Maharashtra Board Class 10 Maths Chapter 6 Statistics Set 6 Solutions

Question 1. Find the correct answer from the alternatives given.
(i) The persons of O - blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O - blood group?
(A) 114°
(B) 140°
(C) 104°
(D) 144°
Answer:
Measure of the central angle = \( \frac{40}{100} \times 360° = 144° \)
(D)
In simple words: The central angle for a category in a pie diagram is found by multiplying its percentage by 360°. For 40% of people with O-blood group, the angle is 144°.

🎯 Exam Tip: Remember the formula for calculating the central angle in a pie diagram: \( \text{Percentage} \times 360° \). This is a fundamental concept for data representation.

Question 1. [Continued]
(ii) Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure of Rs. 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
(A) 2,16,000
(B) 3,60,000
(C) 4,50,000
(D) 7,50,000
Answer:
Measure of the central angle = \( \frac{\text{Expenditure of cement}}{\text{Total expenditure}} \times 360° \)

\( \implies \) Total expenditure = \( \frac{45000 \times 360°}{75°} \) = Rs. 2,16,000
(A)
In simple words: If a part of the expenditure (Rs. 45,000 for cement) corresponds to a certain angle (75°) in a pie chart, you can use the ratio to find the total expenditure by scaling up to 360°.

🎯 Exam Tip: Understanding how to work backwards from a central angle to the total value is crucial. Set up a proportion: \( \frac{\text{Part Value}}{\text{Total Value}} = \frac{\text{Part Angle}}{360°} \).

Question 1. [Continued]
(iii) Cumulative frequencies in a grouped frequency table are useful to find.
(A) Mean
(B) Median
(C) Mode
(D) All of these
Answer:
(B)
In simple words: Cumulative frequencies help determine the position of data points within a distribution, which is essential for calculating the median, as it represents the middle value.

🎯 Exam Tip: Cumulative frequencies are primarily used for finding the median and quartiles, as they indicate how many data points fall below a certain value. They are not directly used for mean or mode.

Question 1. [Continued]
(iv) The formula to find mean from a grouped frequency table is \( \bar{X} = A + \frac{\Sigma f_i u_i}{\Sigma f_i} \times g \)
in the formula \( u_i = \)
(A) \( \frac{x_i + A}{g} \)
(B) \( (x_i - A) \)
(C) \( \frac{x_i - A}{g} \)
(D) \( \frac{A - x_i}{g} \)
Answer:
(C)
In simple words: In the step deviation method for finding the mean, \( u_i \) represents the deviation of the class mark \( x_i \) from the assumed mean \( A \), divided by the class width \( g \).

🎯 Exam Tip: Memorize the formulas for mean, median, and mode using different methods (direct, assumed mean, step deviation). Understanding what each variable in the formula represents is key to accurate application.

Question 1. [Continued]
(v)

The median of the distances covered per litre shown in the above data is in the group
(A) 12 - 14
(B) 14 - 16
(C) 16 - 18
(D) 18 - 20
Answer:
(C)
In simple words: To find the median group, first find the total number of data points, then locate the position of the median. The median group is the one whose cumulative frequency first exceeds or equals this median position.

🎯 Exam Tip: For grouped data, the median lies in the class interval where the cumulative frequency first exceeds \( \frac{N}{2} \), where \( N \) is the total frequency. Calculate cumulative frequencies to find the median class.

Question 1. [Continued]
(vi)

The above data is to be shown by a frequency polygon. The coordinates of the points to show number of students in the class 4 - 6 are.
(A) (4, 8)
(B) (3,5)
(C) (5,8)
(D) (8,4)
Answer:
Class mark = 5
Frequency = 8
\( \therefore \) Co-ordinates of the point = (5, 8)
(C)
In simple words: In a frequency polygon, each class is represented by a point where the x-coordinate is the class mark (midpoint of the class interval) and the y-coordinate is the frequency of that class. For the class 4-6, the class mark is 5 and the frequency is 8, so the point is (5,8).

🎯 Exam Tip: When plotting a frequency polygon, always use the class mark (midpoint of the class interval) for the x-coordinate and the frequency for the y-coordinate. Also, include imaginary classes with zero frequency at both ends to close the polygon.

Question 2. The following table shows the income of farmers in a grape season. Find the mean of their income.

