Frequency Distribution

ICSE Class 9 Mathematics · Solutions for Class 9 RS Aggarwal ICSE Foundation Mathematics · 35 solved questions

Exercise

Marks

1Exercise 15

Define statistics as a subject.

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Statistics is the science which deals with the collection, presentation, analysis and interpretation of numerical data.

2Exercise 15

What are primary data and secondary data ? Which of the two are more reliable and why?

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Primary Data : The data collected by the investigator himself with a definite plan in mind are known as primary data.

Secondary Data : The data collected by someone, other than the investigator are known as secondary data.

Primary data is more reliable than secondary data because primary data collected by the user himself.

3Exercise 15

Fill in the blanks

(i) The difference between the maximum and minimum observations in a data is called the ............ of the data.

(ii) The number of observations in a class-interval is called the ............ of the interval.

(iii) The mid-point of a class-interval is called the ............ of the interval.

(iv) Lower-limit of the class-interval 24-30 is ............

(v) Upper limit of the class-interval 16-20 is ............

(vi) The class-mark of the class-interval is 20-30 is ............

(vii) The class-mark of the class-interval 9.5-19.5 is ............

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(i) Range

(ii) Frequency

(iii) Class mark

(iv) 24

(v) 20

(vi) 25

(vii) 14.5

4Exercise 15

Find the range of the data

(a) 5, 7, 16, 21, 8, 10

(b) 11, 13, 17, 14, 19, 14, 15, 18

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(a) By formula,

Range = Highest value - lowest value

Range = 21 - 5 = 16

Hence, the range of the data = 16.

(b) By formula,

Range = Highest value - Lowest value

Range = 19 - 11 = 8

Hence, the range of the data = 8.

5Exercise 15

The class marks of a frequency distribution are 28, 34, 40, 46, 52. Find the class-size and all the class intervals.

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By formula,

Class Size = Difference between two consecutive class marks

= 34 - 28 = 6.

Lower limit = Class mark - $\frac{\text{Class size}}{2}$

Upper limit = Class mark + $\frac{\text{Class size}}{2}$

Class intervals for class mark 28 :

Lower Limit = 28 - 3 = 25

Upper limit = 28 + 3 = 31

Class = 25 - 31

Class intervals for class mark 34 :

Lower Limit = 34 - 3 = 31

Upper limit = 34 + 3 = 37

Class = 31 - 37

Class intervals for class mark 40 :

Lower Limit = 40 - 3 = 37

Upper limit = 40 + 3 = 43

Class = 37 - 43

Class intervals for class mark 46 :

Lower Limit = 46 - 3 = 43

Upper limit = 46 + 3 = 49

Class = 43 - 49

Class intervals for class mark 52 :

Lower Limit = 52 - 3 = 49

Upper limit = 52 + 3 = 55

Class = 49 - 55.

Hence, class size = 6 and class intervals : 25 - 31, 31 - 37, 37 - 43, 43 - 49, 49 - 55.

6Exercise 15

State which of the following variables are continuous and which are discrete :

(i) Marks obtained by the students of a class in a test.

(ii) Daily maximum temperature of a city.

(iii) I.Q. of a students of a class.

(iv) Weights of players of a Volley-ball team.

(v) Number of car-accidents in a city.

(vi) Distance travelled by a train.

(vii) Time taken by runners in a race.

(viii) Sizes of shoes sold in a shoe-store.

(ix) Number of patients in a hospital per day.

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(i) Marks obtained by the students in a test will always be in whole number or natural number (e.g : 17, 20, ... ) hence, it is discrete variable.

(ii) Temperature can be in any value (e.g : 27⁰ C, 21.5⁰ C). So, it is continuous variable.

(iii) I.Q. of a person can be any number. Hence it is a continuous variable.

(iv) Weight can be measured in any value (e.g : 47.5 kg, 98 kg). Hence it is a continuous variable.

(v) Accidents will always be in whole number (e.g : 2, 5, 15). We cannot say total accidents are 2.5, 7.8. Hence, it is a discrete variable.

(vi) Distance can be of any value (like 78 km, 117.95 km). Hence, it is a continuous variable.

(vii) Time can be measured in any value (like 18.7 sec, 34.7 min). Hence, it is a continuous variable.

(viii) Shoes size always be in whole number or natural number(like 7, 9, 10). We cannot have the shoes size of 8.6, 9.3. Hence, it is a discrete variable.

(ix) Hospitals will count the number of patietns as whole number or natural number(like 12 patients, 19 patients). We cannot say there are 15.6 patients in the hospital. Hence, it is a discrete variable.

