Bhagya

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

PSEB 6th Class English Creative Writing

Write atleast ten sentences about the given pictures:

This is a small village. lt is neat and clean. We can see some huts here and there. These are made of mud and straw. The roofs of the huts are slanting. A small number of people live in this village. Women wear simple clothes like sari. The village is pollution free. A woman is there with a broom in her hand. They collect the garbage and put it in a big basket. A child is sitting under a tree and watching all this. A small girl going somewhere with her mother.

2.

This is a public park. There are trees and plants all around. It is surrounded by a wall. In one part of the park, there is a funny house. This open house is a kind of game for the children. Many children have come to enjoy this game. They go up into it from one side. They come out of it from the other side. They come out slipping on a slide. There is a woman to take care of the playing children. She helps the children to play.

3.

This is a zoo. This is a small world of wild and animals. It is like a mini forest. There is a big gate to enter it. Many people visit it daily. They have to buy tickets. Here, they see lions, tigers, monkeys etc. Lions and tigers are kept in cages. Monkeys move freely from tree to tree. Children seeing the playful monkeys. We can also see some elephants. A visit to a zoo is a happy experience

4.

This is a railway platform. A long train has just arrived. It is standing on the platform. It is standing on the platform. Some passengers are going to board it. Two children are sitting under a tree. There are plates of white marble on the floor. Three people are cleaning them. So the floor is shining. At some distance big buildings can be seen. A tree over looks the train. Thus the scene is very beautiful. Train will leave the station any time. The platform will look deserted.

5.

This is a bus stands A couple is waiting for their bus. They are sitting on a long bench. They have their luggage with them. A man is reading a newspaper. Another man is sitting idly. A bus is standing at its stopage. It is ready to leave. A man and his son are going to board it. Two other persons are also sitting there. We can see their backs. Thus the bits stand presents a lovely look.

6.

This is a large park. Trees are tall and green. There is a small pond. Some ducks are swimming in it. A boy and a girl have come for a picnic. They are watching the ducks. The girl is sitting but the boy is standing. They have a mat before them. Some eatables are lying on the mat. A little bird is sitting near the mat. Green grass and blooming flowers look very charming.

7.

This is a village scene. Some huts can be seen at a distance. Two persons are going to their village. They are going on foot. A bullock cart is going on the road. It is loaded with straw. A man is sitting on the heap (%r) of straw. There is a big tree near the road. It has a platform around it. Some children are playing under the tree. They are playing football.

8.

Here is a small girl. She is wearing a lovely/funny dress. She loves birds and animals. She is surrounded by them. A duck is looking at her. She has a parrot like bird in one hand. A pigeon is sitting on her shoulder. A stag is standing near her. It has long curly horns. We can also see a hare and a squirrel in the picture.

9.

This is a sweet home. It is not very big. lt has a bed and some chairs. The window is covered with a curtain. Some clothes are hanging at the back of the bed. On the bed, a boy and a girl are sitting with an old man. The old man seems to be their grandfather. He loves his grand children very much. He has one arm round the boy’s neck. In his other hand, he is holding a book. He is telling something to his grand children.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

1. There are two cuboidal boxes as shown in the figures. Which box requires the lesser amount of material to make?

Solution:
For 1st cuboidal box:
length (l) = 60 cm, breadth (b) = 40 cm and height (h) = 50 cm
Total surface area of a cuboid
= 2 (lb + bh + lh)
= 2 [(60 × 40) + (40 × 50) + (50 × 60)]
= 2 (2400 + 2000 + 3000)
= 2(7400)
= 14,800 cm²

For 2nd cuboidal box:
l = b = h = 50 cm
Total surface area of a cuboid
= 2 (lb + bh + lh)
= 2 [(50 × 50) + (50 × 50) + (50 × 50)]
= 2(2500 + 2500 + 2500)
= 2 × 7500
= 15,000 cm²
As, total surface area of cuboid (a) is less, so the box (a) requires the less amount of material to make.

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Let us find total surface area of a suitcase (cuboidal in shape).
length (l) = 80 cm, breadth (b) = 48 cm and height (h) = 24 cm
∴ Total surface area of a suitcase = 2 (lb + bh + lh)
= 2 [(80 × 48) + (48 × 24) + (24 × 80)]
= 2(3840 + 1152 + 1920)
= 2 (6912)
= 13,824 cm²
Total surface area of such 100 suitcases = 13,824 × 100
= 13,82,400 cm²
Area of 1 metre trapezium
= length × breadth
= 100 × 96 = 9600 cm²
Tarpaulin required to cover 100 suitcases = \(\frac{1382400}{9600}\) = 144 metres
Hence, 144 m tarpaulin is required to cover 100 suitcases.

3. Find the side of a cube whose surface area is 600 cm².
Solution:
Let the side of the cube be x cm
Total surface area of a cube = 6x²
Total surface area of the cube = 600 cm² (Given)
∴ 6x² = 600
∴ x² = \(\frac {600}{6}\)
∴ x² = 100
∴ x² = 10²
∴ x = 10
Hence, the side of the cube is 10 cm.

4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.

Solution:
For given cabinet:
length (l) = 2 m, breadth (b) = 1 m and height (h) = 1.5 m
Total surface area of the cabinet
= 2 (lb + bh + lh)
= 2 [(2 × 1) + (1 × 1.5) + (2 × 1.5)]
= 2 (2 + 1.5 + 3)
= 2(6.5) = 13 m²
She has not painted bottom. So subtract area of bottom from total surface area of the cabinet.
Area of bottom = l × b = 2 × 1 = 2m²
∴ Painted surface area = (13 – 2) m² = 11 m²
Hence, she has painted 11 m² area of a cabinet.

5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m² of area is painted. How many cans of paint will she need to paint the room?
Solution:
For given wall:
length (l) = 15 m, breadth (b) = 10 m and height (h) = 7 m
∴ Area to be painted
= Area of 4 walls + Area of ceiling
= [2 (l + b) × h] + l × b
= [2(15 + 10) × 7] + 15 × 10
= [2(25) × 7] + 150
= 350 + 150 = 500 m²
Now, one can of paint covers 100 m² area.
∴ Number of paint cans needed
= \(\frac{\text { Area to be painted }}{\text { Area painted by one can }}=\frac{500 \mathrm{~m}^{2}}{100 \mathrm{~m}^{2}}\)
Hence, 5 cans of paint will be needed.

