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Punjab State Board PSEB 6th Class English Book Solutions English Grammar Tenses Exercise Questions and Answers, Notes.

PSEB 6th Class English Grammar Tenses

The form of verb that shows the Time or State of an action is called the Tense.
Verb का जो रुप क्रिया के समय अथवा स्थिति के बारे में बताती है उसे Tense कहते है।

We have three main Tenses in English.
They are-

1. The Present Tense वर्तमानकाल
2. The Past Tense भूतकाल
3. The Future Tense. भविष्यत्काल

Look at the following sentences:

1. Tanu is here today.
2. The girls are happy.
3. Tanu was here yesterday.
4. The girls were happy.
5. Tanu will be here tomorrow.
6. The girls will be happy.

Sentences 1 and 2 show the present state.
We can say they are in the Present Tense.

Sentences 3 and 4 show the past state.
We can say they are in the Past Tense.

Sentences 5 and 6 show the future state.
We can say they are in the Future Tense.

Please Note: 1. We use V₁ for Present Tense.
2. We use V₂ for Past Tense.
3. We use will / shall + V₁ for Future Tense.

Forms of Tenses काल के रूप

Each Tense in English has four different forms:

1. Indefinite
2. Continuous
3. Perfect
4. Perfect Continuous

Thus, there are twelve different Tenses.
They are-

1. Present Indefinite	I write poems.
2. Past Indefinite	I wrote poems.
3. Future Indefinite	I shall write poems
4. Present Continuous	I am writing poems.
5. Past Continuous	I was writing poems.
6. Future Continuous	I shall be writing poems.
7. Present Perfect	I have written poems.
8. Past Perfect	I had written poems.
9. Future Perfect	I shall have written poems.
10. Present Perfect Continuous	I have been writing poems.
11. Past Perfect Continuous	I had been writing poems.
12. Future Perfect Continuous	I shall have been writing poems.

1. Present Indefinite Tense

The Present Indefinite Tense is used to express a universal truth (सर्वमान्य सत्य) or habitual action (आदत की क्रिया)”; as-

1. The baby likes bread.
2. We do our duty.
3. Aman speaks the truth.
4. The earth moves round the sun.

The underlined verbs are all in the Present Indefinite Tense.
For Positive Statements in this tense:
We use V₁ for I, You and a Plural subject; as,

1. You learn your lessons.
2. We pray to God daily
3. I buy milk from this dairy.

We use V₁ + s/es, for third person Singular subject; as,

1. Meena tells lies.
2. My sister cooks delicious food.
3. Mr. Singh teaches us English.

Exercises (Solved)

I. Put each sentence into the plural:

1. A cat eats meat.
2. A dog hates a cat.
3. A writer writes a book.
4. An apple grows on a tree.
Answer:
Cats eat meat.
2. Dogs hate cats.
3. Writers write books
4. Apples grow on trees.

II. Put each sentence into the singular:

1. Houses have roofs.
2. Postmen wear caps.
3. They drink tea out of cups.
4. Classrooms have blackboards.
Answer:
1. A house has a roof.
2. A postman wears a cap.
3. He/She drinks tea out of a cup.
4. A classroom has a blackboard.

Framing Negative Questions

We use do not + V₁ for I, you and a Plural subject; as-
1. They do not learn their lessons.
2. You do not do your homework daily.
3. We do not buy milk from this dairy.

We use does not + V₁ for third person Singular subject; as
1. She does not waste her time.
2. Rita does not cook delicious food.
3. Mr. Singh does not teach us English.

Exercise (Solved)

Rewrite each sentence as a Negative:
1. Ram goes home for lunch.
2. I like coffee.
3. She looks beautiful.
4. These boys run fast.
5. We go for a walk daily.
6. You obey your teachers.
7. He takes care of his health.
8. “They take tea in the evening.
Answer:
1. Ram does not go home for lunch.
2. I do not like coffee.
3. She does not look beautiful.
4. These boys do not run fast.
5. We do not go for a walk daily.
6. You do not obey your teachers.
7. He does not take care of his health.
8. They do not take tea in the evening.

Framing Questions

We use the following sentence pattern:
Do + Plural subject + V₁ + Complement ?
Does + third person Singular subject + V₁ + Complement ?

Exercise (Solved)

Rewrite each sentence as a Question:

1. Owls hoot at night.
2. They work on Sundays.
3. Children play on the road.
4. A postman delivers letters.
5. Farmers grow crops for us.
6. Mosquitoes spread Malaria.
7. She helps her mother in the kitchen.
8. Your brother knows many people in this town.
Answer:
1. Do owls hoot at night ?
2. Do they work on Sundays ?
3. Do children play on the road ?
4. Does a postman, deliver letters ?
5. Do farmers grow crops for us ?
6. Do mosquitoes spread Malaria ?
7. Does she help her mother in the kitchen ?
8. Does your brother know many people in this town?

Framing Negative Questions

We can put not before the main verb or in short form after the helping verb; as-

1. Does Rani not tell lies ?
= Doesn’t Rani tell lies ?

2. Do you not take a bath daily ?
= Don’t you take a bath daily ?

Exercise (Solved)

Rewrite each sentence as a Negative Question:

1. Do cows live on grass ?
Do cows not live on grass ?
(or)
Don’t cows live on grass ?

Note : Do not = Don’t
Does not = Doesn’t

2. She does not like coffee.
3. The sun rises in the east.
4. Do they come here daily ?
5. Does Kusha bring flowers ?
6. We do not pluck the flowers.
7. Does Nitin obey his parents ?
8. He does not drive his car very fast.
9. These boys do not respect their teachers.
Answer:
2. Does she not like coffee ?
Or
Doesn’t she like coffee ?
3. Does the sun not rise in the east ?
4. Do they not come here daily ?
Or
Don’t they come here daily ?
5. Does Kusha not bring flowers ?
6. Do we not pluck the flowers ?
7. Does Nitin not obey his parents ?
8. Does he not drive his car very fast ?
9. Do these boys not respect their teachers ?

2. Past Indefinite Tense

Past Indefinite Tense is used to express an action which took place in the past or was completed before the time of speaking; as-
Past Indefinite Tense का प्रयोग बीते समय (भूतकाल) में हुई क्रिया के लिए किया जाता है जो उसके बारे में बात करने से पहले ही पूरी हो गई हो।

1. Simi liked ice cream.
2. Rohan went to the market.

For Positive Statements in this tense, we use V₁ with all subjects (singular or plural); as-

1. He worked hard.
2. We took milk in the morning today.
3. I bought this book last week.

Exercise (Solved)

Rewrite each sentence using the Past form of the given verbs:

1. Rahul (want) a shirt.
2. Deepa (eat) an ice cream.
3. Nancy (wear) simple clothes.
4. Raj (come) to India in March.
5. They (build) a house in Delhi.
6. The boys (laugh) at the beggar.
7. I (go) to the market with my friend.
8. My mother (buy) a new dress for me.
Answer:
1. Rahul wanted a shirt.
2. Deepa ate an ice cream.
3. Nancy wore simple clothes.
4. Raj came to India in March.
5. They built a house in Delhi.
6. The boys laughed at the beggar.
7. I went to the market with my friend.
8. My mother bought a new dress for me.

Negative Statements

We use did not + V₁ for all subjects (singular or plural); as-

1. He did not work hard.
2. We did not take milk in the morning today.
3. I did not buy this book last week.

Please note that did always take V₁ form with it; it never takes V₂ form.

Exercise (Solved)

Rewrite each sentence as a Negative:

1. Misha told the truth.
Misha did not tell the truth.
2. He took my pen.
3. Tony polished his shoes.
4. She cooked food for me.
5. Rohan respected his teachers.
6. They finished their work in time.
7. The naughty boys broke the glass.
8. Ranjan and his friends went for a picnic.
Answer:
2. He did not take my pen.
3. Tony did not polish his shoes.
4. She did not cook food for me.
5. Rohan did not respect his teachers.
6. They did not finish their work in time.
7. The naughty boys did not break the glass.
8. Ranjan and his friends did not go for a picnic.

