Maharashtra Board Class 11 Maths Part 2 Chapter 1 Partition Values 1.3 Solutions

Question 1. The following table gives the frequency distribution of marks of 100 students in an examination.

Determine D6, Q1, and P85 graphically.
Answer: Solution: To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (20, 9), (25, 21), (30, 44), (35, 75), (40, 85), (45, 93), (50, 100).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक 'कम से कम' ओगीव वक्र को दर्शाता है जो परीक्षा में प्राप्त अंकों के वितरण को प्रदर्शित करता है। इसमें X-अक्ष पर अंकों को (1 सेमी = 10 अंक) और Y-अक्ष पर छात्रों की संचयी आवृत्ति (1 सेमी = 10 छात्र) को दर्शाया गया है। वक्र पर अंकित बिंदु (20,9) से (50,100) तक बढ़ते हुए संचयी आवृत्ति को दर्शाते हैं। Here, N = 100 For D6, \( \frac{6 N}{10} = \frac{6 \times 100}{10} = 60 \)
For Q1, \( \frac{N}{4} = \frac{100}{4} = 25 \)
For P85, \( \frac{85 N}{100} = \frac{85 \times 100}{100} = 85 \)
Therefore, We take the points having Y co-ordinates 60, 25 and 85 on Y-axis. From these points, we draw lines parallel to X-axis. From the points where these lines intersect the curve, we draw perpendiculars on X-axis. X co-ordinates of these points give the values of D6, Q1 and P85.
Therefore, D6 = 32.5, Q1 = 26, P85 = 40
In simple words: This solution involves constructing a less than cumulative frequency table from the given data. Then, a less than ogive curve is plotted using these cumulative frequencies. Finally, D6, Q1, and P85 are determined graphically by finding the corresponding X-axis values for calculated Y-axis positions (60, 25, and 85).

🎯 Exam Tip: When drawing ogive curves, ensure the axes are clearly labeled with appropriate scales. Accuracy in plotting points and drawing perpendiculars is crucial for obtaining correct graphical values for deciles, quartiles, and percentiles.

Question 2. The following table gives the distribution of daily wages of 500 families in a certain city.

Draw a 'less than' ogive for the above data. Determine the median income and obtain the limits of income of central 50% of the families.
Answer: Solution: To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (100, 50), (200, 200), (300, 380), (400, 430), (500, 470), (600, 490) and (700, 500).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ दैनिक मजदूरी के वितरण के लिए एक 'कम से कम' ओगीव वक्र को दर्शाता है। X-अक्ष पर दैनिक मजदूरी (Rs. में, 1 सेमी = 100 दैनिक मजदूरी) और Y-अक्ष पर परिवारों की संचयी संख्या (2 सेमी = 100 परिवार) प्रदर्शित की गई है। वक्र पर अंकित बिंदु (100, 50) से (700, 500) तक बढ़ते हुए संचयी आवृत्ति को दर्शाते हैं, जिससे विभिन्न आय स्तरों पर परिवारों का वितरण स्पष्ट होता है। Here, N = 500 For Q1, \( \frac{N}{4} = \frac{500}{4} = 125 \)
For Q2, \( \frac{N}{2} = \frac{500}{2} = 250 \)
For Q3, \( \frac{3 N}{4} = \frac{3 \times 500}{4} = 375 \)
Therefore, We take the points having Y co-ordinates 125, 250 and 375 on Y-axis. From these points we draw lines parallel to X-axis. From the points where these lines intersect the curve, we draw perpendiculars on X-axis. X-Co-ordinates of these points give the values of Q1, Q2 and Q3.
Therefore, Q1 ~ 150, Q2 ~ 228, Q3 ~ 297
Therefore, Median = 228 50% families lie between Q1 and Q3
Therefore, Limits of income of central 50% families are from Rs. 150 to Rs. 297
In simple words: First, a less than cumulative frequency table is created. Then, a less than ogive curve is plotted using the upper class boundaries and their respective cumulative frequencies. The median is found by identifying the X-value corresponding to N/2 on the Y-axis. The limits of the central 50% of families are determined by finding the X-values corresponding to Q1 (N/4) and Q3 (3N/4).

