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Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 11 Ratio and Proportion Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 11 Ratio and Proportion Ex 11.1

1. Express the following ratios in the simplest form:

Question (i)
12 : 32
Solution:
To express this ratio in the simplest form, we shall have to divide both terms by their H.C.F.

So H.C.F. of 12 and 32 = 4
∴ 12 : 32= (12 ÷ 4) : (32 ÷ 4)
= 3 : 8

Question (ii)
45 : 25
Solution:
To express this ratio in the simplest form, we shall have to divide both terms by their H.C.F.

So H.C.F. of 45 and 25 = 5
∴ 45 : 25 = (45 ÷ 5) : (25 ÷ 5)
= 9 : 5

Question (iiii)
91 : 104
Solution:
To express this ratio in the simplest form, we shall have to divide both terms by their H.C.F.

So H.C.F. of 91 and 104 = 13
∴ 91 : 104 = (91 ÷ 13) : (104 ÷ 13)
= 7 : 8

Question (iv)
60 : 72
Solution:
To express this ratio in the simplest form, we shall have to divide both terms by their H.C.F.

So H.C.F. of 60 and 72 = 12
∴ 60 : 72 = (60 + 12) : (62 ÷ 12)
= 5 : 6

Question (v)
375 : 125.
Solution:
To express this ratio in the simplest form, we shall have to divide both terms by their H.C.F.

So H.C.F. of 375 and 125 = 125
∴ 375 : 125 = (375 ÷ 125) : (125 ÷ 125)
= 3 : 1

2. Write the ratio in the simplest form:

Question (i)
₹ 20 to ₹ 55
Solution:
₹ 20 to ₹ 55
= 20 : 55
= (20 ÷ 5) : (55 ÷ 5)
[Divide both terms by their H.C.F.]
= 4:11

Question (ii)
18 m to 63 m
Solution:
18 m to 63 m
= 18 : 63 [H.C.F. of 18 and 63 = 9]

= (18 ÷ 9): (63 ÷ 9)
= 2 : 7

Question (iii)
40 paise to ₹ 2
Solution:
40 paise to ₹ 2
= 40 paise to 200 paise
(₹ 1 = 100 paise)
= 40 : 200
[H.C.F. of 40 and 200 = 40]

= 1 : 5

Question (iv)
One hour to 36 minutes
Solution:
One hour to 36 minutes
= 60 minutes to 36 minutes
(1 hour = 60 minutes)
= 60 : 36
[H.C.F. of 60 and 36 = 12]

= (60 ÷ 12) : (36 ÷ 12)
= 5 : 3

Question (v)
5 kg to 1200 g.
Solution:
5 kg to 1200 g
= 5000 g : 1200 g
( ∵ 1 kg = 1000 g) [H.C.F. of 5000 and 1200]

= 5000 : 1200
= (5000 ÷ 200) : (1200 ÷ 200)
= 25 : 6

3. Simplify the following ratios:

Question (i)
2 years : 14 months
Solution:
2 years : 14 months
= 24 months : 14 months
(∵ 1 year =12 months)
= 24 : 14 [H.C.F. of 24 and 14 = 2]

= (24 ÷ 2) : (14 ÷ 2)
= 12 : 7

Question (ii)
28 min : 2 hours
Solution:
28 min : 2 hours
= 28 min : 120 min
(∵ 1 hour = 60 min)
= 28 : 120
[H.C.F. of 28 and 120 = 4]

= (28 ÷ 4) : (120 ÷ 4)
= 7 : 30

Question (iii)
125 ml: 21
Solution:
125 ml : 2 l
= 125 ml : 2000 ml
(∵ 1 l = 1000 ml)
= 125 : 2000
[H.C.F. of 125 and 2000 = 125]

= (125 ÷ 125) : (2000 ÷ 125)
= 1 : 16

Question (iv)
4 m 20 cm : 80 cm
Solution:
4 m 20 cm : 80 cm
= 420 cm : 80 cm
(∵ 1 m = 100 cm)
= 420 : 80
[H.C.F. of 420 and 80 = 20]

= (420 ÷ 20) : (80 ÷ 20)
= 21 : 4

Question (v)
3 dozen : 12 pieces.
Solution:
3 dozen : 12 pieces
= 36 pieces : 12 pieces
(∵ 1 dozen =12 pieces)
= 36 : 12
[H.C.F. of 36 and 12 = 12]

= (36 ÷ 12) : (12 ÷ 12)
= 3 : 1

4. Find two equivalent ratios for each given ratio:

Question (i)
4:1
Solution:
4 : 1
4 × 2 : 1 × 2 = 8 : 2
(Multiply both terms by 2)
or 4 × 3 : 1 × 3 = 12 : 3
(Multiply both terms by 3)
∴ 8 : 2 and 12 : 3 are two equivalent ratios

Question (ii)
3:5
Solution:
3 : 5
3 × 2 : 5 × 2 = 6 : 10
(Multiply both terms by 2)
or 3 × 3 : 5 × 3 = 9 : 15
(Multiply both terms by 3)
Thus two equivalent ratios for 3 : 5 are 6 : 10 and 9 : 15

Question (iii)
5 : 12.
Solution:
5 : 12
5 × 2 : 12 × 2 = 10 : 24
(Multiply both terms by 2)
or 5 × 3 : 12 × 3 = 15 : 36
(Multiply both terms by 3)
Thus two equivalent ratios of 5 : 12 are 10 : 24 and 15 : 36

5. The number of boys and girls in a class are 60 and 52 respectively. Find the ratio of number of boys to the number of girls.
Solution:
We have number of boys = 60
and number of girls = 52
Ratio of number of boys to the number of girls = 60 : 52
(Divide both by their H.C.F. = 4)
= 15 : 13

6. Pankaj has 23 pens and 42 pencils. Find the ratio of pens to pencils.
Solution:
Number of pens = 23
Number of pencils = 42
∴ Ratio of pens to pencils = 23 : 42

7. In a year, Harjot earns ₹ 2,80,000 and saves ₹ 60,000. Find the ratio of money:

Question (i)
He saves to the money he spends.
Solution:
Harjot’s income = ₹ 2,80,000
Haijot’s savings = ₹ 60,000
Haijot’s spendings = ₹ 2,80,000 – ₹ 60,000
= ₹ 2,20,000

Ratio of Harjot’s savings of Haijot’s spendings = 60,000 : 2,20,000 (Divide both terms by their H.C.F.
= 20,000)
= 3:11

Question (ii)
He earns to the money he saves.
Solution:
Ratio of Haijot’s income to Haijot’s savings = ₹ 2,80,000 : ₹ 60,000
(Divide both terms by their H.C.F. = 20,000)
= 14 : 3

Question (iii)
He spends to the money he earns.
Solution:
Ratio of Haijot’s spendings to Haijot’s income = 2,20,000 : 2,80,000
(Divide both terms by their H.C.F. = 20,000)
= 11 : 14

8. In a school, there are 175 boys, 205 girls students and 20 teachers. Find the ratio of the number of:

Question (i)
Boys to the number of teachers.
Solution:
Number of boys = 175
Number of girls = 205
Number of teachers = 20
Number of total persons in the school = 175 + 205 + 20 = 400.

Ratio of number of boys to the number of teachers = 175 : 20
(Divide both terms by their H.C.F. = 5)
= 35 : 4

Question (ii)
Girls to the number of boys.
Solution:
Ratio of number of girls to the number of boys = 205 : 175
(Divide both terms by their H.C.F. = 5)
= 41 : 35

Question (iii)
Teachers to the number of total persons in the school.
Solution:
Ratio of number of teachers to the number of total persons in the school = 20 : 400
(Divide both terms by their H.C.F. = 20)
= 1 : 20

9. Out of 144 students in a school, 48 play cricket, 28 play kabaddi, 40 play volleyball and the remaining play kho-kho. Find the ratio of:

Question (i)
Number of students play kabaddi to the number of students play kho- kho.
Solution:
Total number of students in the school = 144
Number of students play cricket = 48
Number of students play kabaddi = 28
Number of students play volley ball = 40
Number of students play kho-kho
= 144 – (48 + 28 + 40)
= 144 – 116
= 28

Ratio of number of students play kabaddi to the number of students play kho-kho
= 28 : 28
= 1 : 1

Question (ii)
Number of students play cricket to the number of students play volleyball.
Solution:
Ratio of number of students play cricket to the number of students play volleyball = 48 : 40
(Divide both terms by their H.C.F. = 8)
= 6 : 5

Question (iii)
Number of students who play kho- kho to the total students of school.
Solution:
Ratio of number of students who play kho-kho to the total students of school = 28 : 144
(Divide both terms by their H.C.F. = 4) = 7 : 36

10. The present age of Kush and Shelly are 22 years and 16 years respectively. Find the ratio of:

Question (i)
Their present ages.
Solution:
Ratio of present age of Kush to the present age of Shelly = 22 : 16
(Divide both terms by their H.C.F. = 2)
= 11 : 8

Question (ii)
Kush’s age to Shelly’s age after 4 years.
Solution:
After four years Kush’s age
= 22 + 4 = 26 years
After four years Shelly’s age
= 16 + 4 = 20 years
Ratio of Kush’s age to Shelly’s age after four years
= 26 : 20
(Divide both terms by their H.C.F. = 2)
= 13 : 10

Question (iii)
Shelly’s age to Kush’s age before 5 years.
Solution:
Ratio of Shelly’s age to Kush’s age before 5 years
= 17 : 11

Question (iv)
Kush’s present age to Shelly’s age after 6 years.
Solution:
Ratio of Kush’s present age to Shelly’s age after 6 years
= 22 : 22
(Divide both terms by their H.C.F. = 22)
= 1 : 1

11. In a pencil box there are 150 pencils. Out of which 40 are red, 60 are black and the rest are blue pencils. Find the ratio of:

Question (i)
Red pencils to the black pencils.
Solution:
Total pencils = 150
Red pencils = 40
Black pencils = 60
Blue pencils = 150 – (40 + 60)
= (150 – 100)
= 50

Ratio of Red pencils to the black pencils = 40 : 60
(Divide both terms by their H.C.F. = 20)
= 2 : 3

Question (ii)
Blue pencils to the total number of pencils.
Solution:
Ratio of Blue pencils to the total number of pencils = 50 : 150
(Divide both terms by their H.C.F. = 50)
= 1 : 3

Question (iii)
Total pencils to the red pencils.
Solution:
Ratio of total pencils to the red pencils = 150 : 40
(Divide both terms by their H.C.F. = 10)
= 15 : 4

