Bhagya – Page 22 – PSEB Solutions

PSEB 7th Class Science Notes Chapter 12 Reproduction in Plants

January 24, 2025January 23, 2025 by Bhagya

This PSEB 7th Class Science Notes Chapter 12 Reproduction in Plants will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 12 Reproduction in Plants

→ Plants reproduce in two ways Sexual reproduction and Asexual reproduction.

→ Asexual reproduction is a method of reproduction by which new plants are bom from a single parent.

→ There are different methods of asexual reproduction like reproduction by sprouting, reproduction by seeds, fragmentation.

→ In the fragmentation reproductive process, the organism is divided into equal parts which grow into two individuals.

→ During sexual reproduction, the male and female reproductive organs of plants produce male gametes and female gametes which fuse together to form zygotes. The zygote develops into a new plant.

→ Sexual reproduction occurs only in flowering plants.

→ Vegetative propagation is a method of reproduction in which new plants grow through organs like roots, stems, or leaves. In this method of reproduction, neither the genitals nor the seed participates.

→ There are also many artificial methods of reproduction in plants. These are grafting, cutting, and burring under the ground.

→ The transfer of mature pollen grains from the anther to the stigma is called pollination. They reach the stigma of the same flower or another flower.

→ Flowerless plants like moss breed through fragmentation, through yeast breed through buds, while fungi and moss breed through spores.

→ The fusion of the male gamete and the female gamete in the ovum is called fertilization.

→ After fertilization of the ovaries, the ovaries develop into fruit and the ovum develops into seeds.

→ In order to move the seeds away from the germinating plants, it is necessary to scatter the seeds so that the seeds can grow into new plants.

→ Reproduction: The ability to live beings to produce new creatures like themselves is called reproduction.

→ Asexual reproduction: A method that does not require seeds to grow new plants. A single parent is required for reproduction.

→ Sexual Reproduction: Sexual reproduction is a method to produce a new organism through the combination of male and female gametes.

→ Vegetative propagation: When a new plant is obtained from any part of the plant except seed, it is called vegetative propagation.

→ Fragmentation: The formation of a new organism by dividing the body of an animal into Two or more parts is called fragmentation.

→ Unisexual Flower: A flower that has only stamens or only pistil is called a unisexual flower.

→ Bisexual flower: A flower in which both stamens and pistil are present is called a bisexual flower.

→ Fertilization: The combination of male and female gametes is called the action of fertilization.

→ Pollination: The transfer of mature pollen from the pollen cell to the stigma is called pollination.

→ Self-pollination: Pollination in flowers, when the pollen grains land on the same flower this action is called self-pollination.

→ Cross-pollination: Pollination action in which the pollen goes from the anther of one flower to the stigma of another flower. This type of pollination is done by two flowers of the same plant or flowers of the same species.

→ Germination of seeds: Reaching the moist soil, the seeds absorb water and swell. As the embryo begins to germinate, the root sprouts sink into the soil, and the stem sprouts up into the air. The leaves come out. This process is called seed germination.

PSEB 7th Class Science Notes Chapter 11 Transportation in Animals and Plants

January 24, 2025January 23, 2025 by Bhagya

This PSEB 7th Class Science Notes Chapter 11 Transportation in Animals and Plants will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 11 Transportation in Animals and Plants

→ All living things need energy for various activities and this energy is received from food.

→ The leaves need water and CO₂ to make food through photosynthesis.

→ In animals, food, oxygen, and water are delivered to every cell in the body, and waste products are transported from the cells to the body’s excretory organs.

→ The movement of substances from one place to another in an organism is called transportation.

→ The circulatory system of developed organisms consists of the heart, blood vessels, and blood, which carry oxygen, carbon dioxide, food, hormones, and enzymes from one part of the body to another.

→ The single-cell organism does not have a circulatory system.

→ Blood contains red blood cells, white blood cells, platelets, and plasma.

→ The red color of the blood is caused by a pigment called Haemoglobin.

→ The heart is a muscular organ that constantly beats like a pump for the circulation of blood.

→ The number of heartbeats per minute is called the pulse rate.

→ The arteries contain oxygenated blood and the veins contain carbon dioxide-rich blood.

→ Cells exchange nutrients, gases, and follicles between blood and tissue fluids.

→ The human excretory system consists of a pair of kidneys, a pair of ureters, a urinary bladder, and a urethra.

→ The kidney produces waste products in the form of urine, lungs in the form of carbon dioxide, and skin in the form of sweat.

→ The blood cells in the human kidneys filter the blood.

→ Dialysis is the process by which a machine removes unwanted substances and excess fluid from the blood.