Answer:
Solution:

\( \therefore \) Mean \( \bar{X} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4410}{84} = 52.5 = (52.5 \times 1000) \) = Rs. 52500
\( \therefore \) The mean of the income of the farmers is Rs. 52,500.
In simple words: The mean income is calculated by summing the products of each class mark and its frequency, then dividing by the total frequency. Here, the average income for farmers is Rs. 52,500.

🎯 Exam Tip: When calculating the mean for grouped data, ensure you correctly find the class mark (midpoint) for each interval. The direct method formula is \( \bar{X} = \frac{\Sigma f_i x_i}{\Sigma f_i} \). Pay attention to units, like "Thousand Rupees" here.

Question 3. The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.

Answer:
Solution:

\( \therefore \) Mean \( \bar{X} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{6540}{100} = 65.4 = (65.4 \times 1000) \) = Rs. 65400
\( \therefore \) The mean of the loans given by the bank is Rs. 65,400.
In simple words: We calculate the mean loan amount by first finding the midpoint (class mark) for each loan range, multiplying it by the number of farm ponds (frequency), summing these products, and then dividing by the total number of farm ponds. The average loan is Rs. 65,400.

🎯 Exam Tip: Pay close attention to the class intervals and ensure accurate calculation of class marks. Double-check your summation of \( f_i x_i \) and \( f_i \) to avoid calculation errors in the mean.

Question 4. The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.

Answer:
Solution:

\( \therefore \) Mean \( \bar{X} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{510000}{120} = 4250 \)
\( \therefore \) The mean of the weekly wages of the workers is Rs. 4250.
In simple words: To find the average weekly wage, we calculate the midpoint of each wage bracket, multiply it by the number of workers in that bracket, sum these values, and then divide by the total number of workers. The average weekly wage is Rs. 4250.

🎯 Exam Tip: Ensure that the class marks are calculated correctly for each interval. The direct method for mean is straightforward but requires careful arithmetic, especially with larger numbers.

Question 5. The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.

Answer:
Solution:

\( \therefore \) Mean \( \bar{X} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{3620}{50} = 72.4 = (72.4 \times 1000) \) = Rs. 72400
\( \therefore \) The mean of the amount of aid given to families is Rs. 72,400.
In simple words: The average aid amount is calculated by first finding the midpoint for each aid category, multiplying it by the number of families, summing these products, and then dividing by the total number of families. The mean aid is Rs. 72,400.

🎯 Exam Tip: Always make sure to consider the units of the data (e.g., "Thousand rupees") in your final answer. Ensure the arithmetic for class marks and products \( f_i x_i \) is precise.

Question 6. The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.

Answer:
Solution:

Here, total frequency = \( \Sigma f_i = N = 250 \)
\( \therefore \frac{N}{2} = \frac{250}{2} = 125 \)
Cumulative frequency which is just greater than (or equal) to 125 is 180.
\( \therefore \) The median class is 220 - 230.
Now, L = 220, f = 80, cf = 100, h = 10
\( \therefore \) Median = \( L + \left[ \frac{\frac{N}{2}-cf}{f} \right] h \)
= \( 220 + \left[ \frac{125-100}{80} \right] 10 \)
= \( 220 + \left[ \frac{25}{80} \right] 10 \)
= \( 220 + \frac{25}{8} \)
= \( 220 + 3.125 \)
= \( 223.125 \approx 223.13 \)
\( \therefore \) The median of the distances is 223.13 km (approx.).
In simple words: To find the median distance, we first identify the median class where the cumulative frequency exceeds half of the total frequency. Then, we apply the median formula using the lower limit of the median class, its frequency, the cumulative frequency of the preceding class, and the class width. The median distance is approximately 223.13 km.

🎯 Exam Tip: Accurately identifying the median class and the values for L, f, cf, and h is critical. Ensure calculations for the median formula are performed step-by-step to avoid errors, especially with fractions and decimals.

Question 7. The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.

Answer:
Solution:

Cumulative frequency which is just greater than (or equal) to 200 is 240.
\( \therefore \) The median class is 20 - 40.
Now,L = 20, f = 100,cf = 140, h = 20
\( \therefore \) Median = \( L + \left[ \frac{\frac{N}{2}-cf}{f} \right] h \)
= \( 20 + \left[ \frac{200-140}{100} \right] 20 \)
= \( 20 + \left[ \frac{60}{100} \right] 20 \)
= \( 20 + 12 \)
= \( 32 \)
\( \therefore \) The median of the prices of different articles is Rs. 32.
In simple words: To find the median price, we identify the class where the cumulative frequency exceeds half of the total frequency. Then, using the median formula with the lower limit of this class, its frequency, the cumulative frequency of the preceding class, and the class width, we calculate the median price, which is Rs. 32.

