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Commission, Brokerage and Discount Class 12 Commerce Maths 2 Chapter 1 Exercise 1.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Ex 1.2 Questions and Answers.

Std 12 Maths 2 Exercise 1.2 Solutions Commerce Maths

Question 1.
What is the present worth of a sum of ₹ 10,920 due six months hence at 8% p.a simple interest?
Solution:
Given, SD = ₹ 10,920
n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
r = 8%
We have,

Thus the present worth is ₹ 10,500

Question 2.
What is the sum due of ₹ 8,000 due 4 months at 12.5% simple interest?
Solution:
Given, PW = ₹ 8,000, n = \(\frac{4}{12}\) year = \(\frac{1}{3}\) year, r = 12.5%
We have,

Thus, the sum due is ₹ 8,333.33

Question 3.
The true discount on the sum due 8 months hence at 12% p.a. is ₹ 560. Find the sum due and present worth of the bill.
Solution:
Given, TD = ₹ 560, n = \(\frac{8}{12}\) year = \(\frac{2}{3}\) year, r = 12%
We have,
TD = \(\frac{\mathrm{PW} \times n \times r}{100}\)
∴ 560 = \(\frac{\mathrm{PW} \times 2 \times 12}{3 \times 100}\)
∴ PW = 560 × \(\frac{25}{2}\) = ₹ 7,000
Now, SD = PW + TD
= 7,000 + 560
= ₹ 7,560

Question 4.
The true discount on a sum is \(\frac{3}{8}\) of the sum due at 12% p.a. Find the period of the bill.
Solution:

8 × n × 12 = 3(100 + n × 12)
96n = 300 + 36n
60n = 300
∴ n = 5
∴ Period of the bill = 5 years.

Question 5.
20 copies of a book can be purchased for a certain sum payable at the end of 6 months and 21 copies for the same sum in ready cash. Find the rate of interest.
Solution:
Given, n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
Let the sum payable be ₹ x
Let the rate of interest be r%
According to given condition,
PW of one book = \(\frac{x}{21}\)
SD of one book = \(\frac{x}{20}\)

Thus, the rate of interest is 10%.

Question 6.
Find the true discount, Banker’s discount, and Banker’s gain on a bill of ₹ 4,240 due 6 months hence at 9% p.a.
Solution:

And, Banker’s Gain (BG) = BD – TD
= 190.80 – 182.58
= ₹ 8.22

Question 7.
The true discount on a bill is ₹ 2,200 and bankers discount is ₹ 2,310. If the bill is due 10 months, hence, find the rate of interest.
Solution:
Given, TD = ₹ 2,200, BD = ₹ 2,310
n = \(\frac{10}{12}=\frac{5}{6}\) year

∴ \(\frac{r}{120}=\frac{1}{20}\)
∴ r = 6%
Thus, rate of interest is 6%

Question 8.
A bill of ₹ 6,395 drawn on 19th January 2015 for 8 months was discounted on 28th February 2015 at 8% p.a. interest. What is the banker’s discount? What is the cash value of the bill?
Solution:
Face value = ₹ 6,395
Date of drawing = 19/01/2015
Period of the bill = 8 months
Nominal Due date = 19/09/2015
Legal due date = 22/09/2015
Date of discounting = 28/02/2015
Now, the unexpired period = Legal due date – Date of discounting
= 22/09/2015 – 28/02/2015
= days (as shown below)

Cash Value = FV – BD
= 6,395 – 313.12
= ₹ 6,621.38

Question 9.
A bill of ₹ 8,000 drawn on 5th January 1998 for 8 months was discounted for ₹ 7,680 on a certain date. Find the date on which it was discounted at 10% p.a.
Solution:
Bankers discount (BD) = FV – cash value
= 8,000 – 7,680
= ₹ 320
Let the unexpired period be x days
∴ BD = \(\frac{\mathrm{FV} \times x \times r}{365 \times 100}\)
∴ 320 = \(\frac{8,000 \times x \times 10}{365 \times 100}\)
∴ x = 146 days
∴ The unexpired days = 146 days
Date of drawing = 05/01/1998
Period of bill = 8 months
Nominal due date = 05/09/1998
Legal due date = 08/09/1998
Thus, the date of discounting is 146 days before the legal due date

∴ Date of discounting of the bill is 15th April 1998

Question 10.
A bill drawn on 5th June for 6 months was discounted at the rate of 5% p.a. on 19th October. If the cash value of the bill is ₹ 43,500, find the face value of the bill.
Solution:
Date of drawing = 5th June
Period of bill = 6 months
Nominal due date = 5th December
Legal due date = 8th December
Date of discounting = 19th October
Rate of interest = 5% p.a.
Let the face value of the bill be ₹ x
The unexpired period

Question 11.
A bill was drawn on 14th April for ₹ 7,000 and was discounted on 6th July at 5% p.a. The Banker paid ₹ 6,930 for the bill. Find the period of the bill.
Solution:
Face value = ₹ 7,000, cash value = ₹ 6,930
∴ Banker’s discount = 7,000 – 6,930 = ₹ 70
Date of drawing = 14/04
Date of discounting = 06/07
Rate of interest = 5%
Let the unexpired period = x days
∴ BD = \(\frac{7,000 \times x \times 5}{365 \times 100}\)
∴ 70 = \(\frac{70 \times x}{73}\)
∴ x = 73 days
∴ Legal due date of the bill is 73 days after the date of discounting.

∴ Legal due date = 17/09
∴ Nominal due date = 14/09
∴ Period of the bill = 5 months

Question 12.
If the difference between true discount and banker’s discount on a sum due 4 months hence is ₹ 20. Find true discount, banker’s discount and amount of bill, the rate of simple interest charged is 5% p.a.
Solution:
Banker’s gain (BG) = Banker’s discount (BD) – True Discount (TD)
∴ BG = ₹ 20
Also, BG = \(\frac{\mathrm{TD} \times n \times r}{100}\)
∴ 20 = \(\frac{\mathrm{TD} \times 4 \times 5}{12 \times 100}\)
∴ 20 = \(\frac{\mathrm{TD}}{60}\)
∴ TD = ₹ 1200
Now, BD = BG + TD
= 20 + 1,200
= ₹ 1,220
Also, BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 1,220 = \(\frac{\mathrm{FV} \times 4 \times 5}{12 \times 100}\)
∴ FV = 1,200 × 60 = ₹ 73,200
∴ Amounting the bill = ₹ 73,200

Question 13.
A bill of ₹ 51,000 was drawn on 18th February 2010 for 9 months. It was encashed on 28th June 2010 at 5% p.a. Calculate the banker’s gain and true discount.
Solution:
Face Value = ₹ 51,000
Date of drawing = 18/02/2010
Period of the bill = 9 months
Nominal due date = 18/11/2010
Legal due date = 21/11/2010
Date of discounting = 28/06/2010
Unexpired period

∴ TD = ₹ 1,000
∴ BG = BD – TD
= 1,020 – 1,000
= ₹ 20

Question 14.
A certain sum due 3 months hence is \(\frac{21}{20}\) of the present worth, what is the rate of interest.
Solution:

Question 15.
A bill of a certain sum drawn on 28th February 2007 for 8 months was encashed on 26th March 2007 for ₹ 10,992 at 14% p.a. Find the face value of the bill.
Solution:
Date drawing = 28/02/2007
Period of the bill = 8 months
Nominal due date = 28/10/2007
Legal due date = 31/10/2007
Date of discounting = 26/03/2007
Cash value = ₹ 10,992
Rate of interest = 14%
Let face value of the bill = ₹ x
Bankers discount = Face value – Cash value = x – 10,992
Also, Banker s discount = \(\frac{F V \times n \times r}{365 \times 100}\)
Where n is the unexpired days

Thus face value of the bill = ₹ 12,000

12 Commerce Maths Textbook Solutions

• Commission, Brokerage and Discount Ex 1.1 Class 12 Commerce Maths Solutions
• Commission, Brokerage and Discount Ex 1.2 Class 12 Commerce Maths Solutions
• Commission, Brokerage and Discount Miscellaneous Exercise 1 Class 12 Commerce Maths Solutions

Commission, Brokerage and Discount Class 12 Commerce Maths 2 Chapter 1 Exercise 1.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Ex 1.1 Questions and Answers.

