Vedanta Excel in Mathematics Teachers' Manual -9 Final

9) In the given figure, ABCD is a rectangle and ADE is a right angle triangle. Find the

size of EC.

Solution:

i) In rectangle ABCD, AB = CD = 50 3 m
p AB
ii) In rt. ‘ed 'ABC; tan45q = b = BC

or, 1 =50BC3 m ? BC = 50 3 m

iii) AD = BC = 50 3 m
[Opposite sides of rectangle ABCD]

iv) In rt.‘ed 'EDA. tan30q = p = ED
b AD
or, 1 = ED
3 50 3 m ? ED = 50 m

Hence, EC = ED DC = 50 m 50 3 m = 136.6 m

Extra Question

1. In the adjoining figure, AB = 6 cm, BC = 8 cm, ‘ABC = 90q, BD

A AC and ‘ABD = T, find the value of cosT and tanT. 4 3 6 cm
5 4
Ans: ,

2. Prove that: 1 tan30q = 1 sin30q 8 cm
1 tan30q 1 sin30q
E

3. In the given figure, ABCD is a rectangle and EAD is a right angled

triangle. Find the length of BE. A 30q D
Ans: 55 cm 50 3cm 5 cm
C
B

A

4. In the equilateral trianlge ABC, AD A BC and AC = x cm. Show that: x cm

sin 60q = 3 and tan 30q = 1.
2 3
BD C

147 Vedanta Excel in Mathematics Teachers' Manual - 9

Unit Statistics

17

Allocated teaching periods 10
Competency

- To collect, present and analyze the data
Learning Outcomes

- To construct the frequency distribution table for the collected data
- To construct the histogram, line graph and pie-chart for the collected data and

solve the related problems
- To introduce the less and more than ogive and construct them
- To find the mean, median, mode and the quartiles of the ungrouped data.
- To collect the real (primary and secondary) data and analyze the data using the

appropriate statistical measure.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define histogram
1. Knowledge (K)

- To define pie-chart

- To recall the formula of finding the mean, median and
quartiles on ungrouped data

- To answer the questions from histogram and line graph
2. Understanding (U)

- To construct the histogram to represent the given groped
data

- To find the mean, median of individual series
- To represent the data through pie-chart

3. Application (A) - To calculate the mean, median and quartiles of discrete
series

- To draw the less and more than ogives and identify the

median from the graph

- To collect the real (primary and secondary) data and

4. High Ability (HA) analyze the data using the appropriate statistical

measure and analyze the data

Vedanta Excel in Mathematics Teachers' Manual - 9 148

Required Teaching Materials/ Resources
Colourful chart-paper, colourful markers, chart paper with required formulae, graph-board,
graph paper, highlighter etc

Pre-knowledge: Frequency distribution table, mean, median etc

Teaching Activities

1. Explain about statistics

2. Discuss on data and types of data with real life examples

3. With example, discuss about the frequency tables

4. Explain with examples about the histogram, its importance and procedures of

constructing histogram

5. Divide the students and tell to draw histograms from the questions given in

exercise and present in class

6. Ask about the pie-chart, its importance and procedures of its construction.

7. Divide the students and tell to show the data in pie-charts from the questions given

in exercise and present in class

8. Discuss about the ogives with examples

9. With appropriate examples, discuss on the central tendencies and their

calculations

10. With discussion, list out the following formulae

(i) Mean

(a) For individual data; Mean ( x ) = 6x
n
6fx
(b) For discrete data; Mean (x) = 6f

(ii) Median

For individual and discrete data; position of median = n + 1 th
2
(iii) Mode

(a) For individual and discrete data; mode = item having highest

frequency

(b) For continuous data; mode = Mode (M0) = fLre+qu2efn1fc1–y–f0of–0f tfh2 eucclass
where L = lower limit of model class, fo =

preceding to model class, f1 = frequency of model class, f2 = frequency
of the class succeeding to model class and c = the width of the class

(iv) Quartiles

n + 1 th
4
(a) For individual and discrete data; position of Q1 = term

n + 1 th
4
(b) For individual and discrete data; position of Q3 =3 term

149 Vedanta Excel in Mathematics Teachers' Manual - 9

Solution of selected problems from Vedanta Excel in Mathematics

1) The pie chart given alongside shows the votes secured by Z
three candidates X, Y and Z in an election. If X secured
5760 votes,

i) how many votes did Z secure? 140°

ii) who secured the least number of votes? How many votes X Y
did he secure? 120°

Solution:

Let the total number of votes secure by three candidates be x.