🎯 Exam Tip: Pay attention to "Less than" type cumulative frequency tables; ensure you correctly convert them to class intervals if needed, and always cross-check your median class identification and formula values.

Question 8. The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.

Answer:
Solution:

Here, the maximum frequency is 60.
The modal class is 250 - 500.
Now, L = 250, \( f_0 \) = 10, \( f_1 \) = 60, \( f_2 \) = 25, h = 250
\( \therefore \) Mode = \( L + \left[ \frac{f_1-f_0}{2f_1-f_0-f_2} \right] h \)
= \( 250 + \left[ \frac{60-10}{2(60)-10-25} \right] 250 \)
= \( 250 + \left[ \frac{50}{120-35} \right] 250 \)
= \( 250 + \left[ \frac{50}{85} \right] 250 \)
= \( 250 + \frac{2500}{17} \)
= \( 250 + 147.06 \)
= \( 397.06 \)
\( \therefore \) The mode of the demand of sweet is 397.06 grams.
In simple words: The mode, which is the most frequent demand, is calculated by identifying the class with the highest frequency (modal class). Then, we apply the mode formula using the lower limit of this class, its frequency, and the frequencies of the classes preceding and succeeding it. The modal demand for sweet is approximately 397.06 grams.

🎯 Exam Tip: The most crucial step in finding the mode is correctly identifying the modal class (the class with the highest frequency) and then assigning the correct values to L, \( f_1 \), \( f_0 \), \( f_2 \), and h. Double-check your arithmetic, especially in the denominator of the formula.

Question 9. Draw a histogram for the following frequency distribution.

Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक हिस्टोग्राम है जो बिजली के उपयोग (यूनिट में) को दर्शाता है और यह भी दिखाता है कि कितने परिवार किस रेंज में आते हैं। X-अक्ष पर 'बिजली का उपयोग (यूनिट)' है और Y-अक्ष पर 'परिवारों की संख्या' है। इसमें 50-70 यूनिट से 150 परिवार, 70-90 यूनिट से 400 परिवार, 90-110 यूनिट से 460 परिवार, 110-130 यूनिट से 540 परिवार, 130-150 यूनिट से 600 परिवार और 150-170 यूनिट से 350 परिवार दर्शाए गए हैं। पैमाने के अनुसार, X-अक्ष पर 1 सेमी = 20 यूनिट और Y-अक्ष पर 1 सेमी = 50 परिवार है।
In simple words: A histogram is a bar graph representing the frequency distribution of continuous data. Each bar represents a class interval, and its height corresponds to the frequency of that class, showing the number of families using electricity within specific unit ranges.

🎯 Exam Tip: When drawing a histogram, ensure that there are no gaps between the bars (as the data is continuous). Label both axes clearly, indicate the scale used, and ensure the heights of the bars accurately reflect the frequencies.

Question 10. In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.

Answer:
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक फ्रीक्वेंसी पॉलीगॉन है जो साड़ी बुनने में लगने वाले दिनों की संख्या और श्रमिकों की संख्या के बीच संबंध दर्शाता है। X-अक्ष पर 'दिनों की संख्या' (वर्ग चिह्न के रूप में) और Y-अक्ष पर 'श्रमिकों की संख्या' है। पॉलीगॉन को बंद करने के लिए काल्पनिक वर्गों (6-8 और 20-22) को शून्य आवृत्ति के साथ शामिल किया गया है। पैमाने के अनुसार, X-अक्ष पर 1 सेमी = 2 दिन और Y-अक्ष पर 1 सेमी = 5 श्रमिक है।
In simple words: A frequency polygon is a line graph that connects the midpoints of the top of each bar in a histogram. It displays the distribution of data by plotting class marks against their frequencies, and typically includes extra points at the ends with zero frequency to close the polygon.

🎯 Exam Tip: To draw a frequency polygon, calculate the class mark (midpoint) for each interval. Plot these class marks against their respective frequencies. Remember to include two extra class intervals (one before the first and one after the last) with zero frequency to create a closed polygon.

Question 11. The time required for students to do a science experiment and the number of students is shown in the following grouped frequency distribution table. Show the information by a histogram and also by a frequency polygon.