7Exercise 15

Define the following terms:

(i) Variable

(ii) Class-interval

(iii) Class-size

(iv) Class-mark

(v) Class-limits

(vi) True class-limits

(Vii) Frequency of a class

(viii) Cumulative frequency of a class

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(i) Variable : A quantity which can take different values is called a variable.

(ii) Class-interval : A range of values into which data is grouped in a frequency distribution.

(iii) Class-size : It is the difference between the true upper limit and the true lower limit of class.

(iv) Class-mark : It is the midpoint( middle value) of a class-interval.

(v) Class-limits : These are the smallest and largest values of the class interval.

(vi) True class-limits : These are the actual limits of a class interval after removing the gap between two consecutive classes.

(vii) Frequency of a class : It is the number of observations that fall within that class-interval.

(viii) Cumulative frequency of a class : It is the sum of the frequencies of all the previous classes and that particular class.

8Exercise 15

Following data gives the number of children in 40 families :

1, 2, 6, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 1, 6, 2, 2

Represent it in the form of a frequency distribution.

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Frequency distribution table :

Number of Children	Tally-Marks	Frequency
0	IIII	4
1	IIII II	7
2	IIII IIII II	12
3	IIII	5
4	IIII II	7
5	II	2
6	III	3
Total		40

9Exercise 15

The marks obtained by 40 students of a class in an examination are given below. Present the data in the form of a frequency distribution using equal class-size, one such class being 10 - 15(15 not included).

3, 20, 13, 1, 21, 13, 3, 23, 16, 13, 18, 12, 5, 12, 5, 24, 9, 2, 7, 18, 20, 3, 10, 12, 7, 18, 2, 5, 7, 10, 16, 8, 16, 17, 8, 23, 21, 6, 23, 15

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Since the given class is 10 - 15 (15 not included)

so, the class size = 5

Since, the maximum marks of the student equals to 24, thus the class intervals will be :

0 - 5, 5 - 10, 10 - 15, 15 - 20, 20 - 25

Frequency distribution table :

Marks (Class Interval)	Tally-Marks	Frequency
0 – 5	IIII I	6
5 – 10	IIII IIII	10
10 – 15	IIII III	8
15 – 20	IIII III	8
20 – 25	IIII III	8
Total		40

10Exercise 15

Construct a frequency table for the following ages (in years) of 30 students using equal class-intervals, one of them being 9-12, where 12 is not included.

18, 12, 7, 6, 11, 15, 21, 9, 8, 13, 15, 17, 22, 19, 14, 21, 23, 8, 12, 17, 15, 6, 18, 23, 22, 16, 9, 21, 11, 16

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Given class is 9 - 12 (12 is not included)

so class size = 12 - 9 = 3

Minimum age = 6

Maximum age = 23

So the class intervals will be

6 - 9, 9 - 12, 12 - 15, 15 - 18, 18 - 21, 21 - 24

Frequency distribution table :

Ages (Class Interval)	Tally-Marks	Frequency
6 – 9	IIII	5
9 – 12	IIII	4
12 – 15	IIII	4
15 – 18	IIII II	7
18 – 21	III	3
21 – 24	IIII II	7
Total		30

11Exercise 15

The weekly wages (in rupees) of 30 workers in a factory given below :

630, 635, 690, 610, 635, 636, 639, 645, 698, 690, 620, 660, 632, 633, 655, 645, 604, 608, 612, 640, 685, 635, 636, 678, 640, 668, 690, 606, 640, 690

Represent the data in the form of a frequency distribution with class size 10.

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Minimum age = 604

Maximum age = 698

Given class size = 10

so we can take the class intervals as

600 - 610, 610 - 620, 620 - 630, 630 - 640, 640 - 650, 650 - 660, 660 - 670, 670 - 680, 680 - 690, 690 - 700

Frequency distribution table :

Class interval	Tally-Marks	Frequency
600 – 610	III	3
610 – 620	II	2
620 – 630	I	1
630 – 640	IIII IIII	9
640 – 650	IIII	5
650 – 660	I	1
660 – 670	II	2
670 – 680	I	1
680 – 690	I	1
690 – 700	IIII	5
Total	30

12Exercise 15

The weights in grams of 50 apples picked at random from a consignment are as follows :

131, 113, 82, 75, 204, 81, 84, 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90, 115, 110, 98, 106, 99, 107, 84, 76, 186, 82, 100, 109, 128, 115, 107, 115, 119, 93, 187, 139, 129, 130, 68, 195, 123, 125, 111, 92, 86, 70, 126

Form the grouped frequency table by dividing the variable range into intervals of equal width of 20 g.

