RS Aggarwal Solutions for Class 10 Maths Chapter 18 Mean, … · 2020. 10. 29. · RS Aggarwal Solutions for Class 10 Maths Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative

RS Aggarwal Solutions for Class 10 Maths Chapter 18-

Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

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Exercise 18A Question 1: Solution: We know that, Mean = Sum of the given observations/ Total number of observations Sum of the given observations = x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = x + x + 2 + x + 4 + x + 6 + x + 8 = 5x + 20 Total number of observations = 5 Therefore, Mean = (5x + 20)/5 Also, Mean = 11 (given) ⇨ 11 = (5x + 20)/5 ⇨ 55 = 5x + 20 or x = 7 Question 2: Solution: Mean of 25 observations = 27 (given) Total observations = 25 Mean = Sum of the given observations/ Total number of observations 27 = (sum of 25 observations)/25 ⇨ sum of 25 observations = 27 x 25 = 675 When each observation is decreased by 7, then New Sum is 675 – 25 x 7 = 500 New Mean = 500/25= 20 Question 3: Solution:

Now, Mean = ∑fixi / ∑fi = 426/80 = 5.325 Question 4:

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Solution:

Mean = ∑fixi/ ∑fi = 1150/40 = 28.75 Question 5: Solution:




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Mean = ∑fixi/ ∑fi = 1980/40 = 49.5 Question 6: Solution:

Mean = ∑fixi/∑fi = 13200/40 = 264 Question 7: Solution:




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Mean = ∑fixi/ ∑fi = 8244/80 = 103.05 Used direct method as it is easy to calculate. Question 8: Solution:

Mean=∑fixi/∑fi Mean = 24 ( given) ⇨ 24 = (270+25p)/(12+p) ⇨ 24(12+p) = 270 + 25p or p = 18 Question 9: Solution:




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Mean=∑fixi/ ∑fi Mean = 18 (given) 18 = (752 + 20f)/(44+f) 792 + 18f = 752 + 20f or f = 20 Question 10: Solution:

Mean=∑fixi/ ∑fi Mean = 54 (given) ⇨ 54 = (2370 + 30p)/(39+p) ⇨ 1053 + 27p = 1185 + 15p or p = 11




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Exercise 18B Question 1: Solution:

Where cf = cumulative frequency Median = I+{h×(N2–cf)f} Here: N = 140 N/2 = 70 cf > 70 is 140 Median class = 45-60 So, l = 45, h = 15, f = 50 and cf = cf of preceding class i.e. 65 Substitute all the value in the above formula, we get Median = 45 + {15 x (70-65)/50} = 45 + 1.5 = 46.5 Therefore, median age of diabetic patients is 46.5 years. Question 2: Solution:




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Where cf = cumulative frequency Median=I+{h×(N2–cf)f} Here: N = 50 N/2 = 25 Here cumulative frequency is 25 Median class = 21-28 So, l = 21, h =7, f = 11 and cf = cf of preceding class i.e. 14 Substitute all the value in the above formula, we get Median =21 + {7x (25-14)/11} = 28 Question 3: Solution:

Where cf = cumulative frequency Median=I+{h×(N2–cf)f} Here: N = 150 N/2 = 75 cf just greater than 75 is 120 Median class = 200-300 So, l = 200, h =100, f = 48 and




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cf = cf of preceding class i.e. 72 Substitute all the value in the above formula, we get Median =200+ {100 x (75-72)/48} = 200 + 6.25 = 206.25 Median of daily wages is Rs. 206.25. Question 4: Solution:

Where cf = cumulative frequency Median=I+{h×(N2–cf)f} Here: N = 49 N/2 = 24.5 cf just greater than 24.5 is 26 Median class = 15-20 So, l = 15, h = 5, f = 15 and cf = cf of preceding class i.e. 11 Substitute all the value in the above formula, we get Median = 15+ {5 x (24.5-11)/15} = 15 + 4.5




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= 19.5 Thus, median of frequency distribution is 19.5. Question 5: Solution:

Where cf = cumulative frequency Median=I+{h×(N2–cf)f} Here: N = 67 N/2 = 33.5 cf just greater than 33.5 is 42 Median class = 125-145 So, l = 125, h = 20, f = 20 and cf = cf of preceding class i.e. 22 Substitute all the value in the above formula, we get Median = 125+ {20 x (33.5-22)/20} = 125 + 11.5 = 136.5 Thus, median of electricity consumed is 136.5.




