ALGEBRA

Adsul Notes

6. o

Statistics - II

Pie - diagram The relative values of items are represented by sector of a circle Since the sectors resemble the slices of pie, therefore it is called a Pie - diagram. The terms related to pie-diagrams are sector and central angle.

o

Steps to draw pie diagram :

ot es

Step 1 : Find out the measure of central angle () for each of the given information. Step 2 : Draw a circle having radius between 4 - 6 cm Step 3 : Draw sector corresponding to measures of central angle () and then lable the different sectors as per given information. Data Measure of central Angle () = × 360º Total

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 140) :

The following data give the number of students using different modes of transport : Mode of transport Number of students

Bicycle

Bus

Walk

Train

Car

140

100

70

40

10

N

1.

Represent the above data using pie diagram. Mode of transport No. of Students Bicycle

140 100

Ad s

Bus

ul

Sol.

Walk

70

Train

40

Car

10

Total

360

(3 marks)

Measure of central angle () 140 × 360º = 140º 360 100 × 360º = 100º 360 70 × 360º = 70º 360 40 × 360º = 40º 360 10 × 360º = 10º 360 360º

Bicycle 140º Bus

100º 70º

10º Car 40º Train

Walk

" Learn from the Best "

1

ALGEBRA 

Adsul Notes

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 140) : Following is the componentwise expenditure per article, draw a pie chart: Component

Expenditure (in Rs.)

Raw material Labour Transportation Packing Taxes Component

(3 marks)

Expenditure

Raw material

800

Labour

300

Transportation

100

Packing

100

Taxes

140

Total

1440

Measure of central angle

800 1440 300 1440 100 1440 100 1440 140 1440

× 360 = 200º × 360 =

75º

× 360 =

25º

× 360 =

25º

× 360 =

35º

360

kin Pac

N

Tran s pora tion

Sol.

800 300 100 100 140

ot es

2.

g

25º 25º 35º

Labour

ul

Taxes

75º

Ad s

200º

Raw materials

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 140) : 3.

Sol.

Area under different crops in a certain village is given below : Represent it by pie diagram (3 marks) Crop Jowar Wheat Sugarcane Vegetables Area in hectare 8000 6000 2000 2000 Crop

Area in hectare

Jowar

8000

Wheat

6000

Sugarcane

2000

Vegetables

2000

Total

" Learn from the Best "

18000

Measure 8000 × 18000 6000 × 18000 2000 × 18000 2000 × 18000

of central angle 360 = 160º 360 = 120º 360 =

40º

360 =

40º 360

2

ALGEBRA



Adsul Notes

Jowar 160º 40º 120º Vegetables Wheat 40º

Su ga rc an e

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 140) : Electricity used by farmers during different parts of a day for irrigation is as follows. Draw pie diagram : (3 marks) Part of day Percentage of electricity used

Morning Afternoon Evening

Night

30

40

20

10

Percentage of Measure of central angle electricity used 30 40 20

10

Ad s

Night

Evening

Total

30 100 40 100 20 100 10 100

N

Part of day

Afternoon

ul

Sol.

Morning

ot es

4.

100

× 360

= 108º

× 360

= 144º

× 360

=

72º

× 360

=

36º 360º

Morning 108º

Afternoon 144º

36º Night 72º Evening

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 140) : 5.

The following table gives information about the monetary investment by some residents in a city : Shares Mutual Read Mode of investment Gold Government funds estate bonds Percentage of residents 10 20 35 30 5

" Learn from the Best "

3

ALGEBRA 

Adsul Notes (3 marks)

Draw pie diagram to represent the data. Sol.

Percentage of residents

Mode of investment Shares

10

Mutual funds

20

Real estate

35

Gold

30

Government bonds

Measure of central angle

10 100 20 100 35 100 30 100 5 100

5

=

36º

× 360

=

72º

× 360

= 126º

× 360

= 108º

× 360

=

100

18º 360º

ot es

Total

× 360

Mutual funds

Shares

72º

36º 126º 18º Governmnet 108º Bonds

N

Real estate

Ad s

ul

Gold

PROBLEM SET - 6 (TEXT BOOK PAGE NO. 176) : 1.

The number of hours, spent by a school boy in different activities in a day is given below. Activity sleep school play home work other Total No. of hours 8 7 2 4 Represent the above information using pie diagram.

Sol.