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Minimum weight = 68 g

Maximum weight = 204 g

Given class width = 20 g

So we can take the class intervals as

60 - 80, 80 - 100, 100 - 120, 120 - 140, 140 - 160, 160 - 180, 180 - 200, 200 - 220

Frequency distribution table :

Weights (in g)	Tally-Marks	Frequency
60 – 80	IIII	5
80 – 100	IIII IIII III	13
100 – 120	IIII IIII IIII II	17
120 – 140	IIII IIII	10
140 – 160	I	1
160 – 180		0
180 – 200	III	3
200 – 220	I	1
Total		50

13Exercise 15

The marks obtained by 35 students in an examination are given below :

370, 290, 318, 175, 170, 410, 378, 405, 380, 375, 315, 305, 325, 275, 241, 288, 261, 355, 402, 380, 178, 253, 428, 240, 210, 175, 154, 405, 380, 370, 306, 460, 328, 440, 425

Form cumulative frequency table with class intervals of length 50.

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Minimum mark = 154

Maximum mark = 460

Given class length = 50

So we can take class intervals as

150 - 200, 200 - 250, 250 - 300, 300 - 350, 350 - 400, 400 - 450, 450 - 500,

Frequency distribution table :

Marks	Tally-Marks	Frequency	Cumulative frequency
150 - 200	IIII	5	5
200 - 250	III	3	8 (5 + 3)
250 - 300	IIII	5	13 (8 + 5)
300 - 350	IIII I	6	19 (13 + 6)
350 - 400	IIII III	8	27 (19 + 8)
400 - 450	IIII II	7	34 (27 + 7)
450 - 500	I	1	35 (34 + 1)
Total		35

14Exercise 15

Construct the cumulative frequency table from the frequency table given below:

Class-Interval	Frequency
0 – 6	7
6 – 12	11
12 – 18	8
18 – 24	14
24 – 30	12

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Cumulative frequency table :

Class-Interval	Frequency	Cumulative Frequency
0 – 6	7	7
6 – 12	11	18 (11 + 7)
12 – 18	8	26 (18 + 8)
18 – 24	14	40(26 + 14)
24 – 30	12	52(40 + 12)
Total	52

15Exercise 15

Construct a frequency distribution table from the following cumulative frequency distribution:

Class-Interval	Cumulative Frequency
0 – 8	8
8 – 16	21
16 – 24	26
24 – 32	33
32 – 40	42

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Frequency = Current C.F - Previous C.F

Frequency distribution table :

Class Interval	Frequency	Cumulative Frequency
0 – 8	8	8
8 – 16	13 (21 - 8)	21
16 – 24	5 (26 - 21)	26
24 – 32	7 (33 - 26)	33
32 – 40	9 (42 - 33)	42

16Exercise 15

Construct a frequency table from the following data :

Age (in years)	Number of Students
Less than 10	6
Less than 20	14
Less than 30	30
Less than 40	52
Less than 50	65
Less than 60	70

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Frequency distribution table :

Class Interval	Frequency	Cumulative Frequency
0 – 10	6	6
10 – 20	8 (14 - 6)	14
20 – 30	16 (30 - 14)	30
30 – 40	22 (52 - 30)	52
40 – 50	13 (65 - 52)	65
50 – 60	5 (70 - 65)	70

17Exercise 15

Convert the following frequency distribution to exclusive form :

Class Interval	Frequency
30 – 34	7
35 – 39	9
40 – 44	13
45 – 49	6
50 – 54	3
55 – 59	10

Use this table to find :

(i) The true class-limits of the fourth class-interval.

(ii) The class-boundaries of the fifth class-interval.

(iii) The class-mark of the third class-interval.

(iv) The class-size of the sixth class-interval.

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Adjustment factor

= $\frac{\text{Lower limit of one class - Upper limit of previous class}}{2} = \frac{35 - 34}{2} = \frac{1}{2}$ = 0.5

Subtract the lower limit of each class by 0.5 and add 0.5 to upper limit of each class.

Class Interval	Exclusive Class Interval	Frequency
30 - 34	29.5 - 34.5	7
35 - 39	34.5 - 39.5	9
40 - 44	39.5 - 44.5	13
45 - 49	44.5 - 49.5	6
50 - 54	49.5 - 54.5	3
55 - 59	54.5 - 59.5	10

(i) The true class limits of the fourth class interval is 44.5 - 49.5.