Answer:
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ही चार्ट पर हिस्टोग्राम और फ्रीक्वेंसी पॉलीगॉन दोनों को दर्शाता है। X-अक्ष पर 'प्रयोग के लिए आवश्यक समय (मिनटों में)' और Y-अक्ष पर 'छात्रों की संख्या' है। हिस्टोग्राम प्रत्येक समय अंतराल के लिए छात्रों की संख्या को आयताकार बारों के माध्यम से दर्शाता है, जबकि फ्रीक्वेंसी पॉलीगॉन इन बारों के शीर्ष के मध्यबिंदुओं को एक रेखा से जोड़कर आवृत्ति वितरण का एक चिकना प्रतिनिधित्व प्रदान करता है। पैमाने के अनुसार, X-अक्ष पर 1 सेमी = 2 मिनट और Y-अक्ष पर 1 सेमी = 2 छात्र है।
In simple words: This solution demonstrates both a histogram and a frequency polygon for the given data. The histogram uses bars to show the frequency within each time interval, while the frequency polygon connects the midpoints of the histogram bars to illustrate the distribution as a line graph.

🎯 Exam Tip: When drawing both a histogram and a frequency polygon on the same graph, first draw the histogram, then mark the midpoints of the tops of the bars. Connect these midpoints with line segments to form the frequency polygon. Remember to extend the polygon to the x-axis by adding imaginary classes with zero frequency.

Question 12. Draw a frequency polygon for the following grouped frequency distribution table.

Answer:
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक फ्रीक्वेंसी पॉलीगॉन है जो रक्तदाताओं की आयु (वर्षों में) और उनकी संख्या के वितरण को दर्शाता है। X-अक्ष पर 'रक्तदाताओं की आयु (वर्षों में)' (वर्ग चिह्न के रूप में) और Y-अक्ष पर 'रक्तदाताओं की संख्या' है। इस पॉलीगॉन को बंद करने के लिए 15-19 और 50-54 जैसे काल्पनिक वर्ग शून्य आवृत्ति के साथ जोड़े गए हैं। पैमाने के अनुसार, X-अक्ष पर 1 सेमी = 5 वर्ष और Y-अक्ष पर 1 सेमी = 5 दाता है।
In simple words: This frequency polygon illustrates the distribution of blood donors by age. Each point on the graph represents the class mark (midpoint of age group) and the corresponding number of blood donors. The line connects these points, creating a visual representation of how donor numbers change across age groups.

🎯 Exam Tip: When class intervals are discontinuous (like 20-24, 25-29), convert them into continuous intervals (e.g., 19.5-24.5, 24.5-29.5) before finding class marks and plotting the frequency polygon to ensure accuracy.

Question 13. The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.

Answer:
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक फ्रीक्वेंसी पॉलीगॉन है जो 150 कस्बों में औसत वर्षा (सेमी में) और कस्बों की संख्या के वितरण को दर्शाता है। X-अक्ष पर 'औसत वर्षा (सेमी में)' (वर्ग चिह्न के रूप में) और Y-अक्ष पर 'कस्बों की संख्या' है। इस पॉलीगॉन में 0-20 और 100-120 के काल्पनिक वर्ग शून्य आवृत्ति के साथ दर्शाए गए हैं ताकि पॉलीगॉन बंद हो सके। पैमाने के अनुसार, X-अक्ष पर 1 सेमी = 10 सेमी वर्षा और Y-अक्ष पर 1 सेमी = 4 कस्बे है।
In simple words: A frequency polygon visually represents the distribution of rainfall data across different towns. By plotting the midpoint of each rainfall interval against the number of towns, and connecting these points, it shows the pattern of average rainfall.

🎯 Exam Tip: When constructing a frequency polygon, accurately determine the class marks and plot them against their corresponding frequencies. To ensure the polygon is closed and properly represents the data, extend it to the x-axis by adding class marks with zero frequency at both ends of the distribution.