→ Diffusion is a process in which a matter moves from a region of higher concentration to a region of low concentration.

→ Dispersion is the process by which a solvent passes through a semi-permeable membrane. And the movement is from low-density solution to high-density solution.

→ Single-cell organism exchanges substances in the external environment from the cell surface.

→ Photosynthesis: The process of formation of carbohydrates (food) by green plants from simple compounds such as carbon dioxide and water in the presence of chlorophyll in the presence of sunlight is called photosynthesis.

→ Dispersion: This is the process in which solvents pass through a semi-permeable membrane from a low concentration solution to a solution with a higher concentration so that the concentration of the solution on both sides becomes equal. This type of transport is for short distances. Plant roots absorb water from the soil by this process.

→ Transpiration: The vaporization of water through the leaves of plants is called transpiration.

→ Transportation: The transfer of food from the leaves to other parts of the plant is called transportation.

→ Phloem: Plant tissues that carry food from the leaves to other parts of the plant, is called Phloem.

→ Arteries: The tubes that carry oxygen-rich blood from the heart to different parts of the body, are called arteries.

→ Veins: The capillaries that carry blood from different parts of the body to the heart, are called veins.

→ Excretion: The process of expelling harmful and waste products from the body is called excretion.

→ Dialysis: The process of removing urea and other toxic substances from the kidneys of the body with the help of an artificial machine is called dialysis.

PSEB 7th Class Science Notes Chapter 10 Respiration in Organisms

January 24, 2025January 23, 2025 by Bhagya

This PSEB 7th Class Science Notes Chapter 10 Respiration in Organisms will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 10 Respiration in Organisms

→ Breathing is a part of the process of Respiration during which the living beings take in (inhale) oxygen and give out (exhale) carbon dioxide into the air.

→ The oxygen we take in when we breathe breaks down glucose into water and carbon dioxide and also releases the energy that is necessary for the existence of living beings.

→ During cellular respiration, glucose is broken down in the cells of an organism.

→ During Aerobic respiration, glucose is broken down in the presence of oxygen.

→ During Anaerobic respiration, the breakdown of glucose occurs in the absence of oxygen.

→ During hard physical activities such as heavy exercise, fast running, and cycling, etc, energy demand is high but oxygen to produce energy is limited so anaerobic respiration takes place.

→ The breathing rate also increases during heavy physical activity.

→ Different organs are present in different organisms for respiration.

→ Lungs expand when oxygen is inhaled and contract again when exhaled.

→ The blood contains haemoglobin which carries oxygen to different parts of the body.

→ In cows, buffaloes, dogs, cats, and other mammals, the respiratory organs are similar to the human respiratory organs and the respiratory function is similar to that of humans.

→ In earthworms, gases in the gut are exchanged through moist skin.

→ In fish, this action takes place through the gills and in insects through the respiratory tract.

→ The breakdown of glucose in plants is similar to that in other organisms.

→ The roots of plants take air from the soil.

→ The leaves have small pores or holes called stomata, through which gases are exchanged.

→ Respiration: It is a biochemical reaction in organisms that involves the oxidation of complex organic foods. This results in the formation of carbon dioxide and water and releases energy.

→ Aerobic Respiration: Respiration that occurs in the presence of oxygen is called aerobic respiration.

→ Anaerobic Respiration: Breathing that occurs in the absence of oxygen is called Anaerobic Respiration.

→ Stomata: A special type of pores are present on the top layer of leaves for the exchange of air and water vapours, called stomata.

→ Respiration: This is a simple mechanical activity in which oxygen-rich air is pulled and goes into the lungs (respiratory organs). This action is called breathing. Carbon dioxide, rich air is released into the atmosphere after respiration called exhalation. This complete process is called Respiration.

→ Breathing: The act of filling the respiratory organs (lungs) with oxygen-rich air from the atmosphere is called breathing.

→ Exhalation: An activity in which carbon dioxide-rich air is expelled out of the lungs.

→ Cellular Respiration: The process that takes place inside a cell in which energy is produced after the chemical decomposition of food is called cellular respiration.

→ Breathing rate: The number of times a person breathes in a minute is called breathing rate.

→ The average person’s breathing rate is 12 to 20 breaths per minute.

→ Gills: These blood vessels are special wing-like organs that are present in some aquatic organisms, such as fish, breathe with help of gills. The water and blood flow in opposite directions which increases the diffusion of oxygen.

PSEB 7th Class Science Notes Chapter 9 Soil

January 24, 2025January 23, 2025 by Bhagya

This PSEB 7th Class Science Notes Chapter 9 Soil will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 9 Soil

→ The top layer of the earth in which crops and plants can grow is called soil.