12. Divide ₹ 175 in ratio 4 : 3 between Preet and Sukhi.
Solution:
Total Amount = ₹ 175
Ratio =4 : 3
Let Preet’s share = 4 x
and Sukhi’s share = 3 x
Acc. to Question
Preet’s share + Sukhi’s share = ₹ 175
⇒ 4x + 3x = 175
⇒ 7x = 175
x = \(\frac {175}{7}\) = 25
∴ Preet’s share = 4x
= 4 × 25
= ₹ 100
Sukhi’s share = 3x
= 3 × 25
= ₹ 75

13. Two numbers are in the ratio 3:7 and their sum is 140. Find the numbers.
Solution:
Sum of numbers = 140
Ratio = 3 : 7
Let first number = 3x
and second number = 4x
Acc. to Question
(First number) + (Second number)
= 140
⇒ 3x + 7x = 140
⇒ 10 x = 140
x = \(\frac {140}{10}\) = 14
∴ First number = 3x
= 3 × 14
= 42
and Second number = 7x
= 7 × 14
= 98

14. The angles of a triangle are in the ratio 1 : 2 : 3. Find the measure of each angle.
Solution:
Sum of the angles of the triangle = 180°
Ratio of angles = 1 : 2 : 3
Let First angle = x
Second angle = 2x
Third angle = 3x
Acc. to Question
Sum of three angles = 180°
x + 2x + 3x = 180°
6x = 180°
x = \(\frac {180°}{6}\) = 30°
∴ First angle = x = 30°
Second angle = 2x
= 2 × 30° = 60°
Third angle = 3x
= 3 × 30° = 90°
Thus, the angles of triangle are 30°, 60°, 90°

15. A pipe of length 4 m 16 cm is cut into two pieces in ratio 3:5. Find the length of each piece of the pipe.
Solution:
Length of pipe = 4 m 16 cm
= 416 cm (1 m = 100 cm)
Ratio of two parts = 3 : 5
Let length of first part = 3x
and length of second part = 5x
Acc. to Question
3x + 5x = 416
8x = 416
x = \(\frac {416}{8}\) = 52

∴ Length of first part = 3x
= 3 × 52
= 156 cm
= 1.56 m.

Length of second part = 5x
= 5 × 52
= 260 cm
= 2.60 m

Punjab State Board PSEB 8th Class English Book Solutions English Reading Comprehension Conversation / Dialogue Based Exercise Questions and Answers, Notes.

PSEB 8th Class English Reading Comprehension Conversation / Dialogue Based

Read the following conversation carefully and answer the questions that follow:

(1) Akram : Why did you not go to village till this time?
Shan : No, I have changed my programme. I do not want to leave comfortable life of the city and lead a very dull and monotonous life in a village.
Akram : You have a very bad unpression about village life. Why ?
Shan : Yes, I am saying right. The village is full of dust and dirt. Many comforts of life are not available in a village. Heaps or garbage can be seen everywhere. People and animals live at the same place. They use the water from the dirty pond. Ignorance prevails everywhere.
Akram : My friend you are mistaken. Many villages have become modern now. There are good schools and hospitals in almost every village. Sanitary system has improved a lot. Electricity has reached in every village which has made life much easier and better.
Shan : But the cities have better facilities, beautiful houses, modern means of communication, fast and comfortable vehicles and many more. The people of cities have comfortable life.
Akram : You are talking superficially. There are crowded houses. A large number of people live in small houses which is injurious to health. Polluted air, dirty streets and stinking drains spread many diseases. The people of cities have no love ‘and sympathy whereas villagers are very sincere. The village life has fresh air, simplicity and love.

Question 1.
Shan thinks that life in a village is:
(a) full of adventures
(b) modern and advanced
(c) dull and monotonous
(d) full of comforts and luxuries
Answer:
(c) dull and monotonous.

Question 2.
What are the major drawbacks of a village life according to Shan ?
(a) lack of sanitation
(b) dirty surroundings
(c) ignorgant people
(d) all of the above.
Answer:
(d) all of the above.

Question 3.
What makes city life better than village life ?
(a) crowded houses
(b) polluted air and stinking drains
(c) better facilities like communication, transportation and high living standard
(d) people who lack love and sympathy.
Answer:
(c) better facilities like communication, transportation and high living standard

Question 4.
Polluted air and stinking drains have made city life:
(a) more comfortable
(b) full of diseases
(c) thrilling and adventurous
(d) none of the above.
Answer:
(b) full of diseases.

Question 5.
What does the above conversation tell us ?
(a) It draws a comparison between life in a village and a city.
(b) Life in cities is better than in villages.
(c) People in villages are uncivilized and ignorant.
(d) Village life is full of discomforts.
Answer:
(a) It draws a comparison between life in a village and a city.

2. Raman : Dad ! You promised to take me to the shopping mall on Saturday.
Daddy : I remember, my child. Finish your breakfast and get ready to go.
Raman : Yah ! You are the best .dad. (Raman finished his breakfast hurriedly and got ready) I am ready.
Daddy : Let’s go.
(While sitting in the car, daddy asked Raman to wear his seat belt.)
Raman : I know daddy that the traffic policeman would challan us if we are not wearing the seat belt.
Daddy (laughs) : We do not wear seat belt for the policeman. We wear it for our safety. It protects driver and passengers from injury during any type of accident.
Raman : Hmm ! (He pulls his seat belt and smiles.)
(At traffic signal, daddy stops the car just on the zebra crossing.)
Raman : Daddy ! We should not stop the vehicle on the zebra crossing. Our teacher told us that zebra crossing is for the safety of pedestrians. Vehicles have to stop before the zebra crossing to let the pedestrians cross the road safely.
Daddy : Very well, Raman.
(Raman starts singing.)
Raman : Red light, Red light, What do you say ?
I say, Stop !
Stop ! right away.
Yellow light, Yellow light,
What do you say ?
I say, wait!
Wait! Right away.
Green light, Green light,
What do you say ?
I say, Go !
Go ! Right away.
(Daddy smiles and moves the car when the signal goes green.)

Question 1.
In the car Daddy asked Raman to
(a) wear his cap
(b) wear his helmet
(c) wear his seat belt
(d) remove his seat belt.
Answer.
(c) wear his seat belt.

Question 2.
The seat belt is meant for:
(a) safety during accidents
(b) alertness on the road
(c) saving us from challan
(d) all the above.
Answer:
(a) safety during accidents

Question 3.
Zebra crosssing is meant for the :
(a) two wheelers to cross the road safely
(b) car drivers to cross the road safely
(c) policeman to control the traffic
(d) pedestrians to cross the road safely.
Answer:
(d) pedestrians to cross the road safely.

Question 4.
Daddy stops his car just:
(a) on the zebra crossing
(b) before the zebra crossing
(c) in the middle of road.
(d) after he crosses the zebra crossing.
Answer:
(a) on the zebra crossing.

Question 5.
Yellow light asks us to:
(a) go
(b) Stop
(c) wait
(d) Look Back
Answer:
(c) wait.

3. Pala said, “Dear, look at the buildings. How tall they are ! Our village is now on the way to advancement”. Beero grumbled, “Advancement! Don’t you realize that our health is at stake ?” “What are you saying ?” asked Pala. “You need to look around you,” said Beero. People have beautiful homes with all kinds of facilities but they are not using them properly.”

“What do you want to say ?” asked Pala. Beero responded, “They are not using their toilets.” “What!” exclaimed Pala. Beero continued, “They defecate in the open near my home. This place stinks ! I am fed up of this unpleasant odour. They are not even afraid of the danger they are going to face.”
Pala said, “What kind of danger, Beero ?”
Beero said, “Diseases ! How can we forget the two children of our village who died of diarrhoea and infection. At least I can’t ! I am surprised how can man be so ignorant about good hygiene practices ?” She continued, “I have decided I will not tolerate it anymore.”
“What will you do ?” asked Pala.

Beero announced, “I will spread awareness among the people about the use of toilet and the advantages of keeping their homes and surroundings clean and healthy. Will you help me ?

“Of course ! A good deed needs no second thought, no permission,” remarked Pala.

Question 1.
The village was on the way to advancement. What was its sign ?
{a) healthy atmosphere
(b) new houses
(c) tall buildings
(d) all the above.
Answer:
(c) tall buildings.

Question 2.
People had beautiful houses but they were not using their:
(a) toilets
(b) drawing rooms
(c) kitchens
(d) store houses.
Answer:
(a) toilets.

Question 3.
The people in the village defecated:
(a) in front of their house
(b) at the back of their house on
(c) in the open near Beero’s home
(d) the roof of their building.
Answer:
(c) in the open near Beero’s home.

Question 4.
Two children in the village had died of:
(a) cholera and malaria
(b) malaria and diarrhoea
(c) typhoid and nausea
(d) diarrhoea and infection.
Answer:
(d) diarrhoea and infection.

Question 5.
Pala and Beero decided to spread awareness among the people about:
(a) the use of toilets
(b) keeping their homes clean
(c) keeping their surroundings clean and healthy
(d) all of these.
Answer:
(d) all of these.

(4) Kamal : Good Morning, Madam !
Madam : Good Morning ! Sit down. What do you want ?
Kamal : I want to get admission in your schoof.
Madam : Which class do you want to take admission in ?
Kamal : I have just passed class seven. I want to take admission in eighth class.
Madam : Where were you studying before ?
Kamal : I studied in Delhi Public School, Ludhiana. Now my father has been transferred to this city.
Madam : What does your father do ?
Kamal : He is a bank manager.
Madam : Okay. You have to fill the admission form first. Attach your School Leaving Certificate with it.
Kamal : Thank you. mon.

Question 1.
Why did Kamal come to the school ?
(a) He wants to apply for a job
(b) He wants to study in the school
(c) He wants to take part in games
(d) None of the above.
Answer:
(b) He wants to study in the school.

Question 2.
In which class does he want to study ?
(a) sixth
(b) seventh
(c) eighth
(d) tenth.
Answer:
(c) eighth.

Question 3.
In which city did he study before ?
(a) Ludhiana
(b) Patiala
(c) Kapurthala
(d) Bathinda.
Answer:
(a) Ludhiana.

Question 4.
Why does he want to change the school ?
(a) He did not like his previous school ?
(b) His father was transferred to another city
(c) He was failed
(d) None of the above.
Answer:
(b) His father was transferred to another city.

Question 5.
What is required to get admission along with the admission from ?
(a) Identity Proof
(b) Detailed Marks Certificate
(c) School Leaving Certificate
(d) Residence Certificate.
Answer:
(c) School Leaving Certificate.