Then, the number of votes secured by x = 5760
120
or, 360 ux = 5760

i) ? x = 17280 = 140 u 17280 = 6720
No. of votes secured by Z 360

ii) No. of votes secured by Y = 360 (120 140) u 17280
= 360
140
360 u 17280 = 4800

Y secured the least number of votes. He secured only 4800 votes.

2) The given pie chart shows the composition of different Polyester
materials in a type of cloth in percentage. 144°

i) Calculate the percentage of each material found in the Cotton Nylon
cloth. 90° 54°

ii) Calculate the weight of each material contained by a Others
bundle of 50 kg of cloth. 72°

Solution:

Materials In Percentage Weight (in kg)
Polyster 40 % of 50 kg = 20 kg
144 u 100 % = 40 %
360

Cotton 90 u 100 % = 25 % 25 % of 50 kg = 12.5 kg
Nylon 360 15 % of 50 kg = 7.5 kg
Other 20 % of 50 kg = 10 kg
54 u 100 % = 15 %
360

72 u 100 % = 20 %
360

3. Draw a ‘less than’ ogive from the data given below.

Data 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 5 8 15 10 6 3

Solution:

Vedanta Excel in Mathematics Teachers' Manual - 9 150

Less than cumulative frequency table.

Marks No. of students (f) Upper limit less than c.f. Coordinates (x, y)
0 – 10 (10, 5)
10 – 20 5 10 5 (20, 13)
20 – 30 (30, 28)
30 – 40 8 20 13 (40, 38)
40 – 50 (50, 44)
50 – 60 15 30 28 (60, 47)

10 40 38

6 50 44

3 60 47

less than ogive

60

Less than cumulative frequencies 50
(60, 47)

40 (50, 44)
(40, 38)

30
(30, 28)

20

10 (20, 13)

(10, 5)
0 10 20 30 40 50 60 70

Upper limits

4. The speeds of vehicles recorded in a highway during 15 minutes on a day is given
in the following table. The average speed of the vehicles was 54 km per hour.

Speed (km/hr) 10 30 50 70 90

No. of vehicles 7 k 10 9 13

(i) Find the value of k. (ii) Find the total number of vehicles counted during the time.

Solution:

Speed (km/hr) No. of vehicles ( f ) fx

10 7 70
30 k 30k
50 10 500
70 9 630
90 13 1170

Total N = 39 + k 6fx = 2370 + 30k
6fx
Now, mean ( x ) = N

2370 + 30k
or, 54 = 39 + k

or, 2106 + 54k = 2370 + 30k
or, 24k = 264

? k = 11

151 Vedanta Excel in Mathematics Teachers' Manual - 9

i) Required value of k is 11.
ii) No. of vehicles counted during the time is 39 k = 39 11 = 50

5. Given that mean in 40 and N = 51, find the missing requencies in the following
data.

x 10 20 30 40 50 60

f 2 3 - 21 - 5

Solution:

Let the missing frequencies be a and b.

x No. of vehicles ( f ) fx

10 2 20

20 3 60

30 a 30a

40 21 840

50 b 50b

60 5 300

Total N = 31 a + b 6fx = 1220 30a + 50b
Now,

N = 51

or, 31 a b = 51 ? a = 20 b .......... (i)

Again, ) = 6fx
mean ( x N

or, 40 = 1220 + 30a 50b

51

or, 2040 = 1220 30a 50b
or, 820 = 30a 50b

or, 82 = 3a 5b ........... (ii)

Substituting the value of a from equation (i) in equation (ii), we get

82 = 3 (20 b) 5b

or, 22 = 2b ? b = 11

Putting the value of b in equation (i), we get

a = 20 11 = 9

Hecne, the missing frequencies are 9 and 11.

6. Find the first quartile (Q1) and third quartile (Q3) from the following distribution.

Age (in year) 22 27 32 37 42

Solution: No. of people 35 42 40 30 24

Cumulative frequency distribution table,

Age (in years) No. of people (f) c.f.

22 35 35
27 42 77
32 40 117
37 30 147
42 24 171

Total N = 171

Vedanta Excel in Mathematics Teachers' Manual - 9 152

Now, th th

Position of (Q1) = N + 1 term = 171 + 1 term = 43th term
4 4

In c.f. column, the c.f. just greater than 43 is 77 and corresponding value is 27.

? The first quartile (Q1) = 27
Also,
Position of median (Q2) = N + 1 th term = 171 + 1 th term = 86th term
2 2

In c.f. column, the c.f. just greater than 86 is 117 and its corresponding value is 32.

