Selina Class 6 ICSE Solutions Archives

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution

November 10, 2023May 18, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution (Including Use of Brackets as Grouping Symbols)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution (Including Use of Brackets as Grouping Symbols)

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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IMPORTANT POINTS

Substitution : The value of an expression depends on the value of its variable (s).
Use of Brackets :
The Symbols —, ( ), { }, [ ] are called brackets.
If an expression is enclosed within a bracket, it is considered a single quantity, even if it is made up of many terms.
Keep in Mind :
- While simplifying an expression containing a bracket, first of all, the terms inside the bracket are operated (combined).
- ( ) is called a small bracket or Parenthesis.
- { } is called a middle bracket or Curly bracket.
- [ ] is called big or square bracket.
- If one more bracket is needed, then we use the bar bracket.
  i.e. a line ———— is drawn over a group of terms.
  Thus, in \(3x+\bar { 4y-5z }\), the line over 4y – 5z serves as the bar bracket and is called Vinculum.

Substitution Exercise 20A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the following blanks, when :
x = 3,y = 6, z = 18, a = 2, b = 8, c = 32 and d = 0.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 1

Solution:

Question 2.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 5
Solution:

Question 3.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 8
Solution:

Question 4.
If a = 3, b = 0, c = 2 and d = 1, find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 10
Solution:

Question 5.
Find the value of 5x² – 3x + 2, when x = 2.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 13

Question 6.
Find the value of 3x³ – 4x² + 5x – 6, when x = -1.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 14

Question 7.
Show that the value of x³ – 8x² + 12x – 5 is zero, when x = 1.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 15

Question 8.
State true and false :
(i) The value of x + 5 = 6, when x = 1
(ii) The value of 2x – 3 = 1, when x = 0
(iii) \(\frac { 2x-4 }{ x+1 }\) = -1,when x = 1
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 16

Question 9.
If x = 2, y = 5 and z = 4, find the value of each of the following :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 18
Solution:

Question 10.
If a = 3, find the values of a² and 2^a.
Solution:
a² = (3)² = 3 x 3 = 9
2^a = (2)³ = 2 x 2 x 2 = 8

Question 11.
If m = 2, find the difference between the values of 4m³ and 3m⁴.
Solution:
4m³ = 4 (2)³ = 4 x 2 x 2 x 2 = 32
3m⁴ = 3 (2)⁴ = 3 x 2 x 2 x 2 x 2 = 48
Now, a difference 3m⁴ – 4m³ = 48 – 32 = 16

Substitution Exercise 20B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Evaluate :
(i) (23 – 15) + 4
(ii) 5x + (3x + 7x)
(iii) 6m – (4m – m)
(iv) (9a – 3a) + 4a
(v) 35b – (16b + 9b)
(vi) (3y + 8y) – 5y
Solution:
(i) (23 – 15) + 4 = 8 + 4 = 12
(ii) 5x + (3x + 7x) = 5x + 10x = 15x
(iii) 6m – (4m – m) = 6m – 3m = 3m
(iv) (9a – 3a) + 4a = 6a + 4a = 10a
(v) 35b – (16b + 9b)= 35b – 25b = 10b
(vi) (3y + 8y) – 5y = 11y – 5y = 6y

Question 2.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 20

Solution:

Question 3.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 25
Solution:

Substitution Exercise 20C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 28
Solution:

(viii) 2t + r – p – q + s = 2t + r – (p + q – s)

Question 2.
Insert the bracket as indicated :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 30
Solution:

Substitution Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the value of 3ab + 10bc – 2abc when a = 2, b = 5 and c = 8.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 32

Question 2.
If x = 2, = 3 and z = 4, find the value of 3x² – 4y² + 2z².
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 33

Question 3.
If x = 3, y = 2 and z = 1; find the value of:
(i) x^y
(ii) y^x
(iii) 3x² – 5y²
(iv) 2x – 3y + 4z + 5
(v) y² – x² + 6z²
(vi) xy + y²z – 4zx
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 34

Question 4.
If P = -12x² – 10xy + 5y², Q = 7x² + 6xy + 2y², and R = 5x² + 2xy + 4y² ; find :
(i) P – Q
(ii) Q + P
(iii) P – Q + R
(iv) P + Q + R
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 35

Question 5.
If x = a² – bc, y = b² – ca and z = c² – ab ; find the value of :
(i) ax + by + cz
(ii) ay – bx + cz
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 37

Question 6.
Multiply and then evaluate :
(i) (4x + y) and (x – 2y); when x = 2 and y = 1.
(ii) (x² – y) and (xy – y²); when x = 1 and y = 2.
(iii) (x – 2y + z) and (x – 3z); when x = -2, y = -1 and z = 1.
Solution:

