Bhagya

Punjab State Board PSEB 8th Class English Book Solutions English Notice Writing Exercise Questions and Answers, Notes.

PSEB 8th Class English Notice Writing

नोटिस (Notice) किसी घटित होने वाली या घटित हो चुकी घटना के बारे में सूचना होती है। इसके द्वारा किसी आदेश, प्रार्थना अथवा चेतावनी पर अमल करने की सूचना दी जाती है।

याद रखने योग्य बातें-

1. नोटिस की भाषा उद्देश्यपूरक होनी चाहिए। इसमें निजीपन नहीं होना चाहिए।
2. दी जाने वाली सूचना पूर्ण होनी चाहिए।
3. इसका प्रस्तुतीकरण संक्षिप्त और स्पष्ट होना चाहिए। नोटिस का उद्देश्य भी स्पष्ट हो।
4. जहां तक हो सके नोटिस में “I’ तथा ‘You’ के प्रयोग से बचें।
5. नोटिस लिखे जाने की तिथि का उल्लेख अवश्य करें।
6. विषय-वस्तु से सम्बन्धित स्थान, समय तथा कार्यक्रम आदि की स्पष्ट जानकारी दें।
7. नोटिस जारी करने वाले व्यक्ति के पद का नाम तथा उसके हस्ताक्षर भी नोटिस में होने चाहिएं।
8. नोटिस दी गई शब्द-सीमा में ही लिखना चाहिए।

Format of a Notice

Notice

Date

Heading / Title
Content

Signatory

Important Notices

1. Notice About Something Found
You have found a purse lying in one of the lawns of your school. Write a notice asking the owner of the purse to contact you.

NOTICE

March 7, 20…..
FOUND ! FOUND ! FOUND!

This is to inform all the students that a purse has been found lying in one of the i school lawns. It is a black leather purse containing some money. The owner should contact the undersigned.

Gulshan Rai
Roll No. 10, VIII A

2. Notice About a Tour
Your school is organising a tour to Delhi and Agra. You are the secretary of Tour Organising Committee. Draft a notice asking the students to give you their names.

NOTICE
JOINING A HISTORICAL TOUR

March 8, 20……

Our school is organising a historical tour to Delhi and Agra. The duration of the tour is six days starting on March 12. Those interested should give their names to the undersigned latest by March 10.

Mohan Lai
(Secretary Tour Organising Committee)

3. Notice About Paper Reading Contest
You are the incharge of Junior Humanities Forum of your school. The Forum is organising a Paper Reading Contest. Draft a notice inviting the participants to give you their names.

NOTICE
PAPER READING CONTEST

March 9, 20….

The Junior Humanities Forum of the school is organising a Paper Reading Contest on March 16 in the school hall. Students of class VI to VIII are eligible to participate. Those interested should give their names to the undersigned before March 12.

Raman
Secretary
(Junior Humanities Forum)

4. Notice for a School Function
You are Shashi Mehta, the Sports Secretary of your school. Your school is organising the Annual Sports meet next week. Write a notice in about 50 words to be put on the school notice board to this effect.

NOTICE
ANNUAL SPORTS MEET

January 15, 20….

The Annual Sports meet of the school is going to be held on 28th and 29th of this month. Students who wish to take part in any event should give their names to the undersigned by 24th of this month positively.

Shashi Mehta
(Sports Secretary)

5. Notice for a Lost Wrist Watch
You have lost a wrist watch in your school. Write a notice about the loss giving the particulars of the watch. Also announce a reward for the finder.

NOTICE
LOST ! LOST ! LOST !

April 10, 20….

This is to inform all students that I lost my wrist watch yesterday in the school during the recess period. It is an H.M.T. watch with a golden case and a golden chain. He/She who happens to find it should contact the undersigned immediately. He (She) will be suitably rewarded.

Aastha Jindal.
R. No. 2, VIII D

6. Notice about a Lost Pen
You have lost your pen somewhere in your school. Write a notice about it.

NOTICE
LOST ! LOST ! LOST !

12 March, 20 …….
A new gel pen has been lost somewhere in the school ground. The pen is of Reynold make with blue colour. The finder is requested to return it to the undersigned or deposit it with the school office.

Kulbir Singh
Roll No. 2
VIII B

7. Books for Sale
Ashok Mathur of Class VIII A has just passed his annual examination. Two of his books are in fairly good condition and he wants to sell them at reduced prices. He puts up a notice on the school notice-board giving all the necessary details. Write this notice in the space below, using not more than 50 words.

NOTICE
BOOKS FOR SALE

March 22, 20

Two VIII Class books fairly in good condition are for sale. The books and their reduced prices are given below :
1. History of India Rs. 50/- only
2. General Science Rs. 60/- only
Those who are interested should contact Ashok Mathur, Class IX A

8. Taking Part in Debate
K.C. Sharma, Senior English teacher, invites applications from the students of VIII class who want to take part in debate to be held at Ludhiana. March 15 is the last date for the applications to reach the undersigned. In this connection, he puts up a notice on the school notice-board. Write this notice in about 40 words.

NOTICE
TAKING PART IN DEBATE

March 10, 20 ………..

Applications are invited from the students of class IX who want to take part in the debate to be held at Ludhiana. The students should apply before 15th March, giving details of their proficiency.
Sd/
K.C. Sharma
Senior English Teacher

9. Shoes From School Red Cross Fund
Anil Kumar Sharma, the school Red Cross teacher, invites applications from the poor students who want to take shoes from the Red Cross. In this connection, he puts up a notice on, the school notice-board. Write this notice in not more than 30 words.

NOTICE
SHOES FROM SCHOOL RED-CROSS FUND

March 2, 20….

Applications are invited from the poor students who want to take shoes from the Red Cross. All applications should reach the undersigned by 15th March.

Anil Kumar Sharma
Red Cross Teacher

10. Selection to the Volleyball Team
Mohan Singh has been selected the captain of the School Volleyball Team. He is to invite applications from the students of class IX for selection to the Volleyball team. So he puts up a notice on the school notice-board in this connection. Write this notice in about 40 words.

NOTICE
SELECTION TO THE VOLLEYBALL TEAM

Applications are invited from the students of class IX for selection to the Volley-ball team. Apply before 15th September, giving details of your proficiency in the game.

Mohan Singh
Captain Volleyball Team

11. Free Yoga Classes
You are the PTI of your school. Write a notice asking the students to enrol for free yoga classes.

NOTICE
FREE YOGA CLASSES

2 April, 20

Attention !
Students interested in attending free yoga classes from 10th April every morning from 6 a.m. to 7 a.m. should contact the undersigned before 7th April.

B.S. Bedi
PTI

12. Notice for returning Library Books
You are the librarian of your school. Write a notice asking the students to return borrowed books before the school closes down for the summer vacation.

NOTICE

May 10, 20……..

ATTENTION !
The school breaks up for the summer vacation next week. All the students who have borrowed any book from the library must return it before the school breaks up for the vacation.

Raman Kumar
Librarian

13. Notice for a Lost Item
You are Kulbir Singh of Class VIII. You have lost your new water bottle. Write the notice that you would like to put up on the school notice-board.

NOTICE
LOST ! LOST ! LOST !

12 March, 20
A new water botde has been lost somewhere in the school garden. The bottle is of Eagle make with blue colour. The finder is requested to return it to the undersigned or deposit it with the school office.

Kulbir Singh
Roll No. 2
VIII B

14. Notice for Blood Donation
You are the Sarpanch of your village. Write a notice inviting adults to donate blood at the blood donation camp to be held at the community centre.

NOTICE
BLOOD DONATION CAMP

The village Panchayat is going to organise a blood donation camp in the community centre on 8th April from 9 a.m. All the adults of the village should come forward and donate blood to save lives of accident victims.

Balwant Singh
Village Sarpanch

15. Notice for a Misplaced Library Book
You have misplaced a library book ‘Panchtantra Tales’. Write a notice that you would like to put up in the classroom.

NOTICE
BOOK MISPLACED

March 10, 20…. .

The book ‘Panchtantra Tales’ has been misplaced somewhere in the classroom. It was borrowed from the library yesterday. The finder is requested to return it to the undersigned or hand it over to the class teacher.

Rajni
Roll No. 5
VIIIA

16. Notice for Sports Participants
You are the sports captain of your school. Write a notice to all participants to submit their names and event in which they are taking part.

NOTICE
ATTENTION ! SPORTS PARTICIPANTS

March 10, 20….

All the participants who are interested to take part in the school sports are requested to give their names to the undersigned before Sunday, the 6th May. They must mention the event they are taking part in.

Raman Kumar
(Sports Captain)

17. Help for Tsunami Victims
You are the Head Boy/Girl of your school. Draft a notice requesting the students to come forward to help the tsunami victims.

NOTICE
HELPING TSUNAMI VICTIMS

March 18, 20
This is to inform all the students of the school that our school is raising a fund to help the tsunami victims. All should come forward with open heart and contribute to the fund as much as possible. Deposit the money with the respective class teachers.

Sham/Suman
Head Boy/Girl

Punjab State Board PSEB 8th Class English Book Solutions English Role Play Writing Exercise Questions and Answers, Notes.

PSEB 8th Class English Role Play Writing

1. Imagine your father is ill. You went to the hospital to get medicine for him and you got late for school. The headmaster asked you the reason for being late. You will play the role of the student and your friend of the headmaster. The beginning of the conversation is given. You will start with the given conversation.

1. Headmaster : Why are you late today ?
Student : I had gone to the hospital, sir.
2. Headmaster : ………………….
Student : ………………….
3. Headmaster : ………………….
Student : ………………….
4. Headmaster : ………………….
Student : ………………….
5. Headmaster : ………………….
Student : ………………….
6. Headmaster : ………………….
Student : ………………….
Answer:
2. Headmaster : Why did you go to the hospital ?
Student : I had gone to get medicine for my father.
3. Headmaster : How long had you to wait for your turn ?
Student : I had to wait there for one hour for my turn.
4. Headmaster : How is your father now ?
Student : He is improving, sir.
5. Headmaster : Who is looking after him now ?
Student : My mother is looking after him now.
6. Headmaster : How do you come to school ?
Student : I come to school on foot.

2. Imagine you are a football player. Your team won a football match yesterday. Your father wants to know about the match. You will play the role of son and the father. The beginning of the conversation is given.

1. Father : Did you play the match yesterday ?
Son : Yes, I played the match yesterday.
2. Father : ………………….
Son : ………………….
3. Father : ………………….
Son : ………………….
4. Father : ………………….
Son : ………………….
5. Father : ………………….
Son : ………………….
6. Father : ………………….
Son : ………………….
Answer:
2. Father : Where was the match played ?
Son : It was played on our school ground.
3. Father : Which team was stronger ?
Son : Both the teams were equally strong.
4. Father : Which team scored the first goal ?
Son : Our team scored the first goal.
5. Father : How many goals were scored ?
Son : Two goals were scored.
6. Father : Did you score any goal ?
Son : Yes, I scored one goal.

3. Suppose a friend (Ramesh) of yours wants to know about your school library. You will play the role of you and Ramesh. The beginning of the conversation is given. You will start with the given conversation.

1. Ramesh : Has your school a big library ?
You : Yes, our school has a big library.
2. Ramesh : ………………….
You : ………………….
3. Ramesh : ………………….
You : ………………….
4. Ramesh : ………………….
You : ………………….
5. Ramesh : ………………….
You : ………………….
6. Ramesh : ………………….
You : ………………….
7. Ramesh : ………………….
You : ………………….
8. Ramesh : ………………….
You : ………………….
9. Ramesh : ………………….
You : ………………….
10. Ramesh : ………………….
You : ………………….
Answer:
2. Ramesh : How many sections has it ?
You : It has three sections—Punjabi, Hindi and English.
3. Ramesh : Who is its incharge ?
You : Mr. Hari Singh is its incharge.
4. Ramesh : How many almirahs are there in the library ?
You : There are fifteen big steel almirahs in the library.
5. Ramesh : How many books are there in the library ?
You : There are more than six thousand books in the library.
6. Ramesh : Are there novels and story books in it ? ‘
You : Yes, there are many novels and story books in it.
7. Ramesh : Has your library any books on general knowledge ?
You : Yes, our library has general knowledge books too.
8. Ramesh : Are there chairs and tables in the library ?
You : Yes, there are chairs and tables in our library.
9. Ramesh : How often does your class go to the library ?
You : Our class goes twice a week to the library.
10. Ramesh : How many books can you take out at a time ?
You : We can take out only one book at a time.

4. You gave some clothes for dry cleaning. Your suit was spoiled. You went to the dry cleaner and made the complaint. Now you will play the role of yourself and dry cleaner. The beginning of the conversation is given. You will start with the given conversation.

1. Dry Cleaner : Do you want to get your suit dry cleaned ?
You : No, I have come to complain about my suit.
2. Dry Cleaner : ………………….
You : ………………….
3. Dry Cleaner : ………………….
You : ………………….
4. Dry Cleaner : ………………….
You : ………………….
5. Dry Cleaner : ………………….
You : ………………….
6. Dry Cleaner : ………………….
You : ………………….
7. Dry Cleaner : ………………….
You : ………………….
8. Dry Cleaner : ………………….
You : ………………….
9. Dry Cleaner : ………………….
You : ………………….
10. Dry Cleaner : ………………….
You : ………………….
Answer:
2. Dry Cleaner : Whom did you give your suit for dry cleaning ?
You : I gave my suit to your servant for dry cleaning.
3. Dry Cleaner : What has gone wrong with your suit ?
You : It has been completely spoiled.
4. Dry Cleaner : When did you buy it ?
You : I bought it last month.
5. Dry Cleaner : How much did you spend on it ?
You : I spent one thousand rupees on it.
6. Dry Cleaner : What can I do to make up your loss ?
You : You should pay me seven hundred rupees.
7. Dry Cleaner : Who is responsible for this damage ?
You : You alone are responsible for this damage.
8. Dry Cleaner : Why should I pay you for this damage ?
You : ‘Because you are the owner of this shop.
9. Dry Cleaner : Can you wait for a few days ?
You : No, I want it immediately.
10. Dry Cleaner : What will you do if I do not pay the money ?
You : If you do not pay the money, I will go to the police station.

5. Imagine you have returned from Delhi after visiting your uncle. Your father asks you a few questions in connection with your visit. Now you will play the role of you and your father. Start you conversation as:

1. Father : Why did you not come back yesterday ?
You : Uncle did not allow me to come yesterday.
2. Father : ………………….
You : ………………….
3. Father : ………………….
You : ………………….
4. Father : ………………….
You : ………………….
5. Father : ………………….
You : ………………….
6. Father : ………………….
You : ………………….
7. Father : ………………….
You : ………………….
Answer:
2. Father : Why did he not allow you ?
You : He is not keeping good health.
3. Father : What is wrong with him ?
You : He is having high blood pressure.
4. Father : Did you request him to visit us ?
You : Yes, I requested him to visit us some time.
5. Father : Did he promise to come to Chandigarh ?
You : Yes, he promised to come to Chandigarh in June.
6. Father : Did you visit any important place in Delhi ?
You : Yes, I visited many places such as Qutab Minar, Red Fort, Appu Ghar, etc.
7. Father : Did you enjoy your visit ?
You : Yes, I enjoyed it very much.

6. Suppose you learn that your friend (Arun) has been involved in an accident and is in the hospital. You go to the hospital and meet his father. Now play the role of yourself and his father. The beginning of the conversation is given. You will start with the given conversation.

1. You : Where is Arun, my friend ?
His Father : He is in the operation theatre.
2. You : ………………….
His Father : ………………….
3. You : ………………….
His Father : ………………….
4. You : ………………….
His Father : ………………….
5. You : ………………….
His Father : ………………….
6. You : ………………….
His Father : ………………….
7. You : ………………….
His Father : ………………….
Answer:
You : What has happened to him ?
His Father : He has a fracture in his left arm.
You : How did all this happen ?
His Father : His scooter was hit by a bus.
You : Was he alone ?
His Father : No, his sister was with him.
You : Is the girl safe ?
His Father : Thank God, she did not receive any injury.
You : Where did the accident occur ?
His Father : It occurred near the Rose Garden.
You : Do you need any kind of help from me ?
His Father : No, thank you. You should only pray for his early recovery.

7. Suppose somebody has stolen your hundred rupees. You go to the Principal of your school to complain. You will play the role of student and the Principal. The beginning of the conversation is given. You will start with the given conversation.

1. Principal : What brings you here ?
Student : Sir, I have come with a problem.
2. Principal : ………………….
Student : ………………….
3. Principal : ………………….
Student : ………………….
4. Principal : ………………….
Student : ………………….
5. Principal : ………………….
Yudent : ………………….
6. Principal : ………………….
Student : ………………….
7. Principal : ………………….
Student : ………………….
Answer:
2. Principal : What is your problem ?
Student : Sir, somebody has stolen my hundred rupees.
3. Principal : Where had you kept it ?
Student : I had kept it in my bag in a book, sir.
4. Principal : Where had you put your bag ?
Student : It was lying in the classroom.
5. Principal : Where had the boys of your class gone ?
Student : All the boys had gone to the ground for P.T.
6. Principal : Did you ask anybody about it ?
Student : Yes sir, I asked a number of my class-fellows.
7. Principal : Have you notified it on the notice-board ?
Student : Not yet sir, but I will put it lip on the notice-board.

8. Suppose you visit your cousin and ask him about his hobby. You will play the role of yourself and cousin. The beginning of the conversation is given. You will start with the given conversation.

1. You : What is your hobby ?
Cousin : Gardening is my hobby.
2. You : ………………….
Cousin : ………………….
3. You : ………………….
Cousin : ………………….
4. You : ………………….
Cousin : ………………….
5. You : ………………….
Cousin : ………………….
6. You : ………………….
Cousin : ………………….
7. You : ………………….
Cousin : ………………….
8. You : ………………….
Cousin : ………………….
9. You : ………………….
Cousin : ………………….
10. You : ………………….
Cousin : ………………….
Answer:
2. You : When do you work in your garden ?
Cousin : I work in my garden after the school time.
3. You : Do you apply manure to the plants ?
Cousin : Yes, I apply manure to the plants.
4. You : What have you grown there ?
Cousin : I have grown roses, pansy and petunia flowers there.
3. You : Do you have any fruit trees there ?
Cousin : Yes, I have some fruit trees there.
6. You : When do you cut the grass of the lawn ?
Cousin : I cut the grass of the lawn on Saturday.
7. You : Where do you bring the plants from ?
Cousin : I bring the plants from Green Nursery.
8. You : Do you have any vine there ?
Cousin : Yes, I have a vine of grapes there.
9. You : When do you take part in a Flower Show ?
Cousin : I take part in a Flower Show in the month of March.
10. You : When can I come to see your garden ?
Cousin : You can come to see my garden on any holiday.

9. Suppose you went to Rose Garden for a picnic. Your younger brother who could not go with you asks you about it. Two boys shall play the role of elder brother and younger brother.The beginning of the conversation is given. You will start with the given conversation.

1. Younger Brother : Where did you go for a picnic ?
Elder Brother : We went to the Rose Garden.
2. Younger Brother : ………………….
Elder Brother : ………………….
3. Younger Brother : ………………….
Elder Brother : ………………….
4. Younger Brother : ………………….
Elder Brother : ………………….
5. Younger Brother : …………………..
Elder Brother : ………………….
6. Younger Brother : ………………….
Elder Brother : ………………….
7. Younger Brother : ………………….
Elder Brother : ………………….
8. Younger Brother : ………………….
Elder Brother : ………………….
9. Younger Brother : ………………….
Elder Brother : ………………….
10. Younger Brother : ………………….
Elder Brother : ………………….
Answer:
2. Younger Brother : How many friends were you ?
Elder Brother : We were a party of ten friends.
3. Younger Brother : How did you go there ?
Elder Brother : We went there on bicycles.
4. Younger Brother : What did you take with you ?
Elder Brother : We took with us a camera, a transistor, a pack of cards, a stove, some milk, tea and sugar.
5. Younger Brother : YHow much time did you take to reach Rose Garden ?
Elder Brother : It took us an hour to reach Rose Garden.
6. Younger Brother : Where did you see children playing merrily ?
Elder Brother : We saw children playing merrily in the garden.
7. Younger Brother : How did you feel there ?
Elder Brother : We felt very happy there.
8. Younger Brother : Where did you sit ?
Elder Brother : We sat on a mat under a shady tree.
9. Younger Brother : What did you do there ?
Elder Brother : We played cards, took snaps, listened to songs and took tea there.
10. Younger Brother : Did you enjoy yourselves ?
Elder Brother : Yes, we really enjoyed ourselves.

10. Suppose Ram has just come out of the Examination Hall after finishing his paper. Another examinee named Sham has also come out. You will play the role of Ram and Sham. Now start your conversation.

1. Ram : How have you done your paper ?
Sham : I have done my paper well.
2. Ram : ………………….
Sham : ………………….
3. Ram : ………………….
Sham : ………………….
4. Ram : ………………….
Sham : ………………….
3. Ram : ………………….
Sham : ………………….
6. Ram : ………………….
Sham : ………………….
7. Ram : ………………….
Sham : ………………….
8. Ram : ………………….
Sham : ………………….
9. Ram : ………………….
Sham : ………………….
10. Ram : ………………….
Sham : ………………….
Answer:
2. Ram : Is the paper easy ?
Sham : It is not so easy.
3. Ram : Which question is difficult ? ‘
Sham : Question No. IV is rather difficult.
4. Ram : What do you think about the answer to this question ?
Sham : I think my answer to this question is also to the point.
5. Ram : From which chapter is the question set ?
Sham : The question is set from Chapter II of the book.
6. Ram : How do you find the question on ‘Fill in the blanks’ ?
Sham : The question on ‘Fill in the blanks’ is confusing.
7. Ram : What do you think about these two questions ?
Sham : These two questions are out of the course, prescribed.
8. Ram : How did you answer these questions ?
Sham : I gave rather poor answers to these questions.
9. Ram : How do you feel on the whole ?
Sham : On the whole I am satisfied with the answers I gave.
10. Ram : How many marks do you hope to get in this paper ?
Sham : I hope to get only 40 per cent marks in this paper.

11. Your friend bought a mobile phone. The mobile phone is not working properly. Now play the role of friend and shop owner. Write the conversation in a dialogue format.

1. Shop owner : Good morning, Sir.
Friend : Good’morning Sir.
2. Shop owner : ………………….
Friend : ………………….
3. Shop owner : ………………….
Friend : ………………….
4. Shop owner : ………………….
Friend : ………………….
5. Shop owner : ………………….
Friend : ………………….
Answer:
2. Shop owner : How can I help you ?
Friend : Sir, the mobile phone I bought yesterday from your shop is not working correctly.
3. Shop owner : Oh ! Please explain your problem in detail.
Friend : It is giving a lot of trouble. There is no ring tones on it. The sound/music/ audio quality is not good either.
4. Shop owner-: OK. Let me check it. Oh ! you are right. Tell me, what to do ?
Friend : It is of no use to me.
5. Shop owner : I understand.
Friend : Please take it back and refund my money.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

Solve the following equations.

Question 1.
\(\frac{8 x-3}{3 x}=2\)
Solution:
\(\frac{8 x-3}{3 x}=2\)
∴ 3x\(\left(\frac{8 x-3}{3 x}\right)\) = 3x (2) (Multiplying both the sides by 3x)
∴ 8x – 3 = 6x
∴ 8x – 6x = 3 [Transposing 6x to LHS and (-3) to RHS]
∴ 2x = 3
∴ \(\frac{2 x}{2}=\frac{3}{2}\) (Dividing both the sides by 2)
∴ x = \(\frac {3}{2}\)

Question 2.
\(\frac{9 x}{7-6 x}=15\)
Solution:
\(\frac{9 x}{7-6 x}=15\)
∴ 9x = 15 (7 – 6x) (Cross multiplication)
∴ 9x = 105 – 90x
∴ 9x + 90x = 105 [Transposing (- 90x) to LHS]
∴ 99x = 105
∴ \(\frac{99 x}{99}=\frac{105}{99}\)(Dividing both the sides by 99)
∴ x = \(\frac {35}{33}\)

Question 3.
\(\frac{z}{z+15}=\frac{4}{9}\)
Solution:
\(\frac{z}{z+15}=\frac{4}{9}\)
∴ z(9) = 4(z + 15) (Cross multiplication)
∴ 9z = 4z + 60
∴ 9z – 4z = 60 (Transposing 4z to LHS)
∴ 5z = 60
∴ \(\frac{5 z}{5}=\frac{60}{5}\) (Dividing both the sides by 5)
∴ z = 12

Question 4.
\(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)
Solution:
\(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)
∴ 5(3y + 4) = -2(2 – 6y) (Cross multiplication)
∴ 15y + 20 = -4 + 12y
∴ 15y – 12y = – 4 – 20 (Transposing 12y to LHS and 20 to RHS)
∴ 3y = -24
∴ \(\frac{3 y}{3}=\frac{-24}{3}\) (Dividing both the sides by 3)
∴ y = (-8)

Question 5.
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
Solution:
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
∴ 3(7y + 4) = -4(y + 2) (Cross multiplication)
∴ 21y + 12 = – 4y – 8
∴ 21y + 4y = – 8 – 12 (Transposing -4y to LHS and 12 to RHS)
∴ 25y = -20
∴ \(\frac{25 y}{25}=\frac{-20}{25}\) (Dividing both the sides by 25)
∴ y = \(\frac {-4}{5}\)

Question 6.
The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Solution:
Age of Hari : Age of Harry
= 5 : 7
Let the present age of Hari be 5x years.
Then, the present age of Harry = 7x years.
After 4 years their ages :
Hari = (5x + 4) years
Harry = (7x + 4) years
∴ (5x + 4) : (7x + 4) = 3 : 4
∴ \(\frac{5 x+4}{7 x+4}=\frac{3}{4}\)
∴ 4(5x + 4) = 3(7x + 4) (Cross multiplication)
∴ 20x + 16 = 21x + 12
∴ 20x – 21x = 12 – 16 (Transposing 21x to LHS and 16 to RHS)
∴ -x = – 4
∴ x = 4 [Multiplying both the sides by (- 1)]
∴ Hari’s present age = 5x = 5 × 4
= 20 years
∴ Harry’s present age = 7x = 7 × 4
= 28 years
Thus, Hari’s present age is 20 years and Harry’s present age is 28 years.

Question 7.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \(\frac {3}{2}\). Find the rational number.
Solution:
Let the numerator be x.
Denominator (8 more than numerator) = x + 8
New numerator = x + 17
(After adding 17)
New denominator = x + 8 – 1
= x + 7
(After decreasing 1)
But new number = \(\frac{x+17}{x+7}\)
But this rational number is \(\frac {3}{2}\)
\(\frac{x+17}{x+7}=\frac{3}{2}\)
∴ 2(x + 17) = 3(x + 7) (Cross multiplication)
∴ 2x + 34 = 3x + 21
∴ 2x – 3x = 21 – 34 (Transposing 3x to LHS and 34 to RHS)
∴ -x = – 13
∴ x = 13 [Multiplying both the sides by (-1)]
∴ Numerator = x = 13
Denominator = x + 8
= 13 + 8
= 21
The rational number = \(\frac {13}{21}\)
Thus, the rational number is \(\frac {13}{21}\).

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 13 Symmetry Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2

1. Draw the reflection of following figures along the dotted line :

Solution:

2. Write ‘yes’ for right reflection and ‘no’ for wrong reflection:

Solution:
(a) Yes
(b) Yes
(c) Yes
(d) Yes.

3. Trace the figures on the graph paper and draw the reflections. The dotted line is the line of symmetry:

Solution:

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 13 Symmetry Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1

1. Classify the figure as symmetrical or non-symmetrical. Also draw the line/ lines of symmetry (if any).

Solution:

2. Which Capital letter of English alphabet have:

Question (i)
No line of symmetry.
Solution:
Capital letters of English alphabet having no line of symmetry.
F, G, J, L, N, P, Q, R, S, Z.

Question (ii)
1 line of symmetry.
Solution:
Capital letters of English alphabet having 1 line of symmetry.
A, B, C, D, E, K, M, T, U, V, W, Y.

Question (iii)
2 lines of symmetry.
Solution:
2 lines of symmetry.
O, X, H, I.

3. Find file numbers of line/lines of symmetry for the following:

Solution:

4. Draw the line (s) of symmetry the following figures:

Question (a)
Rhombus
Solution:
Rhombus: A rhombus has two lines of symmetry.

Question (b)
Scalene Triangle
Solution:
Scalene Triangle: A scalene triangle has no line of symmetry.

Question (c)
Parallelogram
Solution:
Parallelogram: A parallelogram has no line of symmetry.

Question (d)
Rectangle
Solution:
Rectangle: A rectangle has two lines of symmetry.

Question (e)
Square
Solution:
Square: A square has four lines of symmetry.

Question (f)
Regular Pentagon.
Solution:
Regular Pentagon. A regular pentagon has five lines of symmetry.

5. Complete each of the figure using both lines of symmetry:

Solution:

6. Draw a triangle which has:

Question (i)
No line of symmetry
Solution:
No line of symmetry: Scalene triangle has no line of symmetry.

Question (ii)
Exactly one line of symmetry
Solution:
Exactly one line of symmetry: Isosceles triangle has exactly one line of symmetry.

Question (iii)
Exactly three lines of symmetry.
Solution:
Exactly three lines of symmetry. Equilateral triangle has exactly three lines of symmetry.

7. List any three symmetrical objects from your day-to-day life.
Solution:
Glass, Lock Pencil are three symmetrical objects.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

Solve the following linear equations.

Question 1.
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
∴ \(\frac {1}{2}\) [Transposing \(\frac{x}{3}\) to LHS and \(\frac {-1}{2}\) to RHS]
∴ \(\frac{3 x-2 x}{6}=\frac{1 \times 5+1 \times 4}{20}\) [LCM = 6, LCM = 20]
∴ \(\frac{x}{6}=\frac{5+4}{20}\)
∴ \(\frac{x}{6}=\frac{9}{20}\)
∴ \(\frac{x}{6} \times 6=\frac{9}{20} \times 6\) (Multiplying both the sides by 6)
∴ \(\frac {27}{10}\)
∴ x = 2.7

Question 2.
\(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}\) = 21
Solution:
\(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}\) = 21
∴ \(\frac{n \times 6}{2 \times 6}-\frac{3 n \times 3}{4 \times 3}+\frac{5 n \times 2}{6 \times 2}\) = 21
∴ \(\frac{6 n-9 n+10 n}{12}\) = 21
∴ \(\frac {7 n}{12}\) = 21
∴ 7n = 21 × 12 (Multiplying both the sides by 12)
∴ n = \(\frac{21 \times 12}{7}\) (Dividing both the sides by 7)
∴ n = 36

Question 3.
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
(Multipljdng both the sides by 6, the LCM of 3, 6 and 2.)
(6 × x) + (6 × 7) – \(\left(\frac{6 \times 8 x}{3}\right)\) = \(\left(6 \times \frac{17}{6}\right)-\left(\frac{6 \times 5 x}{2}\right)\)
∴ 6x + 42 – 16x = 17- 15x
∴ – 10x + 42 = 17 – 15x
∴ – 10x + 15x = 17 – 42 [Transposing (-15x) to LHS and 42 to RHS]
∴ 5x = – 25
∴ x = – 5 (Dividing both the sides by 5)

Question 4.
\(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:
\(\frac{x-5}{3}=\frac{x-3}{5}\)
∴ 5(x – 5) = 3 (x – 3) (Cross multiplication)
∴ 5x – 25 = 3x – 9
∴ 5x – 3x = 25 – 9 [Transposing 3x to LHS and (-25) to RHS]
∴ 2x = 16
∴ \(\frac{2 x}{2}=\frac{16}{2}\) (Dividing both the sides by 2)
∴ x = 8

Question 5.
\(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
Solution:
\(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
(Multiplying both the sides by 12, the LCM of 4 and 3)
12\(\left(\frac{3 t-2}{4}\right)\) – 12\(\left(\frac{2 t+3}{3}\right)\) = 12 × \(\frac {1}{2}\) – 12t
∴ 3(3t – 2) -4 (2t + 3) = 8 – 12t
∴ 9t – 6 – 8t – 12 = 8 – 12t
∴ t – 18 = 8 – 12t
∴ t + 12t = 8 + 18 [Transposing (-12t) to LHS and (-18) to RHS]
∴ 13t = 26
∴ \(\frac{13 t}{13}=\frac{26}{13}\) (Dividing both the sides by 13)
∴ t = 2

Question 6.
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Solution:
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
(Multiplying both the sides by 6, the LCM of 2 and 3)
6m – 6 \(\left(\frac{m-1}{2}\right)\) = 1 × 6 – 6\(\left(\frac{m-2}{3}\right)\)
∴ 6m – 3(m- 1) = 6 – 2(m-2)
∴ 6m – 3m + 3 = 6 – 2m + 4
∴ 3m + 3 = 10 – 2m
∴ 3m + 2m = 10 – 3 [Transposing (-2 m) to LHS and 3 to RHS]
∴ 5m = 7
∴ \(\frac{5 m}{5}=\frac{7}{5}\) (Dividing both the sides by 5)
∴ m = \(\frac {7}{5}\)

Simplify and solve the following linear equations.

Question 7.
3(t – 3) = 5(2t + 1)
Solution:
3(t – 3) = 5(2t + 1)
∴ 3t – 9 = 10t + 5
∴ 3t – 10t = 5 + 9 [Transposing 10t to LHS and (-9) to RHS]
∴ – 7t = 14
∴ 7t = – 14 [Multiplying both the sides by (-1)]
∴ \(\frac{7 t}{7}=\frac{-14}{7}\) (Dividing both the sides by 7)
∴ t = -2

Question 8.
15 (y – 4) – 2 (y – 9) + 5 (y + 6) = 0
Solution:
15 (y – 4) – 2 (y – 9) + 5 (y + 6) = 0
∴ 15y – 60 – 2y + 18 + 5y + 30 = 0
∴ 15y – 2y + 5y – 60 + 18 + 30 = 0
∴ 15y + 5y – 2y + 18 + 30 – 60 = 0 (Arranging the terms)
∴ 18y – 12 = 0
∴ 18y = 12 [Transposing (- 12) to RHS]
∴ \(\frac{18 y}{18}=\frac{12}{18}\) (Dividing both the sides by 18)
∴ y = \(\frac {2}{3}\)

Question 9.
3(5z – 7) -2 (9z – 11) = 4(8z – 13) – 17
Solution:
3(5z – 7) -2 (9z – 11) = 4(8z – 13) – 17
∴ 15z – 21 – 18z + 22 = 32z – 52 – 17
∴ 15z – 18z – 21 + 22 = 32z + (-52 – 17)
∴ – 3z + 1 = 32z – 69
∴ – 3z – 32z = – 69 – 1 (Transposing 1 to RHS and 32z to LHS)
∴ – 35z = – 70
∴ 35z = 70[Multiplying both the sides by (-1)]
∴ \(\frac{35 z}{35}=\frac{70}{35}\) (Dividing both the sides by 35)
∴ z = 2.

Question 10.
0.25 (4f – 3) = 0.05(10f – 9)
Solution:
0.25 (4f – 3) = 0.05(10f – 9)
0.25 × 4f – 0.25 × 3
= 0.05 × 10f – 0.05 × 9

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 12 Perimeter and Area MCQ Questions

Multiple Choice Questions

Question 1.
The outer boundary of a closed figure is called ………….. .
(a) Perimeter
(b) Region
(c) Area
(d) Curve.
Answer:
(a) Perimeter

Question 2.
Find the perimeter of the given figures:

(a) 30 cm
(b) 31 cm
(c) 32 cm
(d) 33 cm.
Answer:
(b) 31 cm

Question 3.
Perimeter of an equilateral triangle = ………………… .
(a) 3 + Side
(b) Side × Side
(c) Side + Side
(d) 3 × Side.
Answer:
(d) 3 × Side.

Question 4.
Perimeter of Rectangle = …………………… .
(a) 2l + b
(b) 2 (l + b)
(c) l + 2b
(d) l × b.
Answer:
(b) 2 (l + b)

Question 5.
If side of an equilateral triangle is 4 cm then perimeter = ……………….. .
(a) 8 cm
(b) 7 cm
(c) 12 cm
(d) 16 cm.
Answer:
(c) 12 cm

Question 6.
If length and breath of a rectangle are 2.4 cm and 1.9 cm then its perimeter is ………………. .
(a) 4.3 cm
(b) 8.2 cm
(c) 4.2 cm
(d) 8.6 cm.
Answer:
(d) 8.6 cm.

Question 7.
The perimeter of square is 16 cm then its side is ……………………… .
(a) 4 cm
(b) 64 cm
(c) 24 cm
(d) 32 cm.
Answer:
(a) 4 cm

Question 8.
The perimeter of a rectangle is 50 cm and its length is 12 cm then breadth is ……………………….. .
(a) 38 cm
(b) 13 cm
(c) 62cm
(d) 18cm.
Answer:
(b) 13 cm

Question 9.
Two sides of a triangle are 4.8 cm and 3.9 cm. The perimeter of the triangle is 12 cm. Find the third side.
(a) 3.3 cm
(b) 4.3 cm
(c) 20.7 cm
(d) 3.7 cm.
Answer:
(b) 4.3 cm

Question 10.
Samandeep takes 3 rounds of square park side 125 m. Find the distance covered by her.
(a) 1.5 km
(b) 1500 km
(c) 500 m
(d) 375 m.
Answer:
(a) 1.5 km

Question 11.
The measurement of the region enclosed by a closed plane figure is called its ………….. .
(a) Circumference
(b) Curve
(c) Perimeter
(d) Area.
Answer:
(d) Area

Question 12.
If the length of a rectangle is x units and breadth is 5 units then its perimeter is ………….. .
(a) 5x
(b) 2 (x + 5)
(c) 10x
(d) 10 + x
Answer:
(b) 2 (x + 5)

Question 13.
Find the area of the given rectangle whose length is 16 m and breadth is 8 m.
(a) 42 sq.m
(b) 128 sq. m
(c) 72 sq. m
(d) 21 sq. m.
Answer:
(b) 128 sq. m

Question 14.
The area of a rectangle is 144 m . If its breadth is 9 m then find its length.
(a) 16 sq. m
(b) 12 m
(c) 16 m
(d) 18 m.
Answer:
(c) 16 m

Question 15.
1 sq. m = ……………. sq. cm.
(a) 100
(b) 10000
(c) 1000
(d) 1.
Answer:
(b) 10000

Question 16.
Find the area of a square having side 3.6 cm.
(a) 14.4 cm
(b) 12.96 cm
(c) 1.29 sq. cm
(d) 12.96 sq. cm.
Answer:
(d) 12.96 sq. cm.

Question 17.
The perimeter of a square is 68 m. Find its area.
(a) 289 sq.m
(b) 329 sq. m
(c) 279 sq. m
(d) 249 sq.m.
Answer:
(a) 289 sq.m

Question 18.
A marble tile is of side 25 cm by 25 cm. How many tiles will be required to cover a floor of 4 m by 3 m?
(a) 216
(b) 192
(c) 188
(d) 196.
Answer:
(b) 192

Question 19.
What will happen to the area of a square, if side is doubled?
(a) Double
(b) Half
(c) Four times
(d) Nochange.
Answer:
(c) Four times

Question 20.
Find the perimeter of a rectangle whose area is 234 sq. cm and its one side is 13 cm.
(a) 31 cm
(b) 62 cm
(c) 18 cm
(d) 24 cm.
Answer:
(b) 62 cm

Question 21.
How many cm² are in 1 m²?
(a) 1000
(b) 100
(c) 10000
(d) 10.
Answer:
(c) 10000

Question 22.
The distance covered along the boundary forming a closed figure when you go round the figure once is called its:
(a) Length
(b) Perimeter
(c) Breadth
(d) Area.
Answer:
(b) Perimeter

Question 23.
The amount of surface enclosed by a closed figure is called its:
(a) Perimeter
(b) Volume
(c) Area
(d) None of these.
Answer:
(c) Area

Question 24.
The formula to find perimeter of a rectangle is:
(a) Length + Breadth
(b) 2 (Length + Breadth)
(c) Length – Breadth
(d) Length × Breadth.
Answer:
(b) 2 (Length + Breadth)