Question 14. Observe the given pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पाई आरेख है जो सुबह 8 बजे से 10 बजे के बीच एक कस्बे में सिग्नल पार करने वाले वाहनों के विभिन्न प्रकारों का प्रतिशत दर्शाता है। इसमें कारें 30%, टू-व्हीलर 40%, टेम्पो 12%, बसें 8% और ऑटो रिक्शा 10% हैं। यह आरेख दिखाता है कि कौन सा वाहन प्रकार कुल यातायात का कितना हिस्सा है।
(i) Find the central angle for each type of vehicle.
(ii) If the number of two-wheelers is 1200, find the number of all vehicles.
Answer:
Solution:
(i)

(ii) Central angle for two wheelers \( (\theta) = 144° \)
Measure of central angle = \( \frac{\text{Number of two wheelers}}{\text{Total number of vehicles}} \times 360° \)
\( \therefore 144° = \frac{1200}{\text{Total number of vehicles}} \times 360° \)
\( \therefore \) Total number of vehicles = \( \frac{1200 \times 360}{144} = 3000 \)
\( \therefore \) The total number of vehicles is 3000.
In simple words: To find the central angle for each vehicle type, multiply its percentage by 360°. If two-wheelers represent 40% of the total (which corresponds to 144° central angle) and their count is 1200, then the total number of vehicles can be found by setting up a proportion where 144° is to 360° as 1200 is to the total.

🎯 Exam Tip: For pie diagrams, remember that the sum of all percentages is 100% and the sum of all central angles is 360°. Use proportions to find unknown values, either actual numbers from percentages/angles or total values from a part. Always show your calculation steps clearly.

Question 14. Observe the given pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am.
(i) Find the central angle for each type of vehicle.
(ii) If the number of two-wheelers is 1200, find the number of all vehicles.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पाई आरेख है जो सुबह 8 बजे से 10 बजे के बीच एक कस्बे में सिग्नल पार करने वाले विभिन्न प्रकार के वाहनों का प्रतिशत दर्शाता है। आरेख में कार (30%), दो-पहिया वाहन (40%), टेम्पो (12%), बसें (8%), और ऑटो-रिक्शा (10%) शामिल हैं, जो कुल 100% बनाते हैं। प्रत्येक खंड एक विशेष प्रकार के वाहन को प्रदर्शित करता है, जिससे छात्रों को वाहनों के वितरण की कल्पना करने में मदद मिलती है।
Answer: Solution: Measure of central angle \((\theta) = \frac{\text{Number of scores in the components}}{\text{Total number of scores}} \times 360^\circ\)
(i)

(ii) Central angle for two wheelers \((\theta) = 144^\circ\)
Measure of central angle \( = \frac{\text{Number of two wheelers}}{\text{Total number of vehicles}} \times 360^\circ\)
\( \implies 144^\circ = \frac{1200}{\text{Total number of vehicles}} \times 360^\circ\)
\( \implies \text{Total number of vehicles} = \frac{1200 \times 360}{144} = 3000\)
The total number of vehicles is 3000.
In simple words: We calculated the central angle for each vehicle type using their percentage contribution to the total. Then, using the central angle for two-wheelers and their given count, we found the total number of vehicles by setting up a proportion.

🎯 Exam Tip: Remember to express percentages as fractions of 100 when calculating central angles. Ensure all central angles sum up to \(360^\circ\) for verification.

Question 15. The following table shows causes of noise pollution. Show it by a pie diagram.

Answer: Solution: Measure of central angle \((\theta) = \frac{\text{Percentage of components}}{100} \times 360^\circ\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पाई आरेख है जो विभिन्न स्रोतों से ध्वनि प्रदूषण के कारणों को दर्शाता है। सबसे बड़ा हिस्सा यातायात (Traffic) का है जो कुल प्रदूषण का 50% या 180° केंद्रीय कोण बनाता है। उद्योग (Industry) 20% (72°), ट्रेनें (Trains) 11% (40°), निर्माण (Construction) 10% (36°) और विमान उड़ानें (Aircraft take offs) 9% (32°) का योगदान करती हैं। यह आरेख ध्वनि प्रदूषण के विभिन्न कारणों की सापेक्षिक गंभीरता को स्पष्ट रूप से दर्शाता है।
In simple words: To represent noise pollution causes by a pie diagram, we first calculated the central angle for each category based on its percentage contribution. Then, these angles were used to divide a circle into sectors, showing the proportion of each cause.

🎯 Exam Tip: When drawing a pie diagram, always calculate the central angle for each category. Use a protractor for accuracy and label each sector clearly with its category and percentage or angle for easy understanding.