Question 7.
Simplify :
(i) 5 (x + 3y) – 2 (3x – 4y)
(ii) 3x – 8 (5x – 10)
(iii) 6 {3x – 8 (5x – 10)}
(iv) 3x – 6 {3x – 8 (5x – 10)}
(v) 2 (3x² – 4x – 8) – (3 – 5x – 2x²)
(vi) 8x – (3x – \(\bar { 2x-3 }\))
(vii) 12x² – (7x – \(\bar { 3x^{ 2 }+15 }\))
Solution:

Question 8.
If x = -3, find the value of : 2x³ + 8x² – 15.
Solution:

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions

November 10, 2023May 18, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions (Including Evaluation)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions (Including Evaluation)

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Framing Algebraic Expressions Exercise 21 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write in the form of an algebraic expression :
(i) Perimeter (P) of a rectangle is two times the sum of its length (l) and its breadth (b).
(ii) Perimeter (P) of a square is four times its side.
(iii) Area of a square is square of its side.
(iv) Surface area of a cube is six times the square of its edge.
Solution:
(i) Let P be the perimeter and / be the length, and b be the breadth.
P = 2 (l + b)
(ii) Let P be the perimeter and a be the side of the square.
P = 4a
(iii) Let A be the area of the square and a be the sides of the square.
A = (a)²
(iv) Let S be the surface area and a be the edges of the cube.
S = 6a²

Question 2.
Express each of the following as an algebraic expression :
(i) The sum of x and y minus m.
(ii) The product of x and y divided by m.
(iii) The subtraction of 5m from 3n and then adding 9p to it.
(iv) The product of 12, x, y and z minus the product of 5, m and n.
(v) Sum of p and 2r – s minus sum of a and 3n + 4x.
Solution:
(i) x + y – m
(ii) \(\frac { xy }{ m }\)
(iii) 3n – 5m + 9p
(iv) 12xyz – 5mn
(v) p + 2r – s – (a + 3n + 4x)

Question 3.
Construct a formula for the following :
Total wages (₹ W) of a man whose basic wage is (₹ B) for t hours week plus (₹ R) per hour, if he Works a total of T hours.
Solution:
Wages for t hours = ₹ B
Wages for overtime = R(T – t)
=> Total wages = Wages for t hours + wages for overtime of (T – t) hours
=> ₹ W = ₹ B + ₹ R (T – t)

Question 4.
If x = 4, evaluate :
(i) 3x + 8
(ii) x² – 2x
(iii) \(\frac { { x }^{ 2 } }{ 2 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 1

Question 5.
If m – 6, evaluate :
(i) 5m – 6
(ii) 2m² + 3m
(iii) (2m)²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 2

Question 6.
If x = 4, evaluate :
(i) 12x + 7
(ii) 5x² + 4x
(iii) \(\frac { { x }^{ 2 } }{ 8 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 3

Question 7.
If m = 2, evaluate :
(i) 16m – 7
(ii) 15m² – 10m
(iii) \(\frac { 1 }{ 4 } \times { m }^{ 3 }\)
Solution:
16m – 7
= (16 x 2) – 7
= 32 – 7 = 25
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 4

Question 8.
If x = 10, evaluate :
(i) 100x + 225
(ii) 6x² – 25x
(iii) \(\frac { 1 }{ 50 } \times { x }^{ 3 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 5

Question 9.
If a = – 10, evaluate :
(i) 5a
(ii) a²
(iii) a³
Solution:
(i)5a
= 5 x (-10) = -50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 6

Question 10.
If x = – 6, evaluate :
(i) 11x
(ii) 4x²
(iii) 2x³
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 7

Question 11.
If m = – 7, evaluate :
(i) 12m
(ii) 2m²
(iii) 2m³
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 8

Question 12.
Find the average (A) of four quantities p, q, r and s. If A = 6, p = 3, q = 5 and r = 7 ; find the value of s.
Solution:
Given, average of four quantities (A) = 6
and p = 3,q = 5, r = 7 and s = ?
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 9

Question 13.
If a = 5 and b = 6, evaluate :
(i) 3ab
(ii) 6a²b
(iii) 2b²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 10

Question 14.
If x = 8 and y = 2, evaluate :
(i) 9xy
(ii) 5x²y
(iii) (4y)²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 11

Question 15.
If x = 5 and y = 4, evaluate :
(i) 8xy
(ii) 3x²y
(iii) 3y²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 13

Question 16.
If y = 5 and z = 2, evaluate :
(i) 100yz
(ii) 9y²z
(iii) 5y²
(iv) (5z)³
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 14

Question 17.
If x = 2 and y = 10, evaluate :
(i) 30xy
(ii) 50xy²
(iii) (10x)²
(iv) 5y²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 15