(ii) The class-boundaries of the fifth class interval is 49.5 - 54.5.

(iii) The class-mark of the third class interval

= $\frac{39.5 + 44.5}{2}$

= $\frac{84}{2}$

= 42.

Hence, the class mark of third class interval = 42.

(iv) Sixth class interval : 54.5 - 59.5

Class size = Upper limit - lower limit

Class size = 59.5 - 54.5 = 5.

Hence, the class size of sixth class interval is 5.

11 Mark · Multiple choice questions

If in a grouped frequency distribution, the classes are 20 - 30, 30 - 40, 40 - 50, ..., then the observation 40 is included in the class :

(a)30 - 40
(b)40 - 50
(c)50 - 60
(d)20 - 30

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Since the given class-intervals are in exclusive form.

So, in exclusive form lower limit is included but the upper limit is excluded.

So 40 is included in the class-interval, 40-50.

Hence, option 2 is the correct option.

21 Mark · Multiple choice questions

If in a grouped frequency distribution, the class intervals are 20 - 29, 30 - 39, 40 - 49, ...., then the observation 29.4 is included in the class :

(a)20 - 29
(b)30 - 39
(c)40 - 49
(d)29 - 30

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Given class intervals are in inclusive form but the observation is in decimal form (29.4).

So convert the inclusive form into exclusive form.

Gap between the classes = 30 - 29 = 1

Adjustment factor = 0.5

So new class boundaries become

19.5 - 29.5, 29.5 - 39.5, 39.5 - 49.5, ...,

29.4 lies between the class-interval 19.5 - 29.5.

19.5 - 29.5 is the exclusive form of 20 - 29.

Hence, option 1 is the correct option.

31 Mark · Multiple choice questions

The classes of a frequency distribution are 30 -34, 35 - 39, ..., 50 - 54. The lower boundary of the class 35 - 39 is :

(a)30.5
(b)35
(c)34.5
(d)39

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Gap between the classes = 35 - 34 = 1

Adjustment factor = $\frac{1}{2}$ = 0.5

New class boundaries :

29.5 - 34.5, 34.5 - 39.5, ..., 49.5 - 54.5.

Thus lower limit of class 35 - 39 is 34.5

Hence, option 3 is the correct option.

41 Mark · Multiple choice questions

In a grouped frequency distribution, the class interval are 40 - 49, 50 - 59, ..., 80 - 89. The upper boundary of the class (50 - 59) is :

(a)59
(b)59.5
(c)49.5
(d)60

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Gap between the classes = 50 - 49 = 1

Adjustment factor = 0.5

Upper boundary for the class interval (50 - 59) = Upper limit + 0.5

= 59 + 0.5

= 59.5

Hence, option 2 is the correct option.

51 Mark · Multiple choice questions

If 0 - 9, 10 - 19, 20 - 29, ...., are the classes of a grouped frequency distribution, then the width of each class is :

(a)9
(b)9.5
(c)10
(d)9 or 10

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Gap between classes = 10 - 9 = 1

Adjustment factor = 0.5

So the new class boundaries become

-0.5 - 9.5, 9.5 - 19.5, 19.5 - 29.5, ...,

Class width = Upper boundary - Lower boundary

Considering second class :

Class width = 19.5 - 9.5 = 10

Similarly, for rest of the class intervals.

Hence, option 3 is the correct option.

61 Mark · Multiple choice questions

If 10, 15, 20, ... are respectively the mid-value of the classes of a grouped frequency distribution, then the class whole mid-value is 25 is :

(a)20 - 27
(b)22.5 - 27.5
(c)21.5 - 28.5
(d)21 - 28

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Difference between the mid values = 15-10 = 5

So, the class width = 5

For the class whose mid-value 25 is

Lower limit = 25 - $\frac{\text{Class width}}{2}$ = 22.5

Upper limit = 25 + $\frac{\text{Class width}}{2}$ = 27.5

So the Class is 22.5 - 27.5.

Hence, option 2 is the correct option.

71 Mark · Multiple choice questions

The width of each class interval of a grouped frequency distribution is 8. The lower boundary of the class whose mid-value is 10 is :

(a)2
(b)4.5
(c)5.5
(d)6

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Class width = 8

Half of class width = $\frac{8}{2}$ = 4

Mid-value = 10

So, Lower boundary = 10 - 4

= 6

Hence, option 4 is the correct option.