Activity

No. of hrs.

sleep

8

school

7

play

2

home work

4

other

3

Total

24

" Learn from the Best "

3

24 (3 marks)

Measure of central angle ()

8 24 7 24 2 24 4 24 3 24

× 360º

= 120º

× 360º

= 105º

× 360º

=

30º

× 360º

=

60º

× 360º

=

45º 360º

4

ALGEBRA



Adsul Notes

Sleep School

120º 105º 30º

45º 60º

Other

Play Homework

PROBLEM SET - 6 (TEXT BOOK PAGE NO. 176) : Number of students admitted in different faculties of a college are given below : Faculty

ot es

2.

Science Commerce Arts

No. of students

1000

1200

650

Law Home science 450

300

Draw a pie diagram and represent the above data. Faculty

No. of students

Science

1000

1000 × 360 = 100º 3600

1200

1200 × 360 = 120º 3600

ul

Commerce

650

Ad s

Arts

Measure of central angle ()

N

Sol.

650 × 360 = 3600

65º

Law

450

450 × 360 = 3600

45º

Home science

300

300 × 360 = 3600

30º

Total

(3 marks)

3600

360º

Science Commerce 120º

100º

30º Home 65º 45º Science Law Arts

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5

ALGEBRA 

PROBLEM SET - 6 (TEXT BOOK PAGE NO. 176) : Draw a pie diagram to represent the world population given in the following table after determining the value of a : (3 marks) Country

India

Percentage of world population Sol.    

China Russia

15

20

As per given condition, 15 + 20 + a + a + 25 = 60 + 2a = 2a = 2a = a =

100 100 100 – 60 40 20

Percentage of world population

15 100 20 100 20 100 20 100 25 100

India

15

China

20

Russia

20

USA

20

Others

Total

a

25

100

× 360 = 54º

25

× 360 = 72º × 360 = 72º × 360 = 72º × 360 = 90º

100

360º

ul

Total

Other

Measure of central angle ()

N

Country

a

USA

ot es

17.

Adsul Notes

Ad s

China

Russia

72º

72º 72º

U.S.A.

6.

Sol.

India 54º

90º Other

EXERCISE - 6.1 (TEXT BOOK PAGE NO. ) : The following pie diagram represents expenditure on different items in constructing a building. Answer the following questions : (a) Find the expenditure of each of the items if the total construction cost is Rs. 5,40,000. (b) Which is the item with the maximum expenditure ? (c) Which is the item with the minimum (3 marks) expenditure ? Total expenditure = Rs. 540000

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Bricks 50º Cement 75º 100º 45º 90º Steel

Labour

Timber

6

ALGEBRA



Items

Measure of central angle

Cement

75º

Bricks

50º

Labour

100º

Timber

90º

Steel

45º

Total

360º

Adsul Notes

Expenditure (in Rs.)

75 360 50 360 100 360 90 360 45 360

× 540000 = 112500 × 540000 = 75000 × 540000 = 150000 × 540000 = 135000 × 540000 = 67500 540000

ot es

(a) The item with maximum expenditure is labour. (b) The item with minimum expenditure is steel.

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 176) :

The following diagram represents the sectorwise loan amount in crores of Rs. distributed by a bank. From the information Agriculture answer the following questions : 120º (a) If the dairy sector received Rs. 20 40º crores, then find the total loan disbursed. Dairy (b) Find the loan amount for agriculture Industry sector and also for industrial sector. (c) How much additional amount did industrial sector received than agriculture sector. (3 marks) (a) Let the total loan disbursed be x crores The measure of central angle for dairy sector is 40º.  Dairy sector received 20 crores of the total loan i.e. x 40  x = 20  360 20  360  x = 40  x = 180

Ad s

Sol.

ul

N

7.

 Total loan disbursed is Rs. 180 crores. (b) Measure of central angle for agriculture sector is 120º

120  180 = 60 crores 360 Measure of central angle for industrial sector is 200º

 Amount disbursed for agriculture sector =

 Amount disbursed for industrial sector =

200  180 = 100 crores. 360

(c) Amount received by industrial sector = Rs. 100 crores. Amount received by agricultural sector = Rs. 60 crores.  Industrial sector received 100 – 60 = Rs. 40 crores more than agricultural sector.

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7

ALGEBRA 

Adsul Notes

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 178) :

Sol.

