# NCERT Solutions For Class 9 Math CAHPTER- 14 EXERCISE – 14.3

1. A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %):

 S.No. Causes Female fatality rate (%) 1. Reproductive health conditions 31.8 2. Neuropsychiatric conditions 25.4 3. Injuries 12.4 4. Cardiovascular conditions 4.3 5. Respiratory conditions 4.1 6. Other causes 22.0

(i) Represent the information given above graphically.

(ii) Which condition is the major cause of women’s ill health and death worldwide?

(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.

Solution:

(i) The information given in the question is represented below graphically.

(ii) We can observe from the graph that reproductive health conditions is the major cause of women’s ill health and death worldwide.

(iii) Two factors responsible for cause in (ii) are:

• Lack of proper care and understanding.
• Lack of medical facilities.

2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

 S.No. Section Number of girls per thousand boys 1. Scheduled Caste (SC) 940 2. Scheduled Tribe (ST) 970 3. Non SC/ST 920 4. Backward districts 950 5. Non-backward districts 920 6. Rural 930 7. Urban 910

(i) Represent the information above by a bar graph.

(ii) In the classroom discuss what conclusions can be arrived at from the graph.

Solution:

(i) The information given in the question is represented below graphically.

(ii) From the above graph, we can conclude that the maximum number of girls per thousand boys is present in the section ST. We can also observe that the backward districts and rural areas have more number of girls per thousand boys than non-backward districts and urban areas.

3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

 Political party A B C D E F Seats won 75 55 37 29 10 37

(i) Draw a bar graph to represent the polling results.

(ii) Which political party won the maximum number of seats?

Solution:

(i) The bar graph representing the polling results is given below:

(ii) From the bar graph it is clear that Party A won the maximum number of seats.

4. The length of 40 leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table:

 S.No. Length (in mm) Number of leaves 1. 118 – 126 3 2. 127 – 135 5 3. 136 – 144 9 4. 145 – 153 12 5. 154 – 162 5 6. 163 – 171 4 7. 172 – 180 2

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]

(ii) Is there any other suitable graphical representation for the same data?

(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Solution:

(i) The data given in the question is represented in discontinuous class interval. So, we have to make it in continuous class interval. The difference is 1, so taking half of 1, we subtract ½ = 0.5 from lower limit and add 0.5 to the upper limit. Then the table becomes:

 S.No. Length (in mm) Number of leaves 1. 117.5 – 126.5 3 2. 126.5 – 135.5 5 3. 135.5 – 144.5 9 4. 144.5 – 153.5 12 5. 153.5 – 162.5 5 6. 162.5 – 171.5 4 7. 171.5 – 180.5 2

(ii) Yes, the data given in the question can also be represented by frequency polygon.

1. No, we cannot conclude that the maximum number of leaves are 153 mm long because the maximum number of leaves are lying in-between the length of 144.5 – 153.5

5. The following table gives the life times of 400 neon lamps:

 Life Time (in hours) Number of lamps 300 – 400 14 400 – 500 56 500 – 600 60 600 – 700 86 700 – 800 74 800 – 900 62 900 – 1000 48

(i) Represent the given information with the help of a histogram.

(ii) How many lamps have a life time of more than 700 hours?

Solution:

(i) The histogram representation of the given data is given below:

(ii) The number of lamps having a life time of more than 700 hours = 74+62+48 = 184

6. The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

Solution:

The class-marks = (lower limit + upper limit)/2

For section A,

 Marks Class-marks Frequency 0-10 5 3 10-20 15 9 20-30 25 17 30-40 35 12 40-50 45 9

For section B,

 Marks Class-marks Frequency 0-10 5 5 10-20 15 19 20-30 25 15 30-40 35 10 40-50 45 1

Representing these data on a graph using two frequency polygon we get,

From the graph, we can conclude that the students of Section A performed better than Section B.

7The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both the teams on the same graph by frequency polygons.

[Hint: First make the class intervals continuous.]

Solution:

The data given in the question is represented in discontinuous class interval. So, we have to make it in continuous class interval. The difference is 1, so taking half of 1, we subtract ½ = 0.5 = 0.5 from lower limit and add 0.5 to the upper limit. Then the table becomes:

 Number of balls Team A Team B 0.5-6.5 2 5 6.5-12.5 1 6 12.5-18.5 8 2 18.5-24.5 9 10 24.5-30.5 4 5 30.5-36.5 5 6 36.5-42.5 6 3 42.5-48.5 10 4 48.5-54.5 6 8 54.5-60.5 2 10

The data of both the teams are represented on the graph below by frequency polygons.

8. A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

Solution:

The width of the class intervals in the given data is varying.

We know that,

The area of rectangle is proportional to the frequencies in the histogram.

Thus, the proportion of the children per year can be calculated as given in the table below.

 Age(in years) Number of children (frequency) Width of class Length of rectangle 1-2 5 1 (5/1)×1 = 5 2-3 3 1 (3/1)×1 = 3 3-5 6 2 (6/2)×1 = 3 5-7 12 2 (12/2)×1 = 6 7-10 9 3 (9/3)×1 = 3 10-15 10 5 (10/5)×1 = 2 15-17 4 2 (4/2)×1 = 2

Let x-axis = the age of children

y-axis = proportion of children per 1 year interval

9. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Solution:

(i) The width of the class intervals in the given data is varying.

We know that,

The area of rectangle is proportional to the frequencies in the histogram.

Thus, the proportion of the number of surnames per 2 letters interval can be calculated as given in the table below.

 Number of letters Number of surnames Width of class Length of rectangle 1-4 6 3 (6/3)×2 = 4 4-6 30 2 (30/2)×2 = 30 6-8 44 2 (44/2)×2 = 44 8-12 16 4 (16/4)×2 = 8 12-20 4 8 (4/8)×2 = 1

(ii) 6-8 is the class interval in which the maximum number of surnames lie.