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Question 6: Solution:

Where cf = cumulative frequency Median=I+{h×(N2–cf)f} Here: N = 100 N/2 = 50 cf just greater than 50 is 56 Median class = 150-155 So, l = 150, h = 5, f = 22 and cf = cf of preceding class i.e. 34 Substitute all the value in the above formula, we get Median = 150+ {5 x (50-34)/22} = 150 + 3.64 = 153.64 Question 7: Solution: Let the missing frequency be x.




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Median = 24 N/2 = (55+x)/2 = 27.5 + x/2 Median class = 20-30 l = 20, h = 10, f = x and cf = cf of preceding class i.e. 30 Using below formula: Median=I+{h×(N2–cf)f} ⇨ 24 = 20 + {10 x (27.5+ x/2 – 30)/x} ⇨ 24x = 20x + 5x – 25 or x = 25 Missing frequency is 25. Question 8: Solution:




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Here: N = 70 N/2 = 35 Median = 16, Median class = 15-20 So, l = 15, h = 5, f = 15 and cf = cf of preceding class i.e. 24+a We know, Median=I+{h×(N2–cf)f} Substitute all the value in the above formula, we get 16 = 15 + {5x (35-24-a)/15} ⇨ 16 = 15 + (11-a)/3 or a = 8 Now, 70 = 55 + a + b 70 = 55 + 8 + b ⇨ b = 7 So, missing frequencies are 8 and 7. Question 9: Solution:

Here: N = 60 N/2 = 30 Median = 5000, Median class = 4500-5500 So, l = 4500, h = 1000, f = y and cf = cf of preceding class i.e. 5+x We know, Median=I+{h×(N2–cf)f} Substitute all the value in the above formula, we get 5000 = 4500 + {1000 x (30-5-x)/y} ⇨ 500 = 1000 x (25-x)/y




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⇨ 0.5 = (25-x)/y ⇨ x + 05.y = 25 …(1) Again, from table 25 + x + y = 60 ⇨ x + y = 35 …(2) Solving (1) and (2), we have x = 15 and y = 20 Missing frequencies are 15 and 20. Question 10: Solution: Sum of all the frequencies = f1+ 5 + 9 +12 + f2 + 3 + 2 = 40 (given) ⇨ f1+ f2 = 9 …(1) Median is 32.5 lies in 30 – 40, so the median class is 30 – 40 N = 40 N/2 = 20 l = 30, h = 10, f = 12 cf = f1 + 5+9 = f1+14 We know, Median=I+{h×(N2–cf)f} Substitute all the value in the above formula, we get 32.5 = 30 + {10 x (20-f1-14)/12 } ⇨ 32.5 = 30 + (30-5f1)/6 ⇨ 2.5 = (30-5f1)/6 or f1 = 3 From (1) ⇨ 3 + f2 = 9 ⇨ f2 = 6




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Exercise 18C Question 1: Solution: Here: Maximum class frequency is 45. The class corresponding to Maximum class frequency is the modal class. So, Modal class = 30 – 40 Lower limit = l = 30 Modal class size =h = 10 Frequency of class preceding the modal class =f0 = 35 Frequency of the modal class= f1 = 45 Frequency of class succeeding the modal =f2= 25 Mode formula is given by, Mode=I+(f1−f02f1−f0−f2)×h Mode=30+(45−352×45−35−25)×10 = 30 + 100/30 = 33.33 Question 2: Solution: Here: Maximum class frequency = 28 The class corresponding to Maximum class frequency is the modal class. So, Modal class = 40-60 Lower limit = l = 40 Modal class size =h = 20 Frequency of class preceding the modal class =f0 = 16 Frequency of the modal class= f1 = 28 Frequency of class succeeding the modal =f2= 20 Mode formula is given by, Mode=I+(f1−f02f1−f0−f2)×h Mode=40+(28−162×28−16−20)×20 = 40 + 12/20 x 12 = 52 Question 3: Solution: Here: Maximum class frequency = 20 The class corresponding to Maximum class frequency is the modal class. So, Modal class = 160-165 Lower limit = l = 160 Modal class size =h = 5 Frequency of class preceding the modal class =f0 = 8 Frequency of the modal class= f1 = 20 Frequency of class succeeding the modal =f2= 12 Mode formula is given by, Mode=I+(f1−f02f1−f0−f2)×h Mode=160+(20−82×20−8−12)×5