6. Describe how the two figures given are alike and how they are different. Which box has larger lateral surface area?

Solution:
Here, figure (i) is cylinder and figure (ii) is cube.
Similarity: Both have the same height.
Difference: One is the cylinder and the other is a cube.
For cylinder:
Radius = \(\frac{\text { diameter }}{2}=\frac{7}{2}\) = cm
Height = 7 cm
∴ Lateral (curved) surface area of cylinder
= 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 7
= 22 × 7 = 154 cm²

For given cube:
side = 7 cm
Lateral surface area of the cube
= 4l² = 4 × 7²
= 4 × 49 = 196 cm²
Hence, cubical box has a larger area.

7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Solution:
For given cylindrical tank:
radius (r) = 7 m, height (h) = 3 m
Total surface area of tank
= 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 7(7 + 3)
= 44(10) = 440 m²
Hence, 440 m² sheet of metal is required.

8. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Solution:
Lateral surface area of a hollow cylinder = 4224 cm² (Given)
Let length of a rectangular sheet made from hollow cylinder be l cm.
∴ l × b = 4224
∴ 1 × 33 = 4224
∴ l = \(\frac {4224}{33}\) = 128 cm
∴ Length of rectangular sheet =128 cm

Perimeter of rectangular sheet
= 2 (l + b)
= 2 (128 + 33)
= 2 (161) = 322 cm
Hence, perimeter of rectangular sheet is 322 cm.

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road levelled if the diameter of a road roller is 84 cm and length is 1 m.
Solution:
Shape of road roller is cylindrical.

For given cylinder:
Radius (r) = \(\frac{\text { diameter }}{2}\) = \(\frac {84}{2}\) = 42cm
length (height) (h) = 1 m = 100 cm
∴ Curved surface area of a roller
= 2 πrh
= 2 × \(\frac {22}{7}\) × 42 × 100
= 26,400 cm²
Roller covers area of 26,400 cm² in 1 revolution.
∴ Area of covered by roller in 750 revolutions
= 26400 × 750 cm²
= \(\frac{26400 \times 750}{100 \times 100}\) m²
= 1980 m²
Hence, area of road levelled is 1980 m².

10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm.

Company places a lable around the surface of the container (as shown in the figure). If the lable is placed 2 cm from top and bottom, what is the area of the label.
Solution:
For given cylindrical container:
radius (r) = \(\frac {diameter}{2}\) = \(\frac {14}{2}\) = 7 cm
height (h) = 20 cm
Now, label is placed 2 cm away from top and bottom.
∴ Height of label = (20 – 2 – 2) cm
= (20 – 4) cm = 16 cm
For label:
radius (r) = 7 cm, height (h) = 16 cm
∴ Area of cylindrical label = 2πrh
= 2 × \(\frac {22}{7}\) × 7 × 16
= 44 × 16
= 704 cm²
Hence, area of the label is 704 cm².

Punjab State Board PSEB 6th Class English Book Solutions English Bar Graph Reading Comprehension Exercise Questions and Answers, Notes.

PSEB 6th Class English Bar Graph Reading Comprehension

Bar Graph द्वारा विभिन्न प्रकार के आंकड़ों को दंडों (Bars) की सहायता से दर्शाया जाता है। इससे आंकड़ों को समझना, उनका तुलनात्मक अध्ययन करना, उनकी बढ़ती या घटती प्रवृत्ति (Upwards & Downward trends) की जानकारी प्राप्त करना बहुत ही सरल हो जाता है।

How to Read a Bar Graph?

किसी Bar Graph को पढ़ना बहुत ही सरल है। इसके लिए निम्नलिखित विधि अपनाएं।
1. X-Axis अर्थात् क्षैतिज/लेटवीं (Horizontal) line आंकड़ों से जुड़ी items दर्शाती है।
2. Y-Axis अर्थात् लम्बवत् अथवा खड़ी (Vertical) line उनसे संबंधित आंकड़े बताती है। इस रेखा पर सुविधाजनक ढंग से आंकड़ों का विभाजन दिया होता है।
3. इसे पढ़ने के लिए item पर बने दण्ड के ऊपरी तल को सीधी रेखा द्वारा खड़ी लाइन से जोड़ें। वहां दिया गया आंकड़ा ही उसी item की value बताएगा।

Some Important Bar Graphs

1. Study the following Bar graphs and write a short paragraph on each about what it informs.

Paragraph – The graph shows the number of books sold during four months. It is from January to April. In January only two books were sold. In Feb, the number increased. It was three. In March too the equal number of books was sold. April recorded the maximum sale. In this month six books were sold. Thus, the total number of books sold during four months was twelve.

(ii) Total Number of Students in a School

Paragraph – The graph shows the number of students in a school in different years. Their number was maximum (600) in 2002. It was least in 2004 only after two years. It means something must have gone wrong with the school as 500 students left the school.

The school was also popular with the student in 1998 when its strength was 550. Then its strength went on falling for two years. On the whole, the school was a popular one till 2002.

(iii) Number of Corona (Covid-19) Patients in Punjab

Paragraph – The graph informs us about number of Corona (Covid-19) patients in Punjab. In April 2020, it spread very rapidly. Within a week, from April 5 to April 12 the number of Corona patients became six fold. The most alarming phase was April 5 to April 8 when the number jumped from 10 to 40. After that the spread lost its furry and only 20 more patients were added to it during the next four days. On April 12, the total number of patients was 60. The government is of the view that it was the result of the voilation of safety measures. For safey, people must keep distance and use masks. Above all, they should stay at home.

(iv) Read the Bar graph and write down a paragraph on it.

Paragraph – Our government buys wheat from the farmers every year. It is called Buffer stock. It is used in crisis. It is increasing year by year. In 1998 it was only 15 thousand tonnes. It jumped to 25 thousand tonnes next year. During 2002 purchase of wheat was maximum. It also shows the increase in wheat production. The production in 2000 reduced. And, so was the purchase. The next year also saw the same trend of production and purchase.

2. Observe this bar graph which is showing the sale of shirts in a ready-made shop from Monday to Saturday.

Now answer the following questions:
(a) What information does the above bar graph give ?
(b) On which day were the maximum number of shirts sold ? How many shirts were sold that day ?
(c) On which day were the minimum number of shirts sold ?
(d) How many shirts were sold on Thursday ?
Answers:
(a) This bar graph tells us the number of shirts sold in a ready-made shop from Monday to Saturday.
(b) On Saturday the maximum number of shirts were sold. Sixty shirts were sold that day.
(c) On Tuesday, the number of shirts sold was the minimum. It was only ten.
(d) On Thursday thirty-five shirts were sold.
The graph also shows the sale went in increasing after Wednesday.