🎯 Exam Tip: For continuous data, ensure class intervals are correctly formed. When determining quartiles or median from an ogive, precisely locate the N/4, N/2, and 3N/4 values on the Y-axis and draw parallel lines to the X-axis to intersect the ogive, then drop perpendiculars to the X-axis for the corresponding values.

Question 3. From the following distribution, determine the median graphically.

Answer: Solution: To draw an ogive curve, we construct the less than and more than cumulative frequency table as given below:

The points to be plotted for less than ogive are (400, 50), (500, 121), (600, 310), (700, 415), (800, 475), (900, 513) and (1000, 520) and that for more than ogive are (300, 520), (400, 470), (500, 399), (600, 210), (700, 105), (800, 45), (900, 7).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ दैनिक मजदूरी के लिए 'कम से कम' और 'अधिक से अधिक' दोनों ओगीव वक्रों को दर्शाता है। X-अक्ष पर दैनिक मजदूरी (Rs. में, 1 सेमी = 100 रुपये) और Y-अक्ष पर कर्मचारियों की संचयी संख्या (2 सेमी = 100 कर्मचारी) प्रदर्शित की गई है। दोनों वक्र एक बिंदु पर प्रतिच्छेद करते हैं, जो वितरण के माध्यिका मान को ग्राफिक रूप से निर्धारित करने में मदद करता है। From the point of intersection of two ogives, we draw a perpendicular on X-axis. The point where it meets the X-axis gives the value of the median.
Therefore, Median ~ 574
In simple words: To find the median graphically, both a less than and a more than cumulative frequency ogive curve are plotted on the same graph. The X-coordinate of the point where these two ogives intersect represents the median value of the distribution.

🎯 Exam Tip: When calculating the median using two ogives, ensure you accurately plot both curves. The intersection point is key; drawing a perpendicular from this point to the X-axis will give the median, so precision in plotting is vital for correct results.

Question 4. The following frequency distribution shows the profit (in Rs.) of shops in a particular area of the city.

Find graphically
(i) the Unfits of middle 40% shops.
(ii) the number of shops having a profit of fewer than Rs. 35,000 rupees.
Answer: Solution: To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (10, 12), (20, 30), (30, 57), (40, 77), (50, 94), (60, 100).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ प्रति दुकान लाभ के वितरण के लिए एक 'कम से कम' ओगीव वक्र को दर्शाता है। X-अक्ष पर प्रति दुकान लाभ ('000 Rs. में, 1 सेमी = 10 ('000) रुपये) और Y-अक्ष पर दुकानों की संचयी संख्या (1 सेमी = 10 दुकानें) प्रदर्शित की गई है। वक्र पर अंकित बिंदु (10,12) से (60,100) तक बढ़ते हुए संचयी आवृत्ति को दर्शाते हैं। The Middle 40% value lies in between P30 and P70. N = 100 For P30 = \( \frac{30 N}{100} = \frac{30 \times 100}{100} = 30 \)
For P70 = \( \frac{70 N}{100} = \frac{70 \times 100}{100} = 70 \)
Therefore, We take the points having Y co-ordinates 30 and 70 on Y-axis. From these points we draw lines parallel to X-axis. From the points where these lines intersect the curve, we draw perpendiculars on X-axis. X-Co-ordinates of these points give the values of P30 and P70.
Therefore, P30 ~ 20, P70 ~ 36 Limits of middle 40% shops lie between Rs. 20,000 to Rs. 36,000 To find the number of shops having a profit of less than Rs. 35,000, we take the value 35 on the X-axis. From this point, we draw a line parallel to Y-axis, and from the point where it intersects the less than ogive we draw a perpendicular on Y-axis. It intersects the Y-axis at approximately 67.
Therefore, No. of shops having profit less than Rs. 35,000 is 67.
In simple words: To find the middle 40% of shops, P30 and P70 are calculated and identified on the Y-axis of the less than ogive. Perpendiculars are dropped to the X-axis to find the corresponding profit limits. To find the number of shops with profit less than Rs. 35,000, Rs. 35,000 is located on the X-axis, a vertical line is drawn to the ogive, and then a horizontal line to the Y-axis to read the cumulative frequency.

🎯 Exam Tip: When dealing with percentiles, correctly calculate the position (e.g., 30N/100 for P30). Remember to convert profit figures from '000 Rs. to full values when stating final answers (e.g., 20 becomes Rs. 20,000). Visual accuracy in graph plotting directly impacts the correctness of derived values.

