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Unit 7. Statistics
23: Graphical Representation
▶ 24: Measure of Central Tendency (Mean, Median, Quartiles and Mode)
25: Probability
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Solutions for Chapter 24: Measure of Central Tendency (Mean, Median, Quartiles and Mode)
Below listed, you can find solutions for Chapter 24 of CISCE Selina for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई.
Selina solutions for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई 24 Measure of Central Tendency (Mean, Median, Quartiles and Mode) EXERCISE 24 (A) [Pages 368 - 369]
Multiple Choice Type: Choose the correct answer from the options given below.
The mean of 5 variables is 20. If four of them are 15, 25, 10 and 12; the fifth variable is ______.
38
60
48
28
The mean height of 5 students is 140 cm. If height of one student is 156 cm, the mean height of the remaining students is ______.
74 cm
136 cm
16 cm
80 cm
The mean age of eight boys is 16 years. If two more boys, with ages 18 years and 14 years, join them, the resulting mean age is ______.
18 years
14 years
16 years
32 years
The mean value of 15 numbers is 20. If one of these numbers is wrongly taken as 45 instead of 15, the correct mean is ______.
30
15
20
18
The mean of the given frequency distribution is:
| x | 30 | 20 |
| f | 15 | 5 |
\[\frac{30}{15}+\frac{20}{5}\]
\[\frac{30}{15}-\frac{20}{5}\]
\[\frac{550}{20}=27.5\]
\[\frac{50}{20}=2 \cdot 5\]
Marks obtained (in mathematics) by 9 students are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56
- Find the arithmetic mean.
- If marks of each student be increased by 4; what will be the new value of arithmetic mean?
- Find the mean of 7, 11, 6, 5 and 6.
- If each number given in (a) is diminished by 2; find the new value of mean.
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’
The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.
The ages of 40 students are given in the following table :
| Age (in years) | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| Frequency | 2 | 4 | 6 | 9 | 8 | 7 | 4 |
Find the arithmetic mean.
If 69.5 is the mean of 72, 70, ‘x’, 62, 50, 71, 90, 64, 58 and 82, find the value of ‘x’.
The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of ‘f’.
| Height (cm) | 50 | 55 | 58 | 60 | 65 | 70 | 71 |
| No. of plants | 2 | 4 | 10 | f | 5 | 4 | 3 |
From the data given below, calculate the mean wage, correct to the nearest rupee.
| Category | A | B | C | D | E | F |
| Wages (Rs/day) | 50 | 60 | 70 | 80 | 90 | 100 |
| No. of workers | 2 | 4 | 8 | 12 | 10 | 6 |
- If the number of workers in each category is doubled, what would be the new mean wage?
- If the wages per day in each category are increased by 60%; what is the new mean wage?
- If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%, what would be the new mean wage?
The contents of 100 match boxes were checked to determine the number of matches they contained.
| No. of matches | 35 | 36 | 37 | 38 | 39 | 40 | 41 |
| No. of boxes | 6 | 10 | 18 | 25 | 21 | 12 | 8 |
- Calculate, correct to one decimal place, the mean number of matches per box.
- Determine, how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches.
If the mean of the following distribution is 3, find the value of p.
| x | 1 | 2 | 3 | 5 | p + 4 |
| f | 9 | 6 | 9 | 3 | 6 |
In the following table, Σf = 200 and mean = 73. Find the missing frequencies f1, and f2.
| x | 0 | 50 | 100 | 150 | 200 | 250 |
| f | 46 | f1 | f2 | 25 | 10 | 5 |
Find the arithmetic mean (correct to the nearest whole number) by using step-deviation method.
| x | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
| f | 20 | 43 | 75 | 67 | 72 | 45 | 39 | 9 | 8 | 6 |
Find the mean (correct to one place of decimal) by using short-cut method.
|
x |
40 |
41 |
43 |
45 |
46 |
49 |
50 |
|
f |
14 |
28 |
38 |
50 |
40 |
20 |
10 |
Selina solutions for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई 24 Measure of Central Tendency (Mean, Median, Quartiles and Mode) EXERCISE 24 (B) [Pages 374 - 376]
Multiple Choice Type: Choose the correct answer from the options given below.