Std 12 Maths 2 Exercise 1.1 Solutions Commerce Maths

Question 1.
An agent charges a 12% commission on the sales. What does he earn if the total sale amounts to ₹ 48,000? What does the seller get?
Solution:
Rate of commission = 12%
Total sales = ₹ 48,000
Agent’s commission = \(\frac {12}{100}\) × 48,000
= ₹ 5,760
Amount received by the seller = Total sales – commission
= ₹ 8,000 – ₹ 5760
= ₹ 2,240

Question 2.
A salesman receives a 3% commission on sales up to ₹ 50,000 and a 4% commission on sales over ₹ 50,000. Find his total income on the sale of ₹ 2,00,000.
Solution:
Total sales = ₹ 2,00,000
Rate of commission upto ₹ 50,000 = 3%
= \(\frac{3}{100}\) × 50,000
= ₹ 1,500
Rate of commission on the sales over ₹ 50,000 = 4%
Sales over ₹ 50,000 is 2,00,000 – 50,000 = ₹ 1,50,000
Commission on sales over ₹ 50,000 = \(\frac{4}{100}\) × 1,50,000 = ₹ 6,000
His total income = ₹ 1,500 + ₹ 6,000 = ₹ 7,500

Question 3.
Ms. Saraswati was paid ₹ 88,000 as commission on the sale of computers at the rate of 12.5%. If the price of each computer was ₹ 32,000, how many computers did she sell?
Solution:
Total commission = ₹ 88,000
Rate of commission = 12.5%
Let the number of computers sold be x
since price of each computer = ₹ 32,000
Total sales = ₹ 32,000x
Total commission = 12.5% of total sales
88,000 = \(\frac{12.5}{100}\) × 32,000x
= \(\frac{125}{1000}\) × 32,000x
x = \(\frac{88,000}{125 \times 32}\)
x = 22

Question 4.
Anita is allowed 6.5% commission on the total sales made by her, plus, a bonus of \(\frac{1}{2}\)% on the sale over ₹ 20,000. If her total commission amounts to ₹ 3,400. Find the sales made by her.
Solution:
Let the total sales made by Anita be ₹ x
Rate of commission = 6.5% of total sales
= \(\frac{6.5}{100} \times x\)
= \(\frac{65 x}{1,000}\)
= \(\frac{13 x}{200}\)

Question 5.
Priya gets a salary of ₹ 15,000 per month and a commission of 8% on sales over ₹ 50,000. If she gets ₹ 17,400 in a certain month. Find the sales made by her in that month.
Solution:
Let the total sales made by Priya be ₹ x
Salary of Priya = ₹ 15,000
Commission = Total earning – salary
= ₹ 17,400 – ₹ 15,000
= ₹ 2,400
Commission = 8% on the sales over ₹ 50,000
2400 = \(\frac{8}{100}\) (x – 50000)
\(\frac{2,400 \times 100}{8}\) = x – 50,000
30,000 = x – 50,000
30,000 + 50,000 = x
∴ x = ₹ 80,000

Question 6.
The income of the broker remains unchanged though the rate of commission is increased from 4% to 5%. Find the percentage reduction in the value of the business.
Solution:
Let the original value of business be ₹ 100
Original rate of commission = 4%
∴ Original commission = \(\frac{4}{100}\) × 100 = ₹ 4
Let the new value of business be ₹ x
The new rate of commission = 5%
∴ New commission = \(\frac{5}{100}\) × x = \(\frac{x}{20}\)
Given, original income = New income
4 = \(\frac{x}{20}\)
∴ x = ₹ 80
Thus there is 20% reduction in the value of the business.

Question 7.
Mr. Pavan is paid a fixed weekly salary plus commission based on a percentage of sales made by him. If on the sale of ₹ 68,000 and ₹ 73,000 in two successive weeks, he received in all ₹ 9,880 and ₹ 10,180. Find his weekly salary and the rate of commission paid to him.
Solution:
Let the weekly salary of Mr. Pavan be ₹ x and the rate of commission paid to him be y%
Income = Weekly salary + Commission on the sales
∴ 9,880 = x + \(\frac{y}{100}\) × 68,000
i.e. 9,880 = x + 680y …….(1)
Also, 10,180 = x + \(\frac{y}{100}\) × 73,000
i.e 10,180 = x + 730y ………(2)
Subtracting (1) from (2), we get
50y = 300
∴ y = 6
Substituting y = 6 in equation (1)
9,880 = x + 680(6) ‘
∴ 9,880 – 4,080 = x
∴ x = 5,800
Weekly salary = ₹ 5,800
Rate of commission = 6%

Question 8.
Deepak’s salary was increased from ₹ 4,000 to ₹ 5,000. The sales being the same, due to a reduction in the rate of commission from 3% to 2%, his income remained unchanged. Find his sales.
Solution:
Let Deepak’s total sales be ₹ x
Original salary of Deepak = ₹ 4,000
Original rate of commission = 3%
His new salary = ₹ 5,000
New rate of commission = 2%
Original income = New income (given)
4000 + \(\frac{3 x}{100}\) = 5000 + \(\frac{2 x}{100}\)
\(\frac{3 x}{100}-\frac{2 x}{100}\) = 5,000 – 4,000
\(\frac{x}{100}\) = 1000
x = ₹ 1,00,000
∴ His total sales = ₹ 1,00,000

Question 9.
An agent is paid a commission of 7% on cash sales and 5% on credit sales made by him. If on the sale of ₹ 1,02,000 the agent claims a total commission of ₹ 6,420, find his cash sales and credit sales.
Solution:
Total Sales = ₹ 1,02,000
Let cash sales ₹ x
∴ Credit sales = ₹ (1,02,000 – x)
Agent’s commission on cash sales = 7%
= \(\frac{7}{100}\) × x
= \(\frac{7x}{100}\)
Commission on credit sales = 5%
= \(\frac{5}{100}\)(1,02,000 – x)
Given, Total commission = ₹ 6,420
∴ \(\frac{7x}{100}\) + \(\frac{5}{100}\)(1,02,000 – x) = 6420
∴ \(\frac{7x}{100}\) + 5100 – \(\frac{5x}{100}\) = 6,420
∴ \(\frac{2x}{100}\) = 6,420 – 5,100
∴ \(\frac{2x}{100}\) = 1320
∴ x = ₹ 66,000
∴ Cash sales = ₹ 66,000
∴ Credit sales = 1,02000 – 66,000 = ₹ 36,000

Question 10.
Three cars were sold through an agent for ₹ 2,40,000, ₹ 2,22,000 and ₹ 2,25,000 respectively. The rates of the commission were 17.5% on the first, 12.5% on the second. If the agent overall received 14% commission on the total sales, find the rate of commission paid on the third car.
Solution:
Total selling price of three cars = 2,40,000 + 2,22,000 + 2,25,000 = ₹ 6,87,000
Commission on total sales = 14%
= \(\frac{14}{100}\) × 6,87,000
= ₹ 96,180
Selling price of first car = ₹ 2,40,000
Rate of commission = 17.5% = \(\frac{17.5}{100}\) × 2,40,000
∴ Commission on first car = ₹ 42,000
Selling price of second car = ₹ 2,22,000
Rate of commission = 12.5% = \(\frac{12.5}{100}\) × 2,22,000
∴ Commission on second car = ₹ 27,750
Selling price of third car = ₹ 2,25,000
Let the rate of commission be x%
Commission on third car = \(\frac{x}{100}\) × 2,25,000
96,180 – (42,000 + 27,750) = \(\frac{x}{100}\) × 2,25,000
\(\frac{26,430 \times 100}{2,25,000}\) = x
∴ x = 11.75
∴ Rate of commission on the third car = 11.75%