Now write answers in form of a paragraph.
This bar graph tells us the number of shirts sold in a ready made shop from Monday to Saturday. On Saturday, the maximum number of shirts were sold. Sixty shirts were sold on that day. On Tuesday the number of shirts sold was the minimum. It was only ten. On Thursday, thirty five shirts were sold. The graph also shows the sale went in increasing after Wednesday.

3. Observe this bar graph which shows the marks obtained by Aziz in monthly tests in different subjects.

Answer the given questions:
(a) What information does the bar graph give ?
(b) Name the subject in which Aziz scored maximum marks.
(c) Name the subject in which he has scored minimum marks. (d). State the name of the subjects and marks obtained in each of them.
Answers:
(a) This graph informs us about the marks obtained by Aziz in different subjects in monthly tests.
(b) He scored maximum marks in Hindi. He got 80 marks in this subject.
(c) In Social Studies he scored minimum marks. He obtained only forty marks.
(d) The other subjects were English, Mathematics and Science. In these three subjects he obtained 60,70 and 50 marks respectively
Thus we observe that Aziz is poor in Science and Social Studies.

Now write these answers in the form of a short paragraph.
This pargraph informs us about the marks obtaied by Aziz-in different subjects in monthly tests. He scored maximum marks in Hindi. He got 80 marks in this subject. In Social Studies, he scored minimum marks. He obtained only forty marks. The other subjects were English, Mathematics and Science. In these three subjects he obtained 60,70 and 50 marks respectively. Thus we observe that Aziz is poor in Science and Social Studies.

4. Study the following Bar graph and write 7-8 sentences on each about the information it gives. You can also take the help of the given table.

Answer:
(a) The graph tells us about the choice of students for different fruits. The fruits were Banana, Orange, Apple and Guava. Some students made their choice for more than one fruit.
(b) Eight students liked Banana, five students Apple and four students Guava.
(c) Only three students made their choice for orange. As such orange got least number of votes.
(d) Thus Banana was the choice of maximum number of students. Apple and Guava got the second and third position respectively in this choice making contest.

5. Study the given Bar graph carefully and write few sentences on the information it provides.

(a) This bar graph shows the population of India in different census years from 1951 to 2001. The census in India takes place every 10 years. (b) We observe that our population is increasing continuously. (Par).
(c) It was only 36 crores in 1951. It touched the figure 102 crores in 2001.
(d) Thus 76 crore people were added to our population in 50 years only.
(e) It was the maximum increase in population between 1991-2001. The period between 1951-1961 saw the minimum addition.

6. Study the given Bar graph carefully and write a short paragraph on the information it provides.
Runs Scored by a Harmanpreet Kaur in a five O.D.I. matches series.

Paragraph – This bar graphs represents the runs scored by Harmanpreet Kaur in five O.D.I. matches. She began with 35 runs in the first match. In the second match she scored 55 runs. She gave her best performance in the third match by scoring 70 runs. Then she lost her nature torch of the game audience. In fourth match she could score only 10 runs. And in the last match. she made no score as she went for a duck. Thus she failed to impress the audience in the last two matches. On the whole, she scored 170 runs in the series.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Solution:

Now, by drawing the points on the graph
i.e. (120, 12); (140, 26); (160, 34); (180, 40); (200, 50).
We get graph of less than type cumulative frequency.

Scale chosen:
On x-axis 10 units = Rs. 10
On y-axis 10 units = 5 workers.

Question 2.
During the medial check up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verWy the result by using the formula.
Solution:

Now, By drawing the points on the graph i.e., (38, 0); (40, 3); (42, 5); (44, 9); (46, 14); (48, 28) ; (50, 32) ; (52, 35) we get graph of less than type cumulative frequency.

Scale Chosen:
On x-axis, 10 units = 2 kg
On y-axis units = 5 students

From above graph, it is clear that
Median = 46.5 kg ; which lies in class interval 46 – 48.
Now, in the given table
\(\Sigma f_{i}\) = n = 35

∴ \(\frac{n}{2}=\frac{35}{2}\) = 17.5 ; which lies in the interval 46 – 48.

∴ Median class = 46 – 48
So, l = 46; n = 35; f = 14; cf = 14 and h = 2

Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 46 + \(\left\{\frac{\frac{35}{2}-14}{14}\right\}\) × 2

= 46 + \(\left\{\frac{\frac{35-28}{2}}{14}\right\}\)

= 46 + \(\frac{7}{2} \times \frac{1}{14}\) = 46 + \(\frac{1}{2}\)

= 46 + 0.5 = 46.5
From above discussion and graph; it is clear that median is same in both cases. Hence, Median weight of students is 46.5 kg.

Question 3.
The following table gives production yield per hectare of wheat of 1(X) farms of a village.

Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Now, by drawing the points on the graph i.e. (50, 100); (55, 98); (60, 90); (65, 78); (70, 54); (75, 16)
we get graph of more than type cumulative frequency.

Scale chosen:
On x-axis 10 units = 5 kg/ha
On y-axis 10 units = 10 forms

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Solution:

Here, \(\Sigma f_{i}\) = 68 then \(\frac{n}{2}=\frac{68}{2}\) = 34
Which lies in interval 125 – 145
Median class = 125 – 145
So, l = 125; n = 68; f = 20; çf = 22 and h = 20

Using formula, Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h

= 125 + \(\left\{\frac{\frac{68}{2}-22}{20}\right\}\) × 20

=125+ \(\frac{34-22}{20}\) × 20

= 125 + 12 = 137

For mean:

From above data, assumed mean (a) = 135
Width of class (h) = 20
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{7}{68}\) = 0.102
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathrm{X}}\) = 135 + 20 (0.102)
= 135 + 2.04 = 137.04.

For Mode:
In the given data,
Maximum frequency is 20 and it correspond to 125 – 145.
∴ Modal class = 125 – 145
So l = 125; f₁ = 20; f₀ = 13; f₂ = 14and h = 20
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 125 + \(\left(\frac{20-13}{2(20)-13-14}\right)\) × 20

= 125 + \(\frac{7}{40-27}\) × 20

= 125 + \(\frac{140}{13}\)
= 125 + 10.76923
= 125 + 10.77 = 135.77.
Hence. median, mean and mode of given data is 137 units: 137.04 units and 135.77 units.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Solution:

In thegiven data, \(\Sigma f_{i}\) = n = 60
∴ \(\frac{n}{2}=\frac{60}{2}\) = 30
Also, median of the distribution = 28.5 ………….(Given)
which lies in the class interval 20 – 30
Median class = 20 – 30
So, l = 20; f = 20; cf = 5 + x; h = 10
From table, it is clear that 45 + x + y = 60
x + y = 60 – 45 = 15
or x + y = 15 ……………….(1)
Now, using formula, Median = l + {\(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\)

28.5 = 2o + \(\left\{\frac{30-(5+x)}{20}\right\}\)

or 28.5 = 20 + \(\frac{30-5-x}{2}\)

or 28.5 = \(\)

or 2(28.5) = 65 – x
or 57.0 = 65 – x
or x = 65 – 57 = 8
∴ x = 8
Substitute this value of x in (1), we get
8 + y = 15
Hence, values of x and y is 8 and 7.

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.

Solution:

Here, \(\Sigma f_{i}\) = n = 100
then, \(\frac{n}{2}=\frac{100}{2}\) = 50, which lies in the interval 35 – 40
∴ Median class = 35 – 40
So, l = 35; n = 100; f = 33; cf = 45 and h = 5
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

= 35 + \(\left\{\frac{\frac{100}{2}-45}{33}\right\}\) × 5

= 35 + \(\frac{50-45}{33}\) × 5

= 35 + \(\frac{25}{33}\)
= 35 + 0.7575 = 35 + 0.76 (approx.) = 35.76
Hence, median age of given data is 35.76 years.

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

Find the median length of the leaves.
Solution:
Since the frequency distribution is not continuous, so firstly we shall make it continuous.

Here, \(\Sigma f_{i}\) = n = 40
then, \(\frac{n}{2}=\frac{40}{2}\) = 20, which lies in the interval 144.5 – 153.5
∴ Median class = 144.5 – 153.5
So, l = 144.5; f = 12; cf = 17; h = 9
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 144.5 + \(\left\{\frac{20-17}{12}\right\}\) × 9

= 144.5 + \(\frac{3 \times 9}{12}\)
= 144.5 + 225 = 146.75
Hence, median length of the leaves is 146.75 mm.

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

Find the median life time of a lamp.

Solution:

Here, \(\Sigma f_{i}\) = n = 400
∴ \(\frac{n}{2}=\frac{400}{2}\) = 200; which lies in the interval 3000 – 3500.
∴ Median class = 3000 – 3500
So, l = 3000; n = 400; f = 86; cf = 130 and h = 500
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 3000 + \(\left\{\frac{\frac{400}{2}-130}{86}\right\}\) × 500

= 3000 + \(\left(\frac{200-130}{86}\right)\) × 500

= 3000 + \(\frac{70 \times 500}{86}\) + 406.9767441

= 3000 + 406.98 (approx.) = 3406.98
Hence, median life time of a lamp is 3406.98 hours.

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames ? Also, find the modal size of the surnames.
Solution.
For Median:

Here, Here, \(\Sigma f_{i}\) = n = 100
∴ \(\frac{n}{2}=\frac{100}{2}\) = 50, which lies in interval 7 – 10.
∴ Median class = 7 – 10
So, l = 7; n = 100; f = 40; cf = 36 and h = 3
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 7 + \(\left\{\frac{\frac{100}{2}-36}{40}\right\}\) × 3

= 7 + \(\left\{\frac{50-36}{40}\right\}\) × 3

= 7 + \(\frac{14 \times 3}{40}\)

= 7 + \(\frac{21}{20}\) = 7 + 1.05 = 8.05
Hence, the median of letters in the surnames is 8.05.

For Mean:

From above data, Assumed Mean (a) = 8.5
Width of class (h) = 3
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\bar{u}=\frac{-6}{100}\) = – 0.06

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 8.5 + 3 (- 0.06) = 8.5 – 0.18 = 8.32
Hence, mean number of letters in the surnames is 8.32.

For Modal:
In the given data Maximum frequency is 44 and it corresponds to interval 7 — 10
∴ Modal class = 7 – 10
So l = 7; f₁ = 40; f₀ = 30; f₂ = 16 and h = 3
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 7 + \(\left(\frac{40-30}{2(40)-30-16}\right)\) × 3

= 7 + \(\frac{10}{80-46}\) × 3

= 7 + \(\frac{30}{34}\) = 7 + 0.882352941

= 7 + 0.88 (approx.) = 7.88.
Hence. modal size of the surnames is 7.88 letters.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

Here, \(\Sigma f_{i}\) = n = 30
∴ \(\frac{n}{2}=\frac{30}{2}\) = 15; which lies in the interval 55 – 60.
∴ Median class = 55 – 60
So, l = 55; n = 30; f = 6; cf = 13 and h = 5
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 55 + \(\left\{\frac{\frac{30}{2}-13}{6}\right\}\) × 5

= 55 + \(\left\{\frac{15-13}{6}\right\}\) × 5

= 55 + \(\frac{2 \times 5}{6}\)

= 55 + \(\frac{5}{3}\) = 55 + 1.66666
= 55 + 1.67 (approx.) = 56.67
Hence, median weight of the students are 56.67 kg.

Punjab State Board PSEB 6th Class English Book Solutions English Reading Comprehension Conversation / Dialogue Based Exercise Questions and Answers, Notes.

PSEB 6th Class English Reading Comprehension Conversation / Dialogue Based

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

1. Anil : Hello Nimmy, how are you?
Nimmy : Hey Anil, I’m good and how are you doing ?
Anil : Well, I’m quiet bored this vacation. Let us join some dance classes.
Nimmy : That sounds great. I’ll ask my parents and let you know tommorow.
Anil : Sure then we’ll go and fix the timings at Bollywood Dance Centre.
Nimmy : Alright and what about holidays homework ?
Anil : I have finished my work and what about you ?
Nimmy : I will finish it in a day or two.
Anil : Great, see you tomorrow. Bye !
Nimmy : Bye.

1. How Is Anil feeling during the vacation ?
(a) happy
(b) bored
(c) energetic
(d) sad.
Answer:
(b) bored.

2. Whom does Nipimy want to ask about dance classes ?
(a) her brother
(b) her sister
(c) her parents
(d) her friend
Answer:
(c) her parents.

3. Which dance centre do Anil and Nimmy wish to join ?
(a) New Centre
(b) Bollywood Dance Centre
(c) Modem Dance Centre
(d) High Heels Centre.
Answer:
(b) Bollywood Dance Centre

4. Has Anil finished his holidays homework?
(a) no
(b) yes
(c) partially
(d) not at all.
Answer:
(b) yes.

5. Nimmy is planning to finish her work:
(a) tomorrow
(b) in a day or two
(c) in a week
(d) in ten days.
Answer:
(b) in a day or two.

