Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 1.
Calculate D6 and P85 for the following data:
79, 82, 36, 38, 51, 72, 68, 70, 64, 63
Solution:
The given data can be arranged in ascending order as follows:
36, 38, 51, 63, 64, 68, 70, 72, 79, 82
Here, n = 10
D6 = value of 6\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 6\(\left(\frac{10+1}{10}\right)^{\text {th }}\) observation
= value of (6 × 1.1)th observation
= value of (6.6)th observation
= value of 6th observation + 0.6(value of 7th observation – value of 6th observation)
= 68 + 0.6(70 – 68)
= 68 + 0.6(2)
= 68 + 1.2
∴ D6 = 69.2
P85 = value of \(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{100}\right)^{\text {th }}\) observation
= value of (85 × 0. 11)th observation
= value of (9.35)th observation
= value of 9th observation + 0.35(value of 10th observation – value of 9th observation)
= 19 + 0.35(82 – 79)
= 79 + 0.35(3)
= 79 + 1.05
∴ P85 = 80.05

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 2.
The daily wages (in ₹) of 15 labourers are as follows:
230, 400, 350, 200, 250, 380, 210, 225, 375, 180, 375, 450, 300, 350, 250
Calculate D8 and P90.
Solution:
The given data can be arranged in ascending order as follows:
180, 200, 210, 225, 230, 250, 250, 300, 350, 350, 375, 375, 380, 400, 450
Here, n = 15
D8 = value of 8\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 8\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (8 × 1.6)th observation
= value of (12.8)th observation
= value of 12th observation – 0.8(value of 13th observation – value of 12th observation)
= 375 + 0.8(380 – 375)
= 375 + 0.8(5)
= 375 + 4
∴ D8 = 379
P90 = value of 90\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 90\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (90 × 0.16)th observation
= value of (14.4)th observation
= value of 14th observation + 0.4 (value of 15th observation – value of 14th observation)
= 400 + 0.4(450 – 400)
= 400 + 0.4(50)
= 400 + 20
∴ P90 = 420

Question 3.
Calculate 2nd decile and 65th percentile for the following:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q3
Solution:
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q3.1
Here, n = 200
D2 = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{200+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 20.1)th observation
= value of (40.2)th observation
Cumulative frequency which is just greater than (or equal to) 40.2 is 58.
∴ D2 = 120
P65 = value of 65\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 65\(\left(\frac{200+1}{100}\right)^{\text {th }}\) observation
= value of (65 × 2.01)th observation
= value of (130.65)th observation
The cumulative frequency which is just greater than (or equal to) 130.65 is 150.
∴ P65 = 280

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 4.
From the following data calculate the rent of the 15th, 65th, and 92nd house.
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q4
Solution:
Arranging the given data in ascending order.
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q4.1
Here, n = 100
P15 = value of 15
= value of 15\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 15\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (15 × 1.01 )th observation
= value of (15.15)th observation
Cumulative frequency which is just greater than (or equal to) 15.15 is 25.
∴ P15 = 11000
P65 = value of 65\(\left(\frac{n+1}{100}\right)^{\text {th }}\)observation
= value of 65\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (65 × 1.01)th observation
= value of (65.65)th observation
Cumulative frequency which is just greater than (or equal to) 65.65 is 70.
∴ P65 = 14000
P92 = value of 92\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 92\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (92 × 1.01)th observation
= value of (92.92)th observation
Cumulative frequency which is just greater than (or equal to) 92.92 is 98.
∴ P92 = 17000

Question 5.
The following frequency distribution shows the weight of students in a class.
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q5
(a) Find the percentage of students whose weight is more than 50 kg.
(b) If the weight column provided is of mid values then find the percentage of students whose weight is more than 50 kg.
Solution:
(a) Let the percentage of students weighing less than 50 kg be x.
∴ Px = 50
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q5.1
From the table, out of 20 students, 84 students have their weight less than 50 kg.
∴ Number of students weighing more than 50 kg = 120 – 84 = 36
∴ Percentage of students having there weight more than 50 kg = \(\frac{36}{120}\) × 100 = 30%