Question 11.
Swatantra Distributors allows a 15% discount on the list price of the washing machines. Further 5% discount is giver for cash payment. Find the list price of the washing machine if it was sold for the net amount of ₹ 38,356.25.
Solution:
Let the list price of the washing machine be ₹ 100
Trade discount = 15% = \(\frac{15}{100}\) × 100 = ₹ 15
∴ Invoice price =100 – 15 = ₹ 85
Cash discount = 5% = \(\frac{5}{100}\) × 85 = ₹ 4.25
∴ Net price = 85 – 4.25 = ₹ 80.75
Thus if List price is 100 than Net price is 80.75
if List price is x than Net price is 38,356.25.
∴ x = \(\frac{38356.25 \times 100}{80.75}\)
∴ x = ₹ 47,500
The list price of the washing machine is ₹ 47,500

Question 12.
A bookseller received ₹ 1,530 as a 15% commission on the list price. Find the list price of the books.
Solution:
Let the list price of the books be ₹ x
Rate of commission = 15%
Book seller’s commission = ₹ 1,530
∴ \(\frac{15}{100}\) × x = 1,530
∴ x = \(\frac{1,530 \times 100}{15}\)
∴ x = ₹ 10,200

Question 13.
A retailer sold a suit for ₹ 8,832 after allowing an 8% discount on market price and a further 4% cash discount. If he made 38% profit, find the cost price and the market price of the suit.
Solution:
Let the marked price of the suit be ₹ 100
Trade discount = 8% = \(\frac{8}{100}\) × 100 = ₹ 8
Invoice price = 100 – 8 = ₹ 92
Cash discount = 4% = \(\frac{4}{100}\) × 92 = ₹ 3.68
∴ Net price = 92 – 3.68 = ₹ 88.32
Thus if list price is 100 then net price is 88.32, if list price is x then net price is 8,832
∴ x = \(\frac{8,832 \times 100}{88.32}\)
∴ x = ₹ 10,000
The retailer made 38% profit.
Let the CP of the suit be ₹ 100
∴ SP of the suit = 100 + 38 = ₹ 138
Thus if the SP of the suit is ₹ 138 then its CP is ₹ 100
If the SP of the suit is 88.32 then its
CP = \(\frac{88.32 \times 100}{138}\) = ₹ 6400

Question 14.
An agent charges 10% commission plus 2% delcredere. If he sells goods worth ₹ 37,200, find his total earnings.
Solution:
Total sales = ₹ 37,200
Rate of commission = 10%
Agents commission = \(\frac{4}{100}\) × 37200 = ₹ 3720
Rate of delcredere = 2%
Amount of delcredere = \(\frac{2}{100}\) × 37,200 = ₹ 744
Total earning of the agent = ₹ 3,720 + ₹ 744 = ₹ 4,464

Question 15.
A whole seller allows a 25% trade discount and 5% cash discount. What will be the net price of an article marked at ₹ 1600?
Solution:
Marked price of the article = ₹ 1,600
Trade discount = 25%
= \(\frac{25}{100}\) × 1,600
= ₹ 400
∴ Invoice price = 1,600 – 400 = ₹ 1,200
Cash discount = 5%
= \(\frac{5}{100}\) × 1,200
= ₹ 60
∴ Net price = 1,200 – 60 = ₹ 1,140

12 Commerce Maths Textbook Solutions

• Commission, Brokerage and Discount Ex 1.1 Class 12 Commerce Maths Solutions
• Commission, Brokerage and Discount Ex 1.2 Class 12 Commerce Maths Solutions
• Commission, Brokerage and Discount Miscellaneous Exercise 1 Class 12 Commerce Maths Solutions

Measures of Dispersion Class 11 Commerce Maths 2 Chapter 2 Miscellaneous Exercise 2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Questions and Answers.

Std 11 Maths 2 Miscellaneous Exercise 2 Solutions Commerce Maths

Question 1.
Find the range for the following data.
116, 124, 164, 150, 149, 114, 195, 128, 138, 203, 144
Solution:
Here, largest value (L) = 203, smallest value (S) = 114
∴ Range = L – S
= 203 – 114
= 89

Question 2.
Given below the frequency distribution of weekly w ages of 400 workers. Find the range.

Solution:
Here, largest value (L) = 40, smallest value (S) = 10
∴ Range = L – S
= 40 – 10
= 30

Question 3.
Find the range of the following data.

Solution:
Here, upper limit of the highest class (L) = 175, lower limit of the lowest class (S) = 115
∴ Range = L – S
= 175 – 115
= 60

Question 4.
The city traffic police issued challans for not observing the traffic rules:

Find Q.D.
Solution:
The given data can be arranged in ascending order as follows:
24, 36, 40, 58, 62, 80
Here, n = 6
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation
= value of (1.75)th observation
= value of 1st observation + 0.75(value of 2nd observation – value of 1st observation)
= 24 + 0.75(36 – 24)
= 24 + 0.75(12)
= 24 + 9
∴ Q₁ = 33
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 1.75)th observation
= value of (5.25)th observation
= value of 5th observation + 0.25(value of 6th observation – value of 5th observation)
= 62 + 0.25(80 – 62)
= 62 + 0.25(18)
= 62 + 4.5
= 66.5
∴ Q.D. = \(\frac{Q_{3}-Q_{1}}{2}=\frac{66.5-33}{2}=\frac{33.5}{2}\) = 16.75

Question 5.
Calculate Q.D. from the following data.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 35
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{35}{4}\) = 8.75
Cumulative frequency which is just greater than (or equal to) 8.75 is 15.
∴ Q₁ lies in the class 20-30.
∴ L = 20, c.f. = 8, f = 7, h = 10

Question 6.
Calculate the appropriate measure of dispersion for the following data.

Solution:
Since open-ended classes are given, the appropriate measure of dispersion that we can compute is the quartile deviation.
We construct the less than cumulative frequency table as follows:

Here N = 250
Q₁ class class containing \(\left(\frac{N}{4}\right)^{t h}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{250}{4}\) = 62.5
Cumulative frequency which is just greater than (or equal to) 62.5 is 65.
∴ Q₁ lies in the class 35-40.
∴ L = 35, c.f. = 15, f = 50, h = 5

The cumulative frequency which is just greater than (or equal to) 187.5 is 190.
∴ Q₃ lies in the class 45-50.
∴ L = 45, c.f. = 150, f = 40, h = 5

Question 7.
Calculate Q.D. of the following data.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 120
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30
Cumulative frequency which is just greater than (or equal to) 30 is 35.
∴ Q₁ lies in the class 4-6.
∴ L = 4, c.f. = 15, f = 20, h = 2

Cumulative frequency which is just greater than (or equal to) 90 is 90.
∴ Q₃ lies in the class 10-12.
∴ L = 10, c.f. = 72, f = 18, h = 2

Question 8.
Find variance and S.D. for the following set of numbers.
25, 21, 23, 29, 27, 22, 28, 23, 27, 25 (Given √6.6 = 2.57)
Solution:
We prepare the following table for the calculation of variance and S.D.:

Question 9.
Following data gives no. of goals scored by a team in 90 matches.