81 Mark · Multiple choice questions

The width of each class of a frequency distribution is 5. The upper boundary of the class whose class mark is 25 is :

(a)22
(b)22.5
(c)27
(d)27.5

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Class width = 5

Class mark = 25

Half of class width = $\frac{5}{2}$ = 2.5

Upper boundary = 25 + 2.5 = 27.5

Hence, option 4 is the correct option.

91 Mark · Multiple choice questions

Class marks of a distribution are: 47, 52, 57, 62, 67, 72, 77, 82. Class size of the given distribution is :

(a)5
(b)6
(c)7
(d)2

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Class size = Difference between the consecutive class marks.

Class size = 52 - 47 = 5

57 - 52 = 5

Therefore class size = 5

Hence, option 1 is the correct option.

101 Mark · Multiple choice questions

Mid value for the class interval 19.5 - 29.5 is :

(a)14.5
(b)24.5
(c)15.5
(d)25.5

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Class interval = 19.5-29.5

Mid value = $\frac{\text{Lower limit} + \text{Upper limit}}{2}$

= $\frac{19.5 + 29.5}{2}$

= $\frac{49.5}{2}$

= 24.5

Hence, option 2 is the correct option.

1Case study based questions

Case Study

To maintain the health report card in a hospital, the weights of new born babies are recorded on regular basis.

During a week, the weights (in kg) of new born were recorded as below :

2.7, 2.6, 3.1, 2.2, 2.4, 2.9, 3.0, 2.5, 2.8, 2.5, 2.8, 2.9, 2.2, 2.9, 2.7, 2.8, 2.3, 2.8, 2.1, 2.2

Based on the above information, answer the following questions:

(a)Range of the data is :
(a) 1 kg
(b) 0.4 kg
(c) 0.5 kg
(d) 1.1 kg
(b)How many babies are born during the week?
(a) 10
(b) 16
(c) 17
(d) 20
(c)How many babies weigh less than 2.4 kg?
(a) 7
(b) 6
(c) 5
(d) 4
(d)How many babies weigh more than 2.7 kg?
(a) 10
(b) 9
(c) 8
(d) 7
(e)How many babies weigh 2.8 kg?
(a) 4
(b) 5
(c) 6
(d) 3

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1. Range = Highest value - Lowest value

= 3.1 - 2.1

= 1 kg

Hence, option (a) is the correct option.

2. Total babies born during the week = 20

Hence, option (d) is the correct option.

3. Babies weighing less than 2.4 kg are

2.2, 2.2, 2.3, 2.1, 2.2

Total = 5

Hence, option (c) is the correct option.

4. Babies weighing more than 2.7 kg are

2.8, 2.9, 3.0, 2.8, 2.9, 2.9, 2.8, 2.8, 3.1

Total = 9

Hence, option (b) is the correct option.

5. From data,

Babies weighing 2.8 kg = 4.

Hence, option (a) is the correct answer

2Case study based questions

Case Study

The following data gives the weights (in grams) of some marbles picked up randomly from 3 boxes. The data was presented in the tabular form as below:

Weight (in g)	No. of Marbles
50-55	5
55-60	7
60-65	8
65-70	6
70-75	2
75-80	1
80-85	1

Based on the above information, answer the following questions:

(a)Class size of the given distribution is :
(a) 5
(b) 10
(c) 50
(d) 85
(b)Class mark of the 3rd class is :
(a) 60
(b) 65
(c) 125
(d) 62.5
(c)The number of marbles whose weight is more than or equal to 75 g is :
(a) 0
(b) 1
(c) 2
(d) 28
(d)The number of marbles whose weight is less than 70 g is :
(a) 26
(b) 28
(c) 4
(d) 2
(e)The class boundaries of the class 7th class interval are :
(a) 80, 85
(b) 80.5, 85.5
(c) 79.5, 85.5
(d) 80, 85.5

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1. Class size = Upper limit - lower limit

= 55 - 50

= 5.

Hence, option (a) is the correct option.

2. Class mark = $\frac{\text{Lower limit} + \text{Upper limit}}{2}$

= $\frac{60 + 65}{2}$

= $\frac{125}{2}$

= 62.5

Hence, option (d) is the correct option.

3. Marbles weighing more than or equal to 75 g are

75-80 ⟶ 1

80-85 ⟶ 1

Total = 1 + 1 = 2

Hence, option (c) is the correct option.

4. Marbles weighing less than 70 g are

50-55 ⟶ 5

55-60 ⟶ 7

60-65 ⟶ 8

65-70 ⟶ 6

Total = 5 + 7 + 8 + 6 = 26

Hence, option (a) is the correct option.