For Questions

We use the following sentence pattern:

Exercise (Solved)

Rewrite each sentence as a Question:

1. Nancy danced at the party.
Did Nancy dance at the party ?
2. He invited us to dinner.
3. My uncle sent me a gift.
4. Our team won the match.
5. You paid your fees yesterday.
6. We spent our holidays at Shimla.
7. They plucked flowers in the garden.
8. Sonu broke his leg in the accident.
Answer:
2. Did he invite us to dinner ?
3. Did my uncle send me a gift ?
4. Did our team win the match ?
5. Did you pay your fees yesterday ?
6. Did we spend our holidays at Shimla ?
7. Did they pluck flowers in the garden ?
8. Did Sonu break his leg in the accident ?

For Negative Questions

We can put not before the main verb or in short form after the helping verb; as
1. Did she not tell lies ?
= Didn’t she tell lies ?

2. Did you not apply for the post ?
= Didn’t you apply for the post ?

Exercise (Solved)

Rewrite each sentence as a Negative Question:

1. Did the peon ring the bell ?
Did the peon not ring the bell ?
(or)
Didn’t the peon ring the bell ?
2. Did he tell a lie ?
3. We called him a fool.
4. Reeta ate all biscuits.
5. Did she reply your letter ?
6. Your sister painted this picture.
7. Did she finish her work in time ?
8. Did they congratulate you on your success ?
Answer:
2. Did he not tell a lie ?
(or)
Didn’t he tell a lie ?
3. Did we not call him a fool ?
4. Did Reeta not eat all biscuits ?
5. Did she not reply your letter ?
6. Did your sister not paint this picture ?
7. Did she not finish her work in time ?
8. Did they not congratulate you on your success ?

3. Present Continuous Tense

Present Continuous Tense is used to express an action that is going on at the time of speaking; as-

1. Radha is doing her homework.
2. Mona is cooking food in the kitchen.

The italicised words express an action which is going on at present. So we can say these sentences are in the Present Continuous Tense.

Positive Statements (is/am/are + V₁ -ing)

1. I am flying a kite.
2. She is doing her work.
3. They are reading a new lesson.

Negative Statements (is/am/are + not + V₁ -ing)

1. I am not flying a kite.
2. She is not doing her work.
3. They are not reading a new lesson.

Questions
(Helping verb before the Subject)

1. Am I flying a kite ?
2. Is she doing her work ?
3. Are they reading a new lesson ?

Negative Questions

(‘not is used before the main verb or in short form after the helping verb)

1. Am I not flying a kite ?
= An’t I flying a kite ?
(We use an’t in spoken language only.)

2. Is she not doing her work ?
= Isn’t she doing her work ?

3. Are they not reading a new lesson ?
= Aren’t they reading a new lesson ?

Exercises (Solved)

I. Use the Present Continuous Tense to complete each sentence:

1. Mona …………… a test. (take)
2. I ………. my breakfast. (have)
3. The hunter …………. the lion. (kill)
4. The trees …………… their leaves. (shed)
5. The farmers ……….. their fields. (water)
6. The pain in my arm ………… worse. (get)
7. The tailors………… the uniforms. (not make)
Answer:
1. Mona is taking a test.
2. I am having my breakfast.
3. The hunter is killing the lion.
4. The trees are shedding their leaves.
5. The farmers are watering their fields.
6. The pain in my arm is getting worse.
7. The tailors are not making the uniforms.

II. Rewrite each sentence as a question:

1. I am reading a book.
2. She is not doing her work.
3. They are watching a movie.
4. You are not listening to me.
5. We are going for a picnic today.
6. The girls are playing in the park.
7. The boys are not teasing the animals.
Answer:
1. Am I reading a book ?
2. Is she not doing her work ?
3. Are they watching a movie ?
4. Are you not listening to me ?
5. Aren’t we going for a picnic today?
6. Are the girls playing in the park ?
7. Are the boys not teasing the animals ?

4. Past Continuous Tense

The Past Continuous tense is used to express an action which was actually taking place at some particular moment (विशेष क्षण/समय पर) in the past.

Positive Statements
We use was/were + V₁ -ing; as-

1. Harjit was reading a book.
2. They were going to the fair.

Negative Statements
We use was/were + not + V₁ -ing; as-

1. Harjit was not reading a book.
2. They were not going to the fair.

Questions
We put the helping verb before the subject; as-

1. Was Harjit reading a book ?
2. Were they going to the fair ?

Negative Questions
We can put ‘not before the main verb or in short form after the helping verb; as-
1. Was Harjit not reading a book ?
= Wasn’t Harjit reading a book ?

2. Were they not going to the fair ?
= Weren’t they going to the fair ?

Exercises (Solved)

I. Complete each sentence using the Past Continuous Tense:

1. Children ………….. in the bushes. (hide)
2. They ………. through the zoo. (walk).
3. The waiter ……….. the people. (serve)
4. Meera …………. with her friends. (not play)
5. The baby ………. all the morning. (not cry)
6. The dancers ………… on the stage. (not perform)
Answer:
1. Children were hiding in the bushes.
2. They were walking through the zoo.
3. The waiter was serving the people.
4. Meera was not playing with her friends.
5. The baby was not crying all the morning.
6. The dancers were not performing on the stage.

II. Rewrite each sentence as a question:

1. The peon was ringing the bell.
2. We were not going to our village.
3. The boys were wearing red turbans.
4. Hema was not working at that time.
5. The children were playing in the street.
6. The teacher was writing on the blackboard.
7. The little girl was not playing with her doll.
8. Anu and Rosy were not talking to each other.
Answer:
1. Was the peon ringing the bell ?
2. Were we not going to our village ?
3. Were the boys wearing red turbans ?
4. Was Hema not working at that time?
5. Were the children playing in the street ?
6. Was the teacher writing on the blackboard ?
7. Was the little girl not playing with her doll ?
8. Were Anu and Rosy not talking to each other?

Miscellaneous Exercises

I. Use Simple Past form of the given verb to complete each sentence:

1. Did you ………….. this film ? (enjoy)
2. Did Roma …………. this picture ? (paint)
3. Columbus ……….. America in 1492. (discover)
4. She ………. to her village last month. (go)
5. The peon ……….. (not) the bell in time. (ring)
6. The fool didn’t ………. from experience. (learn)
Hints.
1. enjoy
2. paint
3. discovered
4. went
5. did not ring
6. learn.

II. Use Simple Present form of the given verb to complete each sentence:

1. I …….. for a walk daily. (go)
2. The sun …………… in the east. (rise)
3. They ……………. (not) bad workers. (like)
4. Kusha …………. (not) her parents. (obey)
5. Teachers …………… good students. (love)
6. We ………… milk and eggs for breakast.
Hints:
1. go
2. rises
3. do not like
4. does not obey
5. love
6. take

III. Rewrite each sentence in Past Indefinite Tense:

1. The bird flies to its nest.
2. They drink coffee every day.
3. Does he pay his fees regularly ?
4. Do you have milk for breakfast ?
5. Do we not fall ill by over-eating ?
6. You do not finish your work in time.
7. Kusha does not wear simple clothes.
8. Does he not help his friends in need ?
Answer:
1. The bird flew to its nest.
2. They drank coffee every day.
3. Did he pay his fees regularly ?
4. Did you have milk for breakfast ?
5. Did we not fall ill by over-eating ?
6. You did not finish your work in time.
7. Kusha did not wear simple clothes.
8. Did he not help his friends in need ?