(b) The difference between any two consecutive mid values of weight is 5 kg.
The class intervals must of width 5, with 40, 45,….. as their mid values.
∴ The class intervals will be 37.5 – 42.5, 42.5 – 47.5, etc.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q5.2
Here, N = 120
Let Px = 50
The value 50 lies in the class 47.5 – 52.5
∴ L = 47.5, h = 5, f = 29, c.f. = 55
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q5.3
∴ x = 58 (approximately)
∴ 58% of students are having weight below 50 kg.
∴ Percentage of students having weight above 50 kg is 100 – 58 = 42
∴ 42% of students are having weight above 50 kg.

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 6.
Calculate D4 and P48 from the following data:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q6
Solution:
The difference between any two consecutive mid values is 5, the width of class interval = 5
∴ Class interval with mid-value 2.5 is 0 – 5
Class interval with mid value 7.5 is 5 – 10, etc.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q6.1
Here, N = 100
D4 class = class containing \(\left(\frac{4 \mathrm{N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{4 \mathrm{N}}{10}=\frac{4 \times 100}{10}\) = 40
Cumulative frequency which is just greater than (or equal to) 40 is 50.
∴ D4 lies in the class 10 – 15.
∴ L = 10,h = 5, f = 25, c.f. = 25
∴ D4 = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{4 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 10 + \(\frac{5}{25}\) (40 – 25)
= 10 + \(\frac{1}{5}\) (15)
= 10 + 3
∴ D4 = 13
P48 class = class containing \(\left(\frac{48 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{48 \mathrm{~N}}{100}=\frac{48 \times 100}{100}\) = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P48 lies in the class 10 – 15.
∴ L = 10, h = 5, f = 25, c.f. = 25
∴ P48 = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{48 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10 + \(\frac{5}{25}\) (48 – 25)
= 10 + \(\frac{1}{5}\) (23)
= 10 + 4.6
∴ P48 = 14.6

Question 7.
Calculate D9 and P20 of the following distribution.
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q7
Solution:
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q7.1
Here, N = 240
D9 class = class containing \(\left(\frac{9 \mathrm{~N}}{10}\right)^{\mathrm{th}}\) observation
∴ \(\frac{9 \mathrm{~N}}{10}=\frac{9 \times 240}{10}\) = 216
Cumulative frequency which is just greater than (or equal to) 216 is 225.
∴ D9 lies in the class 80 – 100.
∴ L = 80, h = 20, f = 90, c.f. = 135
∴ D9 = \(L+\frac{h}{f}\left(\frac{9 N}{10}-c . f .\right)\)
= 80 + \(\frac{20}{90}\)(216 – 135)
= 80 + \(\frac{2}{9}\)(81)
= 80 + 18
∴ D9 = 98
P20 class = class containing \(\left(\frac{20 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{20 \mathrm{~N}}{100}=\frac{20 \times 240}{100}\) = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P20 lies in the class 40 – 60.
∴ L = 40, h = 20, f = 35, c.f. = 15
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q7.2
∴ P20 = 58.86

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 8.
Weekly wages for a group of 100 persons are given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q8
D3 for this group is ₹ 1100. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 500 – 1000 and class 2000 – 2500 respectively.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q8.1
Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 …..(i)
Given, D3 = 1100
∴ D3 lies in the class 1000 – 1500.
∴ L = 1000, h = 500, f = 25, c.f. = 7 + a
∴ \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 100}{10}=30\)
∴ D3 = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{3 \mathrm{~N}}{10}-\mathrm{c} . \mathrm{f} .\right)\)
∴ 1100 = 1000 + \(\frac{500}{25}\) [30 – (7 + a)]
∴ 1100 – 1000 = 20(30 – 7 – a)
∴ 100 = 20(23 – a)
∴ 100 = 460 – 20a
∴ 20a = 460 – 100
∴ 20a = 360
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18 = 20
∴ 18 and 20 are the missing frequencies of the class 500 – 1000 and class 2000 – 2500 respectively.