2. All friends of Neha come to the party at 6 o’clock in the evening in their best party clothes. Then the teacher enters the room.)

Children : Good evening, madam.
Teacher : Good evening, children. (To Neha) God bless you,
Neha! Here is a birthday gift for you. Have a nice time.
Neha : Thank you, madam.
Mother : Children come here. Now Neha is going to cut the cake.
Chidren : Happy birthday to Neha.
Uncle : Sony, I am late. Happy birthday Neha. Here is a gift for you, a packet of books!
Neha : Thank you, uncle. Thank you very much. That is the nicest gift I have ever received.
Father : Children now have a piece of cake and sweets. Here are small presents for all of you.
(Neha’s mother gives the presents to the children.)
Children : Thank you, uncle, Thank you, aunt for nice gifts.
Neha : Thank you, everyone. Thanks for my birthday presents.

1. Whose birthday party is it ?
(a) Teacher’s
(b) Neha’s
(c) Uncle’s
(d) Father’s
Answer:
(b) Neha’s.

2. Who comes late to the party ?
(a) Neha’s father
(b) Neha’s teacher
(c) Neha’s friends
(d) Neha’s uncle.
Answer:
(d) Neha’s uncle.

3. Uncle brings ……….. as a birthday gift.
(a) cake
(b) some sweets
(c) a packet of books
(d) a beautiful dress.
Answer:
(c) a packet of books

4. Neha’s mother gives children at the end of the party.
(a) small presents
(b) some books
(c) a piece of cake
(d) nothing.
Answer:
(a) small presents.

5. Children are asked to have a piece of cake and ……………
(a) fruit
(b) biscuits
(c) coffee
(d) sweets.
Answer:
(d) sweets.

3. Teacher : Good morning children ! I have some good news for you. We are going to a fair at Ropar tomorrow.
Raj : How are we going ?
Teacher : We shall take a bus to Ropar.
Gurudev : When shall we leave ?
Teacher : We shall leave at 9 o’clock in the morning.
Harpreet : Who will go with us ?
Teacher : Mr. Shanna and Mrs. Singh will go with us.
Mini : Will there be something to eat ?
Teacher : Oh yes, there will be lots of things to eat and many different games to play.
Raj : I will go on the merry-go-round.
Rani : I will go on the giant-wheel.
Ali : Will there be some rides too ?
Teacher : Oh yes ! There will be elephant, camel and hore rides. Some of you may enjoy these rides. You will have a lot of fun.
Nina : What other things shall we see at the fair ?

1. A fair is going to be held at
(a) Amritsar
(b) Patiala
(c) Ropar
(d) Chandigarh.
Answer:
(c) Ropar.

2. Mr. Sharma and ……………. will go with the students.
(a) Mrs. Singh
(b) Mrs. Sharma
(c) Mr. Singh
(d) All the three.
Answer:
(a) Mrs. Singh.

3. Raj wishes to enjoy ……………..
(a) the merry-go-round
(b) the giant-wheel
(c) slides
(d) swings.
Answer:
(a) the merry-go-round.

4. At the fair, the students can enjoy …………..
(a) elephant rides
(b) camel rides
(c) horse rides
(d) all these.
Answer:
(d) all these.

5. The party will leave at……………. in the morning.
(a) 7 o’clock
(b) 9 o’clock
(c) 8 o’clock
(d) 10 o’clock
Answer:
(b) 9 o’clock.

Or
The party will go to Ropar
(a) by train
(b) on bicycles
(c) by bus
(d) on foot.
Answer:
(c) by bus.

4. The sun is out. It is early morning. Amrit and Harpreet are still sleeping.

Mother : Wake up children ! It is six o’clock. You will be late for school.
Amrit is still sleeping but Harpreet is quickly getting up. He is going to the bathroom to brash his teeth and wash his face,
Harpreet : Wake up Amrit, we will be late for school! Look I am washing my face and brushing my teeth.
Amrit : What’s the time ?
Harpreet : It is half past six.
Amrit : Please let me sleep a little longer. Wake me up at half past seven.
Mother : Now get up Amrit. It is seven o’clock. Look Harpreet is finishing his breakfast and dressing for school.
Harpreet : I am leaving for school, it is eight o’ clock. You are late Amrit. You know the school starts at half past eight.

1. It is morning but the children are still ……………….
(a) reading
(b) playing
(c) talking
(d) sleeping.
Answer:
(d) sleeping.

2. Mother wakes up the children at ………………
(a) six o’clock
(b) seven o’clock
(c) eight o’clock
(d) nine o’clock
Answer:
(a) six o’clock.

3. Mother wakes up Amrit and Harpreet because they can be late for
(a) breakfast
(b) school
(c) office
(d) library.
Answer:
(b) school.

4. Harpreet leaves for school at………….
(a) 7 o’clock
(b) half past seven
(c) 8 o’clock
(d) half past eight.
Answer:
(c) 8 o’clock.

5. The school starts at…………………….
(a) eight o’clock
(b) half past eight
(c) seven o’clock
(d) half past seven.
Answer:
(b) half past eight.

5. Neera : Hello Amar ! Where did you go this weekend ?
Amar : I had gone to Amritsar with my mother.
Neera : How lovely ! What did you see there ?
Amar : We visited Sri Darbar Sahib. It is also called the Golden Temple.
Neera : Isn’t it beautiful ?
Amar : Oh yes ! It is very beautiful. The temple is covered in gold. It shines brightly as the sunlight falls on it. We bathed in the holy tank.
Neera : Are there some other places to see ?
Amar: Yes, we also visited the Durgiana Mandir and the Jallianwala Bagh.
Neera : Were there many shops ?
Amar : Yes, there were ! Some shops were big and some small. We bought many things from the market.-
Neera : I will also go there next week with my grandfather.

1. Amar had gone to Amritsar with ……………..
(a) his friends
(b) Neera
(c) his mother
(d) parents.
Answer:
(c) his mother.

2. In Amritsar Amar visited ……………..
(a) Sri Darbar Sahib
(b) Durgiana Mandir
(c) Jalianwala Bagh
(d) All the above.
Answer:
(d) All the above.

3. Sri Darbar Sahib is also called ……………
(a) the Golden Temple
(b) the Durgiana Mandir
(c) the Jalianwala Temple
(d) None of these.
Answer:
(a) the Golden Temple

4. Neera will go to Amritsar with her
(a) parents
(b) brother and sister
(c) grandfather
(d) grandmother.
Answer:
(c) grandfather.

5. At Amritsar, Amar and his mother bathed in
(a) a lake
(b) a canal
(c) the holy tank
(d) a river.
Answer:
(c) the holy tank.

6. Lion (showing a lot of interest): Do you store food in your stomach ?

Camel : No, I don’t. I store it in my hump. I eat a lot of food at one time. Then I don’t need to eat for a fortnight.
Lion : That’s interesting. But what do you eat ?
Camel : I eat leaves, but there are no trees in a desert.
Lion : What do you eat there ?
Camel : There are thorny bushes in the desert. I eat the thorns. The thorns don’t prick my thick tongue.
Lion : How funny!
Camel : Mr. King of die Forest, please come with me to the desert.
Lion : No, I’d better not. I can’t walk on burning sand. I can’t store food and water and I can’t eat thorns. Good-bye and good luck, Mr Ship of the Desert.

1. The camel can live without eating (food) for …………….
(a) a week
(b) A fortnight
(c) a mo nth
(d) three days.
Answer:
(b) a fortnight.

2. The camel stores food in its ………….
(a) stomach
(b) belley
(c) hump
(d) month.
Answer:
(c) hump.

3. In deserts the camel eats ……………
(a) thorns
(b) green grass
(c) small plants
(d) fallen fruit
Answer:
(a) thorns.

4. Who is Mr King of the Forest in the passage?
(a) elephant
(b) camel
(c) lion
(d) none of these.
Answer:
(c) Lion.

5. Mr. Ship of the desert is……………….
(a) camel
(b) lion
(c) burning sand
(d) camel’s hump.
Answer:
(a) camel.

Punjab State Board PSEB 6th Class English Book Solutions English Reading Comprehension Picture / Poster Based Exercise Questions and Answers, Notes.

PSEB 6th Class English Reading Comprehension Picture / Poster Based

Look at the picture carefully and answer the questions that follow

1.

(i) The weather is …………….
(a) sunny
(b) snowy
(c) clear
(d) rainy.
Answer:
(d) rainy.

(ii) There are …………. puddles in the picture.
(a) three
(b) four
(c) two
(d) one.
Answer:
(c) two.

(iii) There are……………… children in the picture.
(a) two
(b) five
(c) three
(d) four.
Answer:
(d) four.

(iv) A ……………. is hiding under a leaf.
(a) snail
(b) frog
(c) bird
(d) fish.
Answer:
(b) frog.

(v) We use umbrellas to save ourselves from …………..
(a) clouds
(b) puddles
(c) trees
(d) rain.
Answer:
(d) rain.

2.

(i) There are …………….. plants under the tree.
(a) thorny
(b) flower
(c) fruit
(d) dry.
Answer:
(b) flower.

(ii) There are…………… huts in the picture.
(a) two
(b) three
(c) many
(d) few
Answer:
(a) two.

(iii) A river flows under the …………
(a) trees
(b) plants
(c) bridge
(d) hills.
Answer:
(c) bridge.

(iv) The weather is ……………….
(a) sunny
(b) snowy
(c) clear
(d) rainy.
Answer:
(c) clear.

(v) It is a …………………. area.
(a) sandy
(b) sea-side
(c) forest
(d) hilly.
Answer:
(d) hilly.

3.

Two children are ………… in their kitchen garden.
(a) walking
(b) working
(c) playing
(d) running.
Answer:
(b) working.

(ii) They are ………… plants.
(a) watering
(b) growing
(c) cleaning
(d) manuring.
Answer:
(a) watering.

(iii) The day is …………..
(a) snowy
(b) cloudy
(c) sunny
(d) rainy.
Answer:
(c) sunny

(iv) A ……………. is flying under a tall tree.
(a) bird
(b) butterfly
(c) fly
(d) kite.
Answer:
(b) butterfly.

(v ) The flowers in the picture look very …………….
(a) faded
(b) beautiful
(c) ugly
(d) muddy.
Answer:
(b) beautiful.

4.

(i) This is a big
(a) school office
(b) consumer store
(c) booking office
(d) library.
Answer:
(d) library.

(ii) Books are lying on the
(a) shelves
(b) roof
(c) tables
(d) floor.
Answer:
(a) shelves.

(iii) In a library, there are books on different ………….
(a) religions
(b) jobs
(c) subjects
(d) problems
Answer:
(c) subjects.

(iv) The reading of books gives us ……………
(a) health
(b) wealth
(c) property
(d) knowledge
Answer:
(d) knowledge.

(v) This library can be given the name of ……………
(a) Entertaining Hut
(b) Knowledge Garden
(c) Knowledge Show
(d) None of these.
Answer:
(b) Knowledge Garden.

5.

(i) The atmosphere in the picture is ………….
(a) dirty
(b) pretty
(c) lovely
(d) charming.
Answer:
(a) dirty.

(ii) We can see a large …………… of garbage.
(a) basket
(b) dustbin
(c) heap
(d) sack.
Answer:
(c) heap

(iii) Many young ………….. are collecting the garbage.
(a) girls
(b) men and women
(c) persons
(d) women
Answer:
(c) persons

(iv) They will …………… it outside the city.
(a) dump
(b) throw
(c) sell
(d) display.
Answer:
(a) dump

(v) At a little distance we can see …………….
(a) a car
(b) a bus
(c) a jeep
(d) an e-rickshaw.
Answer:
(b) a bus.

6.

(i) The weather is …………..
(a) hot
(b) warm
(c) rainy
(d) snowy.
Answer:
(d) snowy.

(ii) There are …………….. children in die picture.
(a) two
(b) three
(c) four
(d) five
Answer:
(a) four

(iii) A dog is running ………….. a boy lying on the snow.
(a) after
(b) towards
(c) near
(d) under
Answer:
(b) towards

(iv) The tree in the picture is a…………….. tree.
(a) Christmas
(b) Diwali
(c) Mango
(d) Neem
Answer:
(a) Christmas.

(v) A girl is putting a cap on ……………….
(a) the dog
(b) the tree
(c) a snow man
(d) his friend
Answer:
(c) a snow man.

Punjab State Board PSEB 6th Class English Book Solutions English Reading Comprehension Unseen Passages Exercise Questions and Answers, Notes.

PSEB 6th Class English Reading Comprehension Unseen Passages

Read the given passages and answer the questions the follow each:

1. It is said that health is wealth. Healthy mind stays in a he%y body. Human body is like a machine. Over-eating and eating junk food harms our body. Ivular exercise keeps us fit and healthy. We must eat at regular intervals. It is our duty to ke, Gur body in good condition. Walking for a few kilometres daily is also a good exercise. If ycr body is in perfect health, you enjoy life. So we all must do regular exercise and stay fit.