IV. Rewrite each sentence in Past Continuous Tense:

1. Isn’t it raining heavily ?
2. We are waiting for the bus.
3. The teacher is teaching the children.
4. I am not living with my aunt these days.
5. They are not going home in the evening.
6. Is the lady knitting a sweater for her son ?
7. Aren’t Anu and Manu playing in the street ?
8. Am I wasting my time in watching Discovery Channel ?
Answer:
1. Wasn’t it raining heavily?
2. We were waiting for the bus.
3. The teacher was teaching the children.
4. I was not living with my aunt these days.
5. They were not going home in the evening.
6. Was the lady knitting a sweater for her son ?
7. Weren’t Anu and Manu playing in the street ?
8. Was I wasting my time in watching Discovery Channel ?

V. Rewrite each sentence in Present Continuous Tense:

1. Do you not speak the truth?
2. The students ask many questions.
3. I sit on the front bench in my class.
4. Does Kamla teach dance and music ?
5. These boys do not respect their elders.
6. She does not play with the poor children.
Answer:
1. Are you not speaking the truth?
2. The students are asking many questions.
3. I am sitting on the front bench in my class.
4. Is Kamla teaching dance and music ?
5. These boys are not respecting their elders.
6. She is not playing with the poor children.

VI. Rewrite each sentence in Past Continuous Tense:

1. The girls did not pluck flowers.
2. Did the peon not ring the bell ?
3. Did Ram break the windowpanes ?
4. My friends talked to me in English.
5. The watchman did not open the gate.
6. He spent all his money in good deeds.
Answer:
1. The girls were not plucking flowers.
2. Was the peon not ringing the bell ?
3. Was Ram breaking the windowpanes.
4. My friends were talking to me in English.
5. The watchman was not opening the gate.
6. He was spending all his money in good deeds.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2

1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Solution:
Area of top surface of a table = \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between the parallel sides
= \(\frac {1}{2}\) × (1.2 + 1) × 0.8
= \(\frac {1}{2}\) × 2.2 × 0.8
= 0.88 m²
Hence, area of top surface of table is 0.88 m².

2. The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Solution:

Let the length of the other parallel side be xcm.
Area of trapezium
= \(\frac {1}{2}\) × (sum of parallel sides) × height
= \(\frac {1}{2}\) × (10 + x) × 4
= (10 + x) × 2
= 20 + 2x
Area of trapezium = 34 cm² (given)
∴ 20 + 2x = 34
∴ 2x = 34 – 20
∴ 2x = 14
∴ x = 7
Hence, length of the other parallel side is 7 cm.

3. Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Solution:
Perimeter of a field = length of fence of a held
∴ AB + BC + CD + DA = 120
∴ AB + 48 + 17 + 40 = 120
∴ AB + 105 = 120
∴ AB = 120 – 105
∴ AB = 15 m
Now, Area of trapezium ABCD = \(\frac {1}{2}\) × (sum of parallel sides)
× perpendicular distance between the parallel sides
= \(\frac {1}{2}\) × (AD + BC) × AB
= \(\frac {1}{2}\) × (40 + 48) × 15
= \(\frac {1}{2}\) × 88 × 15 = 660 m²
Hence, area of the field is 660 m².

4. The diagonal of a quadrilateral shaped field is 24 m and B the perpendiculars dropped on it from c the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Solution:
Area of quadrilateral ABCD
Area of Δ ABD + Area of Δ BCD
= \(\frac {1}{2}\) × BD × AM + \(\frac {1}{2}\) × BD x CN
= \(\frac {1}{2}\) × 24 × 13 + \(\frac {1}{2}\) × 24 × 8
= 12 × 13 + 12 × 8
= 156 + 96
= 252 m²
Hence, area of the field is 252 m².

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
d₁ = 7.5 cm, d₂ = 12 cm
Area of a rhombus = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 7.5 × 12
= 7.5 × 6 = 45 cm².
Hence, area of rhombus is 45 cm².

6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Rhombus is a parallelogram too.
∴ Area of rhombus = base × height
= 5 × 4.8
= 5 × \(\frac {48}{10}\)
= 24 cm²
Now, area of rhombus = \(\frac {1}{2}\) × d₁ × d₂
∴ 24 = \(\frac {1}{2}\) × 8 × d₂
∴ 24 = 4 × d₂
∴ d₂ = \(\frac {24}{4}\) = 6 Cm

Hence, length of the other diagonal of rhombus is 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus-shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹4.
Solution:
The shape of a floor tile is rhombus.
d₁ = 45 cm, d₂ = 30 cm
Area of a tile = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 45 × 30
= 45 × 15
= 675 cm²
Now, floor of a building consists of total 3000 tiles.
∴ Area of floor = Number of tiles × Area of one tile
= 3000 × 675
= 20,25,000 cm²
1cm = \(\frac {1}{100}\)m
∴ 1cm² = \(\frac{1}{100 \times 100}\)m² = \(\frac {1}{10000}\)m²
∴ 20,25,000 cm² = \(\frac{2025000}{10000}\)m²
= 202.5 m²
∴ Cost of polishing the floor = ₹ 4 per m²
∴ Total cost of polishing the floor
= ₹ 4 × 202.5 = ₹ 810
Hence, total cost of polishing the floor is ₹ 810

8. Mohan wants to buy a trapezium-shaped field. Its side along the river is parallel to and twice the sidealong the road. If the area of this field is 10,500 m² and the perpendicular distance between the two parallel sides is 100 m, And the length of the side along the river.

Solution:
The side along the river is parallel to and twice the side along the road.
Let the length of side along the road be x m.
Then, the length of side along the river = 2xm
Area of trapezium
= \(\frac {1}{2}\) × (sum of parallel sides) × height
= \(\frac {1}{2}\) × (x + 2x) × 100
= \(\frac {1}{2}\) × 3x × 100
= 3x × 50 = 150 x
But, area of field = 10500 m² (given)
∴ 150x = 10500
∴ x = \(\frac {10500}{150}\)
∴ x = 70m
∴ 2x = 2 × 70 = 140 m
Hence, the length of the side along the river is 140 m.

9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Solution:
Let us, divide regular octagon into two congruent trapezium and a rectangle. Then add areas of all these three shapes.
Here, height of trapezium = 4 m Length of two parallel sides = 11 m and 5 m respectively.
∴ Area of a trapezium = \(\frac {1}{2}\) × (11 + 5) × 4
= \(\frac {1}{2}\) × 16 × 4
= 8 × 4
= 32 m²
∴ Area of two trapeziums = 2 × 32
= 64m² …….. (i)
Area of a rectangle = length × breadth
= 11 × 5 = 55m² …….. (ii)
∴ Area of regular octagonal platform
= (64 + 55) m² [From (i) and (ii)]
= 119 m²

10. There is a pentagonal-shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution:
Jyoti has divided given pentagon into two congruent trapeziums.
∴ Area of a trapezium = \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between two parallel sides
= \(\frac {1}{2}\) × (15 + 30) × \(\frac {15}{2}\)
= \(\frac {1}{2}\) × 45 × \(\frac {15}{2}\)
= \(\frac {675}{4}\) m²
∴ Area of two trapeziums = 2 × \(\frac {675}{4}\)
= \(\frac {675}{2}\)
= 337.5 m²
Now, Kavita has divided given pentagon into one triangle and the other square.
∴ Area of a triangle = \(\frac {1}{2}\) × b × h
= \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5m² …………….(i)
∴ Area of a square = (side)²
= (15)² = 15 × 15
= 225 m² ……….(ii)
Area of a pentagon
= (112.5 + 225) m² [From (i) and (ii)]
= 337.5 m²
Now, let us use our idea and find another method to find area of a given pentagon. Here, we have divided given pentagon into three triangles.