The mean of given observations is:
| C.I. | 30-40 | 40-50 |
| f | 2 | 3 |
41
27
71
91
For data given in the adjoining table, the mean is:
| C.I. | f | x |
| 0-10 | 5 | |
| 10-20 | 10 | |
| 20-30 | 10 |
17
27
25
60
The mean of observations, given in the adjoining table, is 24, the value of a is:
| x | f |
| 10 | 10 |
| 20 | a |
| 30 | 30 |
20
30
10
25
If the mean of the data given in adjoining table is 20, the relation between x1 and x2 is:
| x | f |
| 10 | x1 |
| 20 | 20 |
| 30 | x2 |
x1 + x2 = 30
x1 − x2 = 15
x1 − x2 = 0
x1 + x2 = 20
The mean of data, represented by given diagram, is:

53
47
42
51
The following table gives the age of 50 student of a class. Find the arithmetic mean of their ages.
| Age-years | 16 – 18 | 18 – 20 | 20 – 22 | 22 – 24 | 24 – 26 |
| No. of students | 2 | 7 | 21 | 17 | 3 |
The following are the marks obtained by 70 boys in a class test:
| Marks | No. of boys |
| 30 – 40 | 10 |
| 40 – 50 | 12 |
| 50 – 60 | 14 |
| 60 – 70 | 12 |
| 70 – 80 | 9 |
| 80 – 90 | 7 |
| 90 – 100 | 6 |
Calculate the mean by:
Short-cut method
The following are the marks obtained by 70 boys in a class test:
| Marks | No. of boys |
| 30 – 40 | 10 |
| 40 – 50 | 12 |
| 50 – 60 | 14 |
| 60 – 70 | 12 |
| 70 – 80 | 9 |
| 80 – 90 | 7 |
| 90 – 100 | 6 |
Calculate the mean by:
Step-deviation method
Find mean by step-deviation method:
| C.I. | 63 – 70 | 70 – 77 | 77 – 84 | 84 – 91 | 91 – 98 | 98 – 105 | 105 – 112 |
| Frequency | 9 | 13 | 27 | 38 | 32 | 16 | 15 |
The mean of the following distribution is `21 1/7`. Find the value of ‘f’.
| C.I. | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 8 | 22 | 31 | f | 2 |
Using the information given in the adjoining histogram, calculate the mean.

If the mean of the following observations is 54, find the value of p.
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency | 7 | p | 10 | 9 | 13 |
The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f1 and f2.
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
| Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
Calculate the mean of the distribution, given below using the short cut method:
| Marks | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 |
| No. of students | 2 | 6 | 10 | 12 | 9 | 7 | 4 |
Selina solutions for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई 24 Measure of Central Tendency (Mean, Median, Quartiles and Mode) EXERCISE 24 (C) [Pages 385 - 387]
Multiple Choice Type: Choose the correct answer from the options given below.
The median of 18, 29, 15, 14 and 21 is ______.
15
18
29
21
The median of 3, 8, 11, 2, 16, 4, 0 and 6 is ______.
6
9
5
8
Numbers 5, 15, 20, x, 28, 30, 35 are in ascending order and have median = 23; then the value of x is ______.
24
29
17.5
23
For numbers 10, 20, 30, 40, 50, 60, 70 and 80; the inter-quartile range is ______.
20 + 60
60 - 30
60 - 20
50 - 10
From the given diagram, the modal class is:

30-40
40-50
50-60
60-70
A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
The weights (in kg) of 10 students of a class are given below:
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.
Find the median of their weights.
The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28.
Find:
- Median
- Lower quartile
- Upper quartile
- Inter-quartile range
The weight of 60 boys are given in the following distribution table:
| Weight (kg) | 37 | 38 | 39 | 40 | 41 |
| No. of boys | 10 | 14 | 18 | 12 | 6 |
Find:
- Median
- Lower quartile
- Upper quartile
- Inter-quartile range
From the following cumulative frequency table, draw ogive and then use it to find:
- Median
- Lower quartile
- Upper quartile
| Marks (less than) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
| Cumulative frequency | 5 | 24 | 37 | 40 | 42 | 48 | 70 | 77 | 79 | 80 |
In a school, 100 pupils have heights as tabulate below:
| Height (in cm) | No. of pupils |
| 121 – 130 | 12 |
| 131 – 140 | 16 |
| 141 – 150 | 30 |
| 151 – 160 | 20 |
| 161 – 170 | 14 |
| 171 – 180 | 8 |
Find the median height by drawing an ogive.
