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# Class 10 SELINA Solutions Maths Chapter 23 - Graphical Representation (Histograms and Ogives)

## Graphical Representation (Histograms and Ogives) Exercise Ex. 23

### Solution 2

(i)

 Class Interval Frequency 0-10 12 10-20 20 20-30 26 30-40 18 40-50 10 50-60 06

Steps of construction:

(a) Taking suitable scales, mark class intervals on x-axis and frequency on y-axis.

(b)Construct rectangles with class intervals as bases and corresponding frequencies as heights.

(ii)

 Class Interval Frequency 10-16 15 16-22 23 22-28 30 28-34 20 34-40 16

Steps of construction:

(a) Taking suitable scales, mark class intervals on x-axis and frequency on y-axis.

(b) Construct rectangles with class intervals as bases and corresponding frequencies as heights.

(iii)

Since the difference between each class marks is 8, therefore, subtract 8/2 = 4 from each class mark to get the lower limit of the corresponding class interval and add 5 to each class mark to get the upper limit.

 Class Marks Class Intervals Frequency 16 12-20 08 24 20-28 12 32 28-36 15 40 36-44 18 48 44-52 25 56 52-60 19 64 60-68 10

Steps of construction:

(a) Convert the class marks into class intervals.

(b) Taking suitable scales, mark class intervals on x-axis and frequency on y-axis.

(c) Construct rectangles with class intervals as bases and corresponding frequencies as heights.

### Solution 3

(i)

The cumulative frequency table for the given distribution is as follows:

 Class Interval Frequency Cumulative frequency 10-15 10 10 15-20 15 25 20-25 17 42 25-30 12 54 30-35 10 64 35-40 8 72

Taking upper limits along X-axis and corresponding cumulative frequencies along Y-axis, mark the points (15,10), (20,25), (25,42), (30,54), (35, 64) and (40, 72).

Join the points marked by a free hand curve.

### Solution 4

(i)

 Marks Obtained No. of students (c.f.) less than 10 8 less than 20 25 less than 30 38 less than 40 50 less than 50 67

Steps Of construction:

(a) Plot the points (10,8), (20, 25), (30, 38), (40, 50) and (50, 67) on the graph.

(b) Join them with free hand to obtain an ogive.

(ii)

 Age in years (less than) Cumulative Frequency 10 0 20 17 30 32 40 37 50 53 60 58 70 65

Steps Of construction:

(a) Plot the points (10, 0), (20, 17), (30, 32), (40, 37), (50, 53), (60, 58) and (70, 65) on the graph.

(b) Join them with free hand to obtain an ogive.

### Solution 5

(a)

 Class Interval Frequency c.f. 8-12 9 9 12-16 16 25 16-20 22 47 20-24 18 65 24-28 12 77 28-32 4 81

(b) Now plot the points (12, 9), (16, 25), (20, 47), (24, 65), (28, 77), (32, 81) and join them to obtain an ogive.

### Solution 6

(i)

 Wages No. Of workers c.f. 6500-7000 10 10 7000-7500 18 28 7500-8000 22 50 8000-8500 25 75 8500-9000 17 92 9000-9500 10 102 9500-10000 8 110

Total = 110

(ii)

Now plot the points (7000,10), (7500,28), (8000,50), (8500,75), (9000,92), (9500,102) and (10000,110) and join them to obtain an ogive.

### Solution 7

(i)

 Height (in cm) No. Of workers c.f. 150-155 6 6 155-160 12 18 160-165 18 36 165-170 20 56 170-175 13 69 175-180 8 77 180-185 6 83

(ii)

We plot the points (155, 6), (160, 18), (165, 36), (170, 56), (175, 69), (180, 77) and (185, 83) on the graph and join them in free hand to obtain an ogive.

### Solution 8

(i)

 Marks (less than) Cumulative frequency Frequency 0-10 7 7 10-20 28 28-7=21 20-30 54 54-28=26 30-40 71 71-54=17 40-50 84 84-71=13 50-60 105 105-84=21 60-70 147 147-105=42 70-80 180 180-147=33 80-90 196 196-180=16 90-100 200 200-196=4 Total 200

(ii)

 Marks (more than) Cumulative frequency Frequency 0-10 100 13 10-20 87 22 20-30 65 10 30-40 55 13 40-50 42 6 50-60 36 5 60-70 31 10 70-80 21 3 80-90 18 11 90-100 7 7 Total 100

### Solution 1(a)

Correct option: (i) a = 16, b = 52 and c = 8

24 + a = 40 a = 16

b = 40 + 12 = 52

b + c = 60

52 + c = 60 c = 8

### Solution 1(b)

Correct option: (ii) 120

Frequency of class 30-50 = 50

Frequency of class 50-70 = 30

Frequency of class 70-90 = 40

Then, cumulative frequency of the class 70-90 = 50 + 30 + 40 = 120

### Solution 1(c)

Correct option: (i)

The adjustment factor is half the difference between the upper limit of the class and the lower limit of the next class.

Note: The adjustment factor is used to convert inclusive classes into exclusive form.

### Solution 1(d)

Correct option: (iii) lower limit of 1st class

The cumulative curve for a frequency distribution starts from the lower limit of 1st class.

### Solution 1(e)

Correct option: (iv) upper limit of last class

The cumulative curve for a frequency distribution terminates at the upper limit of last class.