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= 160 + 12/20 x 5 = 163 The modal height is 163 cm, which implies maximum number of students have model height 163 cm Use Direct Method to find the mean:

= 9670/60 = 161.17 Mean is 161.17 Thus, 161.7 cm is the average height of all the students. Question 4: Solution: Here: Maximum class frequency = 25 The class corresponding to Maximum class frequency is the modal class. So, Modal class = 26-30 Lower limit = l = 26 Modal class size =h = 4 Frequency of class preceding the modal class =f0 = 20 Frequency of the modal class= f1 = 25 Frequency of class succeeding the modal =f2= 22 Mode formula is given by,




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Mode=I+(f1−f02f1−f0−f2)×h Mode=26+(25−202×25−20−22)×4 = 26 + 5/8 x 4 = 28.5 Question 5: Solution: Here: Maximum class frequency = 40 The class corresponding to Maximum class frequency is the modal class. So, Modal class = 1500-2000 Lower limit = l = 1500 Modal class size =h = 500 Frequency of class preceding the modal class =f0 = 24 Frequency of the modal class= f1 = 40 Frequency of class succeeding the modal =f2= 31 Mode formula is given by, Mode=I+(f1−f02f1−f0−f2)×h Mode=1500+(40−242×40−24−31)×10 = 1500 + 16/25 x 500 = 1820




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Exercise 18D Question 1: Solution: Let assumed mean be 35, h = 10.




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Question 5: Solution: Assumed mean A = 145 Class interval = h = 10




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Question 7: Solution: Let assumed mean = 225 and h = 50




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Exercise 18E Question 1: Solution:

Plot points: (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the ‘less than type’ ogive as follows:




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From frequency table, N = 53 N/2 = 26.5 Mark y = 26.5 on y-axis Through A, draw a horizontal line meeting the curve at P. Through P, a vertical line is drawn which meets OX at M. OM = 68 units Hence, median = 68 Question 2: Solution: Plot points on graph: (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the ‘less than type’ ogive as follows:




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From frequency table, N = 80, ⇨ N/2 = 40 At y = 40, Draw a horizontal line meeting the curve at P, AP Draw a vertical line from point P which is parallel to y-axis and meets OX at A’. OA’ = 77 units Median number of wickets = 77 Question 3: Solution:




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Plot points on the graph: (0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the ‘more than type’ ogive as follows:

From frequency table, N = 100 ⇨ N/2 = 50. At y = 50 Draw a horizontal line meeting the curve at P, AP Draw a vertical line from point P which is parallel to y-axis and meets OX at M.




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OM = 47 units Median marks = 47 Question 4: Solution:

Plot the points on the graph: (135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the ‘more than type’ ogive as follows:




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From frequency table, N = 50 ⇨ N/2 = 25 At y = 25 Draw a horizontal line meeting the curve at P, AP Draw a vertical line from point P which is parallel to y-axis and meets OX at M. OM = 151 units Median height = 151 cm Question 5: Solution:

Plot the points on the graph: (140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the ‘more than type’ ogive as follows:




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From frequency table, N = 156 ⇨ N/2 = 78 At y = 78 Draw a horizontal line meeting the curve at P, AP Draw a vertical line from point P which is parallel to y-axis and meets OX at M. OM = 226 units Median consumption of electricity = 226 units

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RS Aggarwal Solutions for Class 10 Maths Chapter 18 Mean, … · 2020. 10. 29. · RS Aggarwal Solutions for Class 10 Maths Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative