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

March 28, 2024March 27, 2024 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

March 28, 2024March 27, 2024 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

March 28, 2024March 27, 2024 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.

PSEB 7th Class Science Notes Chapter 5 Acids, Bases and Salts

March 28, 2024March 27, 2024 by Bhagya

This PSEB 7th Class Science Notes Chapter 5 Acids, Bases and Salts will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 5 Acids, Bases and Salts

→ We eat many foods with different tastes in our daily life.

→ Some foods taste bitter, sour, sweet, and salty.

→ The sour taste of the substance is due to the acid present in them.

→ The word acid is derived from the Latin word ‘acere’, which means sour.

→ Substances that have a bitter taste and are soapy to touch, are called the base.

→ Indicators are substances that give different colours to substances or solutions of acidic and basic nature.

→ They are used to test the acidic or alkaline nature of substances.

→ Turmeric, litmus, and China rose petals (Gudhal) are natural indicators.

→ Neutral solutions do not change the color of litmus red or blue because they are neither acidic nor basic.

→ Phenolphthalein is a synthetic indicator that is prepared in the laboratory.

→ Some acids are strong and some are weak.

→ The reaction between an acid and a base is called Neutralisation.

→ Antacids are used to treat indigestion.

→ Solution of baking soda (Sodium hydrogen carbonate) rubbed to treat the effects of insect bites.

→ The basicity of the soil is eliminated by the use of organic matter.

→ The acidity of the soil is treated with bases like a quick (calcium oxide) or slaked lime (Calcium hydroxide).

→ Waste products from factories should be released into the water only after treatment with basic substances.

→ Acids: Substances that have a sour taste and which react with blue litmus solution to make its color red are called acids.

→ Bases (Alkali): Substances that have a bitter taste and which react with red litmus solution to make it blue are called bases.

→ Neutralisation: The reaction between an acid and an alkali is called neutralisation.

→ Neutral solution: A solution that is neither acidic in nature nor alkaline in nature is called a neutral solution or solution which does not change the color of the indicator is called a neutral solution.

→ Indicators: Indicators are substances that show different colours by reacting with different acids, salts, and bases.

PSEB 7th Class Science Notes Chapter 4 Heat

March 28, 2024March 27, 2024 by Bhagya

This PSEB 7th Class Science Notes Chapter 4 Heat will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 4 Heat

→ The method of determining whether an object is hot or cold by touching it is not reliable.

→ Temperature is the degree of hotness or coldness of an object.

→ The hot or cold state of an object, i.e. the temperature is measured with an instrument called a thermometer.

→ The temperature of a human or living organism is measured with a medical thermometer (or clinical thermometer)

→ A scale is present on a medical thermometer. This scale is either in Celsius [°C] or Fahrenheit [°F] or both.

→ Medical thermometers have a narrow, homogeneous tube of glass with a bulb at the bottom.

→ There is a kink near the bulb of a medical thermometer that prevents the mercury level from falling as a result of its weight.

→ Should clean with antiseptic solution before and after use of a medical thermometer.

→ The scale range of medical thermometers is from 35°C to 42°C.

→ Before using a medical thermometer, make sure that the mercury level is below 35° C.

→ If not, hold the thermometer firmly and shake it to bring a level of mercury below 35°C.

→ The normal temperature of a healthy person is 37°C or 98.4°F.

→ There are other thermometers to measure the temperature of objects.

→ One of these is the Lab thermometer. The lab thermometer ranges from -10°C to 110° C.

→ The temperature of the object should be measured with a lab thermometer when the mercury level in the thermometer becomes stable.

→ The method by which heat is transferred from the hotter end of an object to the cold end by the particles of material of the object is called conduction. Solids are heated by the conduction method.

→ Materials that transmit heat are called conductors.

→ Items made of iron, silver, copper, aluminium are heat conductors.

→ Substances that do not conduct heat well are called insulators or non-conductors, e.g. Wood, plastic, and rubber.

→ Air is not a good conductor of heat.

→ The method of heat transfer in which heat is transmitted by the motion of hot molecules of a substance is called convection.