Area of ∆ a = \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5m² … (i)
Area of ∆ b = \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5 m² ………… (ii)
Area of ∆ c = \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5 m² ……. (iii)
From (i), (ii) and (iii) the area of a given pentagon
= (i) + (ii) + (iii)
= (112.5 + 112.5 + 112.5) m²
= 337.5 m²
Hence, area of the given pentagon is 337.5 m².
[Note : By using such ideas, i.e., dividing any shape into convienient shapes, you can find area of given figure.]

11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Solution:
Here, picture frame is divided into four trapeziums, such that a, b, c and d.
The sides of opposite trapeziums have same measurement so they have equal area.
∴ Area of a = area of c and area of b = area of d

For trapezium a and c:
Length of parallel sides = 24 cm, 16 cm
Height = \(\frac{28-20}{2}=\frac{8}{2}\) = 4 cm
∴ Area of trapezium a
= \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between parallel sides
= \(\frac {1}{2}\) × (24 + 16) × 4
= \(\frac {1}{2}\) × 40 × 4
= 80 cm²
So area of trapezium c = 80 cm².

For trapezium b and d:
Length of parallel sides = 28 cm, 20 cm
Height = \(\frac{24-16}{2}=\frac{8}{2}\) = 4 cm
∴ Area of trapezium b
= \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between parallel sides
= \(\frac {1}{2}\) × (28 + 20) × 4
= \(\frac {1}{2}\) × 48 × 4
= 96 cm²
So area of trapezium d = 96 cm².
Hence, area of each section of frame is as follows.
Area of section a = 80 cm²
Area of section b = 96 cm²
Area of section c = 80 cm²
Area of section d = 96 cm²
To find the total surface area, we find the area of each face and then add them.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 15 Probability Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
When Shyam and Ekta visit a particular shop in the same week. Possible outcomes are:
S = {(T, T) (T, W) (T, Th) (T, F) (T, S) (W, T) (W, W) (W, Th) (W, F) (W, S) (Th, T) (Th, W) (Th, Th) (Th, F) (Th, S) (F, T) (F, W) (F, Th) (F, F) (F, S) (S, T) (S, W) (S, Th) (S, F) (S, S)}

Here T stands For Tuesday
W stands For Wednesday
Th stands For Thursday
F stands For Friday
S stands For Saturday
n(S) = 25
(i) Let A is event that Shyam and Ekta visit the shop on the same day
A = {(T, T) (W, W) (Th, Th) (F, F) (S, S)}
n(A) = 5
Probability that both will visit the shop on same day = \(\frac{5}{25}\)
∴ P(A) = \(\frac{1}{5}\).

(ii) Let B is event that both will visit consecutive days particular shop
B = {(T, W) (W, T) (W, Th) (Th, W) (Th, F) (F, Th) (F, S) (F, S)}
n(B) = 8
∴ Probability that both will visit particular shop on consecutive days = \(\frac{8}{25}\).

(iii) Probability that both will visit the shop on different days = 1 – Probability that both will visit the shop
on same day.
= 1 – \(\frac{1}{5}\) [∵ P \((\bar{A})\) = 1 – P(A)]
= \(\frac{5-1}{5}\)
P \((\bar{A})\) = \(\frac{4}{5}\).

Question 2.
A die, is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The complete table is

Number of all possible out comes = 6 × 6 = 36
(i) Let A is event of getting total as even
A = {2, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 12}
n (A) = 18
∴ Probability of getting an even number = \(\frac{18}{36}=\frac{1}{2}\)
P (Even Number) = \(\frac{1}{2}\).

(ii) Let B is event of getting sum as 6 B = {6, 6, 6, 6)
n(B) = 4
Probability of getting an even number = \(\frac{4}{36}\)
∴ P(B) = \(\frac{1}{9}\).

(iii) Let C is event of getting sum at least 6
C = (6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 12}
n(C) = 15
∴ Probability of getting at least 6 = \(\frac{15}{36}=\frac{5}{12}\)
∴ P(C) = \(\frac{5}{12}\).

Question 3.
A bag contains 5 red balls and some blue balls. lithe probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Number of red balls = 5
Let number of blue balls = x
∴ Total number of balls = 5 + x
According to question,
Probability of drawing blue ball = 2 Probability of Red ball
\(\frac{x}{5+x}=2\left[\frac{5}{5+x}\right]\)
x = 10
∴ Number of blue balls = 10.

Question 4.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
Total number of balls in bag = 12
Number of black balls x
∴ Probability of getting black ball = \(\frac{x}{12}\)
If 6 more balls put in the box then total number of balls in the box = 12 + 6 = 18
Number of black balls = x + 6
Probability of getting black ball = \(\frac{x+6}{18}\)
According to Question,
Probability of drawing black ball = 2
Probability of drawing blackball in first case
\(\frac{x+6}{18}=\frac{2 x}{12}\)

\(\frac{x+6}{3}=\frac{2 x}{2}\)

\(\frac{x+6}{3}\) = x
x + 6 = 3x
6 = 3x – x
6 = 2x
x = 3
∴ Number of black balls = 3.

Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\). Find the number of blue marbles in the jar.
Solution:
Total number of marbles in jar =24
Let number of green marbles = x
∴ Number of blue marbles = 24 – x
P (Green marbles) = \(\frac{x}{24}\)
When a marble is drawn
Probability of drawing green marble = \(\frac{2}{3}\) (Given)
\(\frac{x}{24}=\frac{2}{3}\)
x = \(\frac{24 \times 2}{3}\)
x = 16

∴ Number of green marbles = 16
∴ Number of blue marbles = 24 – x = 24 – 16 = 8.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Question (a)

Solution:
Side of a square field = 60 m
∴ Perimeter of a square field = 4 × side
= 4 × 60 = 240 m
Area of a square field = (side)²
= (60)²
= 60 × 60
= 3600m²

Question (b)

Solution:
Perimeter of a rectangular field = Perimeter of square field
∴ Perimeter of a rectangular held = 240
∴ 2 (length + breadth) = 240
∴ 2 (80 + breadth) = 240
∴ 80 + breadth = \(\frac {240}{2}\)
∴ 80 + breadth =120
∴ breadth = 120 – 80
∴ breadth = 40
Breadth of rectangular field = 40 m
∴ Area of rectangular field = length × breadth
= (80 × 40)
= 3200 m²
Area of square field > Area of rectangular field
Thus, area of square field (a) is larger.

2. Mrs. Kaushik has a square plot with the measurement as shown g in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².

Solution:
Side of the square plot = 25 m
∴ Area of the square plot = (side)²
= (25 × 25) m²
= 625 m²
In square plot, a rectangular-shaped house is to be constructed.
∴ Area of the constructed house
= length × breadth
= (20 × 15) m²
= 300 m²
∴ Area of the garden = Area of square plot – Area of constructed house
= 625 – 300 = 325 m²
Cost of developing garden of 1 m² is ₹ 55
∴ Cost of developing garden of 325 m²
= ₹ (55 × 325)
= ₹ 17,875
Thus, total cost of developing garden is ₹ 17,875.

3. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5+ 3.5) metres].

Solution:
[Note: Here 2 semicircles at the ends of a rectangular garden makes a whole circle. So first find area of a circle and then area of a rectangle. Sum of these two areas is total area. Follow same pattern to find perimeter too. For perimeter of a garden, take only length as rectangle is between two semicircles. Diameter of a circle = Breadth of a rectangle = 7 m]
For semicircle:
∴ Radius = \(\frac{\text { diameter }}{\text { 2 }}\) = \(\frac {7}{2}\)m
Area of circle = πr²
Area of a semicircle = \(\frac {1}{2}\)πr²
∴ Area of 2 semicircles = 2(\(\frac {1}{2}\)πr²)
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)m²
= 38.5 m²
Circumference of two semicircles = 2πr
= 2 × \(\frac {22}{7}\) × \(\frac {7}{2}\)
= 22 m

For rectangle:
length = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
breadth = 7 m
Area of the rectangle = length × breadth
= 13 × 7 = 91 m²
Perimeter of the rectangle
= 2 (length × breadth)
= 2 (13 + 0)
= 2 × 13 = 26 m
∴ Total area of the garden = (38.5 + 91) m²
= 129.5 m²
∴ Perimeter of the garden = (22 + 26) m
= 48 m
Thus, area of the garden is 129.5 m² and the perimeter is 48 m.

