Class 9

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches:
2, 3. 4, 5, 0. 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Answer:
Here, n = 10.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\frac{2+3+4+5+0+1+3+3+4+3}{10}\)
= \(\frac{28}{10}\)
= 2.8
Thus, the mean of the given scores is 2.8 goals.

Arranging the observations in the ascending order, we get:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Since n = 10 is an even number, \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6.

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
= \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
= \(\frac{3+3}{2}\) = 3
Thus, the median of the given scores is 3 goals.
In the given data, observation 3 occurs most frequently (4 times). Hence, the mode of the data is 3 goals.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Answer:
Here, n = 15.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\begin{gathered}
41+39+48+52+46+62+54+40 \\
+96+52+98+40+42+52+60 \\
\hline 15
\end{gathered}\)
= \(\frac{822}{15}\) = 54.8
Thus, the mean of the data is 54.8 marks.
Arranging the observations in the ascending order, we get:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Here, n = 15 is an odd number.
Median M = \(\left(\frac{n+1}{2}\right)\)th observation
= \(\left(\frac{15+1}{2}\right)\)th observation
= 8 th observation
= 52
Thus, the median of the data is 52 marks.
In the given data, observation 52 occurs most frequently (3 times). Hence, the mode of the data is 52 marks.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Answer:
Here, the median = 63 and n = 10.
∴ \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
∴ 63 = \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
∴ 63 = \(\frac{(x)+(x+2)}{2}\)
∴63 × 2 = x + x + 12
∴126 = 2x + 2
∴ 2x = 124
∴ x = 62

Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Answer:
Here, just by simple observation, it is clearly seen that observation 14 occurs most frequently, i.e., 4 times.
Hence, the mode of the data is 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table:

Answer:

Mean X̄ = \(\frac{\Sigma f_{i} x_{i}}{n}\)
= \(\) = \(\frac{3,05,000}{60}\) = 5083.33
Thus, the mean salary is ₹ 5083.33.

Question 6.
Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
Answer:
For the students studying in the same class, usually their level of knowledge and understanding would be more or less equal. There would be a few student having this level low and there would be a few students having this level high. Their level of knowledge and understanding would be reflected in the marks scored by them at an examination. Hence, the mean of marks scored by them at an examination is an appropriate measure of central tendency.

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
If we consider the monthly income of the people of certain region, the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %):

Causes	Female fatality rate (%)
1. Reproductive health conditions	31.8
2. Neuropsychiatric conditions	25.4
3. Injuries	12.4
4. Cardiovascular conditions	4.3
5. Respiratory conditions	4.1
6. Other causes	22.0

(i) Represent the information given above graphically.
Answer:

(ii) Which condition is the major cause of women’s ill health and death worldwide?
Answer:
‘Reproductive health conditions’ is the major cause of womens ill health and death worldwide.

(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Answer:
‘Malnutrition’ and ‘Lack of necessary medical facilities’ can be considered as two other factors which play a major role in female fatality.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:

Section	Number of girls per thousand bays
Scheduled Caste (SC)	940
Scheduled Tribe (ST)	970
Non-SC/ST	920
Backward districts	950
Non-backward districts	920
Rural	930
Urban	910

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Answer:

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) Draw a bar graph to represent the polling results.
Answer:
Seats won by different political parties

(ii) Which political party won the maximum number of seats?
Answer:
Political party: A won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	.5
163-171	4
172-180	2

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous.]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
Answer:
Making the class intervals continuous, we get the following table:

Length (in mm)	Number of leaves
117.5-126.5	3
126.5- 135.5	5
135.5-144.5	9
144.5-153.5	12
153.5- 162.5	5
162.5-171.5	4
171.5-180.5	2

(i) Length of leaves in millimetre

(ii) Yes. The given data can also be represented by ‘Frequency polygon’.

(iii) It is not correct to conclude that the maximum number of leaves are 153 mm long, because even if the frequency of class 145-153 is 12, we do not have the information about the length of each of those 12 leaves individually.