The following pie diagram shows percentage of persons according to blood group. Answer the following questions : (a) Find the measure of central angle for each blood group. (b) Find total number of persons if there are 600 persons of blood group B. (3 marks) Bloodgroup

Percentage

O

45

A

20

B

30

AB

5

O

A

45% 5% AB 20% 30% B

Measure of central angle 45 × 360 = 162º 100 20 × 360 = 72º 100 30 × 360 = 108º 100 5 × 360 = 18º 100 360º

ot es

8.

Ad s

ul

N

Total 100 Let total number of persons be x. No. of persons with group B = 600 persons Percentage of persons with group B = 30 30  x = 600  100 600  100  x = 30  x = 2000  Total no. of persons is 2000. 600  100  x = 30  x = 2000  Total numbers of persons are 2000.

PROBLEM SET - 6 (TEXT BOOK PAGE NO. 177) : 3.

Sol.

The sales due to salesman in week are given below by the pie diagram study the Salesman Salesman diagram and answer the following questions A B if the total sale due to salesman A is 90º 120º Rs. 18000. 80º 70º (a) Find the sales due to each salesman. Salesman (b) Find the salesman with highest sale. Salesman D (c) Find the difference between the highest C sale and the lowest sale. (d) Find the total sale. (3 marks) Let total sales done be Rs. x Sales done due to salesman A = Rs. 18000 Measure of central angle for salesman A = 90º As per the given condition,

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8

ALGEBRA

Adsul Notes



   

90 × x = 18000 360 18000  360 x = 90 x = 18000 × 4 x = 72000 Total sales done is Rs. 72000 Salesman

Measure of central angle

A

90

B

120

C

80

D

70

Total

360

Sales done in Rs. 18000 120 × 72000 = 24000 360 80 × 72000 = 16000 360 70 × 72000 = 14000 360 72000

ot es

(c) The highest sale is done by salesman B and it is Rs. 24000. The difference between highest sale and lowest sale is 24000 – 14000 = Rs. 10000.

PROBLEM SET - 6 (TEXT BOOK PAGE NO. 177) :

N

The following pie diagram represent the number of valid votes obtained by four student who contested for school captain. The total of valid votes polled was 720. Answer the following questions : (a) Who has won the election ? (b) What is the minimum number of votes? Who got it? (c) By how many votes did the winner defect the nearest contestan ? (3 marks)

Sol.

Ad s

ul

4.

Name of the Candidate Raj

Albert Suja Nashima Total

Measure of central angle ()

Albert

Raj

100º 80º 60º 120º Suja Nashima

Number of votes

80º

80 × 720 = 160 360

100º

100 × 720 = 200 360

60º

60 × 720 = 120 360

120º

120 × 720 = 240 360

360º

720

(a) Nashima has won the election. (b) Minimum number of votes is 120 obtained by Suja. (c) Winner Nashima defeated the nearest contestant Albert by 40 votes.

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9

ALGEBRA 

o Step Step Step Step Step Step

Adsul Notes

Steps to draw frequency polygon or freqeuency curve : 1 2 3 4 5 6

: : : : : :

Prepare frequency distribution table with class marks Draw X-axis, Y-axis, choose proper scale Take class marks on X-axis Take frequencies (fi) on Y-axis Plot the points (xi, fi) Plot two additional points with two additional classes one preceding the first class with frequency zero (0) and the other succeeding the last class with frequency zero.

For frequency polygon we join all the points by straight line using scale. For frequency curve we join all the points with smooth free hand curve.

EXERCISE - 6.3 (TEXT BOOK PAGE NO. 151) : Draw histogram and frequency polygon for the following frequency distribution : (5 marks) Class 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 Frequency Sol.

20

30

50

40

ot es

1.

Y

10

N

Scale : On X = axis : 1 cm = 5 units On Y = axis : 1 cm = 5 units

•

ul

50

Ad s

45 40

•

Frequency

35

•

30 25

•

20 15

•

10 5 X

0

Y

" Learn from the Best "

•

5

10

15

20

25

30

•

35

X

Classes

10

ALGEBRA



Adsul Notes

EXERCISE - 6.3 (TEXT BOOK PAGE NO. 152) : 4.

Represent the following data using, histogram and hence draw frequency polygon : (5 marks) No. of words typed 30 - 39 per minute No. of typists

Sol.