4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners.)
Solution:
[Note : To find number of tiles, divide the area of the floor by area of a tile. Let us do it in a simple way. Unit of floor area and tile area should be same.] Here, tile is parallelogram shaped.
So it’s area = base × corresponding height
Area of a floor = 1080 m²
Base of a tile = 24 cm = \(\frac {24}{100}\) m
Corresponding height of a tile = 10 cm = \(\frac {10}{100}\) m
Number of tiles = \(\frac{\text { Area of a floor }}{\text { Area of a title }}\)
= \(\frac{1080}{\frac{24}{100} \times \frac{10}{100}}\)
= \(\frac{1080 \times 100 \times 100}{24 \times 10}\)
= 45,000
Thus, 45,000 tiles are required to cover the given floor.

5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2 πr, where r is the radius of the circle.

Question (a)

Solution:
Here, the shape is semi-circular.
Diameter = 2.8 cm
Radius = \(\frac{\text { Diameter }}{2}=\frac{2.8}{2}\) = 1.4 cm
Circumference of a semicircle = πr
Perimeter of the given figure
= πr + diameter
= (\(\frac {22}{7}\) × 1.4) + 2.8
= 4.4 + 2.8
= 7.2 cm

Question (b)

Solution:
Here, given shape is semicircular at one side, (radius = \(\frac {2.8}{2}\) = 1.4 cm)
So perimeter of semicircular region (circumference) = πr
= \(\frac {22}{7}\) × 1.4
= \(\frac {22}{7}\) × \(\frac {14}{10}\)
= 4.4 cm … (i)
Perimeter of the other portion
= breadth + length + breadth
= (1.5 + 2.8 + 1.5) cm
= 5.8 cm … (ii)
∴ Perimeter of the given figure
= (4.4 + 5.8) cm [from (i) and (ii)]
= 10.2 cm

Question (c)

Solution:
Perimeter of a given part
(semi circular circumference) = πr
= \(\frac {22}{7}\) × 1.4
= 4.4 cm
∴ Perimeter of the given figure
= (4.4 + 2 + 2) cm
= 8.4 cm
Thus, 7.2 cm < 8.4 cm < 10.2 cm.
Thus, the ant would has to take a longer round for food piece (b), as it has a larger perimeter.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:
For mode:
In the given data, Maximum frequency is 23 and it corresponds to the class interval 35 – 45
Modal class = 35 – 45
So, l = 35; f₁ = 23; f₀ = 21; f₂ = 14 and h = 10
Using fonnula, Mode l = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 35 + \(\left[\frac{23-21}{2(23)-21-14}\right]\) × 10
= 35 + \(\frac{2}{46-35}\) × 10
= 35 + \(\frac{20}{11}\) = 35 + 1.8 = 36.8.

For Mean:

From above data,
Assumed mean (a) = 30
Width of class (h) = 10
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{43}{80}\) = 0.5375
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 30 + 10 (0.5375)
= 30 + 5.375 = 35.375 = 35.37
Hence, mode of given data is 36.8 years and mean of the given data is 35.37 years. Also, it is clear from above discussion that average age of a patient admitted in the hospital is 35.37 years and maximum number of patients admitted in the hospital are of age 36.8 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the ¡nodal lifetimes of the components.
Solution:
In the given data.
Maximum frequency is 61 and it corresponds to the class interval 60 – 80.
∴ Model class = 60 – 80
So, l = 60; f₁ = 61 ; f₀ = 52; f₂ = 38 and h = 20
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 60 + \(\left(\frac{61-52}{2(61)-52-38}\right)\) × 20

= 60 + \(\frac{9}{122-52-38}\) × 20

= 60 + \(\frac{9}{32}\) × 20

= 60 + \(\frac{180}{32}\)

= 60 + 5.625 = 65.625
Hence, modal Lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:
For Mode: In the given data.
Maximum frequency is 40, and it corresponds to the class interval 1500 – 2000.
∴ Model class = 1500 – 2000
So, l = 1500; f₁ = 40; f₀ = 24; f₂ = 33 and h = 500
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

= 1500 + \(\left\{\frac{40-24}{2(40)-24-33}\right\}\) × 500

= 1500 + \(\left\{\frac{16}{80-24-33}\right\}\) × 500

= 1500 + \(\frac{16 \times 500}{23}\)

= 1500 + \(\frac{8000}{23}\) = 1500 + 347.83 = 1847.83

For Mean:

From above data,

Assumed Mean (a) = 2750
Length of width (h) = 500
\(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{35}{200}\) = – 0.175
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 2750 + 500 (- 0.175)
= 2750 – 87.50 = 2662.50
Hence, the modal monthly expenditure of family is 1847.83 and the mean monthly expenditure is 2662.50.

Question 4.
The following distribution gives the slate-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
For Mode:
In the given data,
Maximum frequency is 10 and it corresponds to the class interval is 30 – 35.
∴ Modal class = 30 – 35.
So, l = 30; f₁ = 10; f₀ = 9; f₂ = 3 and h = 5
using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 30 + \(\left(\frac{10-9}{2(10)-9-3}\right)\) × 5
= 30 + \(\frac{1}{20-12}\) × 5
= 30 + \(\frac{5}{8}\) = 30 + 0.625 = 30.625 = 30.63 (approx.)

For mean:

From above data, Assumed Mean (a) = 32.5
Width of class (h) = 5
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{23}{35}\) = – 0.65

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 32.5 + 5 (- 0.65)
= 32..5 – 3.25 = 29.25 (approx.)
Hence, mode and mean of given data is 30.63 and 29.25. Also, from above discussion, it clear that states/U.T. have student per teacher is 30.63 and on average, this ratio is 29.25.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches

Find the mode of the data.
Solution:
In the given data,
Maximum frequency is 18 and it corresponds to the class interval 4000 – 5000.
∴ Modal class = 4000 – 5000
So, l = 4000; f₁ = 18; f₀ = 4; f₂ = 9 and h = 1000
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 4000 + \(\left(\frac{18-4}{2(18)-4-9}\right)\) × 1000

= 4000 + \(\frac{14}{36-13}\) × 1000

= 4000 + \(\frac{14000}{23}\) = 4000 + 608.6956

= 4000 + 608.7 = 4608.7 (approx.)
Hence, mode of the given data is 4608.7.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised It in the table given below. Find the mode of the data:

Solution:
In the given data,
Maximum frequency is 20 and it corresponds to the class interval 40 – 50
∴ Modal Class = 40 – 50
So, l = 40; f₁ = 20; f₀ = 12; f₂= 11 and h = 10
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 40 + \(\left(\frac{20-12}{2(20)-12-11}\right)\) × 10

= 40 + \(\frac{8}{40-23}\) × 10

= 40 + \(\frac{80}{17}\) = 40 + 4.70588

= 40 + 4.7 = 44.7 (approx.)
Hence, mode of the given data is 44.7 cars.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 12 Exponents and Powers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Try These : [Textbook Page No. 194]