→ Soil is made up of broken rocks, organic matter, animals, plants, and microorganisms.

→ There are different layers of soil, which can be seen in the soil profile.

→ Soil is made up of both organic and inorganic components.

→ The dead and rotten leaves of plants or the bodies of plants, insects, or dead animals buried in the soil, animal dung, etc. combine to form organic matter called humus.

→ Soil that contains a mixture of organic and inorganic substances is very useful for crops.

→ Depending on the size of the particles, the soil is clayey, sandy, rocky, and loamy.

→ Depending on the chemical nature of the soil, the soil may be acidic, basic, or neutral.

→ Acid soils have a pH of 1 to 6.

→ Alkaline soils have a pH of 8 to 14.

→ Neutral soil has a pH of 7.

→ Ph paper is used to determine the nature of the soil.

→ Black soil contains iron salts and is good for growing cotton.

→ Soil containing sulphur is good for growing onions.

→ Different types of soil are required to grow different types of crops.

→ It takes many years for the formation of the top layer of soil.

→ Removal of topsoil due to floods, winds, storms, and mining is called Erosion.

→ By digging the soil, the animals with their feet loosen the soil and the soil, which has been loosened, gets eroded quickly by wind and water.

→ By planting trees, building check dams, planting grass on the sides of farmland, and building along sides of rivers and canals soil erosion can be prevented.

→ Soil: A mixture of rock/horizontal particles and humus is called soil.

→ Soil Profile: A vertical section through different layers of soil is called the soil profile.

→ Humus: The dead and decaying organisms present in soil are called humus.

→ Soil Moisture: Soil retains water in it, which is called soil moisture.

→ Soil erosion: The removal of the top layer of soil by water, wind, or ice is called Soil erosion.

→ Weathering: It is a method in which soil is formed by the breakdown of rocks by the action of wind, water, and climate.

PSEB 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

January 24, 2025January 23, 2025 by Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

1. Convert the following fractions into percents

(i). $\frac {1}{8}$
Solution:
$\frac {1}{8}$ = $\frac {1}{8}$ × 100
= $\frac {25}{2}$
= 12.5
Thus, $\frac {1}{8}$ = 12.5%

(ii). $\frac {49}{50}$
Solution:
$\frac {49}{50}$ = $\frac {49}{50}$ × 100
= 98
Thus, $\frac {49}{50}$ = 98%

(iii). $\frac {5}{4}$
Solution:
$\frac {5}{4}$ = $\frac {5}{4}$ × 100
= 125
Thus, $\frac {5}{4}$ = 125%

(iv). 1 $\frac {3}{8}$
Solution:
1 $\frac {3}{8}$ = $\frac {11}{8}$ × 100
= $\frac {275}{2}$
= 137 $\frac {1}{2}$
Thus, 1 $\frac {3}{8}$ = 137 $\frac {1}{2}$ %

2. Convert the following percents into fractions in simplest form

(i). 25%
Solution:
25% = $\frac {25}{100}$
= $\frac {1}{4}$

(ii). 150%
Solution:
150% = $\frac {150}{100}$
= $\frac {3}{2}$

(iii). 7 $\frac {1}{2}$
Solution:
7 $\frac {1}{2}$ = $\frac {15}{2}$ × $\frac {1}{100}$
= $\frac {3}{40}$

3. (i) Anita secured 324 marks out of 400 marks. Find the percentage of marks secured by Anita.
(ii) Out of 32 students, 8 are absent from the class. What is the percentage of students who are absent ?
(iii) There are 120 voters, 90 out of them voted. What percent did not vote ?
Solution:
(i) Anita scored 324 marks out of 400 marks.
∴ Percentage of marks secured by Anita = $\left(\frac{324}{400} \times 100\right) \%$ = 50%

(ii) Out of 32 students 8 students are absent
∴ Percentage of students who are absent = $\frac {8}{32}$ × 100% = 25%

(iii) Total voters = 120
Voters who voted = 90
The voters who did not vote = 120 – 90
= 30
Percentage voters who did not vote = $\frac {30}{120}$ × 100%
= 25%

4. Estimate the part of figure which is shaded and hence find the percentage of the part which is shaded.
PSEB 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 1
Solution:
(i) Shaded part = $\frac {2}{4}$ = $\frac {1}{2}$
Percentage of shaded part = $\left(\frac{1}{2} \times 100\right) \%$
= 50%

(ii) Shaded part = $\frac {2}{6}$ = $\frac {1}{3}$
Percentage of shaded part = $\left(\frac{1}{3} \times 100\right) \%$
= 33 $\frac {1}{3}$ %