40 - 49

50 - 59

8

15

2

No. of words typed per minute

Continuous Classes

30 40 50 60 70

29.5 39.5 49.5 59.5 69.5

-

39 49 59 69 79

-

3

Frequency No. of typists 2 8 15 12 3

Y

Scale : On X = axis : 1 cm = 2.5 words On Y = axis : 1 cm = 1 typist

•

ot es

14 13

•

N

12 11

ul

10 9

•

Ad s

No. of typists

12

39.5 49.5 59.5 69.5 79.5

15

60 - 69 70 - 79

8 7 6 5

•

4 3

•

2 1 X

0

•

Y

" Learn from the Best "

29.5

39.5

49.5

59.5

69.5

79.5

•

X 89.5

No. of words typed per minute

11

ALGEBRA 

Adsul Notes

EXERCISE - 6.3 (TEXT BOOK PAGE NO. 152) : 3.

Following is the frequency distribution of customers in a certain year at the departmental store : No. of customers

50 - 100 100 - 150 150 - 200 200 - 250 Total

No. of days

90

98

138

39

(5 marks)

Draw histogram and hence draw frequency curve. Sol.

Y

365

Scale : On X = axis : 1 cm = 25 customers On Y = axis : 1 cm = 10 days

150 140

•

130

ot es

120 110 100

•

•

N

No. of days

90 80

ul

70

Ad s

60 50 40

•

30 20 10

X

0 Y

•

50

100

150

200

250

•

300

No. of customers

X

EXERCISE - 6.3 (TEXT BOOK PAGE NO. 152) : 2.

Represent the following data using frequency curve : Electricity bill in a month (in Rs.) No. of families

" Learn from the Best "

200 - 400 400-600 600 - 800 800 - 1000 362

490

185

63

12

ALGEBRA

Adsul Notes



(5 marks)

Draw histogram and hence draw frequency curve. Sol.

Electricity bill in a month in Rs.

Class mark

200 400 600 800

300 500 700 900

-

400 600 800 1000

362 490 185 63

Y

ot es

Scale : On X = axis : 1 cm = 100 units On Y = axis : 1 cm = 50 families

500

• (500, 490)

450

(300,362)

350

ul

300

•

N

400 No. of families

No. of families

Ad s

250 200

• (700, 185)

150 100

(900, 63)

•

50 (100, 0)

X

0

•

(1100, 0)

200

Y

400

600

800

1000

•

1200

X

Classes (Electricity bill in a month in Rs.)

EXERCISE - 6.3 (TEXT BOOK PAGE NO. 152) : 5.

Draw frequency polygon and frequency curve for the following data on land holding : (5 marks) Area in hectare No. of farmers

" Learn from the Best "

11 20

21 30

31 40

41 50

51 60

61 70

71 80

58

103

208

392

112

34

12

13

ALGEBRA  Sol.

Adsul Notes Area in hectare

Continuous classes Class mark No. of farmers

11 21 31 41 51 61 71

10.5 20.5 30.5 40.5 50.5 60.5 70.5

-

20 30 40 50 60 70 80

-

20.5 30.5 40.5 50.5 60.5 70.5 80.5

15.5 25.5 35.5 45.5 55.5 65.5 75.5

Y

58 103 208 392 112 34 12

Scale : On X = axis : 1 cm = 10 hectres On Y = axis : 1 cm = 25 farmers

375

•

ot es

(45.5, 392)

350 325

N

300 275

ul

250

Ad s

225 200

(35.5, 208)

•

No. of farmers

175 150

(55.5, 112)

•

125 (25.5, 103)

•

100 75

(15.5, 58)

50

• •

25 X

0

Y

" Learn from the Best "

•

(65.5, 34)

•

(5.5,0) 10.5 20.5 30.5

40.5 50.5 60.5 70.5

(75.5, 12) (85.5, 0)

•

80.5 90.5

X

Classes (area in hectres)

14

ALGEBRA



Y

Adsul Notes

Scale : On X = axis : 1 cm = 10 hectres On Y = axis : 1 cm = 25 farmers

400

(45.5, 392) •

375

ot es

350 325 300

N

275

225

ul

(35.5, 208)

200

•

Ad s

No. of farmers

250

175 150 125 100 75

(25.5, 103)

(15.5, 58)

•

(55.5, 112)

•

• •

50

(65.5, 34) (75.5, 12)

25 X

0

• •

(5.5,0) 10.5 20.5 30.5

40.5 50.5 60.5 70.5

•

(85.5, 0)

80.5 90.5

X

Classes (area in hectres)

Y

" Learn from the Best "

15

ALGEBRA 8.



Adsul Notes

From the following graphs, find the median, mode whichever is possible. Y

Y

• •

•

•

23

•

• • •

• • •

X

10

•

20

•

30

•

40

•

50 X

Y

Sol.