5. Aman s father is going to the office. His mother asks his father to pay the electricity bill.
Father : I’m very busy. I have a meeting today.
Mother : Today is the last date to pay the bill.
Father : OK. I will try. (After using his mobile phone) I have paid the bill. Aman has been watching all this and is very curious to know how his father has paid the bill. In the evening, he asks his father about it.
Aman : Papa you did not go to the Electricity office but you paid the bill. How is it possible?
Father : I paid the bill using net banking facility.
Aman : Oh ! What is net-banking facility? Please tell me.
Father : Ok, listen. A bank is a safe place where we can save our money. It receives money from those who want to save it and lends money on interest to those who need it.
Aman : Can we get back our money?
Father : Yes, of course. It depends upon the type of account we choose. From saving account we can withdraw money whenever we need.
Aman : Why should we deposit money in the bank ?
Father : In a bank our money is always safe. A bank also pays us some extra money called interest for our deposit.

Question 1.
What does Aman’s mother ask his father to do?
(a) not to go to the office
(b) to go to the bank
(c) to pay the electricity bill
(d) not to attend the meeting that day.
Answer:
(c) to pay the electricity bill.

Question 2.
The bill must be paid that day because:
(a) it is the last date to pay it
(b) the banks would not be opened the next day
(c) the electricity office would be closed the next day
(d) there was a strike the next day.
Answer:
(a) it is the last date to pay it.

Question 3.
Aman’s father paid the bill through:
(a) a courier
(b) net-banking.
(c) a check
(d) none of these.
Answer:
(b) net-banking.

Question 4.
For saving of our money, a bank is:
(a) a risky place
(b) not a proper place
(c) a place beyond our reach
(d) a safe place.
Answer:
(d) a safe place.

Question 5.
Extra money that a bank pays us on our deposit is called:
(a) principal
(b) principle
(c) saving
(d) interest.
Answer:
(d) interest.

6. One day, Rahim and his father went for a morning walk at 6 o’ clock. Rahim was questioning his father about the things around and enjoying. On the way, he saw beautiful mountains, lush green lands, grazing cows and white ducks swimming in a small pond.
Rahim : Father, I am tired, now.
Father : We can take rest.
(They both sat down under a shady walnut tree. Suddenly, Rahim s eyes fell on a very big watermelon growing in a field nearby.)
Rahim : Which fruit is that father ?
Father : That is a watermelon. It grows on a vine.
Rahim : Walnut is much smaller than the watermelon but the walnut tree is stronger
than the watermelon vine why God did that ?
Father : What do you think ?
Rahim : I think God has made a mistake. The walnut should have grown on a yine and the watermelon on a tree.
Father : Rahim, never doubt God. Whatsoever God has done or does is always wise decision.
(Just then a walnut fell on Rahims head and struck his head sharply.)
Rahim : Ouch! Now I understand. I am glad that walnuts and not watermelons grow on trees. God, the Almighty is, indeed very wise.

Question 1.
What did Rahizn not see on the way?
(a) lush green lands
(b) beautiful mountains
(c) grazing cows
(d) a watermelon growing on a tree
Answer:
(d) A watermelon growing on a tree.

Question 2.
Where was the big watermelon growing’
(a) in a field
(b) on a tree
(c) in a pond
(d) on the mountain.
Answer:
(a) in a field.

Question 3.
What does a walnut grow?
(a) on a tree
(b) on a vine
(c) on a bush
(d) in a pond.
Answer:
(b) a vine.

Question 4.
‘Whose decision is always right?
(a) Rahim’s
(b) Our friend’s decision.
(c) our elders
(d) God’s.
Answer:
(d) God’s.

Question 5.
A watermelon does not grow on a tree because:
(a) it may hurt somebody if it falls down
(b) it does not look nice.
(c) it is very costly
(d) It will be difficult to pide it.
Answer:
(a) it may hurt somebody if it falls down.

7. A building is on fire. The fire started because of a short circuit. Huge flames of fire can be seen coming out of each floor and there is black and thick smoke all around. People can be seen running with buckets full of water. They are trying to put out the fire. But are they successful ? No. The fire is spreading to other buildings around. Let’s see what the people are saying to each other.

Mr. Singh : would you please call the fire service on your telephone ?
Mr. Sharma : I’ve already done so. A fire engine is on the way.
Mr. Singh : Phone all the people living in the building to come out. The police have cordoned off the building. A large crowd has gathered on the site.
Mr. Sharma : Yes, some people have come out, but there are others who are trapped in the building.
Mr. Singh : Can you hear the source of alarm bells ? Oh yes, I can also see fire engine coming at full speed.
Mr. Sharma : What a relief!
Mr. Singh : The firemen are at their task. They can be seen using ladders to bring down the people who are trapped.
Mr. Sharma : The firemen are using hoses to spray water on the fire. Soon the fire will be put out.
Mr. Singh : One of the buildings has been reduced to ashes. Everybody was happy that the fire has been controlled and the other buildings have been saved.
Mr. Sharma : Let us thank the firemen for the wonderful job they have done indeed. They have risked their lives to save the houses and the people.

The firemen feel happy They get into the engines and drive away.
Question 1.
The fire started because of a:
(a) burning match
(b) short circut
(c) a bright lamp
(d) a neglected spark.
Answer:
(b) short circut.

Question 2.
called the fire service?
(a) Mr. Verma
(b) Mr. Singh
(c) Mr Sharma
(d) A policeman.
Answer:
(c) Mr. Sharma

Question 3.
The firemen used hoses to:
(a) bring the trapped people down
(b) to climb the tall building
(c) to prevent people coming near the fire
(d) to spray water on the fire.
Answer:
(d) to spray water on the fire.

Question 4.
How many buildings were reduced to ashes?
(a) one
(b) two
(c) three
(d) four.
Answer:
(a) one.

Question 5.
The firemen were successful in:
(a) controlling the fire
(b) saving the people caught in fire
(c) saving the buildings
(d) all the above.
Answer:
(d) all the above.

Teacher : Happy Birthday to you, Neha.
Neha : Thank you, Madam.
Teacher : Who bought this pretty dress for you ?
Neha : My mother bought it for me.
Teacher : How old are you now, Neha ?
Neha : I am eight years old now.
Teacher : Are you organising a party at home ?
Neha : Yes, Madam, I am holding a tea party in the evening today. You are cordially invited. Please do come.

Teacher : Thank you. Neha I will try to come. Who else have you invited to your party ?
Neha : Madam, all my friends and relatives. I have also got some sweets to distribute among my classmates.
Teacher : (To other children) Let us first sing a Birthday Song for Neha.
Teacher and Children : Happy Birthday to you ! Happy Birthday to you ! Happy Birthday to Dear Neha!
(All friends of Neha come to the party at 6 o’clock in the evening dressed in their best party wear. Then the teacher enters the room.)
Children : Good evening, Madam.
Teacher : Good evening.
Children : (To Neha) God bless you, Neha !
Teacher : Here is a Birthday Gift for you. I wish you many happy returns of the Day!
Neha : Thank you. Madam.
Mother : Children, come here. Now Neha is going to cut the cake.
Children : Happy Birthday to Neha.
Uncle : Sorry I am late. Happy Birthday, Neha. Here is a gift for you: it’s a packet of books.
Neha : Thank you, Uncle. Thank you very much. It’s really a nice gift !
Father: Children, now please do have a piece of cake and sweets.
And here are the return gifts for all of you.
(Neha’s mother gives the presents to the children)
Children : Thank you, Uncle. Thank you, Aunt for these beautiful gifts.
Neha : Thank you, everyone. Thanks for my Birthday gifts.

Question 1.
Whose birthday was it ?
(a) Mother’s
(b) Neha’s
(c) Father’s
(d) Teacher’s
Answer:
(b) Neha’s.

Question 2.
Who sang ‘Birthday Song’ ?
(a) Teacher
(b) Children
(c) Teacher and children
(d) Mother and Father.
Answer:
(c) Teacher and children.

Question 3.
What was the birthday gift of Neha’s uncle ?
(a) a golden wrist watch
(b) a packet of gel-pens
(c) a packet of books
(d) a beautiful dress.
Answer:
(c) a packet of books.

Question 4.
Uncle felt sorry for:
(a) not bringing aunty with him
(b) not bringing some costly gift
(c) not singing Birthday Song
(d) being late.
Answer:
(d) being late.

Question 5.
Children thanked Uncle and Aunt. Who are they ?
(a) Neha’s parents
(b) Neha’s uncle and aunt
(c) Neha’s next door neighbours
(d) None of these
Answer:
(a) Neha’s parents.

9. Teacher : Do you know why I have called you here ? I’ve come to know that most of you start eating your lunch before washing your hands. You should know that this habit will make you fall sick. When you eat with dirty hands, you carry some kinds of germs inside your body.
Students : Sir, is this the only way to keep ourselves healthy ?
Teacher : (smiling) No, there are many other dos and don’ts while we eat. I’ll tell you some of them. These are :
Always eat well-cooked food.
Wash your hands properly before and after taking meals.
Chew your food properly.
Don’t take food more than what you can eat.
Always use clean utensils.
Don’t leave any food in your plate.
Students : Thank you, sir. We’ll follow these.
Teacher : (smiling) You’re welcome. Now go and have your meals peacefully.
(All children queue up to wash their hands.)

Question 1.
What was the teacher’s complaint about ?
(a) eating lunch before washing hands
(b) eating lunch after washing hands.
(c) eating lunch before taking a bath
(d) eating lunch fast.
Answer:
(d) eating lunch before washing hafids.

Question 2.
You may fall sick if you eat with your:
(a) right hand
(b) left hand
(c) dirty hands
(d) wet hands.
Answer:
(c) dirty hands.

Question 3.
What should we not do ? (Pick out two choices)
(a) chewing food properly
(c) using clean utensils
(b) taking food more than we can eat
(d) leaving food in our plate.
Answer:
(b) taking food more than we can eat
(d) leaving food in our plate.

Question 4.
The students thanked the teacher for:
(a) telling them dos and don’ts of healthy food eating
(b) telling them how to cook healthy food
(c) telling them don’ts of food making.
(d) telling them how to wash hands.
Answer:
(a) telling them dos and don’ts of healthy food eating.

Question 5.
To wash their hands, all children:
(a) stood up
(b) sat down
(c) queued up
(d) ran outside
Answer:
(c) queued up.

Punjab State Board PSEB 8th Class English Book Solutions English Reading Comprehension Picture / Poster Based Exercise Questions and Answers, Notes.