(iii) Shaded part = $\frac {5}{8}$
Percentage of shaded part = $\left(\frac{5}{8} \times 100\right) \%$
= $\frac {125}{2}$ %
= 62.5%

5. Convert the following percentages into ratios in simplest form :

(i). 14%
Solution:
14% = 14 × $\frac {1}{100}$
= $\frac {7}{50}$
= 7 : 50

(ii). 1 $\frac {3}{4}$ %
Solution:
1 $\frac {3}{4}$ % = $\frac {7}{4}$ × $\frac {1}{100}$
= $\frac {7}{400}$
= 7 : 400

(iii). 33 $\frac {1}{3}$ %
Solution:
33 $\frac {1}{3}$ % = $\frac {100}{3}$ × $\frac {1}{100}$
= $\frac {1}{3}$
= 1 : 3

6. Express the following ratios as percentages :

(i). 5 : 4
Solution:
5:4 = $\frac {5}{4}$ × 100
= 125%

(ii). 1 : 1
Solution:
1 : 1 = $\frac {1}{1}$ × 100
= 100%

(iii). 2 : 3
Solution:
2 : 3 = $\frac {2}{3}$ × 100
= $\frac {200}{3}$ %
= 66 $\frac {2}{3}$ %

(iv). 9 : 16
Solution:
9 : 16 = $\frac {9}{16}$ × 100
= $\frac {225}{4}$ %
= 56 $\frac {1}{4}$ %

7. Chalk contains calcium, carbon and sand in the ratio 12 : 3 : 10. Find the percentage of carbon in the chalk.
Solution:
Calcium : Carbon : sand
= 12 : 3 : 10
Total of ratios = 12 + 3 + 10
= 25
Percentage of carbon is chalk
= $\frac {3}{25}$ × 100
= 12%

8. Convert each part of the following ratios into percentage :

(i). 3 : 1
Solution:
Total of ratios = 3 + 1 = 4
Percentage of first part = $\frac {3}{4}$ × 100
= 75%
Percentage of second part = $\frac {1}{4}$ × 100
= 25%

(ii). 1 : 4
Solution:
Total of ratios = 1 + 4 = 5
Percentage of first part = $\frac {1}{5}$ × 100
= 20%
Percentage of second part = $\frac {4}{5}$ × 100
= 80%

(iii). 4 : 5 : 6.
Solution:
Total of ratios = 4 + 5 + 6 = 15
Percentage of first part = $\frac {4}{15}$ × 100
= $\frac {8}{3}$ %
= 26 $\frac {2}{3}$ %
Percentage of second part = $\frac {5}{15}$ × 100
= $\frac {100}{3}$ %
= 33 $\frac {1}{3}$ %
Percentage of third part = $\frac {6}{15}$ × 100
= 40%

9. Convert the following percentages to decimals :

(i). 28%
Solution:
28% = $\frac {28}{100}$
0.28

(ii). 3%
Solution:
3% = $\frac {3}{100}$
= 0.03

(iii). 37 $\frac {1}{2}$ %
Solution:
37 $\frac {1}{2}$ %
= $\frac {75}{2}$ × $\frac {1}{100}$ =
= $\frac {3.75}{100}$
= 0.375

10. Convert the following decimals to percentage :

(i). 0.65
Solution:
0.65 = (0.65 × 100)%
= $\left(\frac{65}{100} \times 100\right) \%$
= 65%

(ii). 0.9
Solution:
0.9 = (0.9 × 100)%
= $\left(\frac{9}{10} \times 100\right) \%$
= 90%

(iii). 2.1
Solution:
2.1 = (2.1 × 100)%
= $\left(\frac{21}{10} \times 100\right) \%$
= 210%

11. (i) If 65% of students in a class have a bicycle, then what percent of the students do not have a bicycle ?
(ii) We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what percent are mangoes ?
Solution:
(i) Percentage of student having a bicycle = 65%
Percentage of students do not have a bicycle = (100 – 65)%
= 35%

(ii) Percentage of apples = 50%
Percentage of oranges = 30%
Percentage Of mangoes = (100 – (50% + 30%)
= (100 – 80)%
= 20%

12. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Original population = 25000
Decreased population = 24,500
Decrease in population = (25000 – 24500)
= 500
Percentage decrease = $\frac{\text { Decrease in population }}{\text { Original population }} \times 100 \%$
= $\frac {500}{25000}$ × 100%
= 2%

13. Arun bought a plot for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase ?
Solution:
Original price of plot = ₹ 3,50,000
The increased price of plot = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000.
Percentage of price increase
= $\left(\frac{\text { Increase in price }}{\text { Original price }} \times 100\right)$
= $\frac {20,000}{350000}$ × 100%
= $\frac {40}{7}$ %
= 5 $\frac {5}{7}$

14. Find :

(i). 15% of 250
Solution:
15% of 250 = $\frac {15}{100}$ × 250
= $\frac {375}{10}$
= 37.5

(ii). 25% of 120 litres
Solution:
25% of 120 litres = $\frac {25}{100}$ × 20 litres
= 30 liters.