•

•

X

2

• •6

•8

• •

10

•

12

X

Y

From the first graph histogram we get mode of is 23 units. From the second graph ogive curve we get median of data is 5.8 units.

MCQ’s

ot es

X

Scale : 1 cm = 100 units on the Y-axis

60

X

N

40 30

ul

20 10

Ad s

No. of students

50

0

5

10

Y

15

20

25

30 X

Classes

1.

Which class has least number of students ? (a) 20 - 25 (b) 25 - 30 (c) 5 - 10 (d) 15 - 20

2.

What is the mode ? (a) 17 (c) 12

(b) (d)

10 15

3.

Which two classes have same frequency ? (a) 0 - 5 and 20 - 25 (b) 0 - 5 and 25 - 30 (c) 20 - 25 and 25 - 30 (d) 15 - 20 and 20 - 25

4.

Which is pre modal class ? (a) 5 - 10 (c) 0 - 5

5.

•

5.8

•

•4

•

(b) (d)

10 - 15 15 - 20

In modal class, how many students present ? (a) 60 (b) 55 (c) 50 (d) 65

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16

ALGEBRA 

Adsul Notes Y

•

No. of families

600 500

•

400

•

300

•

200

•

100 X

•

50

100

150

200

250

300

Classes

No. of families in class 150 - 200 is (a) 300 (b) (c) 200 (d)

X

250 none of these

ot es

6.

0 Y

................ class has maximum frequency. (a) 50 - 100 (b) 100 - 150 (c) 150 - 200 (d) none of these

8.

.............. class has minimum frequency. (a) 50 - 100 (b) 150 - 200 (c) 100 - 150 (d) none of these Y

X axis 1 cm = 10 marks Y axis 1 cm = 10 students

(100, 50)

•

Ad s

50

ul

N

7.

40

(80, 36) • (60, 30)

•

30 20 10

(40, 10) • (20, 6)

•

X

9.

10.

11.

20 40 60 0 Y The median of above graph is (a) 25 (c) 100

The classwidth of above graph is (a) 10 (c) 5

80

100

(b) (d)

58 50

(b) (d)

20 not defined

X

The pie diagram sum of all central angles is (a) 90º (b) 100º (c) 180º (d) 360º

" Learn from the Best "

17

ALGEBRA



12.

While drawing histogram krink mark is put on (a) X axis (b) Y axis (c) both X and Y axis (d) Y axis only

13.

Frequency curve is ............... shaped. (a) ball (b) cone (c) tail (d) bell

14.

Using O give curve .............. can be obtained easily. (a) mean and median (b) mean (c) median (d) mode

15.

Frequency can gives idea about ........... . (a) mean (b) mode (c) median (d) mean, mode and median

Adsul Notes

Cement Bricks 50º

110º

Timber

ul

Labour

Steel

45º

N

80º

75º

ot es

Answer the following questions if total construction cost is 5,40,000.

............... is item with the maximum expenditure. (a) Labour (b) Bricks (c) Steel (d) None of these

17.

Expenditure of steel item is ................ . (a) 76500 (b) 67500 (c) 56700 (d) 65700

18.

How much additional expenditure of Timber than cement .......... . (a) 112500 (b) 67500 (c) 121500 (d) none of these

19.

Expenditure of Bricks item in degree is ............. . (a) 50º (b) 80º (c) 75000 (d) none of these

20.

.............. is item with minimum expenditure. (a) Bricks (b) Steel (c) Cement (d) none of these

Ad s

16.

: ANSWERS : 1. 3.

(b) 25 - 30 (a) 0 - 5 and 20 - 25

" Learn from the Best "

2. 4.

(c) 12 (a) 5 - 10

18

ALGEBRA  5. 7. 9. 11. 13. 15. 17. 19.

Adsul Notes

(a) (b) (b) (d) (d) (b) (b) (a)

60 100 - 150 58 360º bell mode 67500 50º

6. 8. 10. 12. 14. 16. 18. 20.

(a) 300 (d) none of these (b) 20 (a) X axis (c) median (d) None of these (b) 67500 (b) Steel

HOTS PROBLEMS (Problems for developing Higher Order Thinking Skill)

Sol.