PSEB 8th Class English Reading Comprehension Picture / Poster Based

Look at the pictures carefully and answer the questions that follow:

Question 1.
What is the purpose of this advertisement ?
(a) to prevent people from using motor vehicle.
(b) to spread awareness about traffic rules.
(c) to stop people from walking on the road.
(d) to secure people of road ancient.
Answer:
(b) to spread awareness about traffic rules.

Question 2.
While on scooter or bike, which thing can help to save our lives:
(a) scarf
(b) cap
(c) helmet
(d) seat belt.
Answer:
(c) helmet

Question 3.
Zebra crossing is meant for:
(a) four wheelers
(b) bikers
(c) cyclists
(d) pedestrians
Answer:
(d) pedestrians

Question 4.
One should stop the vehicle when it is a:
(a) red light
(b) yellow light
(c) green light
(d) none of these
Answer:
(a) red light

Question 5.
Road accidents can be prevented by:
(a) driving within a speed limit
(b) not driving while drinking
(c) obeying the traffic rules
(d) all of the above.
Answer:
(d) all of the above.

Working Together to Keep Our Children Safe.

Question 1.
What is the purpose of this advertisement ?
(a) To make children happy
(b) Teaching children how to drive a bike or a car
(c) Taking chidren to the park
(d) Promoting road safety awareness among children
Answer:
(d) Promoting road safety awareness.

Question 2.
Children should be aware of:
(a) speed limit while driving
(b) road safety rules
(c) parking their vehicles at a safe place
(d) all these.
Answer:
(d) all these.

Question 3.
Parking of vechiles on the roadside can result in:
(a) an accident
(b) theft of the vehicle
(c) traffic jam
(d) all the above.
Answer:
(d) all the above.

Question 4.
For safe driving the driver should have the knowledge of:
(a) signboards on the roadside
(b) his R.C.
(c) the vehicles coming behind him
(d) the condition of his vehicle.
Answer:
(a) signboards on the roadside.

Question 5.
While driving we should:
(a) not drink
(b) not use our mobile
(c) not cross green light
(d) not drive below speed limit.
Answer:
(a) and (b)

Question 1.
What is the theme of the picture ?
(a) The Values and Advantages of Games and Sports
(b) Good Manners
(c) The Value of Reading Books
(d) The hazards of Pollution.
Answer:
(b) Good Manners.

Question 2.
Which of the following is not a good habit ?
(a) helping old people
(b) planting trees
(c) getting up early in the morning
(d) keeping your classroom dirty.
Answer:
(d) keeping your classroom dirty.

Question 3.
Which kind of words ‘please’ and thankyou’ are ?
(a) bad words
(b) harsh words
(c) polite words
(d) difficult words.
Answer:
(c) polite words.

Question 4.
We should wait for our turn by standing in the line.
(a) quietly
(b) uneasily
(c) impatiently
(d) angrily.
Answer:
(a) quietly.

Question 5.
‘Early to bed, early to rise makes a man healthy, and wise.’
(a) dull
(b) poor
(c) wealthy
(d) foolish.
Answer:
(c) wealthy.

Question 1.
The poster tells us that:
(a) India is a land of festivals.
(.b) we celebrate many festivals in India.
(c) festivals of all religions are celebrated in India.
(d) all these.
Answer:
(d) all these.

Question 2.
What is the importance of festivals in our life ?
(a) They give us new energy.
(b) They keep our culture alive.
(c) They entertain us.
(d) All these.
Ans, (d) All these.

Question 3.
Pushkar fair is celebrated:
(a) all over India
(b) in Rajasthan.
(c) in South India
(d) None of these
Answer:

Question 4.
Holi is a festival of:
(a) lights
(b) colours.
(c) praying in mosques
(d) cleaning our houses and shops
Answer:
(b) colours.

Question 5.
Which of the following festivals, in particular, would promote Hindu Muslim unity ?
(a) Diwali and Christmas
(b) Eid and christmas
(c) Pushkar Fair and Christmas
(d) Diwali and Eid.
Answer:
(d) Diwali and Eid.

Question 1.
What is the purpose of this poster/advertisement about ?
(a) Women Backwardness
(b) Women Education
(c) Women Empowerment
(d) Sources of Entertainment for Women.
Answer:
(c) Women Empowerment.

Question 2.
Daughter’s Day gives the message of:
(a) loving daughters only
(b) having daughters only
(c) Beti Bachao Beti Padhao
(d) marry your daughters in their chile
Answer:
(c) Beti Bachao Beti Padhao

Question 3.
Women’s Day is observed on:
(a) 5th September
(b) First sunday of May
(c) 15th September
(d) 8th March.
Answer:
(d) 8th March.

Question 4.
Women feel empowered when they :
(a) use their power to empower others
(b) use their power to belittle others
(c) win elections to rule the country
(d) all these.
Answer:
(a) use their power to empower others

Question 5.
Mother’s Day is celebrated to:
(a) inspire women to become mother soon after their marriage
(b) to honour mothers of the world
(c) to teach uneducated mothers
(d) none of these.
Answer:
(b) to honour mothers of the world

Polio Drops and Healthy Life

Question 1.
The most suitable title for this advertisement is:
(a) Healthy Life
(b) Medication Vs Yoga
(c) Old Age and Yoga
(d) Eating is Better than Yoga.
Answer:
(a) Heatlhy Life.

Question 2.
We should avoid eating:
(a) fruits and vegetables
(b) balanced food
(c) junk food
(d) cooked food.
Answer:
(c) junk food.

Question 3.
Yoga is kind of:
(a) exercise to please Swami Ramdev
(b) diet to grow tall
(c) excercise to keep us fit and healthy
(d) prayer to please god.
Answer:
(c) exercise to keep us fit and healthy.

Question 4.
Which of the following activity is included in a trip to healthy life ?
(a) walking and laughing loudly
(b) crying and yelling
(c) eating food three times a day
(d) taking medicine now and then.
Answer:
(a) walking and laughing loudly.

Question 5.
Polio drops are given to the children of:
(a) two years
(b) Three years
(c) four years
(d) five years.
Answer:
(d) five years.

Question 1.
The best title for this poster is:
(a) Growing and cutting down the trees
(b) Resting and playing under trees
(c) Planting trees in rainy season
(d) Benefits of growing and protecting trees.
Answer:
(d) Benefits of growing and protecting trees.

Question 2.
Trees give us:
(a) fruits
(b) medicines
(c) firewood
(d) all the above.
Answer:
(d) all the above.

Question 3.
Trees serve us by:
(a) giving out oxygen
(b) taking in carbon dioxide
(c) giving us cool shade in summer
(d) all the above.
Answer:
(d) all the above.

Question 4.
Without trees climate would be:
(a) drier and cooler
(b) drier and hotter.
(c) warmer and cooler
(d) drier and hotter wetter and hotter.
Answer:
(b) drier and hotter.

Question 5.
What is our duty towards trees ?
(a) growing more trees and taking proper care of them
(b) cutting down trees only in winter
(c) planting only fruit trees
(d) not to let birds sit in trees.
Answer:
(a) growing more trees and taking proper care of them.

Effects of Noise Pollution

Question 1.
What is purpose of this poster ?
(a) to create awareness against noise pollution.
(b) to use loudspeakers to check noise pollution.
(c) to put hands on ears on hearing a noise
(d) to prevent people from making noise during the day.
Answer:
(a) to create awareness against noise pollution.

Question 2.
Which of the following activity is responsible for noise pollution ?
(a) high volume of loudspeakers
(b) running factories
(c) vehicles running fast on roads
(d) all these.
Answer:
(d) all these.

Question 3.
Too much noise may:
(a) make us deaf
(b) increase the speed of our vehicles
(c) incresae our hearing power
(d) increase our energy to work.
Answer:
(a) make us deaf.

Question 4.
We should not blow horns or ring bells near a hospital because—
(a) it may spoil the medicines
(b) it may disturb the resting patients
(c) the doctors may go on strike
(d) none of these.
Answer:
(b) it may disturb the resting patients.

Question 5.
To avoid noise pollution we should
(a) not blow horns unnecessarily
(b) avoid the use of loudspeakers
(c) not use old vehicles that produce screeching sound.
(d) all these.
Answer:
(d) all these.

Punjab State Board PSEB 8th Class English Book Solutions English Reading Comprehension Unseen Passages Exercise Questions and Answers, Notes.

PSEB 8th Class English Reading Comprehension Unseen Passages

I. Read the given passages and answer the questions that follow:

(1) Trees are as beautiful as they are useful. Wherever they are, they make that place look nice and green. They give us fruits, shade and wood. Birds build nests in their branches. Trees make the whole place like a garden. They are indeed nature’s precious gift to us.

Every tree is a living and breathing creature, like us. But unlike us, it prepares its own food from raw materials such as carbon dioxide, water and sunlight. Also, unlike us, it lacks a well-developed nervous system although it responds to many external stimuli. The, tree breathes through its leaves.

How does the environment affect the growth of the tree ? If there is a lack of water, the roots go down deeper and spread out far and wide, backward and forward, in search of food material. If there are too many trees in one place, they grow higher and higher to reach the sunshine. If there is a strong wind all the time, the tree takes firmer hold of the ground with its roots.

The tree is a strong fighter. It may bend before the wind but it does not always break. It protects itself very well against snow, frost and hail. It can defeat most of its enemies. But human beings defeat the tree every time by cutting it down. Litde do they know that by destroying trees at such a large scale, they are actually destroying themselves.

Question 1.
Trees are natures precious gift to us because:
(a) they provide us food, shade and wood
(b) they provide shelter to the birds
(c) they turn the earth into a beautiful place
(d) all of the above.
Answer:
(d) all of the above.

Question 2.
When there is lack of water, the tree:
(a) grow taller in order to get rain
(b) takes firm hold on the ground
(c) starts breathing through its leaves
(d) sends its roots deep, far and wide.
Answer:
(d) sends its roots deep, far and wide.

Question 3.
Which of the following statement is true for both humans and trees ?
(a) both breathe and grow
(b) both can move and run
(c) both buy their own food
(d) both have a nervous system.
Answer:
(a) both breathe and grow.

Question 4.
Trees fight many enemies but they are not able to defeat :
(a) snow
(b) wind
(c) water scarcity
(d) human beings
Answer:
(d) human beings.

Question 5.
Trees are strong fighters because :
(a) they can adapt themselves to all circumstances
(b) they can kill other trees for their growth
(c) they can defeat all their enemies
(d) they have strong roots and trunk.
Answer:
(a) they can adapt themselves to all circumstances.