5. Class interval of 7th class = 80 - 85

Hence, option (a) is the correct option.

1Assertion-reason questions

Assertion (A) : Classes of the form : 1-10, 11-20, 21-30 etc. are said to be in inclusive form.

Reason (R) : A frequency distribution in which each upper limit as well as lower limit is included is called an inclusive form.

(a)A is true, R is false
(b)A is false, R is true
(c)Both A and R are true
(d)Both A and R are false

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Given class intervals 1-10, 11-20, 21-30 etc shows that gap between the upper limit of one class (10) and the lower limit of the next (11).

Since gap shows that these classes don't overlap, thus they are in inclusive form.

∴ Assertion (A) is true.

We know that,

A frequency distribution in which each upper limit as well as lower limit is included is known as inclusive form.

∴ Reason (R) is true.

Both Assertion and Reason are True.

Hence, option 3 is the correct option.

2Assertion-reason questions

Assertion (A) : In the exclusive form, the upper and lower limits of a class are respectively known as the true upper limit and true lower limit.

Reason (R) : Class size = $\frac{\text{ True lower limit} + \text{ True upper limit}}{2}$

(a)A is true, R is false
(b)A is false, R is true
(c)Both A and R are true
(d)Both A and R are false

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In the exclusive form (e.g., 10 - 20, 20 - 3)there are no gaps between the classes. Because the classes are continuous, therefore, these limits are indeed the true lower limit and true upper limit.

∴ Assertion (A) is true.

We know that,

Class size = Upper limit - Lower limit

∴ Reason (R) is false.

Hence, option 1 is the correct option.

1Competency focussed questions

Age of teachers of a school varies from 24 years to 72 years. Rishabh has to make a grouped frequency distribution table of age of teachers. Which among the following class intervals of age will be most appropriate to make meaningful conclusion in less time and effort?

(a)24 - 26, 26 - 28, 28 - 30, and so on
(b)24 - 30, 30 - 36, 36 - 42, and so on
(c)24 - 48, 48 - 72
(d)none of these

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Age of teachers varies from 24 years to 72 years

So, Range = 72 - 24 = 48

By checking the option (ii)

Class size = 6

Total classes = $\frac{48}{6}$

= 8 classes

∴ 8 class intervals are suitable to check the age of teachers.

Hence, option 2 is the correct option.

2Competency focussed questions

The distribution of marks of students of a class in a test is shown in the table below.

Marks	Number of students
0 - 15	12
15 - 30	19
30 - 45	8
45 - 60	4

Which of the following statements are correct?

(i) We can definitely say that no one scored 60 marks.

(ii) The number of students who got at least 30 marks is 12.

(a)only (i)
(b)only (ii)
(c)both (i) and (ii)
(d)none of these

Show SolutionHide Solution↓

(i) We know that given table is in exclusive form. So in exclusive form generally the upper limit of class is excluded from that group and moved to next one.

However, last class interval of the given table is 45 - 60.

So, we can strongly say that no one scored 60 marks.

∴ Statement (i) is correct.

(ii) The number of students who got at least 30 marks are

30 – 45 ⟶ 8 students

45 – 60 ⟶ 4 students

Total = 8 + 4 = 12.

∴ Statement (ii) is Correct.

Hence, option 3 is the correct option.

3Competency focussed questions

Range of 14, 12, 17, 18, 16 and p is 20. If p>0, then p= ?

(a)10
(b)26
(c)32
(d)35

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Given numbers 14, 12, 17, 18, 16 and p

Maximum Value = 18

Minimum value = 12

Range = Maximum value - Minimum value = 18 - 12 = 6 (excluding p)

But the range given is 20, So p must be either maximum value or minimum value

If p is the maximum value, then

p - 12 = 20

P = 32

If p is the minimum value, then

18 - p = 20

p = -2

Since p > 0

∴ p = 32

Hence, option 3 is the correct option.

4Competency focussed questions

The class mark of a class-interval is 'a'. If its lower limit is a-b, then its upper limit is :

(a)a + b
(b)ab
(c)b-a
(d)2b

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Class mark = a

Lower limit = a − b

Class mark = $\frac{\text{Upper limit + Lower limit}}{2}$

$a = \frac{(a - b) + \text{Upper limit}}{2}$

2a = (a - b) + Upper limit

Upper limit = 2a − (a − b)

= 2a - a + b

= a + b.

Hence, option 1 is the correct option.

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