Among the group of students 10% scored marks below 20, 20% scored marks between 20 and 40, 35% scored marks between 40 and 60, 20% scored marks between 60 and 80. The remaining 30 students scored marks between 80 and 100. (i) Prepare frequency distribution from the above information (ii) Draw histogram and find the mode of marks scored. (5 marks) Let the total no. of students in the group be x

ot es

28.

10 x 100

= 0.1x

No. of students scoring marks between 20 and 40 =

20 x 100

= 0.2x

No. of students scoring marks between 40 and 60 =

35 x 100

= 0.35x

=

ul

N

 No. of students scoring marks less than 20

20 x = 0.2x 100 No. of students scoring marks between 80 and 100 = 30 students As per the given condition, 0.1x + 0.2x + 0.35x + 0.2x + 30 = x 0.85x + 30 = x 0.85x – x = – 30 – 0.15x = – 30 – 30 x = – 0.15 x = 200 0.1x = 0.1 × 200 = 20 0.2x = 0.2 × 200 = 40 0.35x = 0.35 × 200 = 70

Ad s

No. of students scoring marks between 60 and 80 =

       

Marks 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Total

" Learn from the Best "

No. of students 20 40 70 40 30 200

19

ALGEBRA



Y

Adsul Notes

Scale : On X = axis : 1 cm = 10 marks On Y = axis : 1 cm = 5 students

70 60 55 50

40 35 30

ot es

Frequency (No. of students)

45

25 20

N

15 10

0 Y

20

40

60

80

X

100

Classes (Marks obtained)

Ad s

X

ul

5

 The mode of marks obtained is 50 marks. 29. Following is a frequency distribution of marks : Marks 0 - 20 20 - 40 40 - 60 60 - 80 No. of students

Sol.

2

a

56

b

80 - 100

Total

2

100

If a and b are equal find their values. Draw less than cumulative frequency curve and more than frequency curve on the same graph paper and find the median. If 3a = b then find a and b, also draw less than cumulative frequency curve. Hence find the median. (5 marks) As per the condition given to the table we have : 2 + a + 56 + b + 2 = 100  60 + a + b = 100  a + b = 100 – 60  a + b = 40 .....(i) If a = b then, a + a = 40  2a = 40 40  a = 2  a = 20  a = b = 20

" Learn from the Best "

20

ALGEBRA 

Adsul Notes Marks

Frequencies Cumulative Upper frequency boun.

– 20 - 0 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120

0 2 20 56 20 2 0

0 2 22 78 98 100 0

Cumulative frequency

0 20 40 60 80 100 100

Y

100 98 78 22 2

Lower boundaries 0 20 40 60 80

Scale : On X = axis : 1 cm = 10 marks On Y = axis : 1 cm = 10 students

130 120 Cumulative frequency curve less than type.

ot es

100 •

•

•

90

•

•

ul

70 60 50 40 30 20

•

10 X

•

N

80

Ad s

Cumulative frequency (No. of students)

110

0 Y

•

Median of a+b If 3a = a + 3a  4a  a b   b  b

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Median = 47.3

• 20

•

40

60

• 80

Class boundaries (Marks)

•

100

120

X

marks obtained is 47.3 marks. = 40 [From (i)] b then = 40 = 40 = 10 = 3a = 3 (10) = 3

21

ALGEBRA

Adsul Notes



Marks

Cumulative frequency less than type 0 2 12 68 100

Frequency

– 20 - 0 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100

0 2 10 56 30 2

Y

Upper boundaries 0 20 40 60 80 100

Scale : On X = axis : 1 cm = 10 marks On Y = axis : 1 cm = 10 students

120

(80, 98)

100

80

(60, 68) N 2

50

= 50

ul

40 30 20

(40, 12)

•

10

X

•

N

70 60

(100, 100)

ot es

90

(0, 0)

•

•

Ad s

Cumulative frequency (No. of students)

110

Median = 53 marks

(20, 2)

•

0 Y

•

20

40

60

80

100

120

Class boundaries (Marks)

X

 Median of marks obtained is 53 marks. 30.

Following is the frequency distribution with unknown frequencies : Class Frequency

20 - 30 30 - 40 a

2a

40 - 50 3a

50 - 60 Total a

70

Find the value of a, hence find the frequencies. Draw a histogram. (5 marks) Sol. As per the condition given in the table we have, a + 2a + 3a + a = 70  7a = 70 70  a = 7  a = 10

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22

ALGEBRA 

Adsul Notes  2a = 2 (10) = 20  3a = 3 (10) = 30 Classes 20 30 40 50

-

Frequency

30 40 50 60

10 20 30 10

Total

70

Y

Scale : On X = axis : 1 cm = 5 units On Y = axis : 1 cm = 2 units

30 28

ot es

24 22

N

20

16 14

ul

18

Ad s

Cumulative frequency (No. of students)

26

12 10 8 6 4 2

X 32.