(2) Schools all over India celebrate Childrens Day’ on 14th November every year. On this day, our great Prime Minister who had a great love for children was born. His ancestors came down from Kashmir to the rich plains below. Kaul had been his family name; this changed to Kaul-Nehru: and in later years. Kaul was dropped and they became simply Nehrus. Jawahar Lai Nehru was the only son of his prosperous parents. His two sisters were much younger to Jawahar Lai Nehru, And so, he grew up and spent his early years as a lonely child with no companion of his own age. Private tutors were in charge of his education. Then, he went to England and was educated at Harrow and at Trinity College, Cambridge.

Question 1.
Childrens Day is celebrated on:
(a) 15th August
(b) 26th January
(c) 14th November
(d) 30th January.
Answer:
(c) 14th November.

Question 2.
Nehrus ancestors came from:
(a) Delhi
(b) Allahabad
(c) Kashmir
(d) Raibareli.
Answer:
(c) Kashmir.

Question 3.
Jawahar Lai Nehru was educated at:
(a) Wilson College, Mumbai
(b) Harrow and Trinity College, Cambridge
(c) Presidency University, Kalkata
(d) Jesus and Mary College, Delhi.
Answer:
(b) Harrow and Trinity College, Cambridge.

Question 4.
Why is 14th November celebrated as Childrens Day ?
Or
What is the importance of 14th November ?
(a) Pt. Nehru was born on this day.
(b) Mahatma Gandhi was born on this day.
(c) Indira Gandhi was born on this day.
(d) None of the above.
Answer:
(a) Pt. Nehru was born on this day.

Question 5.
Nehru ji belonged to:
(a) a poor family
(b) a family of farmers
(c) a rich / prosperous family
(d) none of these.
Answer:
(c) a rich / prosperous family.

(3) Once a bee felt thirsty. It flew to a pond to drink water. While drinking water, the bee fell into the pond. A dove was sitting on the branch of a tree. It saw all and decided to save the bee’s life. The dove threw a leaf. The bee climbed over the leaf, dried its wings and flew away.

After a few days a hunter came to the forest. He aimed at the dove. Luckily the bee saw the hunter. It flew to the hunter and stung him hard on the hand. The hunter missed his aim. The dove heard the gunshot and flew away. The dove thanked the bee for this timely help.

Question 1.
Where did the bee fly to drink water ?
(a) a canal
(b) a pond
(c) a river
(d) a stream.
Answer:
(b) a pond

Question 2.
What happened to the bee while drinking water ?
(a) It fell from the tree
(b) It fell into the pond
(c) It was shot by the hunter
(d) None of the above.
Answer:
(b) It fell into the pond.

Question 3.
Who saved the bee’s life?
(a) a dove
(b) a hunter
(c) a fish
(d) a tortoise.
Answer:
(a) a dove.

Question 4.
What did the bee do to save the dove’s life?
(a) It killed the hunter
(b) It stung the hunter on the hand
(c) It shouted hard
(d) It did nothing.
Answer:
(b) It stung the hunter on the hand.

Question 5.
What is the moral of the story ?
(a) Do good, have good
(b) Revenge is the best policy
(c) Pride hath a fall
(d) Union is strength.
Answer:
(a) Do good, have good.

(4) Garbage is a great environmental hazard. It comes from various sources-used paper, tiffin packings, plastic bags, ice-cream wrappers, bottle caps, fallen leaves from trees and many more. Garbage makes the premises ugly, unkempt and breeds diseases.

A lot of trash that is thrown away contains material that can be recycled and reused such as paper, metals and glass which can be sent to the nearest recycling centre or disposed of to the junkdealer. It also contains organic matter such as leaves which can enrich land fertility.

A compost pit can be made at a convenient location where the refuse can be placed with layers of soil and and occasional sprinkling of water. This would help decomposition to make valuable manure (fertilizer). This would also prevent pollution that is usually caused by burning such organic waste.

Question 1.
Garbage is a great environmental hazard because it makes the premises:
(a) ugly
(b) unkempt
(c) breed diseases
(d) all these.
Answer:
(d) all these.

Question 2.
What happens to the disposed material at the recycling centre ?
(a) sent back to homes
(b) takes a new shape.
(c) thrown into rivers.
(d) all the above.
Answer:
(b) takes a new shape.

Question 3.
How can we make use of waste organic matter ?
(a) send it to junkdealer
(b) burn it
(c) change it into valuable manure (fertilizer)
(d) all these
Answer:
(c) change it into valuable manure (fertilizer.)

Question 4.
Proper disposal of garbage
(a) spreads pollution
(b) prevents pollution
(c) spoils mineral wealth
(d) none of these
Answer:
(b) prevents pollution.

Question 5.
The organic waste that can be recycled and reused is
(a) paper
(b) glass
(c) metals
(d) all these.
Answer:
(d) all these.

(5) Yoga is the ancient Indian system to keep a person fit in body and mind. It is basically a system of self-treatment. According to the yogic view, diseases, disorders and ailments are the result of some faulty ways of living, bad habits, lack of proper knowledge and unsuitable food. The diseases are thus the resultant state of a short or prolonged malfunctioning of the body system. The root cause of a disease lies in not correcting the mistakes by the same individual. The yogic practice of treatment comprises three steps, namely proper diet, proper yogic practice and proper knowledge of things concerning the self.

Question 1.
The benefit of the system of yoga is:
(a) It keeps a person fit in body and mind.
(b) It is a modern Indian system.
(c) It makes a person religious.
(d) Comprises three steps.
Answer:
(a) It keeps a person fit in body and mind.

Question 2.
What type of system is this basically?
(a) It is a costly treatment
(b) It is a self-treatment.
(c) It avoids bad habits.
(d) All of the above.
Answer:
(b) It is a self-treatment.
Or
Diseases, disorders and ailments are the results of
(a) Some faulty ways of living.
(b) Some normal ways of living
(c) Some cosdy ways of living.
(d) All of the above.
Answer:
(a) Some faulty ways of living.

Question 3.
What is the root cause of diseases ?
(a) Mistakes of the doctors.
(b) Mistakes of the parents.
(c) Mistakes of the governments.
(d) Mistakes of the individual.
Answer:
(d) Mistakes of the individual.

Question 4.
How many steps does yoga practice keep?
(a) Only one step.
(b) Only three steps.
(c) Only two steps.
(d) Only four steps.
Answer:
(b) Only three steps.
Or
Which is the first step of yoga?
(a) Proper yogic practice.
(b) Proper knowledge of things.
(c) Proper diet.
(d) Proper exercise of body.
Answer:
(c) Proper diet.

Question 5.
Which is the third step of yoga?
(a) Proper knowledge.
(b) Proper counselling.
(c) Proper thinking.
(d) None of the above.
Answer:
(a) Proper knowledge.
Or
Whose efforts cure the person ?
(a) The doctor’s efforts.
(b) The yoga experts efforts.
(c) The efforts of the society.
(d) The patients efforts.
Answer:
(d) The patients efforts.

(6) There is an incident which occurred at the examination during my first year at the high school. Mr. Giles, the Education Inspector, had come on a visit of inspection. He had set us five words, to write as a spelling exercise. One of the words was ‘ketde’. I had mis-spelt it. The teacher tried to prompt me with the point of his boot, but I would not be prompted. It was beyond me to see that he wanted me to copy the spelling from my neighbour’s slate for I had thought that the teacher was there to supervise us against copying.

Question 1.
When did the incident occur ?
(a) In the second year.
(b) In the first year.
(c) In the third year.
(d) None of the above.
Answer:
(b) In the first year.
Or
Who was the Education Inspector?
(a) Mr. Gordon
(b) Mr. Graham
(c) Mr. Giles
(d) Mr. George.
Answer:
(c) Mr. Giles

Question 2.
Which exercise was given to write?
(a) Dictation exercise.
(b) Handwriting exercise.
(c) Yoga exercise.
(d) Spelling exercise.
Answer:
(d) Spelling exercise
Or
How many words were given to us ?
(a) Three words.
(b) Five words.
(c) No word was given.
(d) Four words.
Answer:
(b) Five words.

Question 3.
Who tried to prompt the speaker/writer ?
(a) The teacher.
(b) The Inspector.
(c) The students.
(d) All of the above.
Answer:
(a) The teacher.
Or
What mistake had the writer committed ?
(a) A word mistake.
(b) A meaning mistake.
(c) A spelling mistake.
(d) A speaking mistake.
Answer:
(c) A spelling mistake.

Question 4.
What was the point of indication used by the teacher ?
(a) The point of his hand finger.
(b) The point of his right foot finger.
(c) The point of his left hand thumb.
(d) The point of his boot.
Answer:
(d) The point of his boot.

Question 5.
Which word was mis-spelt ?
(a) Kettle.
(b) Catde.
(c) Settle.
(d) Metal.
Answer:
(a) Kettle.

(7) People often curse poverty as a great evil, and it seems to be an accepted belief that if people only had plenty of money, they would be happy and useful and get more out of life. But the reality is that while palaces give a comfortable life, peace and contentment dwell in cottages. I always pity the sons and daughters of rich parents who are attended by servants and governesses. It is because I know how sweet and happy and pure the home of honest poverty is and how loving and united the members of poor families are in common interests. It is for these reasons that so many strong, eminent and self-reliant men have always sprung from poor families.

Question 1.
What do the people often think about the poverty?
(a) It is a curse and great evil.
(b) It is a boon of God.
(c) It is a self-created act.
(d) It is a social evil.
Answer:
(a) It is a curse and great evil.
Or
Who are attended by servants and governesses?
(a) The kings of the world.
(b) The members of the poor families.
(c) The sons and daughters of rich parents.
(d) None of the above.
Answer:
(b) The sons and daughters of rich parents.

Question 2.
Who are happy according to accepted belief ?
(a) People who have no money.
(b) People who have plenty of money.
(c) People who have a higher education.
(d) People who have no higher education.
Answer:
(a) People who have no money.
Or
What is the reality of happy life.
(a) To live in luxurious palaces.
(b) To live in the forests.
(c) To live peaceful and contented life in a hut.
(d) To live in the king’s palaces.
Answer:
(c) To live peaceful and contented life in a hut.

Question 3.
What does the home of poverty provide us?
(a) A life of prosperity.
(b) A sweet, happy and pure home.
(c) A dirty, bad and disturbed life.
(d) A life of dissatisfaction.
Answer:
(b) A sweet, happy and pure home.

Question 4.
Whose members are loving and united ?
(a) Members of poor families.
(b) Members of rich families.
(c) Members of tribal families.
(d) Members of royal families.
Answer:
(a) Members of poor families.
Or
Who have sprung from poor families ?
(a) Weak, cowardly and religious persons.
(b) Educated, rich and royal persons.
(c) Prosperous honoured and noble persons.
(d) Strong, eminent and self-reliant persons.
Answer:
(d) Strong, eminent and self-reliant persons.