Question 16. A survey of students was made to know which game they like. The data obtained in the survey is presented in the given pie diagram. If the total number of students are 1000,
(i) how many students like cricket?
(ii) how many students like football?
(iii) how many students prefer other games?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पाई आरेख है जो 1000 छात्रों के सर्वेक्षण के आधार पर विभिन्न खेलों के प्रति उनकी पसंद को दर्शाता है। आरेख में क्रिकेट (81°), फुटबॉल (63°), हॉकी (54°), कबड्डी (45°), खो-खो (45°) और अन्य खेल (72°) के लिए केंद्रीय कोण दिए गए हैं। यह आरेख स्पष्ट रूप से दिखाता है कि कौन सा खेल कितने छात्रों द्वारा पसंद किया जाता है, जिससे विभिन्न खेलों की लोकप्रियता का दृश्य प्रतिनिधित्व मिलता है।
Answer: Solution: Measure of central angle \((\theta) = \frac{\text{Number of scores in the components}}{\text{Total number of scores}} \times 360^\circ\)
(i) Central angle for cricket \((\theta) = 81^\circ\)
\( \implies 81^\circ = \frac{\text{Students who like cricket}}{1000} \times 360^\circ\)
\( \implies \text{Students who like cricket} = \frac{81 \times 1000}{360} = 225\)
225 students like cricket.
(ii) Central angle for football \((\theta) = 63^\circ\)
\( \implies 63^\circ = \frac{\text{Students who like football}}{1000} \times 360^\circ\)
\( \implies \text{Students who like football} = \frac{63 \times 1000}{360} = 175\)
175 students like football.
(iii) Central angle for other games \((\theta) = 72^\circ\)
\( \implies 72^\circ = \frac{\text{Students who like other games}}{1000} \times 360^\circ\)
\( \implies \text{Students who like other games} = \frac{72 \times 1000}{360} = 200\)
200 students like other games.
In simple words: We used the given central angles for each game and the total number of students to calculate the exact count of students who prefer cricket, football, and other games by rearranging the central angle formula.

🎯 Exam Tip: When interpreting pie diagrams, remember that the central angle is directly proportional to the quantity it represents. Use the formula: quantity = (central angle / \(360^\circ\)) * total quantity, to find specific values.

Question 17. Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of hemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.

Answer: Solution: Total number of women = 180 Measure of central angle \((\theta) = \frac{\text{Number of scores in the components}}{\text{Total number of scores}} \times 360^\circ\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पाई आरेख है जो 180 महिलाओं के चिकित्सा जांच परिणामों को दर्शाता है। आरेख में हीमोग्लोबिन की कमी वाली 50 महिलाओं (100°), मोतियाबिंद वाली 10 महिलाओं (20°), श्वसन संबंधी विकार वाली 25 महिलाओं (50°), और स्वस्थ 95 महिलाओं (190°) का अनुपात दिखाया गया है। यह आरेख गांव की महिलाओं के स्वास्थ्य स्थिति का एक स्पष्ट और संक्षिप्त दृश्य प्रस्तुत करता है।
In simple words: We first calculated the number of healthy women by subtracting the counts of those with specific ailments from the total. Then, for each health category, we determined the central angle by dividing the number of women in that category by the total number of women and multiplying by \(360^\circ\). These angles were then used to draw the pie diagram.

🎯 Exam Tip: Always ensure the sum of all categories matches the total given. For pie diagrams, the sum of all calculated central angles must be \(360^\circ\). Clearly label each sector with its category and value/angle.

Question 18. On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.

Answer: Solution: Total number of trees planted = 120 Measure of central angle \((\theta) = \frac{\text{Number of scores in the components}}{\text{Total number of scores}} \times 360^\circ\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पाई आरेख है जो पर्यावरण दिवस पर स्कूल में छात्रों द्वारा लगाए गए 120 पेड़ों के प्रकारों को दर्शाता है। आरेख में विभिन्न प्रकार के पेड़ - करंज (Karanj - 60°), बेहड़ा (Behada - 84°), अर्जुन (Arjun - 72°), बकुल (Bakul - 66°), और कडुनिंब (Kadunimb - 78°) - उनके केंद्रीय कोणों के साथ प्रदर्शित किए गए हैं। यह आरेख छात्रों के वृक्षारोपण परियोजना में लगाए गए पेड़ों की विविधता और संख्या का एक स्पष्ट दृश्य प्रतिनिधित्व प्रदान करता है।
In simple words: To create the pie diagram, we first calculated the central angle for each type of tree planted by dividing the number of trees of that type by the total number of trees and then multiplying by \(360^\circ\). These calculated angles were then used to draw the sectors of the pie chart.

🎯 Exam Tip: When presenting data with a pie diagram, ensure the categories are mutually exclusive and collectively exhaustive. Always sum up the frequencies and central angles to confirm they match the total and \(360^\circ\) respectively.