→ Heat is transmitted in liquids and gases by the convection method,

→ In coastal areas, the wind that blows from the sea to the coast during the day is called Sea breeze.

→ In coastal areas, the air that blows from the coast to the sea at night is called land breeze.

→ Radiation is the method of transfer of heat due to the emission of radiations by hot objects without a medium.

→ Dark coloured fabrics absorb heat more than light coloured fabrics.

→ So in winter, we wear dark colours and in summer we wear light coloured clothes.

→ Woolen garments keep us warm in winter because of the air trapped in the wool fibres which is a bad conductor of heat.

→ Heat: It is an agent which produces in us the sensation of warmth. This is a kind of energy.

→ Temperature: Degree of hotness or coldness measured on a definite scale.

→ Thermometer: An instrument used to measure the temperature of an object.

→ Celsius Scale: The Celsius scale is a scale for measuring temperature. Sometimes it is also called as the centigrade scale.

→ Resistor: A substance through which heat cannot be transmitted properly is considered to be resistor or resistant to heat.

→ Conduction: This is the method of heat transfer in which heat passes from the hot end of the object to the cold end through the molecules of the material, but the molecules of the object remain fixed in their place.

→ Convection(Liquid or Gas): This is the method of heat transfer in which heat is transferred from the source of heat to the cold part due to the motion of hot molecules and the cold molecules come down to take their place. This method is used for heating liquids and gases.

→ Radiation: This is the method of heat transfer in which heat is transmitted from a hot source or body directly to a cool body without affecting the medium.

→ Sea breeze: Due to the heat of the sun during the day, the soil of the land gets heated very quickly when water molecules of the ocean are not so hot. Therefore, the air near the ground gets warmer and rises up due to being lighter.

→ To replace it, cool air begins to flow from the ocean, causing air currents to flow. The wind that blows from the ocean to the land is called the sea breeze.

→ Land breeze: Due to high heat absorption capacity, water cools later than land, due to which cold air starts flowing towards seawater which is called land breeze.

→ Fahrenheit Scale: The Fahrenheit scale is a scale designed to measure temperature.

PSEB 6th Class Maths MCQ Chapter 1 Knowing Our Numbers

March 27, 2024March 26, 2024 by Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 1 Knowing Our Numbers MCQ Questions

Multiple Choice Questions

Question 1.
The number of digits are:
(a) 9
(b) 10
(c) 8
(d) Infinite.
Answer:
(b) 10

Question 2.
The greatest 4 digit number using 1, 5, 2, 9 once is:
(a) 9215
(b) 9512
(c) 5912
(d) 9521.
Answer:
(b) 9512

Question 3.
The smallest 4 digit number using 2, 0, 3, 7 once is:
(a) 0237
(b) 2037
(c) 7320
(d) 7023.
Answer:
(b) 2037

Question 4.
Which of the following are in ascending order?
(a) 217, 271, 127, 721
(b) 217, 127, 721, 271
(c) 127, 217, 271, 721
(d) 721, 271, 217, 127.
Answer:
(c) 127, 217, 271, 721

Question 5.
The face value of digit 4 in 23468 is:
(a) 4
(b) 400
(c) 40
(d) 468.
Answer:
(a) 4

Question 6.
The place value of digit 2 in 4123 is:
(a) 23
(b) 2
(c) 20
(d) 200.
Answer:
(c) 20

Question 7.
The difference between place value and face value of 5 in 76542 is:
(a) 537
(b) 45
(c) 0
(d) 495
Answer:
(d) 495

Question 8.
5 × 10000 + 3 × 100 + 2 × 10 + 2 = …………..
(a) 5322
(b) 53022
(c) 50322
(d) 53202.
Answer:
(c) 50322

Question 9.
Four lakh two thousand three hundred fifty-one = …………..
(a) 42351
(b) 402351
(c) 420351
(d) 4002351.
Answer:
(b) 402351

Question 10.
How many four-digit numbers are there?
(a) 9999
(b) 9900
(c) 9000
(d) 9990.
Answer:
(c) 9000

Question 11.
Seventeen million twenty-four thousand fifty-four = …………….
(a) 172454
(b) 170024054
(c) 170240054
(d) 17024054.
Answer:
(d) 17024054.