Question 5.
Whose plus points are highlighted in the passage ?
(a) The rich.
(b) The noble,
(c) The honoured.
(d) The poor.
Answer:
(d) The poor.

(8) Books have much value in our life. They are our lifeline and best companion. Everything comes to an end but they live for ever. They never deceive the readers. They help us in difficulties. We get much knowledge and entertainment from them. We get new meanings and beauties in books. By reading books, we just get confidence in life. This world would be quite dark without books. They tell us about people, their culture and profession.

Question 1.
Books have:
(a) no value.
(b) less value.
(c) much value.
(d) different value.
Answer:
(c) much value.
Or
Books are our:
(a) lifeline.
(b) best companion.
(c) companion and enemy
(d) both (a) and (b).
Answer:
(d) both (a) and
(b) lifeline and best companion.

Question 2.
A quality of books is:
(a) They come to an end.
(b) They don’t live for ever.
(c) They live for ever.
(d) All the above.
Answer:
(c) They live for ever.
Or
Books help us:
(a) in trouble.
(b) in difficulties.
(c) Both (a) and (b).
(d) in sorrows.
Answer:
(b) in difficulties.

Question 3.
We get from books:
(a) waste paper
(b) much knowledge.
(c) entertainment.
(d) Both (b) and (c).
Answer:
(d) both (b) and (c) much knowledge and entertainment.
Or
By reading books we get:
(a) confidence in life
(b) difficulties in life
(c) popularity and prosperity
(d) All the above.
Answer:
(a) confidence in life.

Question 4.
Without books the world would become:
(a) quite happy.
(b) quite bright.
(c) quite dark.
(d) Both (a) and (b).
Answer:
(c) quite dark.

Question 5.
Books tell us about:
(a) people.
(b) their culture.
(c) their profession.
(d) All the above.
Answer:
(d) All the above.

(9) Diwali is the greatest festival of Hindus. It is celebrated throughout the world. It is a festival of lights and candles. It comes in the month of October or November every year. On this day people worship Goddess Lakshmi. They put on new clothes and buy sweets. They also give presents to their friends and relatives. Some people gamble on this day which is an evil practice. Children play crackers and fireworks.

Question 1.
Diwali is the greatest festival of
(a) Muslims
(b) the poor
(c) the rich
(d) Hindus
Answer:
(d) Hindus

Question 2.
Diwali is celebrated:
(a) only in India.
(b) throughout Asia
(c) throughout the world.
(d) None of these.
Answer:
(c) throughout the world.
Or
Diwali is a festival of:
(a) lights.
(b) Candles
(c) swings
(d) both (a) and (b)
Answer:
(d) both (a) and (b) lights and candles.

Question 3.
Diwali fells in the month of:
(a) October.
(b) October or November.
(c) December
(d) None of these.
Answer:
(b) October or November.
Or
On this day people worship:
(a) Goddess Kali.
(b) Goddess Lakshmi.
(c) Both (a) and (b).
(d) None of these.
Answer:
(b) Goddess Lakshmi.

Question 4.
On this day people:
(a) buy houses
(b) buy sweets.
(c) give up bad habbits
(d) All the above.
Answer:
(b) buy sweets.
Or
An evil practice related to Diwali is:
(a) drinking
(b) fighting
(c) smoking
(d) gambling.
Answer:
(d) gambling

Question 5.
Who plays cracker and fire works on this day ?
(a) rich people
(b) poor people
(c) Children
(d) Goddess Laxmi
Answer:
(c) Children

(10) Mohan and Sohan were fast friends. Mohan was very selfish and cunning while Sohan was very loyal and dependable. One day they set out on a long journey. Both decided to help each other. While crossing the forest, they saw a bear coming towards them. Mohan at once climbed up a tree. But Sohan did not know how to climb up. He lay on the ground and held his breath. The bear came and took Sohan as dead. After the bear had gone, Mohan came down the tree and asked Sohan what the bear had said in his ear. Sohan replied that the bear had told him never to trust a false friend.

Question 1.
Mohan was:
(a) loyal.
(b) very helpful.
(c) very selfish and cunning.
(d) All the above.
Answer:
(c) very selfish and cunning.

Question 2.
Sohan was very:
(a) selfish.
(b) cunning.
(c) loyal and dependable.
(d) a false friend
Answer:
(c) loyal and dependable.

Question 3.
One day Mohan and Sohan set out:
(a) on a tour.
(b) on a long journey.
(c) on an expedition.
(d) on a short trip.
Answer:
(b) on a long journey.

Question 4.
What did they see in the forest?
(a) An elephant.
(b) A lion.
(c) A tiger.
(d) A bear.
Answer:
(d) A bear.
Or
What did Sohan do?
(a) He ran away.
(b) He lay on the ground.
(c) He held his breath.
(d) Both (b) and (c).
Answer:
(d) both (b) and (c) He lay on the ground and held his breath.

Question 5.
What did the bear do ?
(a) It attacked Sohan.
(b) It killed Sohan.
(c) It took Sohan as dead.
(d) It took Mohan as dead
Answer:
(c) It took Sohan as dead.
Or
The bear had told Sohan not to
(a) trust anybody.
(b) climb up a tree.
(c) go on a journey.
(d) trust a false friend.
Answer:
(d) trust a false friend.

(11) His first ‘Satyagraha in India was in Champaran, in Bihar. The peasants of that district were being cruelly treated by the British indigo planters. Gandhiji left for Champaran to find out the truth. The news that a Mahatma had arrived to inquire into their sufferings attracted thousands of peasants who flocked to Champaran to have has darshan. The Government got alarmed and Gandhiji was asked to leave the district. He refused and was asked to appear before magistrate. Later, the case was withdrawn.Gandhiji lived with the peasants for some time in order to learn about their hard lot. But, he also taught them to be free and to stand on their feet. At last, he succeeded in securing justice for the poor peasants.

Question 1.
Gandhiji left for Champaran to find
(a) out the truth
(b) out the peasants
(c) out the magistrate
(d) none of the above.
Answer:
(a) out the truth.

Question 2.
Who were cruelly treated ?
(a) The British indigo planters
(b) Gandhiji and his followers
(c) The peasants of Champaran
(d) The peasants all over India
Answer:
(c) The peasants of Champaran.

Question 3.
Gandhiji was asked to leave because:
(a) his life was in danger.
(b) the government was alarmed.
(c) he had lost his energy to unite the peasants.
(d) all the above.
Answer:
(b) the government was alarmed.

Question 4.
When refused, Gandhiji was asked to appear before:
(a) the peasants
(b) the public meeting
(c) the indigo planters
(d) the magistrate.
Answer:
(d) the magistrate.

Question 5.
Gandhiji’s first ‘Satyagraha’ was a:
(a) success
(b) failure
(c) false show
(d) poor show
Answer:
(a) success.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 1 Rational Numbers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions

Try These : (Textbook Page No.4)

Question 1.
Fill in the blanks in the following table :

Answer:

[Note : Rational numbers are not closed under division.]
e.g., \(\frac {2}{3}\) ÷ 0 = ? This is not defined. That’s why our answer in the table is ‘No’.

Try These : (Textbook Page No.6)

Question 1.
Complete the following table:

Answer:

Try These : (Textbook Page No.9)

Question 1.
Complete the following table:

Answer:

Think, Discuss and Write : (Textbook Page No.11)

1. If a property holds for rational numbers, will it also hold for integers ? For whole numbers ? Which will ? Which will not ?
Answer:
( i ) Any property which is true for rational numbers is also true for integers except for any integers ‘a’ and ‘b’ (a ÷ b) is not necessarily an integer.
(ii) All properties which are true for rational numbers are also true for whole numbers also except:

• For ‘a’ and ‘b’ being whole numbers (a – b) may not be a whole number.
• For ‘a’ and ‘b’ being whole numbers (b ≠ 0), a ÷ b may not be a whole number.

Try These : (Textbook Page No.13)

1. Find using distributivity :

Question (i).
\(\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}\)
Answer:
\(\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}\)
= \(=\frac{7}{5} \times\left[\frac{-3}{12}+\frac{5}{12}\right]\)
= \(\frac{7}{5} \times\left[\frac{-3+5}{12}\right]\)
= \(\frac{7}{5} \times \frac{2}{12}\)
= \(\frac{7}{5} \times \frac{1}{6}\)
= \(\frac {7}{30}\)

Question (ii).
\(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}\)
Answer:
\(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}\)
= \(\frac{9}{16} \times\left[\frac{4}{12}+\left(\frac{-3}{9}\right)\right]\)
= \(\frac{9}{16} \times\left[\frac{12+(-12)}{36}\right]\) …..(LCM = 36)
= \(\frac{9}{16} \times\left[\frac{0}{36}\right]\)
= \(\frac{9}{16} \times 0\)
= 0.

Try These : (Textbook Page No.17)

1. Write the rational number for each point labelled with a letter:

Question (i).

Answer:

Here, the rational number for-
the point A is \(\frac {1}{5}\)
the point B is \(\frac {4}{5}\)
the point C is \(\frac {5}{5}\) or 1.
the point D is \(\frac {8}{5}\)
the point E is \(\frac {9}{5}\)

Question (ii).

Answer:

Here, the rational number for-
the point F is \(\frac {-2}{6}\) or \(\frac {-1}{3}\).
the point G is \(\frac {-5}{6}\).
the point H is \(\frac {-7}{6}\).
the point I is \(\frac {-8}{6}\) or \(\frac {-4}{3}\).
the point J is \(\frac {-11}{6}\).

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°.
Solution.
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
= \(\frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)}\)
= \(\frac{\sin 18^{\circ}}{\sin 18^{\circ}}\) = 1
[∵ cos (90° – θ) = sin θ]

(ii) \(\frac{\tan 26^{\circ}}{\cos 64^{\circ}}=\frac{\tan 26^{\circ}}{\cot \left(90^{\circ}-26^{\circ}\right)}\)
= \(\frac{\tan 26^{\circ}}{\tan 26^{\circ}}\) = 1
[∵ cot (90°- θ) = tan θ]

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
[∵ cos (90° – 0) = sin O]
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59°
=cosec 31° – sec (90° – 31°)
= cosec 31° – cosec 31°
[∵ sec (90° – θ) = cosec θ].