0 Y

20

40

60

80

Class boundaries (Marks)

100

120

X

The number of Maharashtra Yuvak Mandal collected the following amounts in rupees to help the earthquake affected people. 158, 238, 543, 134, 240, 343, 495, 230, 178, 275, 245, 175, 334, 248, 305, 120, 225, 210, 437, 160, 235, 290, 200, 320, 190, 240, 420, 245, 320, 150, 201, 105, 298, 240, 330, 101, 155, 410, 451, 221.

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23

ALGEBRA

Adsul Notes



Prepare a grouped frequency distribution table of taking classes 100 200, 200 - 300, ...... . Hence prepare a table showing cumulative frequency less than upper limit. (4 marks) Sol.

Classes (Amount in Rs.)

Tally Marks

100 200 300 400

|||| |||| |||| ||||

-

200 300 400 500

Frequency

|||| | |||| |||| || | |

Cumulative frequency less than upper limits

11 17 6 6

11 28 34 40

40 Information from 120 college students was collected with a view to know which vehicle they use to go to college. The data were as follows. Present the information by Pie diagram : (4 marks) Vehicle No. of students Vehicle

Cycle

Moped

Scooter

Car

21

64

23

9

3

No. of students 21

Cycle

64

Moped

23

Scooter

9 3

Ad s

Car

21 120 64 120 23 120 9 120 3 120

N

Bus

Measure of central angle

ul

Sol.

Bus

ot es

33.

× 360 = 63º

× 360 = 192º × 360 = 69º × 360 = 27º × 360 = 9

Moped

Scooter Car

9º

27º

69º

63º

Bus

192º

Cycle

   " Learn from the Best "

24

ALGEBRA

Adsul Notes

S.S.C.

Marks : 30

CHAPTER 6 : Statistics - II SET - A

ALGEBRA

Duration : 1 hr.

Q.1. Atttempt any TWO of the following : The following pie diagram represents expenditure on different items in constructing a building. Answer the following questions : (a) Which is the item with the maximum expenditure ? (b) Which is the item with the minimum expenditure ?

Bricks 50º Cement 75º 100º 45º 90º Steel

Labour

ot es

(i)

2

N

(ii)

Timber

From the following graphs, find the median, mode whichever is possible.

ul

Y

Y

Ad s

• • •

23

•

X

10

•

20

•

30

•

40

•

50 X

Y

(iii)

•

•

• • •

• •

•

•

X

2

• •4

5.8

• •6

•

•

•8

•

•

•

10

12

X

Y

The following table gives profit earned by companies in lacs of Rs. Profit No. of companies

5-10

10-15

15-20

4

28

49

20-25 25-30 17

2

Draw histogram.

" Learn from the Best "

25

ALGEBRA

Adsul Notes

Q.2. Attempt any FOUR of the following : (i)

8

The following data give the number of students using different modes of transport : Mode of transport Number of students

Bicycle

Bus

Walk

Train

Car

140

100

70

40

10

Represent the above data using pie diagram : Draw the histogram to represent the following data, hence find the mode : Daily sales of 01000200030004000Total a store in (`)

1000

2000

3000

4000

5000

2

12

10

4

2

Number of days in a month

The number of hours, spent by a school boy in different activities in a day is given below. Activity sleep school play home work other Total

N

(iii)

ul

No. of hours 8 7 2 4 Represent the above information using pie diagram. (iv)

3

24

Draw the histogram and use it to find the mode for the following frequency distribution : 4000 6000

6000 8000

8000 10000

10000 12000

200

240

300

50

Ad s

House - Rent (in ` per month)

Number of families

(v)

30

ot es

(ii)

The following table gives information about the monetary investment by some residents in a city :

Mode of investment Percentage of residents

Shares funds

Mutual estate

Read bonds

Gold

10

20

35

30

Government 5

Draw pie diagram to represent the data. Q.3. Attempt any FOUR of the following : (i)

20

Draw a pie diagram to represent the world population given in the following table after determining the value of a :

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26

ALGEBRA

Adsul Notes

Country Percentage of

India

world population

15

Other

Total

a

25

100

a

Draw frequency polygon and frequency curve for the following data on land holding : Area in hectare No. of farmers