Find the mode of following data, using a histogram:
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 5 | 12 | 20 | 9 | 4 |
The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:
| Expenditure (Rs) | No. of students |
| 20 – 25 | 4 |
| 25 – 30 | 7 |
| 30 – 35 | 23 |
| 35 – 40 | 18 |
| 40 – 45 | 6 |
| 45 – 50 | 2 |
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
What are his modal marks?
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
What are his median marks?
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
What are his total marks?
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
What are his mean marks?
At a shooting competition the score of a competitor were as given below:
| Score | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of shots | 0 | 3 | 6 | 4 | 7 | 5 |
- What was his modal score?
- What was his median score?
- What was his total score?
- What was his mean score?
Selina solutions for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई 24 Measure of Central Tendency (Mean, Median, Quartiles and Mode) TEST YOURSELF [Pages 387 - 391]
Multiple Choice Type: Choose the correct answer from the options given below.
The mean of numbers in A.P. 2, 4, 6, 8, …, 40 is ______.
`(40+2)/2`
`20/2(2+40)`
`10/2(2+40)`
840
The median of 10, 12, 9, 8, 12, 13, 8, 15 and 12 is ______.
12
`(10+12)/2`
`(12+13)/2`
13
The numbers 10, 12, 14, 16, 17 and x are in ascending order. If the mean and median of these observations are same, the value of x is ______.
16
14
54
21
The median of first six prime numbers is ______.
5
7
6
7.5
The inter quartile range for the given ogive is:

42
32
44
54
The mean age of nine boys is 28 years and if one new boy joins them the mean age increases by one.
Assertion (A): The age of new boy is (29 × 10 − 28 × 9) years.
Reason (R): The age of new boy is (29 − 28) × 10 years.
A is true, R is false.
A is false, R is true.
Both A and R are true and R is the correct reason for A.
Both A and R are true and R is the incorrect reason for A.
Data = 37, 41, 56, 62, 70, 74, 81, 89, 95 and 90.
Assertion (A): Median = 72.
Reason (R): If number of data(n) is odd, the median = `((n+1)/2)^(th)` term.
A is true, R is false.
A is false, R is true.
Both A and R are true and R is the correct reason for A.
Both A and R are true and R is the incorrect reason for A.
| C.I. | 0 - 10 | 10 - 20 | 20 - 30 |
| Frequency | 15 | 25 | b |
| Cumulative frequency | 15 | a | 50 |
Assertion (A): a = 15 + 25 = 40;
b = 50 − a
Reason (R): a + 15 = 25
and b = 50 − 10
A is true, R is false.
A is false, R is true.
Both A and R are true and R is the correct reason for A.
Both A and R are true and R is the incorrect reason for A.
Data: 9, 11, 15, 19, 17, 13 and 7.
Statement (1): For the given data, lower quartile is 11.
Statement (2): For data with n terms, the lower quartile is `((n+1)/4)^(th)` term, if n is odd.
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
The mean of given data is 26.
| C.I. | 0 - 20 | 20 - 40 | 40 - 60 |
| f | 20 | x | 10 |
Statement (1): x = 26.
Statement (2): \[26=\frac{10 \times 20+30 \times x+50 \times 10}{30+x}\]
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
For a given set of data mean = 14 and median = 15.
Statement (1): Mode = 17.
Statement (2): Mode = 3 median − 2 mean.
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m – 1 and median q. Find p and q.
In a malaria epidemic, the number of cases diagnosed were as follows:
| Date July | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Num | 5 | 12 | 20 | 27 | 46 | 30 | 31 | 18 | 11 | 5 | 0 | 1 |
On what days do the mode and upper and lower quartiles occur?
The marks obtained by 120 students in a mathematics test is given below:
| Marks | No. of students |
| 0 – 10 | 5 |
| 10 – 20 | 9 |
| 20 – 30 | 16 |
| 30 – 40 | 22 |
| 40 – 50 | 26 |
| 50 – 60 | 18 |
| 60 – 70 | 11 |
| 70 – 80 | 6 |
| 80 – 90 | 4 |
| 90 – 100 | 3 |
Draw an ogive for the given distributions on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:
- the median
- the number of student who obtained more than 75% in test.
- the number of students who did not pass in the test if the pass percentage was 40.
- the lower quartile.