Question 9.
The weekly profit (in rupees) of 100 shops are distributed as follows:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q9
Find the limits of the profit of middle 60% of the shops.
Solution:
To find the limits of the profit of the middle 60% of the shops, we have to find P20 and P80.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q9.1
Here, N = 100
P20 class = class containing \(\left(\frac{20 \mathrm{N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{20 \mathrm{N}}{100}=\frac{20 \times 100}{100}=20\)
Cumulative frequency which is just greater than (or equal to) 20 is 26.
∴ P20 lies in the class 1000 – 2000.
∴ L = 1000, h = 1000, f = 16, c.f. = 10
∴ P20 = \(L+\frac{h}{f}\left(\frac{20 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 1000 + \(\frac{1000}{16}\) (20 – 10)
= 1000 + \(\frac{125}{2}\) (10)
= 1000 + 625
∴ P20 = 1625
P80 class = class containing \(\left(\frac{80 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{80 \mathrm{~N}}{100}=\frac{80 \times 100}{100}=80\)
Cumulative frequency which is just greater than (or equal to) 80 is 92.
∴ P80 lies in the class 4000 – 5000.
∴ L = 4000, h = 1000, f = 20, c.f. = 72
∴ P80 = \(L+\frac{h}{f}\left(\frac{80 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 4000 + \(\frac{1000}{20}\)(80 – 72)
= 4000 + 50(8)
= 4000 + 400
∴ P80 = 4400
∴ the profit of middle 60% of the shops lie between the limits ₹ 1,625 to ₹ 4,400.

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 10.
In a particular factory, workers produce various types of output units. The following distribution was obtained:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q10
Find the percentage of workers who have produced less than 82 output units.
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.
∴ the class intervals will be 69.5 – 74.5, 74.5 – 79.5, etc.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q10.1
Here, N = 445
Let Px = 82
The value 82 lies in the class 79.5 – 84.5
∴ L = 79.5, h = 5, f = 50, c.f. = 85
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2 Q10.2
∴ 24.72% of workers produced less than 82 output units.

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Question 1.
Compute all the quartiles for the following series of observations:
16, 14.9, 11.5, 11.8, 11.1, 14.5, 14, 12, 10.9, 10.7, 10.6, 10.5, 13.5, 13, 12.6
Solution:
The given data can be arranged in ascending order as follows:
10.5, 10.6, 10.7, 10.9, 11.1, 11.5, 11.8, 12, 12.6, 13, 13.5, 14, 14.5, 14.9, 16
Here, n = 15
Q1 = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of 4th observation
∴ Q1 = 10.9
Q2 = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 4)th observation
= value of 8th observation
∴ Q2 = 12
Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 4)th observation
= value of 12th observation
∴ Q3 = 14

Question 2.
The heights (in cm.) of 10 students are given below:
148, 171, 158, 151, 154, 159, 152, 163, 171, 145
Calculate Q1 and Q3 for the above data.
Solution:
The given data can be arranged in ascending order as follows:
145, 148, 151, 152, 154, 158, 159, 163, 171, 171
Here, n = 10
Q1 = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= value of (2.75)th observation
= value of 2nd observation + 0.75 (value of 3rd observation – value of 2nd observation)
= 148 + 0.75 (151 – 148)
= 148 + 0.75(3)
= 148 + 2.25
∴ Q1 = 150.25
Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 2.75)th observation
= value of (8.25)th observation
= value of 8th observation + 0.25 (value of 9th observation – value of 8th observation)
= 163 + 0.25(171 – 163)
= 163 + 0.25(8)
= 163 + 2
∴ Q3 = 165