Question 2.
Show that:
(i) tan 4 tan 230 tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) L.H.S.
= tan 48° tan 23° tan 42° tan 67°
= tan 48° × tan 23° × tan (90° – 48°) × tan (90° – 23°)
= tan48° × tan 23° × cot48° × cot 23°
= tan 48C × tan 23° × \(\frac{1}{\tan 48^{\circ}}\) × \(\frac{1}{\tan 23^{\circ}}\) = 1
∴ L.H.S. = R.H.S.

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°
= cos 38° × cos (90 – 38°) – sin 38° × sin (90° – 38°)
= cos 38° × sin 38° – sin 38° × cos 38
= 0.
∴ L.H.S. = RH.S.

Question 3.
If tan 2A = cot (A – 18°) where 2A is an acute angle, find the value of A.
Solution:
Given: tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°)
[cot (90° – θ) = tan θ]
⇒ 90°- 2A = A – 18°
⇒ 3A = 108°
⇒A = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
Given that: tan A = cot B
⇒ tan A = tan(90° – B)
[∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B.
⇒ A + B = 90°..

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Given that: sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A — 20°)
[∵ cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20°
⇒ 5A = 110°
⇒ A = 22°.

Question 6.
If A, B and C interior angles of a triangle ABC, then show that: \(\sin \left(\frac{B+C}{2}\right)=\cos \left(\frac{A}{2}\right)\)
Solution:
Since, A, B and C are interior angles of a triangle
∴ A + B + C = 180°
[Sum of three angles of a triangle is 180°]
⇒ B + C = 180° – A
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}\)
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
Taking sin on both sides, we get
⇒ \(\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
[∵ sin (90° – θ) = cos θ].

Question 7.
Express sin 67° + cos 75° in terms of Trigonometric ratios of angles between 0° and 45°.
Solution:
Given that: sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
[∵ sin(90° – θ) = cos θ and cos (90° – θ) = sin θ].

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.6

1. Draw a line XY and point P not lying on XY. Draw a line parallel to XY passing through P with the help of ruler and compasses.
Solution:
Steps of Construction:
1. Draw a line XY and point P not lying on it.

2. Take any point Q, anywhere on line XY.
3. Join PQ.

4. Now take Q as centre, draw arc AB of any radius on XY. Similarly, draw an arc CD of same radius on line segment PQ from point P.

5. Measure arc AB with compasses.
6. Draw an arc equal to radius AB from point C witch intersect CD on E.

7. Join PE and produce it. So, the line l is the required line parallel to XY.

2. Draw a line p parallel to line m passing through a point A which is not lying on line m with the help of set squares.
Solution:
Steps of Construction:
1. Given a line m with point A not lying on it.
2. Place one of the edge of a ruler along the line m and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Hold the ruler firmly, slide the set square along the line m till its vertical side reaches the point A.

5. Firmly hold the set square in this position, take another set square and place it in such a way that one of its edges containing right angle concides with previous set square as shown.
6. Now draw a line p along side of second set square passing through A.

Thus p \(\text { ॥ } \) m passing through A.

3. Given a line AB and the point X is not lying on it. Draw a line parallel to AB passing through X.

Question (i)
By a ruler and compasses
Solution:
By a ruler and compasses:
Let us consider a line AB and point X not lying on it.

Steps of Construction:
1. Take any point, say Y anywhere on line AB.
2. Join XY.

3. Now take Y as centre, draw an arc PQ of any radius on AB. Similarly draw an arc RS of same radius on line segment XY from point X.

4. Measure arc PQ with compasses.
5. Draw an arc equal to radius PQ from point R which intersect RS on T.
6. Join XT and produce it.

So the line m is the required line parallel to AB.

Question (ii)
By set squares.
Solution:
By set squares.

Steps of Construction:
1. Given a line AB with point X not lying on it.
2. Place one of the edge of a ruler along the line AB and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Hold the ruler firmly, slide the set square along the line AB till its vertical side reaches the point X.

5. Firmly hold the set square in this position, take another set square and place it in such a way that one of its edges containing right angle concides with previous set square as shown.
6. Now draw a line l along side of second set square passing through X.

7. Thus l \(\text { ॥ } \) AB passing through X.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.5

1. Draw the following angles in both directions (Left and right) by protractor:

Question (i)
75°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on ray OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 75° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 75°.

If ray OA lies to the left of the centre (midpoint) of the baseline, start reading the angle on the outer scale from 0° and mark 75°. Join OB, then \(\angle AOB\) = 75°.

Question (ii)
110°
Solution:
Steps of Construction:
1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 110° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 110°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 110°. Join OB, then \(\angle AOB\) = 110°.

Question (iii)
62°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-62° base line along OA.

3. Mark a point B on the paper against the mark of 62° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 62°.

If the ray OA lies to the left of the centre (midpoint) of the baseline, start reading the angle on the outer scale from 0° and mark 62°. Join OB, then \(\angle AOB\) = 62°.

Question (iv)
165°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° baseline along OA.

3. Mark a point B on the paper against the mark of 165° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle
\(\angle AOB\) = 165°

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 165°. Join OB, then \(\angle AOB\) = 165°.

Question (v)
170°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on ray OA such that its centre lies on the initial point O and 0-480° base line along OA.

3. Mark a point B on the paper against the mark of 170° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 170°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 170°. Join OB, then \(\angle AOB\) = 170°.

Question (vi)
32°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 32° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 32°.
If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 32°. Join OB, then \(\angle AOB\) = 32°.

Question (vii)
128°
Solution:
Steps of Construction:

1. Draw a ray OA.

Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 128° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle
\(\angle AOB\) = 128°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 128°. Join OB, then \(\angle AOB\) = 128°.

Question (viii)
25°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 25° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 25°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 25°. Join OB, then \(\angle AOB\) = 25°.

Question (ix)
80°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 80° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 80°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 80°. Join OB, then \(\angle AOB\) = 80°.

Question (x)
135°.
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 135° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 135°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 135°. Join OB, then \(\angle AOB\) = 135°.

2. Bisect the following angles by compasses:

Question (i)
48°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 48°.
4. Join OB. Then \(\angle AOB\) = 48°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point
E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 48°.
Measure \(\angle AOB\) and \(\angle AOB\)
\(\angle AOB\) = \(\angle AOB\) = 24°.

Question (ii)
140°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 140°.
4. Join OB. Then \(\angle AOB\) = 140°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 140°.
On measurement \(\angle AOE\) = \(\angle BOE\) = 70°.

Question (iii)
75°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 75°.
4. Join OB. Then \(\angle AOB\) = 75°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 75°.
On measurement
\(\angle AOE\) = \(\angle BOE\) = 37.5°.

Question (iv)
64°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 64°.
4. Join OB. Then \(\angle AOB\) = 64°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 64°.
On measurement
\(\angle AOE\) = \(\angle BOE\) = 32°.

Question (v)
124°.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 124°.
4. Join OB. Then \(\angle AOB\) = 124°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 124°.
On measurement \(\angle AOE\) = \(\angle BOE\) = 62°.

3. Draw an angle of 80° and bisect it in to four equal parts by compasses.
Solution:
1. Draw a line OY of any length.
2. Place the centre of the protractor at O.
3. Starting with 0 mark a point X at 80°.
4. Join OX. Then \(\angle XOY\) = 80°.
5. With O as centre and using compass, draw an arc that cuts both rays of \(\angle XOY\). Name the point of intersection as X’ and Y’.
6. With Y’ as centre, draw an arc whose radius is more than half the length X’Y’.
7. With the same radius and with X’ as a centre, draw another arc which cut the first arc at point C.
8. With O as centre and using compass, draw an arc that cuts both rays of \(\angle COY\). Name the points of intersection as B and A.
9. With A as centre, draw an arc whose radius is more than half the length AB.
10. With the same radius and with B as centre, draw another arc which cuts the first arc at point S.
11. With O as centre and using compass, draw an arc that cuts both rays of \(\angle XOC\) . Name the points of intersection as D and E.
12. With E as centre, draw an arc whose radius is more than half the length DE.
13. With the same radius and with E as centre, draw another arc which bisects the first arc at T. Then OT is the bisector of \(\angle XOC\).
Thus \(\overline{\mathrm{OS}}, \overline{\mathrm{OC}} \text { and } \overline{\mathrm{OT}}\) divide, \(\angle AOB\) = 80° into four equal parts.

4. Draw a right angle and bisect it.
Solution:
1. Draw a ray OB.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point A at 90°.

4. Join OA. Then \(\angle AOB\) = 90°
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\). Name the points of intersection as A’ and B’.
6. With B’ as centre, draw an arc whose radius is more than half of the length B’A’.
7. With the same radius and with A’ as a centre, draw another arc which cuts the first arc at point C. Join OC bisects \(\angle AOB\).

5. Draw the following angles by ruler and compasses:

Question (i)
30°
Solution:
To Construct angle of 30°

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.

3. With C as centre and same radius as before draw another arc cutting the previous arc at E.
4. Join OE and produce it to B. \(\angle AOB\) = 60°.
5. Bisect \(\angle AOB\).
Thus \(\angle AOM\) = \(\angle MOB\) = 30°.

Question (ii)
45°
Solution:
To Construct Angle of 45°:

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.

3. With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.
4. With P and Q as centres and any suitable radius (more than half of PQ) or even the same radius draw arc cutting each other at R.
5. Join OR and produce it to B. Then \(\angle AOB\) = 90°.
6. Bisect \(\angle AOB\).
7. OD is the bisector of \(\angle AOB\).
\(\angle BOD\) = \(\angle DOA\) = 45°.

Question (iii)
135°
Solution:
To Construct Angle of 135°.

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.
3. With C as centre and the same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q and then with Q as centre the same radius cut off the arc again at R.
4. With Q and R as centres and radius more than half of RQ draw arcs cutting each other at L.
5. Join OL and produce it to B. Then \(\angle AOB\) = 150°.
6. Take a point M on the arc where OL intersects the arc.
7. With M and Q as centres and radius more than half of MQ draw arcs cutting each other at N.
8. Join ON and produce it to E. \(\angle AOE\) = 135°.

Question (iv)
180°
Solution:
To Construct Angle of 180°.

Steps of Construction:
1. Draw a line AB and mark a point C on it.
2. Taking C as centre and with any suitable radius, draw an are PQ cutting AB at P and Q.

3. Here \(\angle ACB\) = 180° (It is a straight line).

Question (v)
120°
Solution:
To Construct Angle of 120°:

Steps of Construction:
1. Draw a line segment OA.

2. With O as centre and any suitable radius draw an arc cutting OA at point M.
3. With M as centre and same radius draw an arc which cuts the arc at N and then with N as centre and the same radius cut off the arc again at Q.
4. Join OQ and produce it to B.
Then \(\angle AOB\) = 120°.