Question 25.
The formula to find perimeter of a square is :
(a) 4 × side
(b) 3 × side
(c) 2 × side
(d) 5 × side.
Answer:
(a) 4 × side

Fill in the blanks:

Question (i)
The formula to find area of equilateral triangle is …………….. .
Answer:
3 × side

Question (ii)
The formula to find area of a rectangle is ……………… .
Answer:
Length × Breadth

Question (iii)
The formula to find area of a square is ………………. .
Answer:
Perimeter

Question (iv)
The sum of lengths of all sides of a polygon is called ………………… .
Answer:
Area

Write True/False:

Question (i)
Perimeter of square = 4 × side. (True/False)
Answer:
True

Question (ii)
1 sq. m = 1000 sq. cm. (True/False)
Answer:
False

Question (iii)
The outer boundary of a closed figure is called area. (True/False)
Answer:
False

Question (iv)
Perimeter of a triangle = 3 × side. (True/False)
Answer:
True

Question (v)
Area of rectangle = Length × Breadth. (True/False)
Answer:
True

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

Question 1.
Amina thinks of a number and subtracts \(\frac {5}{2}\) from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let the number be x.
Subtracting \(\frac {5}{2}\) from it, we get, x – \(\frac {5}{2}\)
By multiplying this result by 8,
we get 8 (x – \(\frac {5}{2}\))
According to the condition,
8(x – \(\frac {5}{2}\)) = 3x
∴ 8x – 20 = 3x
∴ 8x = 3x + 20 [Transposing (-20) to RHS]
∴ 8x – 3x = 20 (Transposing 3x to LHS)
∴ 5x = 20
∴ \(\frac{5 x}{5}=\frac{20}{5}\) (Dividing both the sides by 5)
∴ x = 4
Thus, the number though of by Amina is 4.

Question 2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let the smaller number be x
∴ The greater number is 5x
On adding 21 to both the numbers,
we get (5x + 21) and (x + 21)
According to the condition,
5x + 21 = 2 (x + 21)
∴ 5x + 21 = 2x + 42
∴ 5x = 2x + 42 – 21 (Transposing 21 to RHS)
∴ 5x = 2x + 21
∴ 5x – 2x = 21 (Transposing 2x to LHS)
∴ 3x = 21
∴ \(\frac{3 x}{3}=\frac{21}{3}\) (Dividing both the sides by 3)
∴ x = 7
The smaller number x = 7
The greater number = 5x = 5 × 7 = 35
Thus, the required numbers are 7 and 35.

Question 3.
Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Solution:
Let the digit at the units place be x.
Then, the digit at the tens place = (9 – x)
The original number = 10(9 – x) + x
= 90 – 10x + x
= 90 – 9x
On interchanging the digits, the new number =10x + (9 – x)
= 10x + 9 – x
= 9x + 9
According to the condition,
New number = Original number + 27
∴ 9x + 9 = (90 – 9x) + 27
∴ 9x + 9 = 90 – 9x + 27
∴ 9x + 9 = 117 – 9x
∴ 9x = 117 – 9x – 9 (Transposing 9 to RHS)
∴ 9x = 108 – 9x
∴ 9x + 9x = 108 [Transposing (-9x) to LHS]
∴ 18x = 108
∴ \(\frac{18 x}{18}=\frac{108}{18}\) (Dividing both the sides by 18)
∴ x = 6
∴ Original number = 90 – 9x
= 90 – 9(6)
= 90 – 54 = 36
Thus, the original number is 36.

Question 4.
One of the two digits of a two-digit number is three times the other digit. I If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Solution:
Let the digit at units place be x and the digit at tens place be 3x.
The number = 10 (3x) + x
= 30x + x
= 31x
On interchanging the digits, the number – 10x + 3x = 13x
According to the condition,
31x + 13x = 88
44x = 88
∴ \(\frac{44 x}{44}=\frac{88}{44}\) (Dividing both the sides by 44)
∴x = 2
The number = 31x = 31 × 2 = 62
Thus, original number is either 62 or 26.

Question 5.
Saroj’s mother’s present age is six times ; Saroj’s present age. Saroj’s age five years from now will be one third of his mother’s present age. What are their present ages?
Solution:
Let Saroj’s present age be x years and mother’s present age be 6x years
After 5 years:
Saroj’s age will be x + 5 years
Mother’s age will be 6x + 5 years
According to the condition,
\(\frac {1}{3}\) (Mother’s present age) = Saroj’s age after 5 years
∴ \(\frac {1}{3}\)(6x) = x + 5
∴ 2x = x + 5
∴ 2x – x = 5 (Transposing x to LHS)
∴ x = 5
Saroj’s present age = x = 5 years
Mother’s present age = 6x = 6 × 5
= 30 years
Thus, present ages of Saroj and her mother are 5 years and 30 years respectively.

Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11 : 4. At the rate ₹ 100 per metre it will cost the village panchayat ₹ 75,000 to fence the plot. What are the dimensions of the plot ?
Solution :
Length : Breadth = 11 : 4
Let the length be 11x metres.
Then, the breadth = 4x metres.
Perimetre = 2 (length + breadth)
= 2 (11x + 4x)
= 2 (15x) = 30x
Cost of fencing = ₹ 100 × 30x
= ₹3000 x
But, the cost of fencing = ₹ 75,000 (Given)
∴ 3000 x = 75000
∴ \(\frac{3000 x}{3000}=\frac{75000}{3000}\) (Dividing both the sides by 3000)
∴ x = 25
Length = 11x
= 11 × 25
= 275 metres
Breadth = 4x
= 4 × 25
= 100 metres
Thus, the length and breadth of the rectangular plot are 275 metres and 100 metres respectively.

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre.
For every 3 metres of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10 % profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy?
Solution:
Let the length of cloth for trousers be 2x metres
Then, the length of cloth for shirts = 3x metres
Cost of trouser’s cloth = 2x × ₹ 90
= ₹ 180x
Cost of shirt’s cloth = 3x × ₹ 50
= ₹ 150x
10 % profit is made on trouser’s cloth.
If C.E of trouser’s cloth is ₹ 100, then S.E is ₹ 110.
S.E of trouser’s cloth at 10 % profit = ₹\(\frac {110}{100}\) × 180x
= ₹ 198x
12 % profit is made on shirt’s cloth. If C.P of shirt’s cloth is ₹ 100, then S.P is ₹ 112.
S.P of shirt’s cloth at 12 % profit = ₹\(\frac {112}{100}\) × 150x
= ₹ 168x
∴ Total S.P = ₹ 198x + ₹ 168x
= ₹ 366x
But the total S.P. = ₹ 36,600 (Given)
366x = 36600
∴ \(\frac {112}{100}\) (Dividing both the sides by 366)
∴ x = 100
Cloth for trousers = 2x = 2 × 100 = 200
Thus, he bought 200 metres of cloth for trousers.

Question 8.
Half of a herd of deer are grazing | in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Solution:
Let the total number of deer be x.
Number of deer grazing in the field = \(\frac{x}{2}\)
Remaining deer = x – \(\frac{x}{2}\) = \(\frac{x}{2}\)
Deer playing near by = \(\frac {3}{2}\) × (Remaining no. of deer)
= \(\frac {3}{4}\) × \(\frac{x}{2}\)
= \(\frac{3 x}{8}\)
Number of deer drinking water = 9
∴ Total number of deer = \(\frac{x}{2}\) + \(\frac{3 x}{8}\) + 9
∴ \(\frac{x}{2}\) + \(\frac{3 x}{8}\) + 9 = x
\(\frac{x}{2}\) + \(\frac{3 x}{8}\) = x – 9 (Transposing 9 to RHS)
\(\frac{x}{2}\) + \(\frac{3 x}{8}\) -x = -9 (Transposing x to LHS)
\(\frac{4 x+3 x-8 x}{8}\) = -9 (LCM = 8)
∴ \(\frac{-x}{8}\) = -9
∴ \(\frac{-x}{8}\) × 8 = -9 × 8 (Multiplying both the sides by 8)
∴ -x = – 72
∴ x = 72 [∵ Multiplying both sides by (- 1)]
Thus, total number of deer in herd is 72.

Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution:
Let the present age of granddaughter be x years.
Grandfather’s age is 10x years.
Fresent age of daughter + 54 = Grandfather’s age
∴ x + 54 = 10x
∴ 10x = x + 54 (Interchanging the sides)
∴ 10x – x = 54 (Transposing x to LHS)
∴ 9x = 54
∴ \(\frac{9 x}{9}=\frac{54}{9}\) (Dividing both the sides by 9)
∴ x = 6
Granddaughter’s age = x = 6 years
Grandfather’s age = 10x = 10 × 6 = 60 years
Thus, granddaughter’s age is 6 years and grandfather’s age is 60 years.

Question 10.
Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution :
Let the present age of son be x years
Then, the present age of Aman = 3x years
Ten years ago their ages:
Son’s age was (x – 10) years
Aman’s age was (3x – 10) years
5 × (Son’s age 10 years ago) = Aman’s age 10 years ago
∴ 5 (x – 10) = (3x – 10)
∴ 5x – 50 = 3x – 10
∴ 5x = 3x – 10 + 50 [Transposing (-50) to RHS]
∴ 5x = 3x + 40
∴ 5x – 3x = 40 (Transposing 3x to LHS)
∴ 2x = 40
∴ \(\frac{2 x}{2}=\frac{40}{2}\)(Dividing both the sides by 2)
∴ x = 20
Son’s present age = x = 20 years
Aman’s present age = 3x = 3 × 20
= 60 years
Thus, present ages of Aman and his son are 60 years and 20 years respectively.

Punjab State Board PSEB 8th Class English Book Solutions English Voice Messages Writing Exercise Questions and Answers, Notes.

PSEB 8th Class English Voice Messages Writing

1.

Main Point of Voicemail Message

Message

Call from Ramesh Nagar Need two wheeler Not big budget Mobile No …………..

Manju : Hello, this is Manju. Thanks for calling. I’m not at home so please leave a message and I’ll call you back. Preeti : Hello, Manju this is Preeti calling from Ramesh Nagar. Your neighbourer Mr. Mbhan gave me your number. He said you are interested in selling your two wheeler. I was in need of second hand two wheeler. I can’t afford a big budget. Your two wheeler will serve my purpose. I am free on Sunday. Could I come to have a look of .the two wheeler. Thanks a lot. I look forward to hear from you. My Mobile No. is 98102-70 …….

2.

Main Point of Voicemail Message

Message

Business man or service man Location Number of rooms Minimum rent

Gurvinder : Hello, this is Gurvinder. Thanks for calling. I’m not at home so please leave a message and I’ll call you back. Ravinder : Hi, Gurvinder this is Ravinder calling from your society. You uncle Mr. Ramesh Singh gave me your number. He told that you wish to let your house on rent. I have some interested parties with me. They are either business man or service man. Call me back when free. My number is 99012……. Would you please tell me about number of rooms, location of the house, distance of the nearest market, interstate bus stand and railway station. Is there a public park near it ? What would be the minimum rent ? Thanks a lot. I look forward to hear from you.

3.

Main Point of Voicemail Message

Message

Calling from Hi-tech Washing powder, Washing soap etc. Brochure on table Live demo

Rohan : Hello, this is Rohan. Thanks for calling. I’m not at home so please leave a message and I’ll call you back. Rahul : Hi, Rohan this is Rahul calling from Hi-Tech company. Your friend Alka gave me your number. She said you could help me. I wanted to give you some information about our new products like washing powder, washing soap, hair oil etc. I have left the brochure showing the products and their prices on your table. We can give you live demo at your office when you come back. So please call me back. My Mobile No. is 70127-50 Thank you very much. I look forward to hear from you.

4.

Main Point of Voicemail Message

Message

Calling from Dubey Builders Two rooms, three rooms or one room set Fully or semi furnished Visiting card

Jagdish : Hello, this is Jagdish. Thanks for calling. I’m not at home so please leave a message and I’ll call you back. Sunil Dubey : Hi, Jagdish this is Sunil Dubey calling from Dubey Builders Ludhiana. Your nephew Rohan gave me your number. He told that you need a house on rent. We have developed a new colony with two rooms set, three rooms set or single room set houses. They are fully furnished or semi furnished houses. Some are still vacant you can hire one as per your needs. I have left my visiting card with, the receptionist. Please call me when you are back. Thank you. Looking forward to hear from you.

5.

Main Point of Voicemail Message

Message

Deals in two wheelers sale/purchase All companies products Activa in good condition Going on reasonable price Mobile No. 78402-70 ………

Naveen : Hello, this is Naveen. Thanks for calling. I’m not at home so please leave a message and I’ll call you back. Rakesh : Hi, Naveen this is Rakesh calling from Ludhi colony. Your cousin amrit gave me your number. He said you want to purchase a second hand two wheeler. I deal is second hand two wheelers. I have products of almost all companies in my stock, Activa, Splender, Gusto, Hero Glamer, Bajaj Pulser and so on. Could you call me when you are back. I would like to give you the (demo) demonstration of the Activa which is in very good condition and going on reasonable price. My Mobile No. is 78402-70 ……….. . Thanks a lot. I look forward to hear from you.

6.

Main Point of Voicemail Message

Message

Geetanajli from Smith & Smith Decorator Deals in decoration of houses, shops and offices Decoration as per the customers demand.

Sita : Hello, this is Sita. Thanks for calling. I’m not at home so please leave a message and I’ll call you back. Geetanjali : Hi, Sita this is Geetanjali calling from Smith and Smith Decorators. Your cousin Parul gave me your number. She said you want to renovate your office. We deal in the decoration of houses, shops and offices. We can beautifully decorate your sales office from inside and can give it a look of commercial centre from outside to attract the customers attention. Some changes can be incorporated in it to make it easy approachable. Every thing can be discussed in details. Just give me a missed call. My Mobile No. is 88112-30 ………. . Thanks a lot. I look forward to hear from you.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 12 Perimeter and Area Ex 12.2

1. Find the approximate area of each of the following figures by countaing the number of squares – complete, more than half and exactly half.

Solution:
(i) Number of complete squares m = 7
Here we do not have any half square or more than half.
∴ n = 0, p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 7 + 0 + 0
= 7 sq. units

(ii) Number of complete squares m = 2
Number of more than half squares n = 4
Number of half squares p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 4 + 2 + 0
= 6 sq. units.

(iii) Number of complete squares m = 10
Number of more than half squares n = 0
Number of exactly half squares p = 2
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= (10 + 0 + \(\frac {1}{2}\) × 2) sq. units
= (10 + 1) sq. units
= 11 sq. units

(iv) Number of complete squares m = 11
Number of more than half squares n = 0
Number of exactly half squares p = 2
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 11 + 0 + \(\frac {1}{2}\) × 2
= 11 + 1 = 12 sq. units

(v) Number of complete squares m = 10
Number of more than half squares n = 3
Number of exactly half squares p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 10 + 3 + 0
= 13 sq. units

2. Find the area of rectangle whose:

Question (i)
length = 12 cm, breadth = 16 cm
Solution:
Length of rectangle = 12 cm
and Breadth of rectangle = 16 cm.
∴ Area of rectangle = Length × Breadth
= 12 cm × 16 cm
= 192 sq. cm

Question (ii)
length = 25 m, breadth = 18 m
Solution:
Length of rectangle = 25 m
and Breadth of rectangle = 18 m
∴ Area of rectangle = Length × Breadth
= 25 m × 18 m
= 450 sq. m

Question (iii)
length = 2.7 m, breadth = 45 cm
Solution:
Length of rectangle = 2.7 m = 270 cm
and Breadth of rectangle = 45 cm
∴ Area of rectangle = Length × Breadth
= 270 cm × 45 cm
= 12150 sq. cm

Question (iv)
length 4.2 cm, breadth = 1.5 cm
Solution:
Length of rectangle = 4.2 cm
and Breadth of rectangle = 1.5 cm
∴ Area of rectangle = Length × Breadth
= 4.2 cm × 1.5 cm
= 6.3 sq. cm

Question (v)
length = 3.8 mm, breadth = 4 mm.
Solution:
Length of rectangle = 3.8 mm
and Breadth of rectangle = 4 mm
∴ Area of rectangle = Length × Breadth
= 3.8 mm × 4 mm
= 15.2 sq. mm

3. Find the area of the square with side:

Question (i)
19 cm
Solution:
Side of the square = 19 cm
∴ Area of the square = side × side
= 19 cm × 19 cm = 361 sq. cm.