? Median (Q2) = 32
Again,
3 (N + 1) th
Position of (Q3) = = 4 term = 129th term

In c.f. column, the c.f. just greater than 129 is 147 and its corresponding value is 37.
? The third quartile (Q3) = 37

Extra Question

1. Construct a histogram from the data given in the table below.

Marks 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
No. of student 4 7 8 5 3

2. A mobile shop sold the mobile sets of the following brands in a month. Draw a pie chart
to show the data.

Brand No. of mobile sets

Samsung 130
Huawei 120
Oppo 90
Vivo 40
Nokia 20

3. Draw a 'less than' ogive from the data given below.

Wages (in Rs) 400 - 500 500 - 600 600 - 700 700 - 800 800 - 900
No. of workers 43675

4. The mean of the data given below is 17. Determine the value of m.

X 5 10 15 20 25 30

f 2 5 10 m 4 2
[Ans: 7]

5. Find the quartiles from the following distribution.

Height (in cm) 90 100 110 120 130 140
No. of students 20 28 24 40 35 18

[Ans: Q1 = 100, Q2 = 120, Q3 = 130]

153 Vedanta Excel in Mathematics Teachers' Manual - 9

Unit Probability

18

Allocated teaching periods 5

Competency
- To study the probability in daily life and solve the problems using mathematical
structures

Learning Outcomes
- To introduce probability scale and solve the simple problems related to probability

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define probability

1. Knowledge (K)
- To tell the definition of sample space

- To recall the probabilities of certain and impossible
events

- To tell the probability scale

- To find the probability of an event
2. Understanding (U)

- To solve the simple problems on empirical probability

Required Teaching Materials/ Resources
Coin, Dice, cards, spinner etc

Pre-knowledge: Possibility of raining, having birth of son or daughter etc

Teaching Activities

1. Take a coin and ask the following questions
(i) How many faces are there?
(ii) Which face can be surely placed when the coin is tossed?
(iii) Can both faces be shown at once?
(iv) What is the possibility of getting head?
(v) What is the possibility of getting tail?

2. Similarly, take a die and ask the following questions
(i) How many faces are there?
(ii) Which face can be surely placed when the coin is tossed?
(iii) Is it possible to fix all six faces at a roll?

Vedanta Excel in Mathematics Teachers' Manual - 9 154

(iv) What is the possibility of getting 1?

(v) What is the possibility of getting each face separately?

(vi) What is the possibility of getting an odd number?

(vii) What is the possibility of getting a prime number?

3. Also, discuss on the following questions

(i) What is the probability of raining today?

(ii) What is the probability of having daughter from a pregnant woman?

(iii) What is the probability of getting A+ grade by Ram in Mathematics?

4. Discuss on the following terminologies with proper examples

(i) Probability

š It measures the chances of happening or not happening the event

š The numerical measurement of the degree of certainty of the occurrence of events

Example: When a coin is toofssheeda,ditisis215a0n/5d0thchatanocf etatihlaist tahlseoh21ea. d or tail occurs. So, the
probability of occurrence

(ii) Random experiment

š The experiment is an action by which an observation is made.

š The random experiment is an experiment whose outcome cannot be predicted or

determined in advance.

š Examples: tossing a coin, rolling a die, drawing a card from a well-shuffled pack of 52

playing cards etc.

(iii) Sample space

š The results of random experiments are called outcomes.

š The set of all possible outcomes in a random experiment is the sample space.

š Sample space is usually denoted by S.

š Example: While flipping a coin, the possible outcomes are head (H) or tail (T).

?S = {H, T}

(iv) Event

š Any non-empty subset of a sample space S is called an event.

š Example: When a die is rolled, the sample space (S) = {1, 2, 3, 4, 5, 6}

Here, {1}, {2}, {1, 2, 3}, {2, 4, 6} etc are some events.

š ‘S’ itself is the sure event and empty subset I is an impossible event.

(v) Exhaustive and favourable cases

š The number of possible outcomes of a random experiment is the exhaustive cases.

Example: When a coin is tossed twice, then S = {HH, HT, TH, TT} and exhaustive

cases = 4.

š The number of desirable (expected) outcomes in a random experiment is the

favourable cases. Example: When tossing a coin, the favourable cases of getting head

= 1 and that of tail = 1

(vi) Equally likely events

š Two or more events are said to equally likely events, if the chance of occurring any

one event is equal to the chance of occurring other events.