1. Find the multiplicative inverse of the following:

Question (i)
2^-4
Solution:
Multiplicative inverse of 2^-4 = 2⁴

Question (ii)
10^-5
Solution:
Multiplicative inverse of 10^-5 = 10⁵

Question (iii)
7^-2
Solution:
Multiplicative inverse of 7^-2 = 7²

Question (iv)
5^-3
Solution:
Multiplicative inverse of 5^-3 = 5³

Question (v)
10^-100
Solution:
Multiplicative inverse of 10^-100 =10¹⁰⁰

Try These : [Textbook Page No. 194]

1. Expand the following numbers using exponents:
(i) 1025.63
(ii) 1256.249
Solution:

Number	Expanded form
(i) 1025.63	(1 × 1000) + (0 × 100) + (2 × 10) + (5 × 1) + (6 × \(\frac {1}{10}\)) + (3 × \(\frac {1}{100}\)) OR (1 × 103) + (2 × 101) + (5 × 100) + (6 × 10-1) + (3 × 10-2)
(ii) 1256.249	(1 × 1000) + (2 × 100) + (5 × 10) + (6 × 1) + (2 × \(\frac {1}{10}\))+ (4 x \(\frac {1}{100}\)) + (9 × \(\frac {1}{1000}\)) OR (1 × 103) + (2 × 102) + (5 × 101) + (6 × 100) + (2 × 10-1) + (4 × 10-2) + (9 × 10-3)

Try These : [Textbook Page No. 195]

1. Simplify and write in exponential form:

Question (i)
(- 2)^-3 × (- 2)^-4
Solution:
= (-2)^-3+(-4)
= (-2)^-3-4
= (-2)^-7 or \(\frac{1}{(-2)^{7}}\)

Question (ii)
p³ × p^-10
Solution:
= p^3+(-10)
= p^3-10
= (p)^-7 or \(\frac{1}{(p)^{7}}\)

Question (iii)
3² × 3^-5 × 3⁶
Solution:
= 3^2+(-5)+6
= 3^2-5+6
= 3^2+6-5
= 3³

Try These : [Textbook Page No. 199]

1. Write the following numbers in standard form:

Question (i)
0.000000564
Solution:
= \(\frac{564}{1000000000}\)
(The decimal point is shifted to nine places to the right.)
= \(\frac{5.64}{10^{9}}\)
= \(\frac{5.64}{10^{7}}\)
= 5.64 × 10^-7
∴ 0.000000564 = 5.64 × 10^-7

Question (ii)
0.0000021
Solution:
\(\frac{21}{10000000}\)
= \(\frac{2.1 \times 10}{10000000}\)
= \(\frac{2.1}{10^{6}}\)
= 2.1 × 10^-6
∴ 0.0000021 = 2.1 × 10^-6

Question (iii)
21600000
Solution:
= 216 × 100000
= 216 × 10⁵
= 2.16 × 10² × 10⁵
= 2.16 × 10⁷
∴ 21600000 = 2.16 × 10⁷

Question (iv)
15240000
Solution:
= 1524 × 10000
= 1.524 × 1000 × 10000
= 1.524 × 10³ × 10⁴
= 1.524 × 10⁷
∴ 15240000 = 1.524 × 10⁷

2. Write all the facts given in the standard form. Observe the following facts: [Textbook Page No. 198 ]

Question 1.
The distance from the Earth to the Sun is 150,600,000,000 m.
Solution:
150,600,000,000 m
= 1.506 × 10¹¹ m

Question 2.
The speed of light is 300,000,000 m/sec.
Solution:
300,000,000 m / sec
= 3 × 10⁸m/sec

Question 3.
Thickness of Class VII Mathematics book is 20 mm.
Solution:
20 = 2 × 10¹ mm

Question 4.
The average diameter of a Red Blood Cell is 0.000007 m.
Solution:
0.000007 = 7 × 10^-6 m

Question 5.
The thickness of human hair is in the range of 0.005 cm to 0.01 cm.
Solution:
0.005 = 5 × 10^-3cm and
0.01 = 1 × 10^-2 cm

Question 6.
The distance of moon from the Earth is 384,467,000 m (approx).
Solution:
384,467,000 = 3.84467 × 10⁸ m

Question 7.
The size of a plant cell is 0.00001275 m.
Solution:
0.00001275 = 1.275 × 10^-5m

Question 8.
Average radius of the Sun is 695000 km.
Solution:
695000 = 6.95 × 10⁵ km

Question 9.
Mass of propellant in a space shuttle solid rocket booster is 503600 kg.
Solution:
503600 = 5.036 × 10⁵ kg

Question 10.
Thickness of a piece of paper is 0.0016 cm.
Solution:
0.0016 = 1.6 × 10^-3 cm

Question 11.
Diameter of a wire on a computer chip is 0.000003 m.
Solution:
0.000003 = 3 × 10^-6 cm

Question 12.
The height of Mount Everest is 8848 m.
Solution:
8848 = 8.848 × 10³ m

Punjab State Board PSEB 6th Class English Book Solutions English Vocabulary Antonyms Exercise Questions and Answers, Notes.

PSEB 6th Class English Vocabulary Antonyms

An Antonym is a word which is opposite in meaning (विपरीतार्थक) to another word; as-

Word – Antonym
absent – present
active – passive
above – below
accept – reject
before – after
bitter – sweet

blunt – sharp
bold – timid
beautiful – ugly
bright – dim
cheap – costly
clean – dirty
clever – stupid
dark – bright
defeat – victory
difficult – easy
death – life
early – late
empty – full
enemy – friend
far – near
foolish – wise
fresh – stale
good – bad
great – small
happy – sad
high – low
hot – cold
in – out
import – export
increase – decrease
joy – sorrow
Word – Antonym
junior – senior
kind – cruel
lend – borrow
light – heavy
love – hate
long – short
old – young/new
oral – written
night – day
peace – war
poor – rich
profit – loss
right – wrong
shallow – deep
slow – fast
stale – fresh

strong – weak
summer – winter
thick – thin
top – bottom
tall – short
up – down
warm – cool
wide – narrow
wise – silly

Punjab State Board PSEB 6th Class English Book Solutions English Vocabulary Sounds of Animals Exercise Questions and Answers, Notes.

PSEB 6th Class English Vocabulary Sounds of Animals

Animal – Sound
Ass – Brays
Bee – hum
Cat/Kitten – mew
Chicks – cheep
Cows – low, moo

Crows – caw
Cuckoos – cuckoo
Cock – Crow
Deers – bell
Dogs – barks
Dolphins – click
Donkeys – hee-haw
Doves – coo
Ducks – quack
Eagles – scream
Elephants – trumpet, roar
Fly – buzz
Foxes – bark, yelp.
Frogs – croak
Geese – cackle/quack
Grasshoppers – chirp
Goat – baah
Hare – squeak
Horse – neigh
Lions – roar, growl
Monkey – scream, chatter
Owl – hoot
Parrot – talk
Sparrow – chirrup, chirp
Snakes – hiss

Sheep – bleat
Tiger – roar, growl
Tortoise – grunt.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry InText Questions

Think, Discuss and Write (Textbook Page No. 58)

Question 1.
Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, ∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral ? Give reasons for your answer.
Solution:
No, the quadrilateral ABCD cannot be constructed with the given combination of measurements. If the length of side BC or DC is given, then only □ ABCD can be constructed.

Think, Discuss and Write (Textbook Page No. 60)

Question (i).
We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this ?
Solution:
No, any 5 measurements can’t determine a s quadrilateral. To construct a quadrilateral, specific combination of measurements should be needed such as :

• Four sides and one diagonal
• Four sides and one angle
• Three sides and two diagonals
• Two adjacent sides and three angles
• Three sides and two included angles
• Some special properties should be given,

Question (ii).
Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm? Why?
Solution:

Here, to draw a parallelogram BATS, BA = 5 cm, AT = 6 cm and AS = 6.5 cm are given. In parallelogram length of opposite sides are equal. So ST = AB = 5 cm and SB = AT = 6 cm. First we can draw A ASB where SB = 6 cm, AB = 5 cm and AS = 6.5 cm. Then draw ΔATS where AT = 6 cm, ST = 5 cm.
Thus, we can draw a parallelogram from given measurements.