Question 12.
1 Crore = …………….. million.
(a) 1
(b) 10
(c) 100
(d) 1000.
Answer:
(b) 10

Question 13.
Rounded off 7213 to nearest thousands.
(a) 7200
(b) 7000
(c) 7210
(d) 7213.
Answer:
(b) 7000

Question 14.
Rounded off 45553 to nearest hundreds.
(a) 45500
(b) 45550
(c) 45600
(d) 45650.
Answer:
(c) 45600

Question 15.
Solve : (9 – 4) × 6 = …………….. .
(a) 30
(b) 54
(c) 78
(d) 64.
Answer:
(a) 30

Question 16.
Which of the following number does not have symbol in Roman numerals?
(a) 0
(b) 1
(c) 10
(d) 1000.
Answer:
(a) 0

Question 17.
How many symbols are used in Roman Numerals?
(a) 5
(b) 8
(c) 9
(d) 7.
Answer:
(d) 7

Question 18.
Which of the following are meaningless?
(a) LXIX
(b) XC
(c) IL
(d) LI.
Answer:
(c) IL

Question 19.
CLXVI = ………..
(a) 164
(b) 144
(c) 176
(d) 166.
Answer:
(d) 166

Question 20.
XCIX + XLVI = …………….
(a) CVL
(b) CLV
(c) CXLV
(d) CXLIV.
Answer:
(c) CXLV

Question 21.
Using the digits 4, 5, 7 and 0 without repetition which of the following is the smallest four-digit number?
(a) 0457
(b) 4057
(c) 4507
(d) 4075.
Answer:
(b) 4057

Question 22.
Using the digits 2, 8, 7 and 4 without repetition which of the following is the greatest four-digit number?
(a) 2874
(b) 8742
(c) 8472
(d) 8274.
Answer:
(b) 8742

Question 23.
Which is the smallest four digits number made from the digits 3, 8, 7 by using one-digit twice?
(a) 3378
(b) 3783
(c) 3873
(d) 3837.
Answer:
(a) 3378

Question 24.
Make the greatest four-digit number from the digits 9, 0, 5 by using one-digit twice.
(a) 9005
(b) 9905
(c) 9950
(d) 9050.
Answer:
(c) 9950

Question 25.
Take two digits, 2 and 3, from diem make smallest four digit number, using both the digits equal number of time.
(a) 3232
(b) 2323
(c) 3223
(d) 2233.
Answer:
(d) 2233.

Question 26.
Take two digits, 2 and 3 from them make greatest four-digit number, using both the digits equal number of time.
(a) 3232
(b) 3322
(c) 3223
(d) 2323.
Answer:
(b) 3322

Question 27.
The greatest number from 4536, 4892, 4370, 4452 is:
(a) 4536
(b) 4892
(c) 4370
(d) 4452.
Answer:
(b) 4892

Question 28.
Out of 15623, 15073, 15189, 15800 the smallest number is:
(a) 15623
(b) 15073
(c) 15189
(d) 15800.
Answer:
(b) 15073

Question 29.
The ascending order of the numbers 847, 9754, 8320, 571 is:
(a) 847, 9754, 8320, 571
(b) 9754, 8320, 847, 571
(c) 571, 847, 8320, 9754
(d) 571, 8320, 847, 9754.
Answer:
(c) 571, 847, 8320, 9754

Fill in the blanks:

Question 1.
1 lakh = ten thousands.
Answer:
Ten

Question 2.
1 million = ……………… hundred thousand.
Answer:
Ten

Question 3.
1 crore = ……………….. million.
Answer:
Ten

Question 4.
1 crore = …………… ten lakh.
Answer:
Ten

Question 5.
1 million = ……………. lakh.
Answer:
Ten

Write True/False:

Question 1.
The number of digits are 10. (True/False)
Answer:
True

Question 2.
The greatest four-digit number is 1000. (True/False)
Answer:
False

Question 3.
The place value of digit 5 in 3564 is 50. (True/False)
Answer:
False

Question 4.
0 does not have symbol in Roman numbers. (True/False)
Answer:
True

Question 5.
IL is meaningless. (True/False)
Answer:
True

PSEB 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1

March 27, 2024March 26, 2024 by Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