Using a graph paper, draw an ogive for the following distribution which shows a record of the Weight in kilograms of 200 students.
| Weight | Frequency |
| 40 – 45 | 5 |
| 45 – 50 | 17 |
| 50 – 55 | 22 |
| 55 – 60 | 45 |
| 60 – 65 | 51 |
| 65 – 70 | 31 |
| 70 – 75 | 20 |
| 75 – 80 | 9 |
Use your ogive to estimate the following:
- The percentage of student weighting 55 kg or more
- The weight above which the heaviest 30% of the student fall
- The number of students who are
- underweight
- overweight, If 55.70 kg is considered as standard weight.
The distribution, given below, shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.
| Marks obtained | 5 | 6 | 7 | 8 | 9 | 10 |
| No. of students | 3 | 9 | 6 | 4 | 2 | 1 |
The monthly income of a group of 320 employees in a company is given below:
| Monthly income (thousands) | No. of employees |
| 6-7 | 20 |
| 7-8 | 45 |
| 8-9 | 65 |
| 9-10 | 95 |
| 10-11 | 60 |
| 11-12 | 30 |
| 12-13 | 5 |
Draw an ogive of the distribution on a graph paper taking 2 cm = Rs. 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph detemine:
- the median wage.
- number of employee whose income is below Rs. 8,500.
- If salary of a senior employee is above Rs. 11,500, find the number of senior employee in the company.
- the upper quartile.
The mean of following numbers is 68. Find the value of ‘x’. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence, estimate the median.
The marks of 10 students of a class in an examination arranged in ascending order is as follows:
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
The histogram below represents the scores obtained by 25 students in a mathematics mental test. Use the data to:
- Frame a frequency distribution table.
- To calculate mean.
- To determine the Modal class.

For the data given in the following table, find:
| Class-mark | Frequency |
| 15 | 10 |
| 25 | 12 |
| 35 | 14 |
| 45 | 16 |
| 55 | 8 |
- mean
- mode (using graph)
- median (using grap)
For the data given in the following table, find:
| Class-mark | Frequency |
| 35 | 8 |
| 40 | 10 |
| 45 | 12 |
| 50 | 15 |
| 55 | 20 |
| 60 | 15 |
- mean
- mode (using graph)
- median (using grap)
Case-Study Based Question
On average, 80 patients get admitted into a nursing home in a day. The ages of the patients admitted and their number are as given below:
| Age (in years) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| No. of patients | 13 | 23 | 25 | 14 | 5 |
Find:
- The average age for which maximum cases occur.
- The upper limit of modal class.
- The mean of the given data.
Along with an increase in population, electricity consumption per person is also increasing nationwide.
A survey is conducted for 560 families of Mangal Pande Nagar, Meerut. The following table gives the monthly consumption of electricity of these families.
| Monthly consumption (in units) | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
| No. of families | 160 | 120 | 180 | 60 | 40 |
Assuming that no family consumes more than 500 units of electricity, find:
- The mean monthly consumption.
- The modal monthly class.
Solutions for 24: Measure of Central Tendency (Mean, Median, Quartiles and Mode)
![Selina solutions for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई chapter 24 - Measure of Central Tendency (Mean, Median, Quartiles and Mode) Selina solutions for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई chapter 24 - Measure of Central Tendency (Mean, Median, Quartiles and Mode) - Shaalaa.com](/images/concise-mathematics-english-class-10-icse_6:7eb8c97e7ccc4a1c956f7ac8305d25e3.jpg)
Selina solutions for कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई chapter 24 - Measure of Central Tendency (Mean, Median, Quartiles and Mode)
Shaalaa.com has the CISCE Mathematics कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई CISCE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Selina solutions for Mathematics कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई CISCE 24 (Measure of Central Tendency (Mean, Median, Quartiles and Mode)) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.
Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. Selina textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in कन्साइस माठेमटिक्स [इंग्रजी] इयत्ता १० आयसीएसई chapter 24 Measure of Central Tendency (Mean, Median, Quartiles and Mode) are Frequency Distribution and Its Applications, Histograms, Frequency Polygon, Cumulative Frequency and Cumulative Frequency Table, Ogives (Cumulative Frequency Curve), Concepts of Statistics, Frequency Distribution and Its Applications, Histograms, Frequency Polygon, Cumulative Frequency and Cumulative Frequency Table, Ogives (Cumulative Frequency Curve), Concepts of Statistics, Frequency Distribution and Its Applications, Histograms, Frequency Polygon, Cumulative Frequency and Cumulative Frequency Table, Ogives (Cumulative Frequency Curve), Concepts of Statistics.
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