Compute the variance and standard deviation for the above data.
Solution:
We prepare the following table for the calculation of variance and S.D:

Question 10.
Compute the arithmetic mean and S.D. and C.V. (Given √296 = 17.20)

Solution:
We prepare the following table for calculation of arithmetic mean and S.D.:

Question 11.
The mean and S.D. of 200 items are found to be 60 and 20 respectively. At the time of calculation, two items were wrongly taken as 3 and 67 instead of 13 and 17. Find the correct mean and variance.
Solution:
Here, n = 200, \(\bar{x}\) = Mean = 60, S.D. = 20
Wrongly taken items are 3 and 67.
Correct items are 13 and 17.
Now, \(\bar{x}\) = 60

Correct value of \(\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i}\) (sum of wrongly taken items) + (sum of correct items)
= 12000 – (3 + 67) + (13 + 17)
= 12000 – 70 + 30
= 11960
Correct value of mean = \(\frac{1}{n}\) × correct value of \(\sum_{i=1}^{n} x_{i}\)
= \(\frac{1}{200}\) × 11960
= 59.8
Now, S.D. = 20
Variance = (S.D.)² = 20²
∴ Variance = 400
∴ \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^{2}-(\bar{x})^{2}=400\)

Correct value of \(\sum_{i=1}^{n} x_{i}^{2}\)
= \(\sum_{i=1}^{n} x_{i}^{2}\) – (Sum of squares of wrongly taken items) + (Sum of squares of correct items)
= 800000 – (3² + 67²) + (13² + 17²)
= 800000 – (9 + 4489) + (169 + 289)
= 800000 – 4498 + 458
= 795960
∴ Correct value of Variance = (\(\frac{1}{n}\) × \(\sum_{i=1}^{n} x_{i}^{2}\)) – (correct value of \(\bar{x}\))²
= \(\frac{1}{200}\) × 795960 – (59.8)²
= 3979.8 – 3576.04
= 403.76
∴ The correct mean is 59.8 and correct variance is 403.76.

Question 12.
The mean and S.D. of a group of 48 observations are 40 and 8 respectively. If two more observations 60 and 65 are added to the set, find the mean and S.D. of 50 items.
Solution:

Question 13.
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and combined S.D.
Solution:
Let n₁ and n₂ be the number of boys and girls respectively.
Let n = 200, \(\bar{x}_{\mathrm{c}}\) = 65, \(\bar{x}_{1}\) = 70, \(\bar{x}_{1}\) = 62, σ₁ = 8, σ₂ = 10
Here, n₁ + n₂ = n
∴ n₁ + n₂ = 200 …….(i)
Combined mean is given by

∴ 70n₁ + 62n₂ = 13000
∴ 35n₁ + 31n₂ = 6500 ……..(ii)
Solving (i) and (ii), we get
n₁ = 75, n₂ = 125
Combined standard deviation is given by,

Question 14.
From the following data available for 5 pairs of observations of two variables x and y, obtain the combined S.D. for all 10 observations,
where \(\sum_{i=1}^{n} x_{i}=30, \sum_{i=1}^{n} y_{i}=40, \sum_{i=1}^{n} x_{i}^{2}=225, \sum_{i=1}^{n} y_{i}^{2}=340\)
Solution:

Question 15.
The mean and standard deviations of two brands of watches are given below:

Calculate the coefficient of variation of the two brands and interpret the results.
Solution:

Since C.V. (I) > C.V. (II)
∴ the brand I is more variable.

Question 16.
Calculate the coefficient of variation for the data given below. [Given √3.3 = 1.8166]

Solution:

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths

Measures of Dispersion Class 11 Commerce Maths 2 Chapter 2 Exercise 2.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.3 Questions and Answers.

Std 11 Maths 2 Exercise 2.3 Solutions Commerce Maths

Question 1.
The mean and standard deviation of two distributions of 100 and 150 items are 50, 5, and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Solution:


∴ The mean and standard deviation of all 250 items taken together are 44 and √55.6 respectively.

Question 2.
For certain bivariate data, the following information is available.

Obtain the combined standard deviation.
Solution:

Question 3.
Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are: 2, 4, 6, 8, 10. (Given: √8 = 2.8284)
Solution:

Question 4.
Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Solution:
Given, \(\bar{x}\) = 25, σ = 5
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
= 100 × \(\frac{5}{25}\)
= 20%

Question 5.
A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variation of 2.5%. What is the standard deviation of their height?
Solution:
Given, n = 65, \(\bar{x}\) = 150.4, C.V. = 2.5%
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
∴ 2.5 = 100 × \(\frac{\sigma}{150.4}\)
∴ \(\frac{2.5 \times 150.4}{100}\) = σ
∴ σ = 3.76
∴ the standard deviation of students’ height is 3.76.

Question 6.
Two workers on the same job show1 the following results:

(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Solution:

(i) Since, C.V. (P) < C.V.(Q)
∴ Worker P is more consistent regarding the time required to complete the job.

(ii) Since, \(\bar{p}\) > \(\bar{q}\)
i.e., the expected time for completing the job is less for worker Q.
∴ Worker Q seems to be faster in completing the job.

Question 7.
A company has two departments with 42 and 60 employees respectively. Their average weekly wages are ₹ 750 and ₹ 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Solution:

(i) Since, \(\bar{x}_{1}>\bar{x}_{2}\)
i.e., average weekly wages are more for the first department.
∴ the first department has a larger bill.
(ii) Since, C.V. (1) < C.V. (2)
∴ the second department is less consistent.
∴ the second department has larger variability in wages.

Question 8.
The following table gives the weights of the students of class A. Calculate the coefficient of variation (Given √8 = 0.8944)

Solution:

Question 9.
Compute coefficient of variation for team A and team B. (Given: √2.5162 = 1.5863, √2.244 = 1.4980)

Which team is more consistent?
Solution:
Let f₁ denote no. of goals of team A and f₂ denote no. of goals of team B.


Since C.V. of team A > C.V. of team B.
∴ team B is more consistent.

Question 10.
Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?

Solution:

Since C.V. (S) > C.V. (M)
∴ The subject statistics show higher variability in marks.

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths

Measures of Dispersion Class 11 Commerce Maths 2 Chapter 2 Exercise 2.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.2 Questions and Answers.

Std 11 Maths 2 Exercise 2.2 Solutions Commerce Maths

Question 1.
Find the variance and S.D. for the following set of numbers.
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
Solution:
Given data: 7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
The tabulated form of the above data is as given below:

We prepare the following table for the calculation of variance and S. D.

Question 2.
Find the variance and S.D. for the following set of numbers.
65, 77, 81, 98, 100, 80, 129
Solution:

Question 3.
Compute the variance and standard deviation for the following data:

Solution:
We prepare the following table for the calculation of variance and S.D.:

Question 4.
Compute the variance and S.D.

Solution:
We prepare the following table for the calculation of variance and S.D.:

Question 5.
The following data gives the age of 100 students in a school. Calculate variance and S.D.

Solution:
We prepare the following table for the calculation of variance and S.D:

Question 6.
The mean and variance of 5 observations are 3 and 2 respectively. If three of the five observations are 1, 3, and 5, find the values of the other two observations.
Solution:


∴ (x₄ – 4)(x₄ – 2) = 0
∴ x₄ = 4 or x₄ = 2
From (i), we get
x₅ = 2 or x₅ = 4
∴ The two numbers are 2 and 4.

Question 7.
Obtain standard deviation for the following data:

Solution:
We prepare the following table for the calculation of standard deviation.

Question 8.
The following distribution was obtained by change of origin and scale of variable X.