š Example: While tossing a coin, the change of occurring head and tail is equal.

155 Vedanta Excel in Mathematics Teachers' Manual - 9

(vii) Mutually exclusive events
š Two or more events in a sample space are called mutually exclusive if the occurrence

of one event excludes the occurrence of other.
š Two events A and B of a sample space S are mutually exclusive if AªB =I.
š Example: While tossing a coin, the occurrence of head excludes the occurrence of tail

or HªT = I
(viii)Dependent and independent events
š Two or more events are said to be dependent if the occurrence of one event affects the

occurrence of the other events.
š Example: While drawing a marble in successive trials from a bag containing 3 green

and 6 blue marbles without replacement, getting any one coloured ball in the first trial
affects to draw another ball in the second trail.
š Two or more events are said to be independent if the occurrence of one event does not
affect the occurrence of the other events.
š Example: While tossing a coin twice or more, the occurrence of any one event in the
first toss does not affect the occurrence of any events in other trials.
5. Explain the probability of an event (E) in an exhaustive case/sample space (S) as

P(E) = Favourable number of cases = n(E)
Exhaustive number of cases n(S)

and P'(E) = 1 – P(E)

6. Show the coin, die and pack of 52 playing cards, and make clear about the facts on
coin, dice and playing cards.

7. Discuss about the probabilities of different events including certain and impossible
events

8. With examples, make the students discover probability scale
9. Divide the students into groups and give 1/1 coin to each group. Tell them to flip the

coin 20/20 times and record the outcomes.

Outcomes Head(H) Tail (T)

Frequency ... ....

10. Then ask to find the probability of getting head and getting tail.

11. With above experiment, discuss about the empirical probability.

Solution of selected questions from Excel in Mathematics

1. In a class of 40 students, 3 boys and 5 girls wear spectacles. If a teacher called one of

the students randomly in the office, find the probability that this student is wearing

the spectacle.

Solution:

Here, total number of students, n (S) = 40

Favourable number of cases of wearing glasses, n (E) = 3 + 5 = 8

Now, P (E) = n(E) = 8 = 1
Hence, n(S) 40 5
1
the probability that the student wearing spectacle is 5 .

Vedanta Excel in Mathematics Teachers' Manual - 9 156

2. There are 40 students in a class with roll numbers from 1 to 40. The roll number of

Bhurashi is 18. If a teacher calls one student with roll number exactly divisible by 3

to do a problem on blackboard, what is the probability that Bhurashi will be selected?

Also, find the probability that she will not be selected.

Solution:

Here,

Set of roll numbers which are exactly divisible by 3, (S) = {3, 6, 9, 12, 15, 18, 21, 24, 27,

30, 33, 36, 39} ?n (S) = 13

Favourable number of cases, n (E) = 1

Now, P (E) = n(E) = 1 Thus, the probability that Bhurashi can be selected is 113.
n(S) 13
113= 12
Hence, the probability that Bhurashi will not be selected, P’ (E) = 1 – 13 .

3. A card is drawn at random from a deck of 52 cards, what is the probability that the

card

(i) is an ace (ii) is an ace of spade (iii) is a black ace

Solution:

Here, total number of cards, n (S) = 52

(i) Favourable number of cases of getting ace, n (E1) = 4
nn((ES1)) 4 1
?P (E1) = = 52 = 13

Thus, the probability of getting an ace is 113.
(ii) Favourable number of cases of getting ace of spade, n (E2) = 1
n(E) 1
?P (E2) = n(S) = 52

Thus, the probability of getting an ace of spade is 1 .
52
(iii) Favourable number of cases of getting black ace, n (E3) =2
nn((ES3)) 2 1
?P (E3) = = 52 = 26

Thus, the probability of getting a black ace is 216.
4. A card is drawn at random from a well shuffled pack of 52 cards. Find the probability

that the card will be black or face cards.

Solution:

Here, total number of cards, n (S) = 52

Favourable number of cases of getting black cards = 26

Favourable number of cases of getting faced cards = 12

Favourable number of cases of getting black faced cards = 6

Total number of favourable cases, n(E) = 26 + 12 – 6 = 32

Now, P (E) = n(E) = 32 = 8
n(S) 52 13
8
Thus, the probability that the card will be black or faced card is 13 .

157 Vedanta Excel in Mathematics Teachers' Manual - 9

5. One card is drawn at random from the number cards numbered from 10 to 21. Find

the probability that the card may be prime or even numbered card.