1. Find the ratio of :

Question (i).
₹ 5 to 50 paise
Solution:
To find ratio of ₹ 5 to 50 paise.
Firstly convert both the quantities into same units.
₹ 1 = 100 paise
₹ 5 = 5 × 100 paise = 500 paise.
So, ratio of 500 paise to 50 paise
= \(\frac{500}{50}=\frac{10}{1}\)
= 10 : 1
Hence, required ratio is 10 : 1

Question (ii).
15 kg to 210 g
Solution:
To find ratio of 15 kg to 210 g
Firstly, convert both the quantities into same unit
1 kg = 1000 g
15 kg = 15 × 1000 g = 15000 g.
So, ratio of 15000 to 210 g
= \(\frac{15000}{210}=\frac{500}{7}\)
= 500 : 7
Hence, required ratio is 500 : 7

Question (iii).
4 m to 400 cm
Solution:
To find ratio of 4 m to 400 cm
Firstly convert both the quantities into same unit.
1 m = 100 cm
4 m = 4 × 100 cm = 400 cm
So, ratio of 400 cm to 400 cm
= \(\frac{400}{400}=\frac{1}{1}\)
= 1 : 1
Hence, required ratio is 1 : 1

Question (iv).
30 days to 36 hours
Solution:
To find ratio of 30 days to 36 hours
1 day = 24 hours
30 days = 30 × 24 hours
= 720 hours
So, ratio of 720 hours to 36 hours
= \(\frac{720}{36}=\frac{20}{1}\)
= 20 : 1
Hence required ratio is 20 : 1

2. Are the ratios 1: 2 and 2 :3 equivalent ?
Solution:
To check this we need to know whether
1 : 2 and 2 : 3 are equal
First convert the ratios into fractions
1 : 2 is written as \(\frac {1}{2}\)
2 : 3 is written as \(\frac {2}{3}\)
Now to change these fraction into like fraction.
Make denominator of both the fraction same.
\(\frac{1}{2}=\frac{1}{2} \times \frac{3}{3}=\frac{3}{6}\) and \(\frac{2}{3}=\frac{2}{3} \times \frac{2}{2}=\frac{4}{6}\)
4 > 3
\(\frac{4}{6}>\frac{3}{6}\)
Hence 1 : 2 and 2 : 1 are not equivalent

3. If the cost of 6 toys is ₹ 240, find the cost of 21 toys.
Solution:
The more the number of toys one purchases, the more is the amount to be paid.
Therefore, there is a direct proportion between the number of toys and the amount to be paid.
Let x be number of toys to be purchased
∴ 6 : 240 : : 21 : x
\(\frac{6}{240}=\frac{21}{x}\)
x = \(\frac{21 \times 240}{6}\) = ₹ 840
Thus cost of 21 toys is ₹ 840.

4. The car that I own can go 150 km with 25 litres of petrol. How far can it go with 30 litres of petrol ?
Solution:
More km ↔ More petrol
So, the consumption of petrol and the distance travelled by a car is a case of direct proportion.
Let the distance travelled be x km.
∴ 150 : 25 : : x : 30
\(\frac{150}{25}=\frac{x}{30}\)
x = \(\frac{150 \times 30}{25}\)
x = 180
Thus, it can go 180 km is 30 litres of petrol.

5. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students ?
Solution:
The more the number of students, the more is the number of computer needed.
More students ↔ More will be the computed needed.
There is a direct proportion between the number of students and the number of computers needed. Let computer needed will be x.
6 : 3 : : 24 : x
\(\frac{6}{3}=\frac{24}{x}\)
6 × x = 24 × 3
x = \(\frac{24 \times 3}{6}\) = 12
Thus, 12 computers will be needed.

PSEB 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.5

March 27, 2024March 26, 2024 by Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.5

1. Which of the following are meaningless:

Question (a)
IC
Solution:
Since I can be subtracted only from V and X not from C
So IC is meaningless.

Question (b)
VD
Solution:
VD, since V cannot be subtracted
So VD is meaningless.