Question 5. The following is the frequency distribution of overtime (per week) performed by various workers from a certain company. Determine the values of D2, Q2, and P61 graphically.

Answer: Solution: To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (8, 4), (12, 12), (16, 28), (20, 46), (24, 66) and (28, 80) Here, N = 80
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ विभिन्न श्रमिकों द्वारा किए गए ओवरटाइम के वितरण के लिए एक 'कम से कम' ओगीव वक्र को दर्शाता है। X-अक्ष पर ओवरटाइम (घंटों में, 1 सेमी = 4 घंटे) और Y-अक्ष पर श्रमिकों की संचयी संख्या (1 सेमी = 10 श्रमिक) प्रदर्शित की गई है। वक्र पर अंकित बिंदु (8,4) से (28,80) तक बढ़ते हुए संचयी आवृत्ति को दर्शाते हैं। For D2, we have to consider \( \frac{2 N}{10} = \frac{2 \times 80}{10} = 16 \)
For Q2, we have to consider \( \frac{N}{2} = \frac{80}{2} = 40 \)
and for P61, we have to consider \( \frac{61 N}{100} = \frac{61 \times 80}{100} = 48.8 \)
Therefore, We consider the values 16, 40 and 48.8 on the Y-axis. From these points, we draw the lines which are parallel to the X-axis. From the points where they intersect the less than ogive, we draw perpendiculars to X-axis. The values at the foot of perpendiculars represent the values of D2, Q2, and P61 respectively.
Therefore, D2 ~ 13, Q2 ~ 19, P61 ~ 20.5
In simple words: This solution requires constructing a less than cumulative frequency table from the given overtime data. Then, an ogive curve is plotted. Decile 2 (D2), Quartile 2 (Q2, which is also the median), and Percentile 61 (P61) are determined graphically by finding their respective positions on the Y-axis and then reading the corresponding values on the X-axis from the ogive.

🎯 Exam Tip: When calculating D-values, Q-values, and P-values, correctly apply the formulas for their positions (e.g., 2N/10, N/2, 61N/100). Drawing precise horizontal lines from the Y-axis to the ogive and then vertical lines to the X-axis is critical for accurate graphical determination of these measures.

Question 6. Draw ogive for the following data and hence find the values of D1, Q1, and P40.

Answer: Solution: N = 100 To draw the less than ogive we have to plot the points (10, 4), (20, 6), (30, 24), (40, 46), (50, 67), (60, 86), (70, 96), (80, 99), (90, 100).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ छात्रों द्वारा प्राप्त अंकों के वितरण के लिए एक 'कम से कम' ओगीव वक्र को दर्शाता है। X-अक्ष पर अंकों को (1 सेमी = 10 अंक) और Y-अक्ष पर छात्रों की संचयी संख्या (1 सेमी = 10 छात्र) प्रदर्शित की गई है। वक्र पर अंकित बिंदु (10,4) से (90,100) तक बढ़ते हुए संचयी आवृत्ति को दर्शाते हैं, जो अंकों के वितरण को ग्राफिक रूप से प्रस्तुत करता है। For D1, we have to consider \( \frac{N}{10} = \frac{100}{10} = 10 \)
For Q1, we have to consider \( \frac{N}{4} = \frac{100}{4} = 25 \)
For P40, we have to consider \( \frac{40 N}{100} = \frac{40 \times 100}{100} = 40 \)
Therefore, We consider the values 10, 25 and 40 on the Y-axis. From these points we draw lines parallel to X-axis. From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis. The values at the foot of perpendicular represent the values of D1, Q1 and P40 respectively.
Therefore, D1 ~ 22, Q1 ~ 30.5, P40 ~ 37
In simple words: A less than ogive curve is drawn by plotting the upper limits of marks against their cumulative frequencies. Decile 1 (D1), Quartile 1 (Q1), and Percentile 40 (P40) are then found graphically by identifying their positions on the Y-axis, drawing horizontal lines to the ogive, and then vertical lines down to the X-axis to read the corresponding mark values.

🎯 Exam Tip: When the data is already in 'less than' cumulative frequency form, directly plot the upper class boundaries against the given frequencies. Ensure precise calculation of the positions for D1, Q1, and P40 on the Y-axis for accurate graphical derivation.