If it is given that mean and variance are 59.5 and 413 respectively, determine actual class intervals.
Solution:
Here, Mean = \(\bar{x}\) = 59.5, and
Var(X) = σ² = 413
Let x_i be a mid value of class and
d = \(\frac{x-a}{h}\), where a is assumed mean and h is class width.
We prepare the following table for calculation of mean and variance of d_i.


Now, Var(X) = h². Var(D)
∴ 413 = h² × 4.13
∴ h² = 100
∴ h = 10
Substituting h = 10 in (i), we get
-0.1 × 10 + a = 59.5
∴ -1 + a = 59.5
∴ a = 59.5 + 1
∴ a = 60.5
We prepare the following table to determine actual class intervals for corresponding values of d_i.

∴ The actual class intervals are 15.5 – 25.5, 25.5 – 35.5, …….., 95.5 – 105.5

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths

Measures of Dispersion Class 11 Commerce Maths 2 Chapter 2 Exercise 2.1 Answers Maharashtra Board

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Std 11 Maths 2 Exercise 2.1 Solutions Commerce Maths

Question 1.
Find range of the following data:
575, 609, 335, 280, 729, 544, 852, 427, 967, 250
Solution:
Here, largest value (L) = 967, smallest value (S) = 250
∴ Range = L – S
= 967 – 250
= 717

Question 2.
The following data gives the number of typing mistakes done by Radha during a week. Find the range of the data.

Solution:
Here, largest value (L) = 21, smallest value (S) = 10
∴ Range = L – S
= 21 – 10
= 11

Question 3.
Find range for the following data:

Solution:
Here, upper limit of the highest class (L) = 72, lower limit of the lowest class (S) = 62
∴ Range = L – S
= 72 – 62
= 10

Question 4.
Find the Q. D. for the following data.
3, 16, 8, 15, 19, 11, 5, 17, 9, 5, 3.
Solution:
The given data can be arranged in ascending order as follows:
3, 3, 5, 5, 8, 9, 11, 15, 16, 17, 19
Here, n = 11
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{11+1}{4}\right)^{\text {th }}\) observation
= value of 3rd observation
∴ Q₁ = 5
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{11+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 3)th observation
= value of 9th observation
= 16
∴ Q.D.= \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)
= \(\frac{16-5}{2}\)
= \(\frac{11}{2}\)
= 5.5

Question 5.
Given below are the prices of shares of a company for the last 10 days. Find Q.D.:
172, 164, 188, 214, 190, 237, 200, 195, 208, 230.
Solution:
The given data can be arranged in ascending order as follows:
164, 172, 188, 190, 195, 200, 208, 214, 230, 237
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)
= 172 + 0.75(188 – 172)
= 172 + 0.75(16)
= 172 + 12
= 184
∴ 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)
= 214 + 0.25(230 – 214)
= 214 + 0.25(16)
= 214 + 4
= 218
∴ Q.D. = \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)
= \(\frac{218-184}{2}\)
= \(\frac{34}{2}\)
= 17

Question 6.
Calculate Q.D. for the following data.

Solution:
Since the given data is arranged in ascending order, we construct less than cumulative frequency table as follows:

Here, n = 30
Q₁ = value of \(\left(\frac{\mathrm{n}+1}{4}\right)^{\mathrm{th}}\) observation
= value of \(\left(\frac{30+1}{4}\right)^{\text {th }}\) observation
= value of (7.75)th observation
Cumulative frequency which is just greater than (or equal to) 7.75 is 11.
∴ Q₁ = 25
Q₃ = value of \(\left[3\left(\frac{\mathrm{n}+1}{4}\right)\right]^{\mathrm{th}}\) observation
= value of \(\left[3\left(\frac{30+1}{4}\right)\right]^{\text {th }}\) observation
= value of (3 × 7.75)th observation
= value of (23.25)th observation
Cumulative frequency which is just greater than (or equal to) 23.25 is 27.
∴ Q₃ = 29
∴ Q.D. = \(\frac{Q_{3}-Q_{1}}{2}\)
= \(\frac{29-25}{2}\)
∴ Q.D. = 2

Question 7.
Following data gives the age distribution of 240 employees of a firm. Calculate Q.D. of the distribution.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 240
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{N}{4}=\frac{240}{4}\) = 60
Cumulative frequency which is just greater than (or equal to) 60 is 70.
∴ Q₁ lies in the class 25 – 30.
∴ L = 25, c.f. = 30, f = 40, h = 5

Cumulative frequency which is just greater than (or equal to) 180 is 180.
∴ Q₃ lies in the class 35-40.
∴ L = 35, c.f. = 130, f = 50, h = 5

Question 8.
Following data gives the weight of boxes. Calculate Q.D. for the data.

Solution:

Here, N = 60
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{60}{4}\) = 15
Cumulative frequency which is just greater than (or equal to) 15 is 26.
∴ Q₁ lies in the class 14 – 16.
∴ L = 14, c.f. = 10, f = 16, h = 2

Cumulative frequency which is just greater than (or equal to) 45 is 58.
∴ Q₃ lies in the class 18 – 20.
∴ L = 18, c.f. = 40, f = 18, h = 2

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths

Partition Values Class 11 Commerce Maths 2 Chapter 1 Miscellaneous Exercise 1 Answers Maharashtra Board

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

Std 11 Maths 2 Miscellaneous Exercise 1 Solutions Commerce Maths

Question 1.
The data gives the number of accidents per day on a railway track. Compute Q₂, P₁₇, and D₇.
4, 2, 3, 5, 6, 3, 4, 1, 2, 3, 2, 3, 4, 3, 2
Solution:
The given data can be arranged in ascending order as follows:
1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6
Here, n = 15
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 = 3
P₁₇ = value of 17\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 17\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (17 × 0.16)th observation
= value of (2.72)th observation
= value of 2nd observation + 0.72 (value of 3rd observation – value of 2nd observation)
= 2 + 0.72 (2 – 2)
∴ P₁₇ = 2
D₇ = value of 7\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 7\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (7 × 1.6)th observation
= value of (11.2)th observation
= value of 11th observation + 0.2(value of 12th observation – value of 11th observation)
= 4 + 0. 2(4 – 4)
∴ D₇ = 4

Question 2.
The distribution of daily sales of shoes (size-wise) for 100 days from a certain shop is as follows:

Compute Q₁, D₂, and P₉₅.
Solution:
By arranging the given data in ascending order, we construct the less than cumulative frequency table as given below:

Here, n = 100
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation
= value of (25.25)th observation
Cumulative frequency which is just greater than (or equal) to 25.25 is 27.
∴ Q₁ = 3
D₂ = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{100+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 10.1)th observation
= value of (20.2)th observation
Cumulative frequency which is just greater than (or equal) to 20.2 is 27.
∴ D₂ = 3
P₉₅ = value of 95\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 95\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (95 × 1.01)th observation
= value of (95.95)th observation
The cumulative frequency which is just greater than (or equal) to 95.95 is 100.
∴ P₉₅ = 8

Question 3.
Ten students appeared for a test in Mathematics and Statistics and they obtained the marks as follows:

If the median will be the criteria, in which subject, the level of knowledge of the students is higher?
Solution:
Marks in Mathematics can be arranged in ascending order as follows:
23, 23, 25, 25, 32, 35, 36, 37, 38, 42
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
Median = value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 32 + 0.5 (35 – 32)
= 32 + 0.5(3)
= 32 + 1.5
= 33.5
Marks in Statistics can be arranged in ascending order as follows:
22, 23, 26, 28, 29, 32, 34, 36, 45, 50
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 29 + 0.5(32 – 29)
= 29 + 0.5(3)
= 29 + 1.5
= 30.5
∴ Median marks for Mathematics = 33.5 and
Median marks for Statistics = 30.5
∴ The level of knowledge in Mathematics is higher than that of Statistics.