(iii). 4% of 12.5
Solution:
4% of 12.5 = $\frac {4}{100}$ × $\frac {125}{10}$
= $\frac {5}{10}$
= 0.5

(iv). 12% of ₹ 250.
Solution:
12% of ₹ 250 = ₹ $\frac {12}{100}$ × 250
= ₹300

15. Multiple Choice Questions :

Question (i).
The ratio 2 : 3 expressed as percentage is
(a) 40%
(b) 60%
(c) 66 $\frac {2}{3}$ %
(d) 33 $\frac {1}{3}$ %
Answer:
(c) 66 $\frac {2}{3}$ %

Question (ii).
If 30% of x is 72, then x is equal to
(a) 120
(b) 240
(c) 360
(d) 480
Answer:
(b) 240

Question (iii).
0.025 when expressed as a percent is
(a) 250%
(b) 25%
(c) 4%
(d) 2.5%
Answer:
(d) 2.5%

Question (iv).
In a class, 45% of students are girls. If there are 22 boys in the class, then the total number of students in the class is
(a) 30
(b) 36
(c) 40
(d) 44
Answer:
(c) 40

Question (v).
What percent of $\frac {1}{7}$ is $\frac {2}{35}$ ?
(a) 20%
(b) 25%
(c) 30%
(d) 40%
Answer:
(d) 40%

PSEB 7th Class Science Notes Chapter 8 Winds, Storms and Cyclones

January 24, 2025January 23, 2025 by Bhagya

This PSEB 7th Class Science Notes Chapter 8 Winds, Storms and Cyclones will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 8 Winds, Storms and Cyclones

→ The air around us exerts pressure.

→ Moving air is called wind.

→ Very strong wind lowers the pressure.

→ Air expands on heating and contracts on cooling.

→ Hot air is lighter than cold air.

→ The wind moves from high-pressure areas to low-pressure areas.

→ Wind speed is measured with an Anemometer.

→ The direction of wind speed is measured by the wind vane.

→ Wind currents (movement) are caused by the uneven heating of the earth.

→ Monsoon winds are filled with moisture (water vapours) and bring rain.

→ Cyclones are destructive.

→ A cyclone crossed the coast of Orissa on October 18, 1999.

→ Cyclones have higher wind speeds.

→ A cyclone is a very strong whirlwind that revolves around very low-pressure areas.

→ A hurricane is a storm with strong winds blowing through, in a funnel-shaped cloud.

→ The loud noise produced during lightning is called thunder.

→ Heavy rain with strong winds is called a storm.

→ Hurricanes in the United States and typhoons in Japan are cyclones.

→ Tornadoes are dark cone-like clouds that form between the earth’s crust and the sky.

→ All kinds of natural disasters like cyclones, tornadoes, etc. destroy trees, wires, and communication systems.

→ Special policies are adopted during disasters.

→ A cyclone warning is given 48 hours in advance with the help of satellite and radar.

→ And self-help is the best help. So it would be helpful to plan your safety in advance and take precautionary measures before any cyclone actually strikes.

→ Wind: Fast-moving air is called wind.

→ Monsoon winds: The winds that come from the sea and carry water are called monsoon winds.

→ Tornado: Dark coloured cone-like clouds whose cone structure is from sky to earth is called Tornado.

→ Cyclone: A violent wind that moves in a circle causing a storm is called a cyclone.

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.3

January 24, 2025January 23, 2025 by Bhagya

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
Answer:
The remainder theorem states that when polynomial p (x) of degree greater than or equal to 1 is divided by linear polynomial x – a, the remainder is p (a).
Here. p(x) = x³ + 3x² + 3x + 1

(i) x + 1
Answer:
Divisor g (x) = x + 1.
Comparing x + 1 with zero, we get x = – 1.
Then, remainder
= p(- 1)
= (- 1)³ + 3(- 1)² + 3(- 1) + 1
= – 1 + 3 – 3 + 1
= 0

(ii) x – $\frac{1}{2}$
Answer:
Divisor g (x) = x – $\frac{1}{2}$
x – $\frac{1}{2}$ = 0 gives x = $\frac{1}{2}$
Then, remainder
= p $\left(\frac{1}{2}\right)$
= $\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1$
= $\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$
= $\frac{27}{8}$