Choose the correct option:

(i) Health is ……………
(a) money
(b) wealth
(c) gold
(d) nothing
Answer:
(b) wealth.

(ii) Human body is like a …………….
(a) scooter
(b) machine
(c) engine
(d) furniture.
Answer:
(b) machine.

(iii) When should we eat ?
(a) every time
(b) at regular in;rvals
(c) after long hours
(d) not eat in thtmorning.
Answer:
(b) at regular intervals.

(iv) What is our duty ?
(a) to eat junk food
(b) to eat all the tine
(c) to keep our body fit
(d) to get up late in the morning.
Answer:
(c) To keep our body fit.

(v) If your body is in perfect health
(a) you enjoy life
(b) you do not enjoy mything
(c) everything sounds dull
(d) you feel tired all tie time.
Answer:
(a) you enjoy life.

2. Golu did not like to read that much. One day, Golu’s mother bought him a nev book and Golu loved it. His favourite character in the book was a dog named Sheru. Golu read the whole book in one day. He read three pages before school, five pages after school and two pages before he went to bed. Golu loves reading now.

Choose the correct option:

(i) Golu did not like to:
(a) playgames
(b) watch t.v.
(c) read much
(d) talk to his mother.
Answer:
(c) read much.

(ii) Who bought a new book for Golu ?
(a) his father
(b) his mother
(c) his sister
(d) Sheru.
Answer:
(b) his mother.

(iii) Sheru was the name of:
(a) a dog
(b) Golu’s friend
(c) Golu’s neighbour
(d) none of these.
Answer:
(a) a dog.

(iv) Golu read three page’s of the book:
(a) after he went to bed
(b) before he went to bed
(c) before school
(d) after school.
Answer:
(c) before school

(v) Golu liked:
(a) the book
(b) the dog
(c) neither of these two
(d) both the book and the dog.
Answer:
(d) both the book and the dog.

3. Deenu was so involved r a his dream that he forgot he was carrying a tin on his head. He started running and slipped. The t: in fell down and the oil spilled on the ground. He felt very upset. He came back to the grocer w ith a sad heart and narrated the incident. The grocer scolded him and said. ‘ ‘You have caused m e a loss of rupees five hundred.”

Choose the correct option:

(i) Deenu was invo lved in:
(a) his story
(b) his dream
(c) a grocer
(d) shopping.
Answer:
(b) his dream.

(ii) In the tin, there was:
(a) milk
(b) ghee
(c) sugar
(d) oil.
Answer:
(d) oil

(iii) Why did the grocer scold Deenu ?
(a) because he stole his money.
(b) because of hi$ loss
(c) because he told a lie
(d) none of the above.
Answer:
(b) because of his loss.

(iv) Deem went to the grocer:
(a) happily
(b) cheerfully
(c) With a sad heart
(d) dreaming.
Answer:
(c) with a sad heart.

(v) Deenu felt very upset because:
(a) he fell down
(b) the oil spilled on the ground
(c) the oil made his clothes dirty
(d) he got hurt.
Answer:
(b) the oil spilled on the ground.

4. Shravan is the month of rains. After a long spell of summer, the weather becomes cool and pleasant. There is greenery all around. New plants start growing. All men, women and children begin to feel happy and cheerful. The festival of Teej in this month expresses their feelings of joy. People put on new and colourful clothes on this day. Girls sing folk songs and enjoy swinging.

Choose the correct option:

(i) Shravan is the month of:
(a) fairs
(b) festival after festival
(c) hot summer days
(d) rains
Ans.
(d) rains.

(ii) The weather becomes cool and pleasant in the month of:
(a) Barisakh
(b) Ashad
(c) Shravan
(d) Kartik.
Answer:
(c) Shravan.

(iii) What brings greenery all around ?
(a) rains
(b) summer heat
(c) cold winter
(d) all these.
Answer:
(a) rains.

(iv) The girls sing:
(a) fake songs
(b) folk songs
(c) new songs
(d) film songs
Answer:
(b) folk songs.

(v) When do girls enjoy swinging ?
(a) on Diwali
(b) on Lohri
(c) on Holi
(d) on Teej.
Answer:
(d) On Teej.

5. It was the 15th August last Monday. This day is celebrated as a national festival in India. It was on this day in the year 1947 that we won our Independence from the British rule after a long struggle. This year we celebrated this festival in our school with a great zeal. We had been working hard for a week to beautify our school. We prepared colourful banners and placards with patriotic slogans written on them.

Choose the correct option:

(i) We won our freedom from the British rule in:
(a) 1950
(b) 1952
(c) 1947
(d) 1974.
Answer:
(c) 1947.

(ii) We celebrate 15th August as a:
(a) social festival
(b) religious festival
(c) national festival
(d) none of these.
Answer:
(c) national festival

(iii) How did we win our freedom ?
(a) after a short period of struggle.
(b) after a long period of struggle
(c) without any struggle
(d) after a bloody struggle.
Answer:
(b) after a long period of struggle.

(iv) This year we celebrated the Independence Day
(a) with a great zeal
(b) without any preparation
(c) in a simple manner
(d) with the help of students
Answer:
(a) with a great zeal.

(v) Patriotic slogans were written on ………………….
(a) die walls
(b) walls and banners
(c) playcards
(d) both banners and placards
Answer:
(d) both banners and placards.

6. After the Headmaster’s speech, a cultural programme was organised. The children sang patriotic songs. Some recited poems related to the freedom movement. Some students spoke about the great patriots of India such as Jhansi ki Rani. Tantya Tope, Sardar Bhagat Singh, Neta Ji Subhash Chander Bose, Mahatma Gandhi and the other great sons and daughters of our motherland.

Choose the correct option:

(i) There was a cultural programme:
(a) before the Headmaster’s speech
(b) after the Headmaster’s speech.
(c) after the poems were recited
(d) after the speech of the students.
Answer:
(b) after the Headmaster’s speech.

(ii) What did the cultural programme include ?
(a) patriotic songs
(b) poems related to freedom movement
(c) speeches about the great patriots of India
(d) all the above.
Answer:
(d) all the above.

(iii) Mahatma Gandhi was:
(a) a common man
(b) an ordinary person
(c) a great scholar
(d) a great patriot.
Answer:
(d) a great patriot.

(iv) A great daughter of India mentioned in the passage is:
(a) Jhansi ki Rani
(b) Rani of motherland
(c) Tantya Tope
(d) None of the above.
Answer:
(a) Jhansi ki Rani.

