Exercise-12.1
Maths - Exercise
Q1. 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.N. | 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 |
(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:
- Graph as Shown
- Which condition is the major cause of women’s ill health and death worldwide? From the table and graph, Reproductive health conditions stand out as the major cause, accounting for the highest female fatality rate of 31.8% among women aged 15 to 44 worldwide. This means that issues related to reproductive health contribute the most to illness and death in this age group.
-
Any two factors which play a major role in reproductive health conditions being the major cause
With guidance from your teacher, you may identify these two important factors:
- Limited Access to Healthcare and Education: In many parts of the world, women do not have adequate access to reproductive health services. This includes lack of proper antenatal care, safe delivery facilities, family planning, and information about reproductive rights, meaning preventable complications can turn fatal.
- Social and Economic Barriers: Early marriages, poor nutrition, lack of empowerment, and poverty can lead to increased vulnerability to health problems like maternal mortality, childbirth complications, and infections. Societal norms and taboos often prevent women from seeking timely help for reproductive issues.
Q2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.
| Section | Number of girls per thousand boys |
|---|---|
| Scheduled Caste (SC) | 940 |
| Scheduled Tribe (ST) | 970 |
| Non SC/ST | 920 |
| Backward districts | 950 |
| Non-backward districts | 920 |
| Rural | 930 |
| Urban | 910 |
- Represent the information above by a bar graph.
- In the classroom discuss what conclusions can be arrived at from the graph.
Solution:
- Conclusions and observations from the bar graph (for classroom discussion) Looking at the graph, several key insights emerge:
- Scheduled Tribe (ST) communities have the highest ratio of girls to boys (970 per 1000), which is encouraging and above all other sections.
- Urban areas report the lowest ratio (910 per 1000), indicating that cities tend to have fewer girls per thousand boys compared to rural areas (930 per 1000).
- The ratio is consistently lower (920-950) for Non SC/ST groups and Non-backward districts, suggesting a trend that might be linked to socioeconomic factors or access to resources.
- Backward districts (950) and Scheduled Castes (940) show better ratios than their Non-backward and Non SC/ST counterparts, hinting that certain government interventions, support, or cultural practices could be positively impacting the girl-to-boy ratio there.
- Overall, while most sections hover close to or below 950 girls per thousand boys, none reach perfect parity, and urban areas lag behind, suggesting the need for more awareness and targeted measures.
Summary for Discussion:
- Urbanization may not always mean better gender balance.
- Tribal and backward districts might benefit from supportive policies or local customs that favour girls.
- The data is useful for identifying which communities need more attention regarding gender equality.
Q3. 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 |
- Draw a bar graph to represent the polling results.
- Which political party won the maximum number of seats?
Solution:
- Bar graph representing the polling results at given
- Political party 'A' won the maximum number of 75 seats
Q4. The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:
| Length (in mm) | Number of leaves |
|---|---|
| 118 - 126 | 3 |
| 127 - 135 | 5 |
| 136 - 144 | 9 |
| 145 - 153 | 12 |
| 154 - 162 | 5 |
| 163 - 171 | 4 |
| 172 - 180 | 2 |
- Draw a histogram to represent the given data.
- Is there any other suitable graphical representation for the same data?
- Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
Solution:
- To draw histogram, First make the class intervals
continuous
Class Interval = 9Length (in mm) Number of leaves 117.5 - 126.5 3 126.5 - 135.5 5 135.5 - 144.5 9 144.5 - 153.5 12 153.5 - 162.5 5 162.5 - 171.5 4 171.5 - 180.5 2
Based on these data Histogram is drawn. -
Is there any other suitable graphical representation for the same data?
Yes, apart from the histogram, other suitable graphical representations are:
-
Frequency Polygon:
Plot points at the mid-value of each interval for corresponding frequency; connect these points with straight lines. This gives a visual sense of how frequencies rise and fall. -
Bar Graph:
If you treat the intervals as discrete categories (not ideal for continuous data), you can use a bar graph, but a histogram remains superior here.
-
Frequency Polygon:
- Is it correct to conclude that the maximum number of leaves is 153 mm long? Why?
No, it is not correct to conclude that the maximum number of leaves are exactly 153 mm long.