Question 4.
In the frequency distribution of families given below, the number of families corresponding to expenditure group 2000 – 4000 is missing from the table. However, the value of the 25th percentile is 2880. Find the missing frequency.

Solution:
Let x be the missing frequency of expenditure group 2000 – 4000.
We construct the less than cumulative frequency table as given below:

Here, N = 75 + x
Given, P₂₅ = 2880
∴ P₂₅ lies in the class 2000 – 4000.
∴ L = 2000, h = 2000, f = x, c.f. = 14
∴ P₂₅ = L + \(\frac{h}{f}\left(\frac{25 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 2880 = 2000 + \(\frac{2000}{x}\left(\frac{75+x}{4}-14\right)\)
∴ 2880 – 2000 = \(\frac{2000}{x}\left(\frac{75+x-56}{4}\right)\)
∴ 880x = 500(x + 19)
∴ 880x = 500x + 9500
∴ 880x – 500x = 9500
∴ 380x = 9500
∴ x = 25
∴ 25 is the missing frequency of the expenditure group 2000 – 4000.

Question 5.
Calculate Q₁, D₆, and P₁₅ for the following data:

Solution:
Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
∴ the class intervals will be 0 – 50, 50 – 100, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 160.
Q₁ lies in the class 100 – 150.
∴ L = 100, h = 50, f = 80, c.f. = 80
∴ Q₁ = L + \(\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 100 + \(\frac{50}{80}\)(125 – 80)
= 100 + \(\frac{5}{8}\)(45)
= 100 + 28.125
= 128.125
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 500}{10}\) = 300
Cumulative frequency which is just greater than (or equal) to 300 is 410.
∴ D₆ lies in the class 200 – 250.
∴ L = 200, h = 50, f = 150, c.f. = 260
∴ D₆ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{6 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 200 + \(\frac{50}{150}\)(300 – 260)
= 200 + \(\frac{1}{3}\)(40)
= 200 + 13.33
= 213.33
P₁₅ class = class containing \(\left(\frac{15 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{15 \mathrm{~N}}{100}=\frac{15 \times 500}{100}\) = 75
Cumulative frequency which is just greater than (or equal) to 75 is 80.
∴ P15 lies in the class 50 – 100.
∴ L = 50, h = 50, f = 70, c.f. = 10
∴ P₁₅ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{15 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 50 + \(\frac{50}{70}\) (75 – 10)
= 50 + \(\frac{5}{7}\) (65)
= 50 + \(\frac{325}{7}\)
= 50 + 46.4286
= 96.4286
∴ Q₁ = 128.125, D₆ = 213.33, P₁₅ = 96.4286

Question 6.
Daily income for a group of 100 workers are given below:

P₃₀ for this group is ₹ 110. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 50 – 100 and class 200 – 250 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 ……(i)
Given, P30 = 110
∴ P30 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 25, c.f. = 7 + a
\(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
∴ P₃₀ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{30 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 110 = 100 + \(\frac{50}{25}\) [30 – (7 + a)]
∴ 110 – 100 = 2(30 – 7 – a)
∴ 10 = 2(23 – a)
∴ 5 = 23 – a
∴ a = 23 – 5
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18
∴ b = 20
∴ 18 and 20 are the missing frequencies of the class 50 – 100 and class 200 – 250 respectively.

Question 7.
The distribution of a sample of students appearing for a C.A. examination is:

Help C.A. institute to decide cut-off marks for qualifying for an examination when 3% of students pass the examination.
Solution:
To decide cut-off marks for qualifying for an examination when 3% of students pass, we have to find P97.
We construct the less than cumulative frequency table as given below:

Here, N = 1100
P97 class = class containing \(\left(\frac{97 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{97 \mathrm{~N}}{100}=\frac{97 \times 1100}{100}\) = 1067
Cumulative frequency which is just greater than (or equal) to 1067 is 1100.
∴ P₉₇ lies in the class 500 – 600.
∴ L = 500, h = 100, f = 130, c.f. = 970
∴ P₉₇ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{97 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 500 + \(\frac{100}{130}\)(1067 – 970)
= 500 + \(\frac{10}{13}\) (97)
= 500 + 74.62
= 574.62 ~ 575
∴ the cut off marks for qualifying an examination is 575.

Question 8.
Determine graphically the value of median, D₃, and P₃₅ for the data given below:

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 (15, 8), (20, 22), (25, 30), (30, 55), (35, 70), (40, 84), (45, 90).

N = 90
For median, consider \(\frac{\mathrm{N}}{2}=\frac{90}{2}\) = 45
For D3, consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 90}{10}\) = 27
For P35, consider \(\frac{35 \mathrm{~N}}{100}=\frac{35 \times 90}{100}\) = 31.5
∴ We take the values 45, 27 and 31.5 on the Y-axis and draw lines from these points parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendicular on the X-axis.
Foot of the perpendicular represent the values of median, D3 and P35 respectively.
∴ Median ~ 29, D₃ ~ 23.5, P₃₅ ~ 26

Question 9.
The I.Q. test of 500 students of a college is as follows:

Find graphically the number of students whose I.Q. is more than 55 graphically.
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 (30, 41), (40, 93), (50, 157), (60, 337), (70, 404), (80, 449), (90, 489), (100, 500)

To find the number of students whose I.Q. is more than 55, we consider the value 55 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point this line intersects the less than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of students whose I.Q. is less than 55.
∴ The foot of perpendicular ~ 244
∴ No. of students whose I.Q. is less than 55 ~ 244
∴ No. of Students whose I.Q. is more than 55 = 500 – 244 = 256

Question 10.
Draw an ogive for the following distribution. Determine the median graphically and verify your result by a mathematical formula.

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 (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).

N = 60
∴ \(\frac{\mathrm{N}}{2}=\frac{60}{2}\) = 30
∴ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis.
From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis.
The foot perpendicular gives the value of the median.
∴ Median ~ 164.67
Now, let us calculate the median from the mathematical formula.
∴ \(\frac{\mathrm{N}}{2}\) = 30
The median lies in the class interval 160 – 165.
∴ L = 160, h = 5, f = 15, c.f. = 16
Median = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{2}-\mathrm{c} . \mathrm{f} .\right)\)
= 160 + \(\frac{5}{15}\) (30 – 16)
= 160+ \(\frac{1}{3}\) × 14
= 160 + 4.67
= 164.67

Question 11.
In a group of 25 students, 7 students failed and 6 students got distinction and the marks of the remaining 12 students are 61, 36, 44, 59, 52, 56, 41, 37, 39, 38, 41, 64. Find the median marks of the whole group.
Solution:
n = 25
Median = \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}\) = 13th observation
We have been stated that 7 students failed (assuming passing marks on 35) and 6 students got distinction (assuming distinction as 70+), and the marks of the remaining 12 students (who will be situated between the two groups mentioned above, if arranged in ascending order), we have,
F, F, F, F, F, F, F, 36, 37, 38, 39, 41, 41, 44, 52, 56, 59, 61, 64, D, D, D, D, D, D
∴ median = 13th observation = 41.

Question 12.
The median weight of a group of 79 students is found to be 55 kg. 6 more students are added to this group whose weights are 50, 51, 52, 59.5, 60, 61 kg. What will be the value of the median of the combined group if the lowest and the highest weights were 53 kg and 59 kg respectively?
Solution:
n = 79
Median = 55kg
Lowest observation = 53 kg
Flighest observation = 59 kg
6 new students are added to the group having weights in Kg as follows:
50, 51, 52, 59.5, 60, 61
From the above, we see that of the 6 new students, 3 have weights which are below the lowest weight of the earlier group and 3 have weights which are above the highest weight of the earlier group.
∴ the median remains the same
∴ median = 55 kg.