(v) What was the first item of the day?
(a) recitation of poems
(b) patriotic songs
(c) the Headmaster’s speech
(d) a cultural programme.
Answer:
(c) The Headmaster’s speech.

7. Vineet, a student of class VIII was also in the same bus. He swung into action. The exit door near the engine was all on fire. He broke open the emergency door and started pushing the children out through this door. The driver and the conductor saw Vineet, struggling alone. They rushed to help him. By the time the last student was rescued the entire bus had turned into a ball of fire. It was Vineet’s timely action that averted a big tragedy. He saved twenty seven lives. However, he himself suffered some bums during this heroic deed.

Choose the correct option:

(i) Vineet was the student of:
(a) Class VI
(b) Class VII
(c) Class Vin
(d) none of these.
Answer:
(c) Class VIII.

(ii) Which door was on fire ?
(a) entrance door
(b) front door
(c) exit door
(d) emergency dpor.
Answer:
(c) exit door.

(iii) The children were taken out through the ……………….
(a) main door
(b) exit door
(c) frontdoor
(d) emergency door.
Answer:
(d) emergency door.

(iv) Who helped Vineet ?
(a) The driver and the teacher
(b) The conductor and the children
(c) Both the driver and the conductor
(d) The teachers and the children.
Answer:
(c) Both the driver and the conductor.

(v) What happened to Vineet during the heroic deed ?
(a) He suffered some bums.
(b) His face was badly wounded.
(c) All his clothes were burnt.
(d) He came out safe and sound.
Answer:
(a) He suffered some bums.

8. One day the librarian said to him. “Do you really read the books or just return them without reading ?” Narendra said, “Of course, Sir, I thoroughly read all the books borrowed from the library. You can ask me any question from these books.” The librarian took out a book and asked him the questions. Narendra answered all the questions correctly. The librarian was amazed at his sharp memory.

Choose the correct option:

(i) Narendra answered correctly:
(a) two questions
(b) three questions
(c) all the questions
(d) no question.
Answer:
(c) all the questions.

(ii) Who borrowed books from the library ?
(a) librarian
(b) Narendra
(c) Surendra
(d) Narendra’s friends.
Answer:
(b) Narendra.

(iii) The librarian thought that Narendra:
(a) returned the books after reading them.
(b) returned the books after spoiling them.
(c) returned the books without reading them.
(d) all these.
Answer:
(c) returned the books without reading them.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 5 Data Handling Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2

1. A survey was made to find the type of music that a certain group of young people liked in a city. Given pie chart shows the findings of this survey.

From this pie chart answer the following:

Question (i).
If 20 people liked classical music, how many young people were surveyed ?
Solution:
Let the required number of total young people surveyed = x
∴ 10% of x = 20
∴ \(\frac {10}{100}\) × x = 20
∴ x = \(\frac{20 \times 100}{10}\)
∴ x = 200
200 young people were surveyed.

Question (ii).
Which type of music is liked by the maximum number of people ?
Solution:
Maximum number of people like the light music.

Question (iii).
If a cassette company were to make 1000 CD’s, how many of each type would they make ?
Solution:
Total number of CD’s = 1000
(a) Number of CD’s for semi classical music = 20 % of 1000
= \(\frac {20}{100}\) × 1000 = 200
(b) Number of CD’s for classical music = 10 % of 1000
= \(\frac {10}{100}\) × 1000 = 100
(c) Number of CD’s for folk music = 30 % of 1000
= \(\frac {30}{100}\) × 1000 = 300
(d) Number of CD’s for light music = 40 % of 1000
= \(\frac {40}{100}\) × 1000 = 400

2. A group of 360 people were asked to vote for their favourite season from the ? three seasons rainy, winter and summer:

Question (i).
Which season got the most votes ?
Solution:
Winter season got the most votes (150).

Question (ii).
Find the central angle of each sector.
Solution:
Total Votes = 90 + 120 + 150 = 360
∴ Central angle of the sector corresponding to :
Summer season = \(\frac {90}{360}\) × 360° = 90°
Rainy season = \(\frac {120}{360}\) × 360° = 120°
Winter season = \(\frac {150}{360}\) × 360° = 150°

Question (iii).
Draw a pie chart to show this information.
Solution:
Draw three radi in such a way that it makes angles of 90°, 120° and 150° at the centre.

3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Colours	Number of people
Blue	18
Green	9
Red	6
Yellow	3
Total	36

Solution:
Central angle of the sector corresponding to :
(i) The blue colour = \(\frac {18}{36}\) × 360°
= 18 × 10° = 180°
(ii) The green colour = \(\frac {9}{36}\) × 360° = 90°
(iii) The red colour = \(\frac {6}{36}\) × 360° = 60°
(iv) The yellow colour = \(\frac {3}{36}\) × 360° = 30°
Thus, the required pie chart is given below.

4. The given pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.

Question (i).
In which subject did the student score 105 marks ?
[Hint: For 540 marks, the central angle = 360°. So for 105 marks, what is the central angle ?]
Solution:
Toted marks = 540
∴ Central angle corresponding to 540 marks = 360°
∴ Central angle corresponding to 105 marks = \(\frac {360}{540}\) × 105° = 70°
The sector having central angle 70° is corresponding to Hindi.
Thus, the student scored 105 marks in Hindi.

Question (ii).
How many more marks were obtained by the student in Mathematics than in Hindi?
Solution:
The central angle corresponding to the sector of Mathematics = 90°
∴ Marks scored in Mathematics = \(\frac{90^{\circ}}{360^{\circ}}\) × 540 = 135
30 (135 – 105) more marks were obtained by the student in Mathematics than in Hindi.

Question (iii).
Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
[Hint:Just study the central angles.]
Solution:
The sum of the central angles for Social Science and Mathematics = 65° + 90° = 155°
The sum of the central angles for Science and Hindi = 80° + 70° = 150°
∴ Marks obtained in Social Science and Mathematics is more than the marks scored in Science and Hindi.

5. The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart:

Solution:
Central angle of the sector representing:
(i) Gujarati language = \(\frac {40}{72}\) × 360°
= 40 × 5° = 200°
(ii) English language = \(\frac {12}{72}\) × 360°
= 12 × 5°= 80°
(iii) Urdu language = \(\frac {9}{72}\) × 360°
= 9 × 5° = 45°
(iv) Hindi language = \(\frac {7}{72}\) × 360°
= 7 × 5° = 35°
(v) Sindhi language = \(\frac {4}{72}\) × 360°
= 4 × 5° = 20°
The required pie chart is as follows.

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.