Question (c)
XCVII
Solution:
XCVII = XC + VII = 90 + 7 = 97
So XCVII is meaningful.

Question (d)
IVC
Solution:
Since IV cannot be subtracted from C.
So IVC is meaningless.

Question (e)
XM.
Solution:
Since X can be subtracted only from L and C not from M.
So XM is meaningless.

2. Write the following in Hindu Arabic Numerals:

Question (a)
XXV
Solution:
XXV = X + X + V
= 10 + 10 + 5 = 25

Question (b)
XLV
Solution:
XLV = XL + V = 40 + 5 = 45

Question (c)
LXXIX
Solution:
LXXIX = L + X + X + IX
= 50 + 10 + 10 + 9 = 79

Question (d)
XCIX
Solution:
XCIX = XC + IX = 90 + 9 = 99

Question (e)
CLXIX
Solution:
CLXIX = CL + X + IX = 40 + 10 + 9 = 59

Question (f)
DCLXII
Solution:
DCLXII = D + C + L + X + II
= 500 + 100 + 50 + 10 + 2 = 662

Question (g)
DLXIX
Solution:
DLXIX = D + L + X + IX
= 500 + 50 + 10 + 9
= 569

Question (h)
DCCLXVI
Solution:
DCCLXVI = D + CC + L + X + VI
= 500 + 200 + 50 + 10 + 6 = 766

Question (i)
CDXXXVIII
Solution:
CDXXXVIII = CD + XXX + VIII
= 400 + 30 + 8 = 438

Question (j)
MCCXLVI = M + CC + XL + VI
= 1000 + 200 + 40 + 6
= 1246

3. Express each of the following as Roman numerals:

Question (i)
(a) 29
(b) 63
(c) 94
(d) 99
(e) 156
(f) 293
(g) 472
(h) 638
(i) 1458
(j) 948
(k) 199
(l) 499
(m) 699
(n) 299
(o) 999
(p) 1000
Solution:
(a) 29 = 20 + 9 = XXIX
(b) 63 = 60 + 3 = LXIII
(c) 94 = 90 + 4 = XCIV
(d) 99 = 90 + 9 = XCIX
(e) 156 = 100 + 50 + 6 = CLVI
(f) 293 = 200 + 90 + 3 = CCXCffl
(g) 472 = 400 + 70 + 2 = CDLXXII
(h) 638 = 600 + 30 + 8 = DCXXXVIII
(i) 1458 = 1000 + 400 + 50 + 8 = MCDLVIII
(j) 948 = 900 + 40 + 8 = CMXLVIH
(k) 199 = 100 + 90 + 9 = CXCIX
(l) 499 = 400 + 90 + 9 = CDXCIX
(m) 699 = 600 + 90 + 9 = DCXCIX
(n) 299 = 200 + 90 + 9 = CCXCIX
(o) 999 = 900 + 90 + 9 = CMXCIX
(p) 1000 = M.

PSEB 7th Class Maths MCQ Chapter 7 Congruence of Triangles

March 27, 2024March 26, 2024 by Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 7 Congruence of Triangles MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 7 Congruence of Triangles MCQ Questions

Multiple Choice Questions :

Question 1.
If ∠ABC ≅ ΔDEF, then which of the following statement is correct ?
(a) ∠A = ∠D
(b) ∠A = ∠E
(c) ∠B = ∠D
(d) ∠C = ∠E
Answer:
(a) ∠A = ∠D

Question 2.
If ΔABC ≅ ΔDEF, then which of the following statement is correct ?
(a) AB = EF
(b) BC = DE
(c) BC = EF
(d) AB = EF
Answer:
(c) BC = EF

Question 3.
Which of the following is congruent ?
(a) Shaving blades of the same company
(b) Sheets of the same letter pad.
(c) Biscuits of the same packet.
(d) All of above three are congruent
Answer:
(d) All of above three are congruent

Question 4.
Two line segments are congruents :
(a) Their shapes are same
(b) Direction is same
(c) Lengths are same
(d) All of the above.
Answer:
(c) Lengths are same