Question 5.
The following table gives the life times of 400 neon lamps:

Life time (in hours)	Number of lamps
300 – 400	14
400 – 500	56
500 – 600	60
600 – 700	86
700 – 800	74
800 – 900	62
900 – 1000	48

(i) Represent the given information with the help of a histogram.
Answer:

(ii) How many lamps have a life time of 700 hours or more than 700 hours ?
Answer:
The-frequencies of classes 700-800, 800-900 and 900-1000 are 74, 62 and 48 respectively.
Hence, the life time of 184 (74 + 62 + 48) lamps is 700 hours or more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Answer:
To draw the frequency polygons of both the sections, we find the class marks of each class and prepare the following tables:

Section A

Marks	Class mark	Frequency
0-10	5	3
10-20	15	9
20-30	25	17
30-40	35	12
40-50	45	9

Section B

Marks	Class mark	Frequency
0-10	5	5
10-20	15	19
20-30	25	15
30-40	35	10
40-50	45	1

Comparing the performance of both the sections from the frequency polygons, we observe that the performance of students of section A is better than the performance of students of section B.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Number of balls	Team A	Team B
1-6	2	5
7-12	1	6
13-18	8	2
19-24	9	10
25-30	4	5
31-36	5	6
37-42	6	3
43-48	10	4
49-54	6	8
55-60	2	10

Represent the data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Answer:

Number of runs made by Team A and Team B in first 60 balls.

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows:

Age (in years)	Number of children
1-2	5
2-3	3
3-5	6
5-7	12
7-10	9
10-15	10
15-17	4

Draw a histogram to represent the data above.
Answer:

Children of various age groups playing in a park

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of letters	Number of surnames
1-4	6
4-6	30
6-8	44
8-12	16
12-20	4

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Answer:

(i) Information regarding the number of surnames having given number of letters
Answer:

(ii) Write the class interval in which the maximum number of surnames lie.
Answer:
The maximum number of surnames lie in the class interval 6-8.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2

Question 1.
A park, in the shape of a quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Answer:

In ∆ BCD, ∠C = 90°
∴ BD² = BC² + CD²
= (12)² + (5)²
= 144 + 25
= 169
= (13)²
∴ BD = 13 m

In ∆ BCD, a = 5 m, b = 12 m and c = 13 m
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{5+12+13}{2}\) = \(\frac{30}{2}\) = 15 m
Then, s – a = 15 – 5 = 10m,
s – b = 15 – 12 = 3m and
s – c = 15 – 13 = 2 m.

Area of ∆ BCD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 10 \times 3 \times 2}\) m²
= \(\sqrt{900}\) m²
= 30 m²

Note: ∆ BCD is a right triangle.
∴ Area of ∆ BCD = \(\frac{1}{2}\) × BC × CD
= \(\frac{1}{2}\) × 12 × 5 = 30 m²

Now, in ∆ ABD, a = 9 m, b = 13 m arid c = 8 m
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{9+13+8}{2}\) = \(\frac{30}{2}\) = 15 m
Then,
s – a = 15 – 9 = 6m,
s – b = 15 – 13 = 2m and
s – c = 15 – 8 = 7 m.

Area of ∆ ABD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 6 \times 2 \times 7}\) m²
= \(\sqrt{5 \times 3 \times 3 \times 2 \times 2 \times 7}\) m²
= 6 √35 m²
= 35.5 m² (approx.)
Then, the area of park in the shape of quadrilateral ABCD
= Area of ∆ BCD + Area of ∆ ABD
= (30 + 35.5) m² (approx.)
= 65.5 m² (approx.)
Thus, the area of the park is 65.5 m² (approx.)