(iii)

20

USA

11 20

21 30

31 40

41 50

51 60

61 70

71 80

58

103

208

392

112

34

12

Following data represent the age wise distribution of employees in office : Age in years

25 30

30 35

4

16

Number of employees

35 - 40 - 45 40 45 50

50 55

55 60

8

3

ot es

(ii)

China Russia

19

28

22

(iv)

Represent the following data using, histogram and hence draw frequency polygon : 40 - 49

2

Ad s

No. of typists

30 - 39

ul

No. of words typed per minute

(v)

N

Draw less than type cumulative frequency curve. Hence find the median age of employees.

8

50 - 59

60 - 69 70 - 79

15

12

3

Draw more than type cumulative frequency curve for the following data on time required to do a certain job : Time required (in hours)

Number of workmen

above above above above above above above 170 50 70 90 110 130 150 100

98

90

60

35

21

2

Best Of Luck 

" Learn from the Best "

27

ALGEBRA

Adsul Notes

S.S.C.

Marks : 30

CHAPTER 6 : Statistics - II SET - B

ALGEBRA

Duration : 1 hr.

Q.1. Atttempt any TWO of the following : The following diagram represents the sectorwise loan amount in crores of Rs. distributed by a bank. From the information answer the following questions : If the dairy sector received Rs. 20 crores, then find the total loan disbursed.

Agriculture 120º 40º

ot es

(i)

2

Ad s

(iii)

Dairy

N

The following pie diagram represent the number of valid votes obtained by four student who contested for school captain. The total of valid votes polled was 720. Answer the following question : What is the minimum number of votes? Who got it?

ul

(ii)

Industry

The sales due to salesman in week are given below by the pie diagram study the diagram and answer the following questions if the total sale due to salesman A is Rs. 18000. Find the total sale.

Albert

60º 120º Suja Nashima

Salesman Salesman A B 120º 90º 80º 70º Salesman C

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Raj

100º 80º

Salesman D

28

ALGEBRA

Adsul Notes

Q.2. Attempt any FOUR of the following : (i)

Following is the componentwise expenditure per article, draw a pie chart: Component

Expenditure (in Rs.)

Raw material Labour Transportation Packing Taxes (ii)

800 300 100 100 140

Represent the following data using histogram : 140-144

145-149

150-154

155-159

ot es

Height of students (cm.) Number of students

2

12

10

4

Number of students admitted in different faculties of a college are given below : Faculty

Science Commerce Arts

No. of students

1000

1200

N

(iii)

8

650

Law

Home science

450

300

Draw a pie diagram and represent the above data. Represent the following data by histogram :

ul

(iv)

18 20

20 22

22 24

24 26

26 28

Total

Number of weeks

4

8

22

12

6

52

Ad s

(v)

Price of sugar per kg (in `)

Electricity used by farmers during different parts of a day for irrigation is as follows. Draw pie diagram : Part of day

Percentage of electricity used

Morning

Afternoon

Evening

Night

30

40

20

10

Q.3. Attempt any FOUR of the following : (i)

20

Following data represent the age wise distribution of employees in office : Age in years Number of employees

25 30

30 35

4

16

35 - 40 - 45 40 45 50 19

28

22

50 55

55 60

8

3

Draw : less than type cumulative frequency curve. Hence find the median age of employees.

" Learn from the Best "

29

ALGEBRA

(ii)

Adsul Notes

Represent the following data using frequency curve : Electricity bill in a month (in Rs.)

200 - 400

400-600

600 - 800

800 - 1000

362

490

185

63

No. of families

Draw histogram and hence draw frequency curve. (iii)

Represent the following data using frequency curve : Electricity bill in a month (in Rs.)

200 - 400

400-600

600 - 800

800 - 1000

362

490

185

63

No. of families

(iv)

Represent the following data using, histogram and hence draw frequency polygon : No. of words typed per minute

30 - 39 2

40 - 49

50 - 59

8

15

60 - 69 70 - 79 12

3

N

No. of typistsx (v)

ot es

Draw histogram and hence draw frequency curve.

At customer service centre time required to serve customers is tabulated below : 31 - 40 41- 50 51- 60 61 - 70 71 - 80 81-90

ul

Time required (in minutes)

12

Ad s

No. of customers

18

27

20

17

6

Draw less than cumulative frequency curve.

Best Of Luck 

" Learn from the Best "

30