Question 13.
The median of the following incomplete table is 92. Find the missing frequencies:

Solution:
Let a and b be the missing frequencies of class 50 – 70 and class 110 – 130 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 54 + a + b
Since, N = 80
∴ 54 + a + b = 80
∴ a + b = 26 …..(i)
Given, Median = Q₂ = 92
∴ Q₂ lies in the class 90 – 110.
∴ L = 90, h = 20, f = 20, c.f. = 24 + a
\(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 80}{4}\) = 40
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
∴ 92 = 90 + \(\frac{20}{20}\) [40 – (24 + a)
∴ 92 – 90 = 40 – 24 – a
∴ 2 = 16 – a
∴ a = 14
Substituting the value of a in equation (i), we get
14 + b = 26
∴ b = 26 – 14 = 12
∴ 14 and 12 are the missing frequencies of the class 50 – 70 and class 110 – 130 respectively.

Question 14.
A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

estimate the number of defective tubes in the central batch.
Solution:
To find the number of defective tubes in the central batch, we have to find Q₂.
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 Less than 4.5, 4.5 – 9.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 251
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 251}{4}\) = 125.5
Cumulative frequency which is just greater than (or equal to) 125.5 is 180.
∴ Q₂ lies in the class 9.5 – 14.5.
∴ L = 9.5, h = 5, f = 84, c.f. = 96
∴ Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 9.5 + \(\frac{5}{84}\) (125.5 – 96)
= 9.5 + \(\frac{5}{84}\) × 29.5
= 9.5 + \(\frac{147.5}{84}\)
= 9.5 + 1.76
= 11.26

Question 15.
In a college, there are 500 students in junior college, 5% score less than 25 marks, 68 scores from 26 to 30 marks, 30% score from 31 to 35 marks, 70 scores from 36 to 40 marks, 20% score from 41 to 45 marks and the rest score 46 and above marks. What are the median marks?
Solution:
Given data can be written in tabulated form as follows:

Since the given data is not continuous, we have to convert it into the 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 Less than 25.5, 25.5 – 30.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 500}{4}\) = 250
Cumulative frequency which is just greater than (or equal to) 250 is 313.
∴ Q₂ lies in the class 35.5 – 40.5.
∴ L = 35.5, h = 5, f = 70, c.f. = 243
∴ Median = Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 35.5 + \(\frac{5}{70}\) (250 – 243)
= 35.5 + \(\frac{1}{14}\) (7)
= 35.5 + 0.5
= 36

Question 16.
Draw a cumulative frequency curve more than typical for the following data and hence locate Q₁ and Q₃. Also, find the number of workers with daily wages
(i) Between ₹ 170 and ₹ 260
(ii) less than ₹ 260

Solution:
For more than ogive points to be plotted are (100, 200), (150, 188), (200, 160), (250, 124), (300, 74), (350, 49), (400, 31), (450, 15), (500, 5)

Here, N = 200
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{200}{4}\) = 4
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 200}{4}\) = 150
We take the points having Y co-ordinates 50 and 150 on Y-axis.
From these points, we draw lines which are parallel to X-axis.
From the points of intersection of these lines with the curve, we draw perpendicular on X-axis.
X-Co-ordinates of these points gives the values of Q₁ and Q₃.
Since X-axis has daily wages more than and not less than the given amounts.
∴ Q₁ = Q₃ and Q₃ = Q₁
∴ Q₂ ~ 215 , Q₃ ~ 348

(i) To find the number of workers with daily wages between ₹ 170 and ₹ 260,
Take the values 170 and 260 on X-axis. From these points, we draw lines parallel to Y-axis.
From the point where they intersect the more than ogive, we draw perpendiculars on Y-axis.
The points where they intersect the Y-axis gives the values 178 and 114.
∴ Number of workers having daily wages between ₹ 170 and ₹ 260 = 178 – 114 = 64

(ii) To find the number of workers having daily wages less than ₹ 260, we consider the value 260 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point where the line intersects the more than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of workers having daily wages of more than 260.
The foot of perpendicular ~ 114
∴ No. of workers whose daily wages are more than ₹ 260 ~ 114
∴ No. of workers whose daily wages are less than ₹ 260 = 200 – 114 = 86

Question 17.
Draw ogive of both the types for the following frequency distribution and hence find the median.

Solution:

For less than given points to be plotted are (10, 5), (20, 10), (30, 18), (40, 30), (50, 46), (60, 61), (70, 71), (80, 79), (90, 84), (100, 86)
For more than given points to be plotted are (0, 86), (10, 81), (20, 76), (30, 68), (40, 56), (50, 40), (60, 25), (70, 15), (80, 7), (90, 2)

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.

Question 18.
Find Q1, D6 and P78 for the following data:

Solution:
Since the given data is not in the form of a continuous frequency distribution, we have to convert it into that form by subtracting 0.025 from the lower limit and adding 0.025 to the upper limit of each class interval.
∴ the class intervals will be 7.975 – 8.975, 8.975 – 9.975, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 50
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{50}{4}\) = 12.5
Cumulative frequency which is just greater than (or equal) to 12.5 is 15.
∴ Q₁ lies in the class 8.975 – 9.975.
∴ L = 8.975, h = 1, f = 10, c.f. = 5
Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 8.975 + \(\frac{1}{10}\) (12.5 – 5)
= 8.975 + 0.1(7.5)
= 8.975 + 0.75
= 9.725
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 50}{10}\) = 30
Cumulative frequency which is just greater than (or equal) to 30 is 35.
∴ D6 lies in the class 9.975 – 10.975.
∴ L = 9.975, h = 1, f = 20, c.f. = 15
D₆ = L + \(\frac{h}{f}\left(\frac{6 N}{10}-\text { c.f. }\right)\)
= 9.975 + \(\frac{1}{20}\) (30 – 15)
= 9.975 + 0.05(15)
= 9.975 + 0.75
= 10.725
P₇₈ class = class containing \(\left(\frac{78 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
\(\frac{78 \mathrm{~N}}{100}=\frac{78 \times 50}{100}\) = 39
Cumulative frequency which is just greater than (or equal) to 39 is 45.
∴ P₇₈ lies in the class 10.975 – 11.975.
∴ L = 10.975, h = 1, f = 10, c.f. = 35
∴ P₇₈ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{78 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10.975 + \(\frac{1}{10}\) (39 – 35)
= 10.975 + 0.1(4)
= 10.975 + 0.4
= 11.375

Question 19.

For the above data, find all quartiles and number of persons weighing between 57 kg and 72 kg.
Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 111
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{111}{4}\) = 27.75
Cumulative frequency which is just greater than (or equal) to 27.75 is 39.
∴ Q₁ lies in the class 50 – 55.
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 50 + \(\frac{5}{20}\) (27.75 – 19)
= 50 + \(\frac{1}{4}\) × 8.75
= 50 + 2.1875
= 52.1875
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 N}{4}=\frac{2 \times 111}{4}\) = 55.5
Cumulative frequency which is just greater than (or equal) to 55.5 is 69.
∴ Q₂ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 30, c.f. = 39
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
= 55 + \(\frac{5}{30}\) (55.5 – 39)
= 55 + \(\frac{1}{6}\) × 16.5
= 55 + 2.75
= 57.75
Q₃ class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 111}{4}\) = 83.25
Cumulative frequency which is just greater than (or equal) to 83.25 is 89.
∴ Q₃ lies in the class 60 – 65.
∴ L = 60, h = 5, f = 20, c.f. = 69
∴ Q₃ = L + \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 60 + \(\frac{5}{20}\) (83.25 – 69)
= 60 + \(\frac{1}{4}\) × 14.25
= 60 + 3.5625
= 63.5625
In order to find the number of persons between 57 kg and 72 kg,
We need to find x in P_x, where P_x = 57 kg and y in P_y, where P_y = 72 kg
Then (y – x) would be the % of persons weighing between 57 kg and 72 kg
P_x = 57
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{x \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 57
∴ 55 + \(\frac{5}{30}\) (1.11x – 39) = 57
∴ \(\frac{1}{6}\) (1.11x – 39) = 2
∴ 1.11x – 39 = 12
∴ 1.11x = 51
∴ x = 45.95
∴ P_y = 72
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{y \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 72
∴ 70 + \(\frac{5}{8}\) (1.11y – 99) = 72
∴ 0.625(1.11y – 99) = 2
∴ 1.11y – 99 = 3.2
∴ 1.11y = 102.2
∴ y = 92.07
∴ % of people weighing between 57 kg and 72 kg = 92.07 – 45.95 = 46.12 %
∴ No. of people weighing between 57 kg and 72 kg = 111 × 46.12% = 51.1932 ~ 51