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Answer:

In ∆ ABC, a = 3 cm; b = 4 cm and c = 5 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{3+4+5}{2}\)
= \(\frac{12}{2}\) = 6 cm
Then,
s – a = 6 – 3 = 3 cm,
s – b = 6 – 4 = 2 cm and,
s – c = 6 – 5 = 1 cm.
Area of ∆ ABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{6 \times 3 \times 2 \times 1}\) cm²
= 6 cm²
Note: Proving that ∆ ABC is a right triangle, Area of ∆ ABC = \(\frac{1}{2}\) × 3 × 4 = 6 cm² can be obtained easily.
In ∆ ACD, a = 4 cm; b = 5 cm and c = 5 cm

Area of quadrilateral ABCD
= Area of ∆ ABC + Area of ∆ ACD
= (6 + 9.2) cm² (approx.)
= 15.2 cm² (approx.)

Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in the given figure, s Find the total area of the paper used. ;

Answer:
The sides of the triangle in part 1 measure 5 cm, 5 cm and 1 cm.
∴ a = 5 cm, b = 5 cm and c = 1 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{5+5+1}{2}\) = \(\frac{11}{2}\) cm
Area of part 1
= Area of triangle

The length and breadth of rectangle in part II are 6.5 cm and 1 cm respectively.
Area of part II = Area of rectangle
= length × breadth
= (6.5 × 1) cm²
= 6.5 cm²
For the trapezium in part III, the parallel sides measure 1 cm and 2 cm, while both the non-parallel sides measure 1 cm each.

Drawing DM ⊥ AB and CN ⊥ AB. we get
AM = BM = \(\frac{2-1}{2}\) = \(\frac{1}{2}\) cm.
In ∆ DMA, ∠M = 90°
Area of trapezium ABCD
= \(\frac{1}{2}\) × Sum of parallel sides X Distance between parallel sides
= \(\frac{1}{2}\) × (AB + CD) × DM
= \(\frac{1}{2}\) × (2 + 1) × \(\frac{\sqrt{3}}{2}\)cm²
= \(\frac{1}{2}\) × 3 × \(\frac{\sqrt{3}}{2}\) cm²
= 1.3 cm² (approx.)
For the right triangle in part IV the sides forming the right angle measure 6 cm and 1.5 cm.
Area of right triangle in part IV.
= \(\frac{1}{2}\) × Product of sides forming the right angle
= \(\frac{1}{2}\) × 6 × 1.5 cm²
= 4.5 cm²
The right triangle in part V is congruent to the right triangle in part IV.
∴ Area of right triangle in part V = 4.5 cm²
Now, total area of the paper used
= Areas of figures in part I to part V
= (2.5 + 6.5 + 1.3 + 4.5 + 4.5) cm²
= 19.3 cm²

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer:
In the given triangle, a = 26 cm, b = 28 cm and c = 30 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{26+28+30}{2}\) = \(\frac{84}{2}\) = 42 cm
Area of triangle

= 7 × 6 × 4 × 2 cm²
= 336 cm²
The area of the triangle and the area of the parallelogram are equal.
∴ Area of the parallelogram = 336 cm²
∴ Base × Corresponding altitude = 336 cm²
∴ 28 cm × Corresponding altitude = 336 cm²
∴ Corresponding altitude = \(\frac{336}{28}\) cm
∴ Corresponding altitude = 12 cm
Thus, the height of the parallelogram is 12 cm.

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Answer:

Rhombus ABCD in the given figure represents the field.
A diagonal of a rhombus divides it into two congruent triangles.
∴ Area of rhombus ABCD = 2 × Area of ∆ ABC
In ∆ ABC, a = 30 m; b = 30 m; and c = 48 m.
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{30+30+48}{2}\) = \(\frac{108}{2}\) = 54 cm
Area of ∆ ABC

= 3 × 6 × 24 m²
= 432 m²
Now, area of the field
= area of rhombus ABCD
= 2 × area of ∆ ABC
= 2 × 432 m²
= 864 m²
Now, area of grass field available for 18 cows to graze = 864 m²
∴ Area of grass field available for 1 cow to graze = \(\frac{864}{18}\) m² = 48 m²
Thus, each cow gets 48 m² of grass field to graze.