Question 7. The following table shows the age distribution of heads of the families in a certain country. Determine the third, fifth, and eighth decile of the distribution graphically.

Answer: Solution: To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (35, 46), (45, 131), (55, 195), (65, 270), (75, 360), (85, 400).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ परिवारों के प्रमुखों की आयु वितरण के लिए एक 'कम से कम' ओगीव वक्र को दर्शाता है। X-अक्ष पर आयु (वर्षों में, 1 सेमी = 10 वर्ष) और Y-अक्ष पर परिवारों की संचयी संख्या (मिलियन में, 1 सेमी = 50 परिवार) प्रदर्शित की गई है। वक्र पर अंकित बिंदु (35,46) से (85,400) तक बढ़ते हुए संचयी आवृत्ति को दर्शाते हैं। N = 400 For D3, we have to consider \( \frac{3 N}{10} = \frac{3 \times 400}{10} = 120 \)
For D5, we have to consider \( \frac{5 N}{10} = \frac{5 \times 400}{10} = 200 \)
For D8, we have to consider \( \frac{8 N}{10} = \frac{8 \times 400}{10} = 320 \)
Therefore, We consider the values 120, 200 and 320 on Y-axis. From these points we draw the lines parallel to X-axis. From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis. The foot of perpendicular represent the values of D3, D5 and D8.
Therefore, D3 ~ 44, D5 ~ 55.5 and D8 ~ 70
In simple words: A less than cumulative frequency table is created from the age distribution data. A less than ogive curve is then plotted. Decile 3 (D3), Decile 5 (D5), and Decile 8 (D8) are determined graphically by locating their respective positions on the Y-axis, drawing horizontal lines to the ogive, and then dropping perpendiculars to the X-axis to find the corresponding age values.

🎯 Exam Tip: When dealing with age distribution or any continuous data, ensure the cumulative frequency table is accurately constructed, especially for "Under" or "Above" class types. Calculating the correct positions (3N/10, 5N/10, 8N/10) on the Y-axis is vital for graphically determining deciles.

Question 8. The following table gives the distribution of females in an Indian village. Determine the median age graphically.

Answer: Solution: To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (10, 175), (20, 275), (30, 343), (40, 391), (50, 416), (60, 466), (70, 489), (80, 497), (90, 499), (100, 500).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक भारतीय गाँव में महिलाओं के आयु वितरण के लिए 'कम से कम' ओगीव वक्र को दर्शाता है। X-अक्ष पर आयु समूह (1 सेमी = 10 इकाइयाँ) और Y-अक्ष पर महिलाओं की संचयी संख्या (हजारों में, 1 सेमी = 100 महिलाएँ) प्रदर्शित की गई है। वक्र पर अंकित बिंदु (10,175) से (100,500) तक बढ़ते हुए संचयी आवृत्ति को दर्शाते हैं। N = 500 For median we have to consider \( \frac{N}{2} = \frac{500}{2} = 250 \)
Therefore, We consider the value 250 on Y-axis. From this point, we draw a line parallel to X-axis. From the point it intersects the less than ogive, we draw a perpendicular to X-axis. The foot perpendicular represents the value of the median.
Therefore, Median ~ 17.5
In simple words: To find the median age graphically, a less than cumulative frequency table is constructed from the given age distribution of females. An ogive curve is then plotted. The median is determined by locating the N/2 value on the Y-axis, drawing a horizontal line to the ogive, and then a vertical line to the X-axis to read the corresponding age.

🎯 Exam Tip: When determining the median from an ogive, ensure the N/2 position on the cumulative frequency (Y) axis is accurately identified. The precision of the horizontal and vertical lines drawn from this point to the ogive and then to the X-axis will directly affect the accuracy of the graphical median.

.. We consider the value 250 on Y-axis. From this point, we draw a line parallel to X-axis. From the point it intersects the less than ogive, we draw a perpendicular to X-axis. The foot perpendicular represents the value of the median.
.: Median ~ 17.5
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक 'कम संचयी बारंबारता वक्र' या ओजाइव वक्र को दर्शाता है। X-अक्ष 'आयु समूह' (Age group) को दर्शाता है, जबकि Y-अक्ष 'महिलाओं की संख्या' (No. of females-in '000) को दर्शाता है। वक्र एक बढ़ती हुई रेखा है जो विभिन्न आयु समूहों के लिए संचयी बारंबारता को दर्शाती है, जिसका उपयोग ग्राफिक रूप से माध्यिका जैसे मानों को निर्धारित करने के लिए किया जाता है। In simple words: To find the median graphically, we locate the N/2 value (250) on the Y-axis, draw a horizontal line to the ogive, and then a vertical line down to the X-axis; the point on the X-axis is the median.