Solution:

Here, Sample space (S) = {10, 11, ..., 21} ?n (S) = (21 – 10) + 1 = 12

Favourable number of cases of prime numbers = 4

Favourable number of cases of even numbers = 6

Total number of favourable cases, n (E) = 4 + 6 = 10

Now, P (E) = n(E) = 10 = 5
n(S) 12 6
Hence, the probability that the card may be prime or even numbered card is 5 .
6

6. Glass tumblers are packed in cartons, each containing 12 tumblers. 200 cartons were
examined for broken glasses and the results are given in the table below:

No. of broken glasses 0 1 2 3 4 More than 4

Frequency 164 20 9 4 2 1

If one carton is selected at random, what is the probability that:

(i) it has no broken glass?

(ii) it has broken glasses less than 3?

(iii) it has broken glasses more than 1?

(iv) it has broken glasses more than 1 and less than 4?

Solution:

Here, total number of cartons, n (S) = 200

(i) The number of cartons that have no broken glass = n (E1) = 164
nn((ES1)) 164 41
P (E1) = = 200 = 50 = 0.82

(ii) The number of cartons that have less than 3 broken glasses

n (E2) = 164+ 20 + 9 = 193
nn((ES2)) 193
? P (E2) = = 200 = 0.965

(iii) The number of cartons that have more than 1 broken glasses

n (E3) = 9+4+2+1 = 16
nn((ES3)) 16
? P (E3) = = 200 = 0.08

(iv) The number of cartons that have more than 1 and less than 4 broken glasses, n (E4) =
9+4= 13

P (E4) = nn((ES4)) = 13 = 0.065
200

7. From a pack of playing cards, two cards are taken, which are not hearts. They are not
replaced, and the remaining cards are shuffled. What is the probability that the next
card drawn is heart?

Solution:
Here, total number of cards, n (S) = 52 – 2 = 50
Favourable number of cases of getting the cards of hearts, n (E) = 4

Vedanta Excel in Mathematics Teachers' Manual - 9 158

Now, P (E) = n(E) = 4 = 2
n(S) 50 25

8. Three athletes A, B and C are to run a race. B and C have equal chances of winning,

but A is twice as likely to win as either. Find the probability of each athlete winning.

Solution:

Here, let the probability of winning the race by B be x

According to question, probability of winning the race by C = x and the probability of

winning the race by A = 2x 1
4
Now, P (A) + P (B) + P (C) = 1 or, 2x +x+x=1 1 =or12, 4x = 1 ?x =
A = 2x = 2 × 4 and the
Hencwei,nthneinpgrtohbeabrailcietyboyfBwainnndinCgisth14e race by probability of
each.

9. A man has 3 pairs of black socks and 2 pair of brown socks. If he dresses hurriedly

in the dark, find the probability that

(i) the first sock he puts on is brown.

(ii) the first sock he puts on is black.

(iii) after he put on a black sock, he will then put on another black sock.

(iv) that after he has first put on a brown sock, the next sock will also be brown.

Solution:

Here, total number of socks, n (S) = 2 (3 + 2) = 10

(i) The number of brown socks = n (E1) = 4 ? P (E1) = nn((ES1)) = 4 = 2
10 5

(ii) The number of black socks = n (E2) = 6 ? P (E2) = nn((ES2)) = 6 = 3
10 5
(iii) After putting on a black sock, the number of black socks for the next trial = n (E3) = 5
and total number of socks, n (S) = 10 – 1 = 9

? P (E3) = nn((ES3)) = 4 = 5
50 9
(iv) After putting on a brown sock, the number of brown socks for the next trial = n (E4) =
3 and total number of socks, n (S) = 10 – 1 = 9

? P (E4) = nn((ES4)) = 3 = 1
9 3
Extra Questions

1. Define sample space. What is the probability of a certain event?

2. A die is rolled once. What is the probability that the digit turn off is a prime number.

[Ans: 1 ]
2
3. A bag contains a dozen of identical balls. Among them, 3 are red, 5 are greed and the

rest are white. If a ball is randomly drawn, what is the probability of getting the white

ball? [Ans: 1 ]
3
4. A card is drawn at random from a well shuffled pack of 52 cards. Find the probability

that the card will be black king or red queen. [Ans: 113]

5. Three athletes A, B and C are to run a race. If A is twice as likely to win B and B is thrice

as likely to win C. Find the probability of each athlete winning. [Ans: 3 , 3 , 1 ]
5 10 10

159 Vedanta Excel in Mathematics Teachers' Manual - 9