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Answer:
Out of 10 triangular pieces, 5 are dark coloured and 5 are light coloured.
For each triangle, a = 20 cm, b = 50 cm and c = 50 cm

Hence, the total area of 5 dark coloured cloth pieces = 5 × 200 √6 cm² = 1000 √6 cm²
Similarly, the total area of 5 light coloured cloth pieces = 5 × 200 √6 cm² = 1000 √6 cm²

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it?

Answer:
Let us name the square part as ABCD and the triangular part as CMN.
Suppose the length of square ABCD is xcm.
∴ In ∆ ABD, AB = AD = x cm and ∠A = 90°
The length of hypotenuse BD is given to be 32 cm.
AB² + AD² = BD² (Pythagoras’ theorem)
∴ (x)² + (x)² = (32)²
∴ 2x² = 1024
∴ x² = 512
∴ x = √512
∴ x = \(\sqrt{256 \times 2}\)
∴ x = 16√2
Thus, the length of each side of square ABCD is 16 √2 cm.
Area of part I = Area of ∆ ABD
= \(\frac{1}{2}\) × AB × AD (∠A is a right angle.)
= \(\frac{1}{2}\) × 16 √2 × 16 √2 cm²
= 256 cm²
Area of part II = Area of A BCD
= \(\frac{1}{2}\) × BD × CD (∠A is a right angle.)
= \(\frac{1}{2}\) × 16 √2 × 16 √2 cm²
= 256 cm²
Note: Here, area of square ABCD can easily be found as below:
Area of square ABCD = \(\frac{(\text { Hypotenuse })^{2}}{2}\)
= \(\frac{(32)^{2}}{2}\)
= \(\frac{1024}{2}\)
= 512 cm²
To find the area of part III, we find the area of ∆ CMN.
In ∆ CMN, a = 6 cm, b = 8 cm and c = 6 cm.
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{6+8+6}{2}\) = \(\frac{20}{2}\) = 10 cm

Area of part III
= Area of ∆ CMN

= 8 × 2.24 cm² (approx.)
= 17.92 cm² (approx.)

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see the given figure). Find the cost of polishing the tiles at the rate of 50 p per cm².

Answer:
For each of 16 triangular tiles,
a = 9 cm; b = 28 cm and c = 35 cm

= 88.2 cm² (approx.)
∴ Area of 16 tiles = 16 × 88.2 cm²
= 1411.2 cm²
50 paise = ₹ 0.50
Cost of polishing 1 cm² region = ₹ 0.50
∴ Cost of polishing 1411.2 cm² region
= ₹ (1411.2 × 0.50)
= ₹ 705.60

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Answer:

In the given figure, trapezium ABCD represents the field in which AB || CD,
AB = 25 m, BC = 14 m, CD = 10 m and DA = 13 m.
Through C, draw a line parallel to DA to intersect AB at E.
In quadrilateral AECD, AE || CD and DA || CE
∴ AECD is a parallelogram.
∴ CE = DA = 13 m and AE = CD = 10 m
Now, BE = AB – AE = 25 – 10 = 15 m
In ∆ CEB, a = 13 m; b = 15 m and c = 14 m

In ∆ CEB, draw CM ⊥ BE.
Area of ∆ CEB = \(\frac{1}{2}\) × BE × CM
∴ 84 m² = \(\frac{1}{2}\) × 15 m × CM
∴ CM = \(\frac{84 \times 2}{15}\) m
∴ CM = 11.2 m
Area of parallelogram AECD
= Base × Corresponding altitude
= AE × CM
= 10 × 11.2 m²
= 112 m²
Hence, area of the field
= Area of trapezium ABCD
= Area of ∆ CEB + Area of parallelogram AECD
= 84 m² + 112 m²
= 196 m²
Note: After finding CM = 11.2m, the area of . the field can also be found as below:
Area of the field
= Area of trapezium ABCD
= \(\frac{1}{2}\) × sum of parallel sides × distance between parallel sides
= \(\frac{1}{2}\) × (AB + CD) × CM
= \(\frac{1}{2}\) × (25 + 10) × 11.2 m²
= \(\frac{1}{2}\) × 35 × 11.2 m²
= 196 m²