(iii) x
Answer:
Divisor g (x) = x.
x = 0 gives x = 0.
Then, remainder = p (0)
= (0)³ + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1
= 1

(iv) x + π
Answer:
Divisor g (x) = x + π.
x + π = 0 gives x = – π.
Then, remainder
= p(- π)
= (- π)³ + 3(- π)² + 3(- π) + 1
= – π³ + 3π² – 3π + 1

(v) 5 + 2x
Answer:
Divisor g(x) = 5 + 2x.
5 + 2x = 0 gives x = – $\frac{5}{2}$
Then, remainder
= P $\left(-\frac{5}{2}\right)$
= $\left(-\frac{5}{2}\right)^{3}+3\left(-\frac{5}{2}\right)^{2}+3\left(-\frac{5}{2}\right)+1$
= $-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1$
= $\frac{-125+150-60+8}{8}$
= $-\frac{27}{8}$

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Answer:
Here, p (x) = x³ – ax² + 6x – a and divisor
g (x) = x – a.
x – a = 0 gives x = a.
Then, remainder = p (a)
= (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
= 5a

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
Here, p (x) = 3x³ + 7x and divisor g (x) = 7 + 3x.
7 + 3x = 0 gives x = – $\frac{7}{3}$ .
Then, remainder = p $\left(-\frac{7}{3}\right)$
= $3\left(-\frac{7}{3}\right)^{3}+7\left(-\frac{7}{3}\right)$
= $-\frac{343}{9}-\frac{49}{3}$
= $\frac{-343-147}{9}$
= – $\frac{490}{9}$ ≠ 0
Since the remainder is not zero when
p (x) = 3x³ + 7x is divided by 7 + 3x, it is clear that 7 + 3x is not a factor of 3x³ + 7x.

PSEB 7th Class Science Notes Chapter 7 Weather, Climate and Adaptations of Animals to Climate

January 24, 2025January 23, 2025 by Bhagya

This PSEB 7th Class Science Notes Chapter 7 Weather, Climate and Adaptations of Animals to Climate will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 7 Weather, Climate and Adaptations of Animals to Climate

→ The weather of a place changes day by day and week by week.

→ The weather depends on temperature, humidity, and rainfall.

→ Humidity is a measure of water vapour in the air.

→ Indian Meteorological Department of weather forecasting daily collects statistical data of heat, wind speed at various places and makes weather predictions.

→ Atmospheric conditions in terms of temperature, humidity, rainfall, wind speed, etc. of a place are called the weather of that place.

→ The weather can change in an instant.

→ Factors on which the weather depends, are called elements of weather.

→ Special high-low thermometers are used to measure the temperature.

→ The highest temperature of the day is usually in the afternoon and the lowest is normal usually in the morning.

→ All changes in the weather are caused by the sun.

→ In winters, the length of the day is shorter and the night is earlier.

→ The length of the weather of a place is based on the data collected at that place is called the climate of that place.

→ The climate of different places is different. It changes from hot and dry to hot and humid.

→ Energy reflected and absorbed by the earth’s surface, ocean and atmosphere play important roles in determining weather at any place.

→ Climate has a great impact on living things.

→ Animals are adapted to the conditions in which they live.

→ The Polar Regions are located near the poles, such as the North Pole and the South Pole.

→ Canada, Greenland, Iceland, Norway, Sweden, Finland and Alaska, and Siberian areas of Russia in America, are the Polar Regions

→ Tropical rainforests are found in India, Malaysia, Indonesia, Brazil, the Republic of Congo, Kenya, Uganda, and Nigeria. The Polar Regions have a cold climate.

→ Rainfall is measured by an instrument called the rain gauge.

→ Penguins and polar bears live in Polar Regions.

→ The Polar Regions are covered with white ice.

→ The white hairs on the polar bear’s body help protect it and catch prey.

→ Penguins are well-known animals found in Polar Regions. It is also white and merges well with the background.

→ In addition to polar bears and penguins, many other animals are found in Polar Regions.

→ Many fishes can live in cold water.

→ The climate of the Subtropical Regions is generally warmer, as these areas are closer to the equator.

→ Temperatures in these regions vary from 15°C to 40°C.

→ In areas near the equator, the length of night and day are approximately equal throughout the year.

→ Weather: Everyday changes in the atmosphere in terms of temperature, humidity, rainfall, wind speed, etc., is called the weather of that place.

→ Climate: The average weather pattern taken over a long time, say 25 years is called the climate of the place.

→ Adaptation: The special characteristics of plants and organisms, that is, the nature that enables them to live in a habitat, is called adaptation.

→ Migration: Moving from one place to another to avoid harsh climatic conditions by birds and animals is called migration.