Question (ii)
24 mm
Solution:
Side of square = 24 mm
∴ Area of square = side × side
= 24 mm × 24 mm
= 576 sq. mm

Question (iii)
3.5 cm
Solution:
Side of the square = 3.5 cm
Area of square = side × side
= 3.5 cm × 3.5 cm
= 12.25 sq. cm

Question (iv)
2.6 cm
Solution:
Side of the square = 2.6 cm
Area of square = side × side
= 2.6 cm × 2.6 cm
= 6.76 sq. cm

Question (v)
8.2 cm.
Solution:
Side of the square = 8.2 cm
Area of the square = side × side
= 8.2 cm × 8.2 cm
= 67.24 sq. cm.

4. The area of a rectangle is 216 sq. cm and its length is 12 cm. Find its breadth.
Solution:
The area of the rectangle = 216 sq. cm.
Length of the rectangle = 12 cm
Breadth of the rectangle = \(\frac{\text { Area }}{\text { Length }}\)
= \(\frac {216}{12}\) cm
= 18 cm.

5. The area of a rectangle is 225 sq. m and its breadth is 9 in. Find its length.
Solution:
The area of the rectangle = 225 sq. m
and Breadth of the rectangle = 9 m
∴ Length of the rectangle = \(\frac{\text { Area }}{\text { Breadth }}\)
= \(\frac {225}{9}\) m
= 25 m

6. The length and breadth of a ground are 32 m and 24 m. Find the cost of levelling the ground at the rate of ₹ 3 per sq. m.
Solution:
Length of ground = 32 m
and Breadth of ground = 24 m
Area of the ground = Length × Breadth
= 32 m × 24 m
= 768 sq. m.
Levelling cost of 1 sq. m = ₹ 3
Levelling cost of 768 sq. m = ₹ 3 × 768
= ₹ 2304

7. Find the perimeter of a rectangle whose area is 324 sq. cm and its one side is 36 cm.
Solution:
Area of rectangle = 324 sq. cm
One side of rectangle = 36 cm
∴ Other side of rectangle = \(\frac{\text { Area }}{\text { One side }}\)
= \(\frac {324}{36}\) = 9 cm
Perimeter of rectangle
= 2 × (1st side + 2nd side)
= 2 × (36 + 9) cm
= 2 × 45 cm
= 90 cm

8. The perimeter of a square field is 100 m. Find its area.
Solution:
The perimeter of square field = 100 m
∴ 4 × side of square = 100 m
∴ Side of a square = \(\frac {100}{4}\) = 25 m
Hence area of square field = side × side
= 25 m × 25 m
= 625 sq. m.

9. Area of a rectangle of length 20 cm is 340 sq. cm. Find its perimeter.
Solution:
Area of rectangle = 340 sq. cm
and Length of rectangle = 20 cm
Breadth of rectangle = \(\frac{\text { Area }}{\text { Length }}\)
= \(\frac {340}{20}\) cm = 17 cm
Perimeter of rectangle
= 2 × (Length + Breadth)
= 2 × (20 + 17) cm
= 2 × 37 cm
= 74 cm

10. A marble tile measure 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6 m?
Solution:
Area of the wall = 4 m × 6 m
= 400 cm × 600 cm
= 240000 sq. cm
Area of a marble tile = 15 cm × 20 cm
= 300 sq. cm
Number of tiles required to cover wall
= \(\frac{\text { Area of wall }}{\text { Area of tile }}=\frac{240000}{300}\) = 800
Hence number of tiles required to cover wall = 800

11. Find the cost of levelling the square field of side 75 m at rate of ₹ 5 per square metre.
Solution:
Side of the square field = 75 m
Area of square field = side × side
= 75 m × 75 m
= 5625 sq. m
Cost of levelling 1 sq. m = ₹ 5
Cost of levelling 5625 sq. m = ₹ 5 × 5625
= ₹ 28125

12. How many stamps of size 2 cm × 1.5 cm can be pasted on a sheet of paper of size 6 cm × 12 cm?
Solution:
Area of the sheet of paper
= 6 cm × 12 cm = 72 sq. cm
Area of one stamp = 2 cm × 1.5 cm = 3 sq cm
Number of stamps pasted on sheet of paper
= \(=\frac{\text { Area of paper sheet }}{\text { Area of stamp }}=\frac{72}{3}\) = 24.

13.

Question (i)
What will happen to the area of a square if its side is trebled (tripled)?
Solution:
Let side of the square = x cm
∴ Area of the square = (x × x) sq. m
Now, if the side is trebled, then side of new square = 3x cm
∴ Area of the new square
= [(3x) × (3x)] sq. m
= (3 × 3 × x × x) sq. m
= 9 (x × x) sq. cm
= 9 × (Area of original square)
∴ If side is trebled, then area becomes 9 times of original area.

Question (ii)
What will happen to the area of a rectangle if its length is halved and breadth is doubled?
Solution:
l cm and b cm be the length and breadth of the rectangle respectively.
∴ Area of rectangle = l × b
Now, If its length is halved and breadth is doubled.
∴ New length = \(\frac {1}{2}\) l
and new breadth = 2b
Thus area of new rectangle = length × breadth
= \(\frac {1}{2}\) × l × 2b
= \(\frac {1}{2}\) × 2 × (l × b)
= (l × b) = original area
Area will remain the same.

Question (iii)
What will happen to the area of a square if its side is halved?
Solution:
Let side of the square = x cm
∴ Area of the square = (x × x) sq. cm
Now, if side is halved, then
side of new square = \(\frac {1}{2}\)x cm
∴ Area of the new square sq.cm


Hence, if side is halved, then the area becomes one-fourth times original area.

14. Find the area of the following figures by splitting it into rectangles and squares:

Question (i)

Solution:
The given figure can be divided into 2 parts

Rectangle A of size 8 cm × 3 cm
Rectangle B of size 4 cm × 2 cm
∴ Area of rectangle A
= 8 cm × 3 cm = 24 sq. cm
and Area of rectangle B
= 4 cm × 2 cm
= 8 sq. cm
⇒ Total area of the figure
= 24 sq. cm + 8 sq. cm
= 32 sq. cm

Question (ii)

Solution:
The given figure can be divided into 2 parts

Rectangle A of size 10 cm × 2 cm
Rectangle B of size 8 cm × 2 cm
∴ Area of rectangle A
= 10 cm × 2 cm = 20 sq. cm
and Area of rectangle B
= 8 cm × 2 cm = 16 sq. cm
Total Area of the figure
= (20 + 16) sq. cm = 36 sq. cm

Question (iii)

Solution:
The given figure can be divided into 3 parts

Rectangle A of size 12 cm × 3 cm
Rectangle B of size 3 cm × 2 cm
Rectangle of size 12 cm × 3 cm
Area of rectangle A
= 12 cm × 3 cm = 36 sq. Cm
Area of rectangle B
= 3 cm × 2 cm = 6 sq. cm
Area of rectangle C
= 12 cm × 3 cm = 36 sq. cm
Total area of the figures
= (36 + 6 + 36) sq. cm
= 78 sq. cm

Question (iv)

Solution:
The given figure can be divided into 3 parts

Rectangle A of size 13 × 3 units
Rectangle B of size 5 × 3 units
Rectangle C of size 5 × 3
Area of rectangle A
= 13 × 3 = 39 sq. units
Area of rectangle B
= 5 × 3 = 15 sq. units
Area of rectangle C
= 5 × 3 = 15 sq. units
Hence total area of the figure
= (39 + 15 + 15) sq. units
= 69 sq. units Ans.

15. Fill in the blanks:

Question (i)
1 square metre = ……….. sq. cm.
Solution:
10000

Question (ii)
1 square cm = ………. sq. mm.
Solution:
100

Question (iii)
Area of Rectangle = …………. × ……………
Solution:
length, breadth

Question (iv)
Length = ………….. ÷ breadth.
Solution:
Area

Question (v)
Area of square = …………. × ……………
Solution:
side × side.