Question 20.
For the following data showing weights of 100 employees, find the maximum weight of the lightest 25% of employees.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 100
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
Cumulative frequency which is just greater than (or equal) to 25 is 29.
∴ Q₁ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 15, c.f. = 14
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 55 + \(\frac{5}{15}\) (25 – 14)
= 55 + \(\frac{1}{3}\) × 11
= 55 + 3.67
= 58.67
∴ Maximum weight of the lightest 25% of employees is 58.67 kg.

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths

Partition Values Class 11 Commerce Maths 2 Chapter 1 Exercise 1.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.3 Questions and Answers.

Std 11 Maths 2 Exercise 1.3 Solutions Commerce Maths

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

Determine D₆, Q₁, and P₈₅ graphically.
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).

Here, N = 100
For D₆, \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 100}{10}\) = 60
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₈₅, \(\frac{85 \mathrm{~N}}{100}=\frac{85 \times 100}{100}\) = 85
∴ 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.
∴ D₆ = 32.5, Q₁ = 26, P₈₅ = 40

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.
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).

Here, N = 500
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
For Q₂, \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 500}{4}\) = 375
∴ 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.
∴ Q₁ ~ 150, Q₂ ~ 228, Q₃ ~ 297
∴ Median = 228
50% families lie between Q₁ and Q₃
∴ Limits of income of central 50% families are from ₹ 150 to ₹ 297

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

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).

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.
∴ Median ~ 574

Question 4.
The following frequency distribution shows the profit (in ₹) 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 35,000 rupees.
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).

The Middle 40% value lies in between P₃₀ and P₇₀.
N = 100
For P₃₀ = \(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
For P₇₀ = \(\frac{70 \mathrm{~N}}{100}=\frac{70 \times 100}{100}\) = 70
∴ 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.
∴ P₃₀ ~ 20, P₇₀ ~ 36
Limits of middle 40% shops lie between ₹ 20,000 to ₹ 36,000
To find the number of shops having a profit of less than ₹ 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.
∴ No. of shops having profit less than ₹ 35,000 is 67.

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.

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

For D₂, we have to consider \(\frac{2 \mathrm{~N}}{10}=\frac{2 \times 80}{10}\) = 16
For Q₂, we have to consider \(\frac{\mathrm{N}}{2}=\frac{80}{2}\) = 40
and for P₆₁, we have to consider \(\frac{61 \mathrm{~N}}{100}=\frac{61 \times 80}{100}\) = 48.8
∴ 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.
∴ D₂ ~ 13, Q₂ ~ 19, P₆₁ ~ 20.5

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

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).

For D₁, we have to consider \(\frac{\mathrm{N}}{10}=\frac{100}{10}\) = 10
For Q₁, we have to consider \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₄₀, we have to consider \(\frac{40 \mathrm{~N}}{100}=\frac{40 \times 100}{100}\) = 40
∴ 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.
∴ D₁ ~ 22, Q₁ ~ 30.5, P₄₀ ~ 37

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.

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).

N = 400
For D₃, we have to consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 400}{10}\) = 120
For D₅, we have to consider \(\frac{5 \mathrm{~N}}{10}=\frac{5 \times 400}{10}\) = 200
For D₈, we have to consider \(\frac{8 \mathrm{~N}}{10}=\frac{8 \times 400}{10}\) = 320
∴ 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.
∴ D₃ ~ 44, D₅ ~ 55.5 and D₈ ~ 70

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

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).

N = 500
For median we have to consider \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
∴ 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

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).

N = 120
For Q₁ and Q₃ we have to consider
\(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30
\(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 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 Q₂.
∴ Q₁ ~ 1025 and Q₃ ~ 1248
∴ the limits of the weight of the middle 50% of fishes lie between 1025 to 1248.

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).

N = 200
For D₃, \(\frac{3 \mathrm{N}}{10}=\frac{3 \times 200}{10}\) = 60
For P₆₅, \(\frac{65 \mathrm{N}}{100}=\frac{65 \times 200}{100}\) = 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

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths

Partition Values Class 11 Commerce Maths 2 Chapter 1 Exercise 1.2 Answers Maharashtra Board

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

Std 11 Maths 2 Exercise 1.2 Solutions Commerce Maths

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

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:

Solution:
We construct the less than cumulative frequency table as given below:

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

Question 4.
From the following data calculate the rent of the 15th, 65th, and 92nd house.

Solution:
Arranging the given data in ascending order.

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.

(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

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:

Here, N = 120
Let P_x = 50
The value 50 lies in the class 47.5 – 52.5
∴ L = 47.5, h = 5, f = 29, c.f. = 55

∴ 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.

Question 6.
Calculate D₄ and P₄₈ from the following data:

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:

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.

Solution:
We construct the less than cumulative frequency table as given below:

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

∴ P₂₀ = 58.86

Question 8.
Weekly wages for a group of 100 persons are given below:

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:

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:

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:

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.

Question 10.
In a particular factory, workers produce various types of output units. The following distribution was obtained:

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:

Here, N = 445
Let P_x = 82
The value 82 lies in the class 79.5 – 84.5
∴ L = 79.5, h = 5, f = 50, c.f. = 85

∴ 24.72% of workers produced less than 82 output units.

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths

Partition Values Class 11 Commerce Maths 2 Chapter 1 Exercise 1.1 Answers Maharashtra Board

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

Std 11 Maths 2 Exercise 1.1 Solutions Commerce Maths

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

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.

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:

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.

Question 5.
Calculate all the quartiles for the following frequency distribution:

Solution:
We construct the less than cumulative frequency table as given below:

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:

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:

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.

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.

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:

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:

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:

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

Question 9.
Following is grouped data for the duration of fixed deposits of 100 senior citizens from a certain bank:

Calculate the limits of fixed deposits of central 50% senior citizens.
Solution:
We construct the less than cumulative frequency table as given below:

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.

Solution:
Let x be the missing frequency of the class 1550 – 1750.
We construct the less than frequency table as given below:

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

Maharashtra State Board 11th Commerce Maths

• Partition Values Ex 1.1 11th Commerce Maths
• Partition Values Ex 1.2 11th Commerce Maths
• Partition Values Ex 1.3 11th Commerce Maths
• Partition Values Miscellaneous Exercise 1 11th Commerce Maths
• Measures of Dispersion Ex 2.1 11th Commerce Maths
• Measures of Dispersion Ex 2.2 11th Commerce Maths
• Measures of Dispersion Ex 2.3 11th Commerce Maths
• Measures of Dispersion Miscellaneous Exercise 2 11th Commerce Maths
• Skewness Ex 3.1 11th Commerce Maths
• Skewness Miscellaneous Exercise 3 11th Commerce Maths