Question (iii).
Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm ? Why?
Solution:

Here, to draw a rhombus ZEAL, ZE = 3.5 cm and EL = 5 cm are given. All of rhombus are equal to each other. So ZE = EA = AL = LZ = 3.5 cm. Moreover, diagonal EL = 5 cm Is given.
So We know all necessary measurements to draw rhombus. Yes, we can draw a rhombus ZEAL.

Question (iv).
A student attempted to draw a quadrilateral PLAY where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm, but could not draw it. What is the reason?
[Hint: Discuss it using a rough sketch. ]
Solution :
Here, to draw a quadrilateral PLAY,
PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm are given.

Now, let us look at measurements of
ΔPLY. PL + PY = 3 cm + 2 cm = 5 cm while YL = 6 cm.
We know that the sum of the lengths of any two sides of triangle is always greater than the length of the third side.
So point P cannot be determined even after constructing ΔLAY. Thus, a student failed s due to this reason.

Think, Discuss and Write (Textbook Page No. 62)

Question 1.
In the above example, can we draw the quadrilateral by drawing ΔABD first and then find the fourth point C ?
Solution:
Since, the measurement of AB is not given, we cannot draw ΔABD, so question does not arise to find the foruth point C.

Thus, we cannot draw the quadrilateral

Question 2.
Can you construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm ? Justify your answer.
Solution:
No, the quadrilateral PQRS cannot be constructed as in ΔQSP, SQ + PQ ≯ SP

Think, Discuss and Write (Textbook Page No. 64)

Question 1.
Can you construct the above quadrilateral MIST if we have 100° at M instead of 75° ?
Solution:
Yes. The quadrilateral MIST can be constructed with ∠M = 100° instead of 75°.
[Note : Only size of quadrilateral is changed.]

Question 2.
Can you construct the quadrilateral PLAN if PL = 6 cm, LA = 9.5 cm, ∠P = 75°, ∠L = 150° and ∠A = 140° ?
[Hint: Recall angle sum property.]
Solution:
Here, ∠P + ∠L + ∠A + ∠N
= 75° + 150° + 140° + ∠N
= 365° + ∠N
∴ Construction of quadrilateral PLAN is not possible as according to angle sum property. The sum of all the angles of a quadrilateral is 360°. Here, 365° + N > 360°.

Question 3.
In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct as in the example above?
Solution :
In a parallelogram, the opposite sides are parallel and of equal length. Here, we know the lengths of adjacent sides, so measures of the angles are not needed.
If we know the length of a diagonal the quadrilateral can be drawn.

Think, Discuss and Write (Textbook Page No. 66)

Question 1.
In the above example, we first drew BC. Instead, what could have been be the other starting points?
Solution :
Instead of drawing BC, we can start with \(\overline{\mathrm{AB}}\) or \(\overline{\mathrm{CD}}\).

Question 2.
We used some five measurements to draw quadrilaterals so far. Can there be different sets of five measurements (other than seen so far) to draw a quadrilateral ? The following problems may help you in answering the question.
(i) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and ∠B = 80°.
(ii) Quadrilateral PQRS with PQ = 4.5 cm, ∠P = 70°, ∠Q = 100°, ∠R = 80° and ∠S = 110°.
Construct a few more examples of your own to find sufficiency/insufficiency of the data for construction of a quadrilateral.
Solution:
(i) Here, four sides and one angle are given. So given data is sufficient to construct quadrilateral ABCD.
(ii) We cannot locate the points R and S with the help of given data. So given data is insufficient to construct quadrilateral PQRS.

Few examples of sufficient data to construct quadrilaterals :

• Quadrilateral PQRS in which RS = 6 cm, QR = 5 cm, PQ = 5 cm, ∠Q = 135°, ∠R = 90°. (three sides and two angles)
• Quadrilateral ABCD in which AB = 5 cm, BC = 4 cm, ∠B = 60°, ∠A = 90° and ∠C = 135°. (two sides and three angles)

Try these (Textbook Page No. 67)

Question 1.
How will you construct a rectangle PQRS if you know only the lengths PQ and QR?
Solution:

Each angle of a rectangle is a right angle. Rectangle has opposite sides of equal lengths. Here, PQ is given, So PQ = RS.
ΔPQR can be drawn using PQ, QR and ∠Q = 90°.
ΔQRS can be drawn using QR, RS and ∠R = 90°.
Thus, the required rectangle PQRS can be constructed.

Question 2.
Construct the kite EASY if AY = 8 cm, EY = 4 cm and SY = 6 cm. Which properties of the kite did you use in the process ?
Solution:

Diagonals intersect each other at right angle. ? In kite pairs of consecutive sides are of equal lengths.
While constructing, E cannot be located.
∴ Kite EASY cannot be constructed
(∵ For Δ EYA, the sum of lengths of two sides EY + EA (4 + 4) is not greater than length of third side AY (8).

Punjab State Board PSEB 7th Class English Book Solutions English Message Writing Exercise Questions and Answers, Notes.

PSEB 7th Class English Message Writing

Type – I

1. Read the following telephonic conversation between Kavita and Karan. Karan will not be able to meet Avik. He leaves a message for him. Write this message by using not more than 50 words.

Kavita : Hello ! Hello ! I am Kavita from Indore. Can I speak to Avik? I am his sister.
Karan : Hello ! Kavita I am Karan, Avik’s colleagues. Avik is on leave today. Can I take a message ?
Kavita : Yes, Karan, I am coming to Mumbai tomorrow. Ask him to pick me up at the
airport. I have an interview for the post of Scientist at NPL on the day after tomorrow.
Karan : Which flight are you coming on ?
Kavita : It is the Jetline flight which arrives there at 7.15 p.m. I am bringing along with me that big box which contains his books. I hope it won’t be any trouble for you coming.
Karan : Not at all, I will leave a message on his table. Okay, Kavita.
Kavita : Thank you, Karan. Bye.

Message

Avik

Today, there was a telephonic message from your sister, Kavita. She is coming to Mumbai tomorrow as she has some interview here. She is coming by the Jetlnte flight which will arrive here at 7.15 p.m. She says she will be bringing with her a big box containing your books. Please pick her up at the airport.

Karan

2. Read the following telephone conversation which took place when Suresh was staying with his uncle. Write the message from Suresh to his maid, using not more than 50 words.

Seshu : Hello ! Hello ! This is Sheshu from Lucknow. Can I speak to Mrs. Rao, please ? I am a friend of his son, Madhav.
Suresh : This is Suresh Rao. My uncle is not here at the moment. We heard about the earthquake. Is Madhav all right ?
Seshu : Yes, yes. He’s okay now. He had a bad fall during the earthquake and he broke his left leg. It was a multiple fracture, but there’s nothing to worry about now.
Suresh : Is he in hospital ?
Sheshu : Yes, he’s at the Tata Memorial Hospital here. Would you please inform his family ?
Suresh : Of course I will.

Message

Dear Uncle

There was a telephonic call for you from one Mrs. Sheshu. He is our Madhav’s friend, from Lucknow. He had a fall during the earthquake and he broke his leg. He got a multiple fracture and has been in the Tata Memorial Hospital there. But he added that there was nothing to worry.

Suresh

3. Here is telephonic talk between Gurbani and Jaspreet. Gurbani give her a message , Write the message on behalf of Gurmeet not more than 50 words.