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Question 3.
The monthly consumption of electricity (in units) of families in a certain locality is given below:
205, 201, 190, 188, 195, 172, 210, 225, 215, 232, 260, 230
Calculate electricity consumption (in units) below which 25% of the families lie.
Solution:
To find the consumption of electricity below which 25% of the families lie, we have to find Q1.
Monthly consumption of electricity (in units) can be arranged in ascending order as follows:
172, 188, 190, 195, 201, 205, 210, 215, 225, 230, 232, 260.
Here, n = 12
Q1 = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{12+1}{4}\right)^{\text {th }}\) observation
= value of (3.25)th observation
= value of 3rd observation + 0.25 (value of 4th observation – value of 3rd observation)
= 190 + 0.25(195 – 190)
= 190 + 0.25(5)
= 190 + 1.25
= 191.25
∴ the consumption of electricity below which 25% of the families lie is 191.25.

Question 4.
For the following data of daily expenditure of families (in ₹), compute the expenditure below which 75% of families include their expenditure.
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q4
Solution:
To find the expenditure below which 75% of families have their expenditure, we have to find Q3.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q4.1
Here, n = 100
Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 25.25)th observation
= value of (75.75)th observation
Cumulative frequency which is just greater than (or equal to) 75.75 is 87.
∴ Q3 = 650
∴ the expenditure below which 75% of families include their expenditure is ₹ 650.

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Question 5.
Calculate all the quartiles for the following frequency distribution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q5
Solution:
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q5.1
Here, n = 300
Q1 = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation
= value of (75.25)th observation
Cumulative frequency which is just greater than (or equal to) 75.25 is 90.
∴ Q1 = 2
Q2 = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 75.25)th observation
= value of (150.50)th observation
∴ Cumulative frequency which is just greater than (or equal to) 150.50 is 185.
∴ Q2 = 3
Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 75.25)th observation
= value of (225.75)th observation
Cumulative frequency which is just greater than (or equal to) 225.75 is 249.
∴ Q3 = 4

Question 6.
The following is the frequency distribution of heights of 200 male adults in a factory:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q6
Find the central height.
Solution:
To find the central height, we have to find Q2.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q6.1
Here, N = 200
Q2 class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 200}{4}\) = 100
Cumulative frequency which is just greater than (or equal to) 100 is 156.
∴ Q2 lies in the class 165 – 170.
∴ L = 165, h = 5, f = 64, c.f. = 92
Q2 = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 165 + \(\frac{5}{64}\) (100 – 92)
= 165 + \(\frac{5}{64}\) × 8
= 165 + \(\frac{5}{8}\)
= 165 + 0.625
= 165.625
∴ Central height is 165.625 cm.

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Question 7.
The following is the data of pocket expenditure per week of 50 students in a class. It is known that the median of the distribution is ₹ 120. Find the missing frequencies.
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q7
Solution:
Let a and b be the missing frequencies of class 50 – 100 and class 150 – 200 respectively.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q7.1
Here, N = 25 + a + b
Since, N = 50
∴ 25 + a + b = 50
∴ a + b = 25 …..(i)
Given, Median = Q2 = 120
∴ Q2 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 15, \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 50}{4}\) = 25
∴ Q2 = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
∴ 120 = 100 + \(\frac{50}{15}\) [25 – (7 + a)]
∴ 120 – 100 = \(\frac{10}{3}\) (25 – 7 – a)
∴ 20 = \(\frac{10}{3}\) (18 – a)
∴ \(\frac{60}{10}\) = 18 – a
∴ 6 = 18 – a
∴ a = 18 – 6 = 12
Substituting the value of a in equation (i), we get
12 + b = 25
∴ b = 25 – 12 = 13
∴ 12 and 13 are the missing frequencies of the class 50 – 100 and class 150 – 200 respectively.