PSEB 7th Class Science Notes Chapter 6 Physical and Chemical Changes

January 24, 2025January 23, 2025 by Bhagya

This PSEB 7th Class Science Notes Chapter 6 Physical and Chemical Changes will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 6 Physical and Chemical Changes

→ Change is the tendency of life. There are many changes in our daily life.

→ There are two types of changes:

Physical changes
Chemical changes

→ There is always a reason for the change.

→ Some changes can be controlled and some others cannot be controlled.

→ No new matter is formed in physical change.

→ Chemical changes usually cannot be reversed.

→ The properties of new substances produced in a chemical change are completely different (new).

→ Changes can be classified based on their similarities.

→ The properties of a substance such as size, measure, color, state are called its physical properties.

→ The change that takes place in the physical properties of a substance is called a physical change.

→ The magnesium strip (ribbon) is bums with bright white light.

→ When Carbon dioxide is passed through lime water, it becomes milky.

→ Chemical change produces sound, light, heat, smell, gas, color, and so on.

→ Burning is a chemical change in which there is always an outflow of heat.

→ There is a layer of Ozone in the atmosphere.

→ For the occurrence of rust both oxygen and water are required.

→ In the galvanization process, a layer of zinc is deposited on the iron.

→ Iron can be saved from rust by applying paint.

→ Large crystals can be obtained from a saturated solution of a substance by crystallisation method.

→ Physical changes: Changes in which only the physical properties of matter change and no new matter is created are called physical changes. Example: Salt solution in water.

→ Chemical changes: Changes that involve the formation of new substances with new properties are called chemical changes. Example: Burning of coal.

→ Rust: The process in which iron gets covered with a layer of brownish substance in the presence of moist air is called rust.

→ Galvanization: The process of depositing zinc on iron to protect it from corrosion is called Galvanization.

→ Crystallisation: The process of obtaining large size crystals of a soluble substance is called crystallisation.

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.2

January 24, 2025January 23, 2025 by Bhagya

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at (i) x = 0, (ii) x = – 1 and (iii) x = 2.
Answer:
Here, p (x) 5x – 4x² + 3
(i) The value of polynomial p (x) at x = 0 is given by
p(0) = 5(0) – 4(0)² + 3
= 3

(ii) The value of polynomial p (x) at x = – 1 is given by
p(- 1) = 5(- 1) – 4 (- 1)² + 3
= – 5 – 4 + 3
= – 6

(iii) The value of polynomial p (x) at x = 2 is given by
p(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3
= – 3

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p (y) = y² – y + 1
Answer:
p(y) = y² – y + 1
∴ p(0) = (0)² – (0) + 1 = 1
∴ P(1) = (1)² – (1) + 1 = 1 – 1 + 1 = 1
∴ p(2) = (2)² – (2) + 1 = 4 – 2 + 1 = 3

(ii) p (t) = 2 + t + 2t² – t³
Answer:
p(t) = 2 + t + 2t² – t³
∴ p(0) = 2 + 0 + 2(0)² – (0)³ = 2
∴ p (1) = 2 + (1) + 2 (1)² – (1)³
= 2 + 1 + 2 – 1
= 4
∴ p(2) = 2 + (2) + 2(2)² – (2)³
= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x³
Answer:
p(x) = x³
∴ p (0) = (0)³ = 0
∴ p ( 1) = (1)³ = 1
∴ p (2) = (2)³ = 8

(iv) p(x) = (x – 1) (x + 1)
Answer:
p(x) = (x – 1) (x + 1)
∴ p(0) = (0 – 1) (0 + 1) = (- 1) × 1 = – 1
∴ p(1) = (1 – 1) (1 + 1) = 0 × 2 = 0
∴ p(2) = (2 – 1)(2 + 1) = 1 × 3 = 3

Question 3.
Verify whether the following are zeros of the polynomial, indicated against them:
(i) p(x) = 3x + 1, x = – $\frac{1}{3}$
Answer:
Here, p (x) = 3x + 1
Then, p $\left(-\frac{1}{3}\right)$ = 3 $\left(-\frac{1}{3}\right)$ + 1 = – 1 + 1 = 0
Hence, – $\frac{1}{3}$ is a zero of polynomial
p(x) = 3x + 1

(ii) p(x) = 5x – π, x = $\frac{4}{5}$
Answer:
Here, p(x) = 5x – π,
Then, p $\left(\frac{4}{5}\right)$ = 5 $\left(\frac{4}{5}\right)$ – π = 4 – π ≠ 0
Hence, $\frac{4}{5}$ is not a zero of polynomial
p(x) = 5x – π

(iii) p(x) = x² – 1, x = 1, – 1
Answer:
Here, p(x) = x² – 1
Then, p(1) = (1)² – 1 = 1 – 1 = 0 and
p(- 1) = (- 1)² – 1 = 1 – 1 = 0.
Hence, 1 and – 1 both are zeroes of polynomial p(x) = x² – 1.