Reason:
The class interval 145-153 mm contains the highest frequency (12 leaves), meaning 12 leaves fall somewhere between 145 mm and 153 mm in length. We do not know the exact lengths of these leaves they could be anywhere within this group, not necessarily all at the upper limit (153 mm). The data tells us that this interval, not a single value, is the most common.
Q5. The following table gives the lifetimes 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 |
- Represent the given information with the help of a histogram.
- How many lamps have a lifetime of more than 700 hours?
Solution:
184 lamps have a lifetime of more than 700 hours
Q6. The following table gives the distribution of students of two sections according to the marks obtained by them:
| Section A | Section B | ||
|---|---|---|---|
| Marks | Frequency | Marks | Frequency |
| 0-10 | 3 | 0-10 | 5 |
| 10-20 | 9 | 10-20 | 19 |
| 20-30 | 17 | 20-30 | 15 |
| 30-40 | 12 | 30-40 | 10 |
| 40-50 | 9 | 40-50 | 1 |
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.
Graph representing Marks of Students
Q7. The runs scored by two teams, A and B on the first 60 balls in a cricket match are given below
| Number of balls | Team A | Team B |
|---|---|---|
| 1-6 | 2 | 5 |
| 7-12 | 1 | 6 |
| 13-18 | 8 | 2 |
| 19-24 | 9 | 10 |
| 25-30 | 4 | 5 |
| 31-36 | 5 | 6 |
| 37-42 | 6 | 3 |
| 43-48 | 10 | 4 |
| 49-54 | 6 | 8 |
| 55-60 | 2 | 10 |
Solution:
Make the Class Interval continuous| 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 |
Q8. A random survey of the number of children of various age groups playing in a park was found as follows:
| Age (in years) | Number of children |
|---|---|
| 1-2 | 5 |
| 2-3 | 3 |
| 3-5 | 6 |
| 5-7 | 12 |
| 7-10 | 9 |
| 10-15 | 10 |
| 15-17 | 4 |
Solution:
Here is this table, we can see that the class interval is not regular. So we will scale down the height of rectangle so that the area of the rectangle correctly shows the frequency.The formula used to find the adjusted height for each interval \[\text{Height}=\left(\frac{f}{\text{class width}}\right)\times \text{min. class width}\] Read more...
| Age Interval | \(f\) | class Width | Height |
|---|---|---|---|
| 1-2 | 5 | 1 | \(\frac{5}{1}\times{1}=5\) |
| 2-3 | 3 | 1 | \(\frac{3}{1}\times{1}=3\) |
| 3-5 | 6 | 2 | \(\frac{6}{2}\times{1}=3\) |
| 5-7 | 12 | 2 | \(\frac{12}{2}\times{1}=6\) |
| 7-10 | 12 | 3 | \(\frac{12}{3}\times{1}=4\) |
| 10-15 | 9 | 5 | \(\frac{9}{5}\times{1}=1.8\) |
| 15-17 | 4 | 2 | \(\frac{4}{2}\times{1}=2\) |
Q9. 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:
| Number of letters | Number of surnames |
|---|---|
| 1-4 | 6 |
| 4-6 | 30 |
| 6-8 | 44 |
| 8-12 | 16 |
| 12-20 | 4 |
- Draw a histogram to depict the given information.
- Write the class interval in which the maximum number of surnames lie
Solution:
Here is this table, we can see that the class interval is not regular. So we will scale down the height of rectangle so that the area of the rectangle correctly shows the frequency.The formula used to find the adjusted height for each interval \[\scriptsize\text{Height}=\left(\frac{f}{\text{class width}}\right)\times \text{min. class width}\] Read more...
| Number of letters | \(f\) | class Width | Height |
|---|---|---|---|
| 1-4 | 6 | 3 | \(\frac{6}{3}\times{2}=4\) |
| 4-6 | 30 | 2 | \(\frac{30}{2}\times{2}=30\) |
| 6-8 | 44 | 2 | \(\frac{44}{2}\times{2}=44\) |
| 8-12 | 16 | 4 | \(\frac{16}{4}\times{2}=8\) |
| 12-20 | 4 | 8 | \(\frac{4}{8}\times{2}=1\) |