Question 8.
The following is the distribution of 160 workers according to the wages in a certain factory:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q8
Determine the values of all quartiles and interpret the results.
Solution:
The given table is a more than cumulative frequency.
We transform the given table into less than cumulative frequency.
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q8.1
Here, N = 160
∴ Q1 class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{160}{4}\) = 40
Cumulative frequency which is just greater than (or equal to) 40 is 69.
∴ Q1 lies in the class 10000 – 11000
∴ L = 10000, h = 1000, f = 46, c.f. = 23
Q1 = \(L+\frac{h}{f}\left(\frac{N}{4}-\text { c.f. }\right)\)
= 10000 + \(\frac{1000}{46}\) (40 – 23)
= 10000 + \(\frac{1000}{46}\) (17)
= 10000 + \(\frac{17000}{46}\)
= 10000 + 369.57
= 10369.57
Q2 class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 160}{4}\) = 80
Cumulative frequency which is just greater than (or equal to) 80 is 103.
∴ Q2 lies in the class 11000 – 12000.
∴ L = 11000, h = 1000, f = 34, c.f. = 69
∴ Q2 = \(L+\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
= 11000 + \(\frac{1000}{34}\)(80 – 69)
= 11000 + \(\frac{1000}{34}\)(11)
= 11000 + \(\frac{11000}{34}\)
= 11000 + 323.529
= 11323.529
Q3 class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 160}{4}\) = 120
Cumulative frequency which is just greater than (or equal to) 120 is 137.
∴ Q3 lies in the class 12000 – 13000.
∴ L = 12000, h = 1000, f = 34, c.f. = 103
∴ Q3 = \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 12000 + \(\frac{1000}{34}\) (120 – 103)
= 12000 + \(\frac{1000}{34}\) (17)
= 12000 + \(\frac{1000}{2}\)
= 12000 + 500
= 12500
Interpretation:
Q1 < Q2 < Q3

Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Question 9.
Following is grouped data for the duration of fixed deposits of 100 senior citizens from a certain bank:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q9
Calculate the limits of fixed deposits of central 50% senior citizens.
Solution:
We construct the less than cumulative frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q9.1
To find the limits of fixed deposits of central 50% senior citizens, we have to find Q1 and Q3.
Here, N = 100
Q1 class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{100}{4}\) = 25
Cumulative frequency which is just greater than (or equal to) 25 is 35.
∴ Q1 lies in the class 180 – 360.
∴ L = 180, h = 180, f = 20, c.f. = 15
∴ Q1 = \(L+\frac{h}{f}\left(\frac{N}{4}-c . f .\right)\)
= 180 + \(\frac{180}{20}\) (25 – 15)
= 180 + 9(10)
= 180 + 90
∴ Q1 = 270
Q3 class = class containing \(\left(\frac{3 \mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{N}}{4}=\frac{3 \times 100}{4}\) = 75
Cumulative frequency which is just greater than (or equal to) 75 is 90.
∴ Q3 lies in the class 540 – 720.
∴ L = 540, h = 180, f = 30, c.f. = 60
∴ Q3 = \(L+\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 540 + \(\frac{180}{30}\) (75 – 60)
= 540 + 6(15)
= 540 + 90
∴ Q3 = 630
∴ Limits of duration of fixed deposits of central 50% senior citizens is from 270 to 630.

Question 10.
Find the missing frequency given that the median of the distribution is 1504.
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q10
Solution:
Let x be the missing frequency of the class 1550 – 1750.
We construct the less than frequency table as given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1 Q10.1
Here, N = 199 + x
Given, Median (Q2) = 1504
∴ Q2 lies in the class 1350 – 1550.
∴ L = 1350, h = 200, f = 100, c.f. = 63,
\(\frac{2 \mathrm{~N}}{4}=\frac{199+x}{2}\)
∴ Q2 = \(L+\frac{h}{f}\left(\frac{2 N}{4}-c . f .\right)\)
∴ 1504 = 1350 + \(\frac{200}{100}\left(\frac{199+x}{2}-63\right)\)
∴ 1504 – 1350 = 2\(\left(\frac{199+x-126}{2}\right)\)
∴ 154 = 199 + x – 126
∴ 154 = x + 73
∴ x = 81