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
Answer:
Here, p (x) = (x + 1) (x – 2)
Then, p(- 1) = (- 1 + 1) (- 1 – 2) = 0 × (-3)= 0
and p (2) = (2 + 1) (2 – 2) = 3 × O = O.
Hence, – 1 and 2 both are zeros of polynomial p(x) = (x + 1) (x – 2).

(v) p(x) = x², x = 0
Answer:
Here, p(x) = x²
Then, p(0) = (0)² = 0
Hence, 0 is a zero of polynomial p (x) = x².

(vi) p(x) = lx + m, x = – $\frac{n}{l}$
Answer:
Here, p (x) = lx + m
Then, p $\left(-\frac{m}{l}\right)$ = l $\left(-\frac{m}{l}\right)$ + m = – m + m = 0
Hence, $-\frac{m}{l}$ is a zero of polynomial
p(x) = lx + m.

(vii) p(x) = 3x² – 1, x = – $\frac{1}{\sqrt{3}}$ , $\frac{2}{\sqrt{3}}$
Answer:
Here, p(x) = 3x² – 1
Then, p $\left(-\frac{1}{\sqrt{3}}\right)$ = 3 $\left(-\frac{1}{\sqrt{3}}\right)^{2}$ – 1
= 3 $\left(\frac{1}{3}\right)$ – 1 = 1 – 1 = 0
and p $\left(\frac{2}{\sqrt{3}}\right)$ = 3 $\left(\frac{2}{\sqrt{3}}\right)^{2}$ – 1
= 3 $\left(\frac{4}{3}\right)$ – 1 = 4 – 1 = 3 ≠ 0
Hence, – $\frac{1}{\sqrt{3}}$ is a zero of polynomial p (x) = 3x² – 1, but $\frac{2}{\sqrt{3}}$ is not a zero of polynomial p(x) = 3x² – 1.

(viii) p (x) = 2x + 1, x = $\frac{1}{2}$
Answer:
Here, p(x) = 2x + 1
Then, p $\left(\frac{1}{2}\right)$ = 2 $\left(\frac{1}{2}\right)$ + 1 = 1 + 1 = 2 ≠ 0
Hence, $\frac{1}{2}$ is not a zero of polynomial
p(x) = 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Answer:
To find the zero of polynomial p (x) = x + 5,
we solve the equation p (x) = 0.
∴ x + 5 = 0
∴ x = – 5
Thus, – 5 is the zero of polynomial
p(x) = x + 5.

(ii) p(x) = x – 5
Answer:
To find the zero of polynomial p(x) = x – 5,
we solve the equation p (x) = 0.
∴ x – 5 = 0
∴ x = 5
Thus, 5 is the zero of polynomial
p(x) = x – 5.

(iii) p (x) = 2x + 5
Answer:
To find the zero of polynomial p(x) = 2x + 5,
we solve the equation p (x) = 0.
∴ 2x + 5 = 0
∴ 2x = – 5
Thus, – $\frac{5}{2}$ is the zero of polynomial
p(x) = 2x + 5.

(iv) p (x) = 3x – 2
Answer:
To find the zero of polynomial p (x) = 3x – 2,
we solve the equation p (x) = 0.
∴ 3x – 2 = 0
∴ 3x = 2
Thus, $\frac{2}{3}$ is the zero of polynomial
p(x) = 3x – 2.

(v) p(x) = 3x
Answer:
To find the zero of polynomial p (x) = 3x.
we solve the equation p (x) = 0.
∴ 3x = 0
∴ x = 0
Thus, 0 is the zero of polynomial P(x) = 3x.

(vi) p(x) = ax, a ≠ 0
Answer:
To find the zero of polynomial p (x) = ax,
a ≠ 0, we solve the equation p (x) = 0.
∴ ax = 0
∴ x = 0 (∵ a ≠ 0)
Thus, 0 is the zero of polynomial
p(x) = ax, a ≠ 0.

(vii) p(x) = cx + d, c ≠ 0, C, d are real numbers.
Answer:
To find the zero of polynomial
p(x) = cx + d, c ≠ 0, c, d are real numbers, we solve the equation p (x) = 0.
∴ cx + d = 0
∴ cx = – d
∴ x = – $\frac{d}{c}$
Thus, – $\frac{d}{c}$ is the zero of polynomial p (x) = cx + d, c ≠ 0, c, d are real numbers.