30 | 30-34 | 34-38 | 38-42 |
Frequency | 8 | 6 | 11 | 20 | 25 | 22 | 10 | 4 |
Solution:
Here:
Maximum class frequency = 25
The class corresponding to Maximum class frequency is the modal class.
So, Modal class = 26-30
Lower limit = l = 26
Modal class size =h = 4
Frequency of class preceding the modal class =f0 = 20
Frequency of the modal class= f1 = 22
Frequency of class succeeding the modal =f2= 12
Mode formula is given by,
\(Mode = I +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h\) \(Mode = 26 +\left ( \frac{25-20}{2\times 25-20-22} \right )\times 4\)= 26 + 5/8 x 4
= 28.5
Question 5: Given below is the distribution of total household expenditure of 200 manual workers in a city:
Expenditure (in Rs) | 1000-1500 | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 |
No. of manual workers | 24 | 40 | 31 | 28 | 32 | 23 | 17 | 5 |
Find the expenditure done by maximum number of manual workers.
Solution:
Here:
Maximum class frequency = 40
The class corresponding to Maximum class frequency is the modal class.
So, Modal class = 1500-2000
Lower limit = l = 1500
Modal class size =h = 500
Frequency of class preceding the modal class =f0 = 24
Frequency of the modal class= f1 = 40
Frequency of class succeeding the modal =f2= 31
Mode formula is given by,
\(Mode = I +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h\) \(Mode = 1500 +\left ( \frac{40-24}{2\times 40-24-31} \right )\times 10\)= 1500 + 16/25 x 500
= 1820
Exercise 18D
Question 1: Find the mean, mode and median of the following frequency distribution.
Class | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
Frequency | 4 | 4 | 7 | 10 | 12 | 8 | 5 |
Solution:
Let assumed mean be 35, h = 10.
Question 2: Find the mean, mode and median of the following data:
Class | 0−20 | 20−40 | 40−60 | 60−80 | 80−100 | 100−120 | 120−140 |
Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
Solution:
Question 3: Find the mean, median and mode of the following data:
Class | 0−50 | 50−100 | 100−150 | 150−200 | 200−250 | 250−300 | 300−350 |
Frequency | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
Solution:
Question 4: Find the mode, median and mean for the following data:
Marks obtained | 25−35 | 35−45 | 45−55 | 55−65 | 65−75 | 75−85 |
Number of students | 7 | 31 | 33 | 17 | 11 | 1 |
Solution:
Question 5: A survey regarding the heights (in cm) of 50 girls of a class was conducted and the following data was obtained:
Height (in cm) | 120-130 | 130-140 | 140-150 | 150-160 | 160-170 | Total |
Number of girls | 2 | 8 | 12 | 20 | 8 | 50 |
Find the mean, median and mode of the above data.
Solution:
Assumed mean A = 145
Class interval = h = 10
Question 6: The following table gives the daily income of 50 workers of a factory:
Daily income (in Rs | 100-120 | 120-140 | 140-1 | 140-180 | 180-200 |
Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean, mode and median of the above data.
Solution:
Question 7: The table below shows the daily expenditure of food of 30 households in a locality:
Find the mean and median daily expenditure on food.
Solution:
Let assumed mean = 225 and h = 50
Exercise 18E
Question 1: Find the median of the following data by making a ‘less than ogive’.
Marks | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 | 90−100 |
Number of students | 5 | 3 | 4 | 3 | 3 | 4 | 7 | 9 | 7 | 8 |
Solution:
Plot points: (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the ‘less than type’ ogive as follows:
From frequency table, N = 53
N/2 = 26.5
Mark y = 26.5 on y-axis
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 68 units
Hence, median = 68
Question 2: The given distribution shows the number of wickets taken by the bowlers in one-day international cricket matches:
Number of wickets | Less than 15 | Less than 30 | Less than 45 | Less than 60 | Less than 75 | Less than 90 | Less than 105 | Less than 120 |
Number of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |
Draw a ‘less than type’ ogive from the above data. Find the median.
Solution:
Plot points on graph:
(15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the ‘less than type’ ogive as follows:
From frequency table, N = 80, ⇨ N/2 = 40
At y = 40,
Draw a horizontal line meeting the curve at P, AP
Draw a vertical line from point P which is parallel to y-axis and meets OX at A’.
OM = 77 units
Median number of wickets = 77
Question 3: Draw a ‘more than’ ogive for the data given below which gives the marks of 100 students.
Marks | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
Number of students | 4 | 6 | 10 | 10 | 25 | 22 | 18 | 5 |
Solution:
Plot points on the graph:
(0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the ‘more than type’ ogive as follows:
From frequency table, N = 100 ⇨ N/2 = 50.
At y = 50
Draw a horizontal line meeting the curve at P, AP
Draw a vertical line from point P which is parallel to y-axis and meets OX at M.
OM = 47 units
Median marks = 47
Question 4:The height of 50 girls of class X of a school are recorded as follows:
Height (in cm) | 135−140 | 140−145 | 145−150 | 150−155 | 155−160 | 160−165 |
Number of girls | 5 | 8 | 19 | 12 | 14 | 2 |
Draw a ‘more than type’ ogive for the above data.
Solution:
Plot the points on the graph:
(135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the ‘more than type’ ogive as follows:
From frequency table, N = 50 ⇨ N/2 = 25
At y = 25
Draw a horizontal line meeting the curve at P, AP
Draw a vertical line from point P which is parallel to y-axis and meets OX at M.
OM = 151 units
Median height = 151 cm
Question 5: The monthly consumption of electricity (in units) of some families of a locality is given in the following frequency distribution:
Monthly consumption (in units) | 140−160 | 160−180 | 180−200 | 200−220 | 220−240 | 240−260 | 260−280 |
Number of families | 3 | 8 | 15 | 40 | 50 | 30 | 10 |
Prepare a ‘more than type’ ogive for the following frequency distribution.
Solution:
Plot the points on the graph:
(140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the ‘more than type’ ogive as follows:
From frequency table, N = 156 ⇨ N/2 = 78
At y = 78
Draw a horizontal line meeting the curve at P, AP
Draw a vertical line from point P which is parallel to y-axis and meets OX at M.
OM = 226 units
Median consumption of electricity = 226 units
R S Aggarwal Solutions For Class 10 Maths Chapter 18 Exercises:
Get detailed solutions for all the questions listed under below exercises:
R S Aggarwal Solutions for Chapter 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Topics:
In this chapter students will study important concepts on Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive as listed below:
- Mean of grouped data – Direct Method
- Mean of grouped data – Assumed-Mean Method
- Mean of grouped data – Step derivation method
- Mathematical derivation of assumed mean formula
- Mathematical derivation of step derivation method
- Mean for an inclusive series
- Median for grouped data
- Mode of Grouped Data
- Measures of central tendency
- Cumulative Frequency Curve
Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Topics:
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