Question 5.
Out of two congruent angles measure of one angle is 70°, then the measure of other angle is :
(a) 70°
(b) 110°
(c) 90°
(d) 140°
Answer:
(a) 70°

Question 6.
When we write ∠A = ∠B then we really means :
(a) A = B
(b) m∠A = m∠B
(c) A and B are in the same direction
(d) A and B are of same shape
Answer:
(b) m∠A = m∠B

Question 7.
When we write ΔABC ≅ ΔDEF, we means:
(a) AB = DE
(b) BC = EF
(c) AC = DF
(d) All of the above
Answer:
(d) All of the above

Question 8.
If ΔABC ≅ ΔQPR, then which of the following statement is correct ?
(a) ∠A = ∠P
(b) ∠B = ∠R
(c) ∠B = ∠P
(d) ∠B = ∠Q
Answer:
(c) ∠B = ∠P

Fill in blanks :

Question 1.
When we write ∠A = ∠B, there we really means.
Answer:
m∠A = m∠B

Question 2.
Two lines segment are equal if their length are …………….
Answer:
equal

Question 3.
…………….. symbol is used to denote congruence between two figure.
Answer:
≅

Question 4.
The figures having same shape and size are called …………….. figures.
Answer:
congruent

Question 5.
…………….. stands for right angle hypotenuse side.
Answer:
RHS

Write True or False

Question 1.
Shaving blades of the same company are congruent. (True/False)
Answer:
True

Question 2.
Sheets of the same letter pad are congruent. (True/False)
Answer:
True

Question 3.
Two line segments are congruent if their shapes are same. (True/False)
Answer:
False

Question 4.
AAA is one of the laws of congruency of triangles. (True/False)
Answer:
False

Question 5.
Biscuits of the same packet are congruent. (True/False)
Answer:
True

PSEB 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.4

March 27, 2024March 26, 2024 by Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.4

1. Simplify each of following:

Question (a)
13 × 104
Solution:
13 × 104 = (10 + 3) (104)
= 10 × 104 + 3 × 104
= 10 × (100 + 4) + 3 × (100 + 4)
= 10 × 100 + 10 × 4 + 3 × 100 + 3 × 4
= 1000 + 40 + 300 + 12
= 1352

Question (b)
102 × 105
Solution:
102 × 105 = (100 + 2) × 105 = 100 × 105 + 2 × 105
= 100 × (100 + 5) + 2 × (100 + 5)
= 100 × 100 + 100 × 5 + 2 × 100 + 2 × 5
= 10000 + 500 + 200 + 10
= 10710

Question (c)
6 × 107
Solution:
6 × 107 = 6 × (100 + 7)
= 6 × 100 + 6 × 7
= 600 + 42
= 642

Question (d)
16 × 106
Solution:
16 × 106 = (10 + 6) × 106 = 10 × 106 + 6 × 106
= 10 × (100 + 6) + 6 × (100 + 6)
= 10 × 100 + 10 × 6 + 6 × 100 + 6 × 6
= 1000 + 60 + 600 + 36 = 1696

Question (e)
201 × 205
Solution:
201 × 205 = (200 + 1) × 205 = 200 × 205 + 1 × 205
= 200 × (200 + 5) + 1 × (200 + 5)
= 200 × 200 + 200 × 5 + 1 × 200 + 1 × 5
= 40000 + 1000 + 200 + 5 = 41205

Question (f)
22 × 102
Solution:
22 × 102 = (20 + 2) × 102 = 20 × 102 + 2 × 102
= 20 × (100 +-2) + 2 × (100 + 2)
= 20 × 100 + 20 × 2 + 2 × 100 + 2 x 2
= 2000 + 40 + 200 + 4 = 2244

Question (g)
6 × (4 + 3)
Solution:
6 × (4 + 3) = 6 × 4 + 6 × 3
= 24 + 18 = 42

Question (h)
(17 – 9) × 3
Answer:
(17 – 9) × 3
= 17 × 3 – 9 × 3
= 51 – 27 = 24

Question (i)
(20 + 4) ÷ 2
Solution:
= 20 ÷ 2 + 4 ÷ 2
= 10 + 2 = 12