🎯 Exam Tip: Accurately calculating N/2 and drawing precise lines parallel to axes from the ogive intersection point is crucial for correctly determining median values in graphical methods.

Question 9. Draw ogive for the following distribution and hence find graphically the limits of the weight of middle 50% fishes.

Solution: Since the given data is not continuous, we have to convert it into the continuous form by subtracting 5 from the lower limit and adding 5 to the upper limit of every class interval. To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (895, 8), (995, 24), (1095, 44), (1195, 69), (1295, 109), (1395, 115), (1495, 120).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक 'कम संचयी बारंबारता ओजाइव वक्र' प्रस्तुत करता है। X-अक्ष 'मछलियों का वजन' (Weight of fishes in gms) को दर्शाता है, जबकि Y-अक्ष 'मछलियों की संख्या' (No. of fishes (less than c.f.)) को दर्शाता है। वक्र दर्शाता है कि कैसे मछलियों का संचयी वजन विभिन्न भार सीमाओं में बढ़ता है, जो ग्राफिक रूप से चतुर्थक मानों जैसे Q1 और Q3 को निर्धारित करने में सहायता करता है।

N = 120 For Q1 and Q3 we have to consider \( \frac{N}{4} = \frac{120}{4} = 30 \)
\( \frac{3N}{4} = \frac{3 \times 120}{4} = 90 \) For finding Q1 and Q3 we consider the values 30 and 90 on the Y-axis. From these points, we draw the lines which are parallel to X-axis. From the points where these lines intersect the less than ogive, we draw perpendicular on X-axis. The feet of perpendiculars represent the values Q1 and Q3.
.: Q1 ~ 1025 and Q3 ~ 1248
.: the limits of the weight of the middle 50% of fishes lie between 1025 to 1248. In simple words: To find the middle 50% limits graphically, we first calculate the positions for Q1 (25th percentile) and Q3 (75th percentile) using N/4 and 3N/4, then find their corresponding X-axis values on the ogive.

🎯 Exam Tip: Remember to convert discontinuous data into continuous data before constructing the ogive, as this ensures accurate graphical representation and calculation of quartile values.

Question 10. Find graphically the values of D3 and P65 for the data given below:

Solution: Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval. To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (69.5, 20), (79.5, 60), (89.5, 110), (99.5, 160), (109.5, 180), (119.5, 190), (129.5, 200).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक 'कम संचयी बारंबारता ओजाइव वक्र' प्रस्तुत करता है। X-अक्ष 'छात्रों के IQ' (I.Q. of students) को दर्शाता है, जबकि Y-अक्ष 'छात्रों की संख्या' (No. of students) को दर्शाता है। वक्र दर्शाता है कि कैसे छात्रों का संचयी IQ विभिन्न IQ श्रेणियों में बढ़ता है, जिसका उपयोग ग्राफिक रूप से दशमक (D3) और प्रतिशतक (P65) मानों को निर्धारित करने में सहायता करता है।

N = 200 For D3, \( \frac{3N}{10} = \frac{3 \times 200}{10} = 60 \) For P65, \( \frac{65N}{100} = \frac{65 \times 200}{100} = 130 \)
.: We take the values 60 and 130 on the Y-axis. From these points we draw lines parallel to X-axis and from the points where these lines intersect less than ogive, we draw perpendiculars on X-axis. The foot of perpendiculars represents the median of the values, D3 and P65.
.: D3 = 79.5, P65 = 93.5 In simple words: To graphically find D3 and P65, we calculate their cumulative frequency positions (60 for D3, 130 for P65), locate these on the Y-axis, draw horizontal lines to the ogive, and then vertical lines down to the X-axis to read the corresponding IQ values.

🎯 Exam Tip: For deciles and percentiles, accurately calculating the Nth/10 (for deciles) or Nth/100 (for percentiles) position on the Y-axis and then tracing to the ogive and X-axis is key to finding the correct graphical values.