Gurbani : Hi ! Gurmeet.
Jaspreet : Sorry, I’m not Gurmeet. I’m her elder sister Jaspreet. Can I know who is calling ?
Gurbani : I Gurbani, her friend, Is it not her contact number ? I have some urgent message for her.
Jaspreet : Yes, it is but she has gone to the market to by some fruit and her mobile, is with me.
Gurbani : Would you please convey my message to her ? .
Jaspreet : But she is not coming back for about two hours. I am also going.to the hospital to see one of our neighbours. Would you gave me the message. I’ll leave it on her table before I go.
Gurbani : Sure ! Please note our family is going to Hazoor Sahib on Sunday. She can accompnay as if her parents allow. It will be a good company for me.
Jaspreet : All right, Don’t worry. The message will reach her.
Gurbani : Thank you very much.

Message

22.06.2020
Dear Gurmeet,

There was a call from your friend Gurbani in your absence. Their family is going to Hazoor Sahib on Sunday. You can accompany them if our parents permit. She will be feeling good in your company. Talk to her for confirmation.

Jaspreet.

4. Read the conversation between Mrs. Singh and the Principal of G.S.S. Modern School. Write the message on behalf of the Principal that he will send to the Preeti’s class-teacher.

Mrs. Singh : Hello ! Is that G.S.S. Modem School ?
Principal : Yes, what do you want ?
Mrs. Singh : I would like to speak to the school Principal.
Principal : yes, speaking. What can I do for you.
Mrs. Singh : My daughter, Preeti is a student of VII A of your school. Today was the last day for the payment of her school fee. I have deposited it the school account.
Principal : What is the problem in that ? It was your duty.
Mrs. Singh : Madam, she was worried about fine before she left for school and looked sad. Send this message to her in class, to make her tension free.
Principal : Of course ! The message will be sent to her through her class teacher.
Mrs. Singh : Thank you, Madam.

Message

12.05.2020
Dear Preeti

You looked worried and sad before you left for school. It was natural because your school fee was not paid and you could be fined or punished in some other way for that. Now, don’t take any tension as your fee has been deposited.

Mummy

5. Read the following telephonic conversation between Ravinder and Ranjit about to leave for her coaching soon and his mother is not at home for the moment. Write this message on behalf of Ranjit.

Ravinder : Hello, is that Amit ?
Ranjit : No I am her elder brother, Ranjit speaking May I know who is speaking ?
Ravinder : I am Ravinder speaking. I wanted to speak to your brother for an important message.
Ranjit : He is not at home now. He has just gone to visit one of two friends. I am also leaving for my office. If there is some message I shall give him.
Ravinder : OK, then Please tell Amit all about me. We are to play a friendly cricket match tomorrow morining. I am one of the players in his team front. I will not be able to take part in it as I am suddenly fell ill and the doctor has advised me complete rest. He can take any other player with him.

Message

23 March, 2020
Dear Amit

In your absence there was a telephone from your friend, Ravinder. He is one of the players of your cricket team for the tomorrow match.

But he will not be able to come, as he has suddenly fallen ill and the doctor has advised him complete rest. You can take any other friend with you.

Ranjit

6. Here is telephonic talk between Gurbani and Jaspreet Gurmeet gives her message for her brother. Write the message on behalf of Gurmeet not more than 50 words.

Gurmeet : Hi ! Gurmeet.
Jaspreet : Sorry, I’m not Gurmeet. I am his elder sister, Jaspreet. Can I know who is calling ?
Gurbani : I am Gurbani, Bedi, her tutor. Is it not his contact number ? I have some urgent message for her.
Jaspreet : Yes it is. But she has gone to the market to buy fruit and her mobile is with me.
Gurbani : Would you please, convey my message to him
Jaspreet : But he is not coming back for about two hours, I’m also going to the hospital to see one of our neighbours. Would you give me the message. I’ll leave it on his table before I go.
Gurbani : Sure ! Please note this. I shall not be able ‘to come for coaching as I have sprained my ankle. Therefore she should not write for me in the evening.
Jaspreet : All right. Don’t worry the message will reach him.
Gurbani : Thank you very much.

Message

April 10, 2020
Dear Sarbjit

There was a call from you tutor when you were not at home. She will not be coming today for coaching because she has sprained her ankle. Therefore don’t wait for her in the evening.

Mother/Mummy

7. Read the following telephone conversation between kamal and Hardeep from a hospital. Hardeep wants to talk to kamlesh sharma but she is at present not at home. She will be back after an hour. Thinking yourself as Kamal and using the telephone conversation as the subject, write a message to Kamlesh Sharma in not more than 50 words.

Hardeep : Is that Kamlesh Sharma ?
Kamal : No, we are her tanents.
Hardeep : May I speak to Mrs Sharma ? I have to talk to her urgently.
Kamal : She is not here at present. She has gone out and will return after an hour.
Hardeep : Then, would you give a message to her as soon as she returns ?
Kamal : Yes, of course. But I am also going to my friend’s home for his birthday party this evening. However, you needn’t worry I shall leave your message on table for her. What’s it ?
Hardeep : Kindly tell her that her grandmother has met with an accident and she is at present, in the civil hospital here in Khanna. She has broken her leg and is in plaster. Now she is feeling easy. Please tell kamlesh sharma to reach the hospital at her earliest with her husband.
Kamal : O.M.G. Please worry not. I shall leave the message for her before I leave Hardeep. Thank you very much.
Kamal : By, Who speaks on the other side ?
Hardeep : I am her neighbour, Hardeep Sodhi.

Message

May 5, 2020.
Dear Sharma Aunty

There was a telephone call for you from Khanna. I am sorry to inform that your grandmother has broken her leg and is in the Civil Hospital there. Don’t worry she is feeling easy now though she is still in plaster. Reach the hospital with your husband at your earliest.

Kamal

8. Read the following telephonic conversation between Ravinder and Shilpa. Shilpa is about to leave for her coaching within five minutes and her sister is not at home for the moment. Write this message conveyed on behalf of Shilpa.

Mrs. Ravinder : Hello, is that Jaspreet ?
Shilpa : No, I am her sister, Shilpa speaking. May I know who speaks on the other side ?
Mrs. Ravinder : I am Mrs. Ravinder, her friend Sonum’s mother speaking from bus stand. I wanted to speak to your sister urgently.
Shilpa : She is not at home now aunty. She has just gone to her college if there is some message I shall give her.
Mrs. Ravinder : Ok, then please note down I have returned from Amritsar. I have brought some holy books for her. I wanted her to collect the packet at bus stand as I am already late for home. However, she can collect it from there in the evening today urgently. I will not be availabe for the night as we have to attend a marriage party.
Shilpa : You needn’t worry aunty. As I am also going to the Gurudwara, I shall leave this message on table before leave.
Mrs. Ravinder : Thank you very much.

Message

March 5, 2020
Didi

There was a telephone from Mrs. Ravinder, Sonum’s mother. She has bought some holy books for you from Amritsar. Collect the packet from her home in the evening today, urgenlty as she will not be available at night as they are going to attend a marriage party.

Shilpa.

TYPE – II

1. You want to send a message to your niece on her birthday as you are unable to attend it because of an urgent meeting in the office. Write the S.M.S. is not more than 50 words.

May 15, 2020
Dear Vani

Many many happy returns of the day. Stay blessed and in high spirits. Don’t mind my absence as I am unable to attend your birthday party due to an urgent meeting in the office. You will soon receive a lovely gift from me through courier.

Your loving uncle
Gurnarn

2. You were to attend the marriage of your friend but you suddenly fell ill that night. Send your friend an S.M.S. informing your friend about your disability to reach and giving him congratulations and expressing your good wishes for the wedding couple.

Dear Madhur

Congratulations on the wedding of your elder brother. Please pardon my absence as I am unable to attend the marriage ceremony because of sudden illness. I had got my briefcase ready to take a bus, but I was forced it lie in bed. Let me convey my hearty wishes for the happy life of the wedding couple.

Sharan