Question (vi)
75°.
Solution:
To Construct Angle of 75°.

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at C.
3. With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.
4. With P and Q as centres and radius more than half of PQ draw arcs cutting each other at R.
5. Join OR and produce it to B. \(\angle AOB\) = 90°.
6. Bisect \(\angle AOB\).
7. OD is the bisector of \(\angle AOB\).
\(\angle BOD\) = \(\angle DOA\) = 45°.
8. Again draw OE bisector of \(\angle DOB\).
Thus angle \(\angle EOA\) = 75°.

6. Draw an angle of 30° by protractor and bisect it by a ruler and compasses.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 30°.
4. Join OB. Then \(\angle AOB\) = 30°.
5. With O as centre and using compasses draw an arc that cuts both rays of \(\angle AOB\), with the point intersection as C and D.
6. With C as centre, draw an arc whose radius is more than half of the length of CD.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, it bisect \(\angle AOE\).

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 15 Probability MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
When a balanced die is thrown, the probability of getting 3 is …………….. .
A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{6}\)
Answer:
D. \(\frac{1}{6}\)

Question 2.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a king is …………………. .
A. \(\frac{1}{52}\)
B. \(\frac{1}{26}\)
C. \(\frac{1}{13}\)
D. 1
Answer:
C. \(\frac{1}{13}\)

Question 3.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a card other than picture cards is ……………….. .
A. \(\frac{4}{13}\)
B. \(\frac{10}{13}\)
C. \(\frac{3}{13}\)
D. \(\frac{1}{13}\)
Answer:
B. \(\frac{10}{13}\)

Question 4.
When an unbiased coin is tossed thrice, the probability of receiving three heads is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{8}\)
Answer:
A. \(\frac{1}{8}\)

Question 5.
When three unbiased coins are tossed simultaneously, the probability of receiving exactly one tail is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
Answer:
D. \(\frac{3}{8}\)

Question 6.
When a balanced die is thrown, the probability of receiving an even number is ………………… .
A. \(\frac{1}{6}\)
B. \(\frac{5}{6}\)
C. \(\frac{1}{2}\)
D. \(\frac{1}{4}\)
Answer:
C. \(\frac{1}{2}\)

Question 7.
When a balanced die is thrown, the probability of receiving a prime number is ……………….. .
A. \(\frac{2}{3}\)
B. \(\frac{3}{4}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{2}\)
Answer:
D. \(\frac{1}{2}\)

Question 8.
When two balanced dice are thrown simultaneously, the probability of getting the total of numbers on dice as 9 is ………………. .
A. \(\frac{1}{9}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{12}\)
Answer:
A. \(\frac{1}{9}\)

Question 9.
Out of 100 days, the forecast predicted by the wheather department proved to be true on 20 days. Chosen any one day from these 100 days, the probability that the forecast proved to be false is ………………… .
A. \(\frac{1}{3}\)
B. \(\frac{1}{4}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
Answer:
D. \(\frac{4}{5}\)

Question 10.
The probability of a month of January having 5 Sundays is ………………….. .
A. \(\frac{2}{7}\)
B. \(\frac{3}{7}\)
C. \(\frac{5}{7}\)
D. \(\frac{1}{7}\)
Answer:
B. \(\frac{3}{7}\)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Answer:
The batswoman played 30 balls. Hence, the total number of trials = 30. If the event that she did not hit a boundary is denoted by A, then. the number of trials when event A occured is 30 – 6 = 24.
∴ p(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{24}{30}\)
= \(\frac{4}{5}\)
Thus, the probability that she did not hit a boundary is \(\frac{4}{5}\).

Question 2.
1500 families with 2 children were selected randomly, and the following data were s recorded:

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl. Also check whether the sum of these probabilities is 1.
Answer:
Here, the total number of families is 1500.
Hence, the total number of trials = 1500

(i) Let event A denote the event that the family chosen at random is having 2 girls.
Then, the number of trials when event A occured is 475.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{475}{1500}\)
= \(\frac{19}{60}\)

(ii) Let event B denote the event that the family chosen at random is having 1 girl.
Then, the number of trials when event B occured is 814.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{814}{1500}\)
= \(\frac{407}{750}\)

(iii) Let event C denote the event that the family chosen at random Is having no girl.
Then, the number of trials when event C occured is 211.
∴ p(C) = \(\frac{\text { No. of trials in which event } \mathrm{C} \text { occured }}{\text { The total number of trials }}\)
= \(\frac{211}{1500}\)
Now,
P(A) + P(B) + P(C) = \(\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\)
= \(\frac{475+814+211}{1500}\)
= \(\frac{1500}{1500}\)
= 1

Question 3.
Refer to sum no. 5 of “Sums to Enrich ‘Remember’” in chapter 14. Find the probability that a student of the class was born in August.
Answer:
From the Bar graph in the sum which is referred here, we get the following information:

Total number of students = 40 and the number of students born in August = 6.
Hence, if event A denotes the event that a student of the class is born in August, then the number of trials when event A occured is 6 and the total number of trials is 40.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{6}{40}\)
= \(\frac{3}{20}\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Answer:
Here, the total number of trials = 200. If event A denotes the event that 2 heads come up, then the number of trials when event A occured is 72.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{72}{200}\)
= \(\frac{9}{25}\)

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below :

Suppose a family is chosen. Find the probability that the family chosen is ( i ) earning ? 10000- ? 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not .own any vehicle.
(iv) earning ₹ 13000 – ₹ 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Answer:
Here, the total number of families is 2400. Hence, the total number of trials = 2400

(i) Let event A denote the event that the family is earning ₹ 10000 – ₹ 13000 per month and owning exactly 2 vehicles.
Then, the number of trials when event A occured = 29.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{29}{2400}\)

(ii) Let event B denote the event that the family is earning ₹ 16000 or more per month and owning exactly 1 vehicle.
Then, the number of trials when even B occured = 579.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{579}{2400}\)
= = \(\frac{193}{800}\)

(iii) Let event C denote the event that the family is earning less than ₹ 7000 per month and does not own any vehicle.
Then, the number of trials when event C occured = 10.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{10}{2400}\)
= = \(\frac{1}{240}\)

(iv) Let event D denote the event that the family is earning ? 13000 -? 16000 per month and is owning more than 2 vehicles. Then, the number of trials when event D occured = 25.
∴ P(D) = \(\frac{\text { No. of trials in which event D occured }}{\text { The total number of trials }}\)
= \(\frac{25}{2400}\)
= \(\frac{1}{96}\)

(v) Let event E denote the event that the family is owning not more than 1 vehicle, i.e., 1 vehicle or no vehicle.
Then, the number of trials when event E occured.
= 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
∴ P(E) = \(\frac{\text { No. of trials in which event E occured }}{\text { The total number of trials }}\)
= \(\frac{2062}{2400}\)
= \(\frac{1031}{1200}\)

Question 6.
Refer to table 7 of sum no. 7 in “Sums to Enrich ‘Remember’” in chapter 14.
(i) Find the probability that a student obtained less than 20 marks in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Answer:
According to the table referred here, the total number of students = 90.
Hence, the total number of trials = 90.
(i) According to the same table, the number of students who obtained less than 20 marks in the mathematics test is 7. So, if the event that a student obtained less than 20 marks in mathematics test is called event A, then the number of trials when event A occured is 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{90}\)

(ii) Let event B denote the event that a student obtained 60 or more marks. Then, , according to the same table, the number of trials when event B occured = 15 + 8 = 23.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{23}{90}\)

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

Opinion	Number of students
Like	135
Dislike	65

Find the probability that a student chosen at random
(i) Likes statistics,
(ii) Does not like it.
Answer:
Here, the total number of students = 200.
Hence, the total number of trials = 200.

(i) Let event A denote the event that a student likes statistics.
Then, the number of trials when event A occured = 135
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{135}{200}\)
= \(\frac{27}{40}\)

(ii) Let event B denote the event that a student does not like statistics. Then, the number of trials when event B occured = 65.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{65}{200}\)
= \(\frac{13}{40}\)

Question 8.
Refer to sum no. 2, Exercise 14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work ?
(ii) more than or equal to 7km from her place of work ?
(iii) within \(\frac{1}{2}\)km from her place to work?
Answer:
The total number of observations in the question referred here is 40.
Hence, the total number of trials = 40.

(i) Let event A denote the event that the distance between her residence and the place of work is less than 7 km. Then there are 9 such observations, viz., 5, 3, 2, 3, 6, 5, 6, 2, 3.
Hence, the number of trials when event A occured = 9.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{9}{40}\)

(ii) Let event B denote the event that the said distance is 7 km or more than 7 km. Then, all the remaining 31(40-9) observations refer to event B.
Hence, the number of trials when event B occured = 31
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{31}{40}\)

(iii) Let event C denote the event that the engineer lives within \(\frac{1}{2}\) km from her place of work. There is no observation which is \(\frac{1}{2}\) or less than \(\frac{1}{2}\).
Hence, the number of trials when event C occured = 0.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{0}{40}\)
= 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer:
Note: Students should do this Activity themselves.

Question 10.
Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her / him is divisible by 3 ? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Answer:
Note: Students should do this Activity themselves.

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg) :
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer:
The total number of bags = 11.
Hence, the total number of trials = 11.
Let event A denote the event that a bag contains more than 5 kg of flour.
There are 7 bags weighing more than 5 kg.
Their weights (in kg) are 5.05, 5.08, 5.03, 5.06, 5.08, 5.04 and 5.07. Hence, the number of trials when event A occured = 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{11}\)

Question 12.
In sum no. 5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Answer:
In sum no. 5, Exercise 14.2, total number of days is 30.
Hence, the total number of trials = 30.
In the table prepared there, we see that the frequency of class 0.12 – 0.16 is 2.
Hence, during 2 days the concentration of sulphur dioxide (in ppm) was in the interval 0.12 – 0.16.
Let event A denote the event that the concentration of sulphur dioxide (in ppm) is in the interval 0.12 – 0.16.
Hence, the number of trials when event A occured = 2.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{2}{30}\)
= \(\frac{1}{15}\)

Question 13.
In sum no. 1, Exercise 14.2, you were asked) to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Answer:
In sum no. 1, Exercise 14.2, the total number of students is 30.
Hence, the total number of trials = 30.
Let event A denote the event that a student has blood group AB. The number of students having blood group AB is 3.
Hence, the number of trials when event A occured = 3.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{3}{30}\)
= \(\frac{1}{10}\)