Question 18.
If m = 3 and n = 7, evaluate :
(i) 12mn
(ii) 5mn²
(iii) (10m)²
(iv) 4n²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 16

Question 19.
If a = -10, evaluate :
(i) 3a – 2
(ii) a² + 8a
(iii) \(\frac { 1 }{ 5 }\) x a2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 17

Question 20.
If x = -6, evaluate :
(i) 4x – 9
(ii) 3x² + 8x
(iii) \(\frac { { x }^{ 2 } }{ 2 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 18

Question 21.
If m = -8, evaluate :
(i) 2m + 21
(ii) m² + 9m
(iii) \(\frac { { m }^{ 2 } }{ 4 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 20

Question 22.
If p = -10, evaluate :
(i) 6p + 50
(ii) 3p² – 20p
(iii) \(\frac { { p }^{ 2 } }{ 50 }\)
Solution:
(i) 6p + 50
= (6 x p) + 50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 21

Question 23.
If y = -8, evaluate :
(i) 6y + 53
(ii) y²+ 12y
(iii) \(\frac { { y }^{ 3 } }{ 4 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 22

Question 24.
If x = 2 and 7 = -4, evaluate :
(i) 11xy
(ii) 5x²y
(iii) (5y)²
(iv) 8x²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 24

Question 25.
If m = 9 and n = -2, evaluate
(i) 4mn
(ii) 2m²n
(iii) (2n)³
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 25

Question 26.
If m = -8 and n = -2, evaluate :
(i) 12mn
(ii) 3m²n
(iii) (4n)²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 27

Question 27.
If x = -5 and y = -8, evaluate :
(i) 4xy
(ii) 2xy²
(iii) 4x²
(iv) 3y²
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 28

Question 28.
Find T, if T = 2a – b, a = 7 and b = 3.
Solution:
T = 2a – b, a = 1 and b = 3
Put the value of a = 1, and b = 3 in above equation
T = (2 x 7) -3
T = 14 – 3 = 11
T = 11

Question 29.
From the formula B = 2a² – b², calculate the value of B when a = 3 and b = -1.
Solution:
B = 2a² – b²
Put the values of a = 3 and b = -1 in above equation
B = 2 x (3)² – (-1)²
B = 18 – 1
B = 17
Value of B is = 17

Question 30.
The wages ₹ W of a man earning ₹ x per hour for t hours are given by the formula W = xt. Find his wages for working 40 hours at a rate of ₹ 39.45 per hour.
Solution:
T = 40 hours
x = ₹ 39.45
W = xt = 40 x 39.45
W = ₹ 1578

Question 31.
The temperature in Fahrenhiet scale is represented by F and the tempera¬ture in Celsius scale is represented by C. If F = \(\frac { 9 }{ 5 }\) x C + 32, find F when C = 40.
Solution:
F = \(\frac { 9 }{ 5 }\) x C + 32
Given, C = 40
F = \(\frac { 9 }{ 5 }\) x 40 + 32 = 9 x 8 + 32
F = 104°

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 3 Numbers in India and International System

November 10, 2023May 16, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 3 Numbers in India and International System (With Comparison)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 3 Numbers in India and International System (With Comparison)

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Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

Place -Value (Local Value) : The place – value (Local-value) of a digit depends upon the position, it occupies in the number:
For example :
In the number 6453 the place value of 6 is 6 thousand = 6 x 1000 = 6000
the place value of 4 is 4 hundred = 4 x 100 = 400
the place value of 5 is 5 ten = 5 x 10 = 50
the place value of 3 is 3 one = 3 x 1=3
Face -Value (True-Value) : Each digit in a number has a fixed value, regardless to its position in the number
For Example :
In the above number 6453
Face of value of 6 is 6, 4 is 4, 5 is 5 and 3 is 3
An Abstract Number : Which does not refer to any particular unit e.g., 3, 5, 9.
A Concrete Number : Which refers to particular unit e.g., 3 boys, 5 girls, 9 rooms etc.
The smallest (least) one digit Number is 1.
The largest (greatest) one digit number is 9.

Numbers in India and International System Exercise 3 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write the following numerals using Indian System words or International system (as required in words) :
(i) 4, 35, 342 = …………..
(ii) 36, 71, 430 = …………..
(iii) 4, 28, 30, 004 = …………..
(iv) 75, 132, 684 = ……………
(v) 815, 906 = …………….
(vi) 5, 420, 700 = ……………..
Solution:
(i) 4, 35, 342 = four lakh, thirty five thousand, three hundred forty two
(ii) 36, 71, 430 = Thirty six lakh, seventy one thousand four hundred thirty
(iii) 4, 28, 30, 004 = four crore, twenty eight lakh, thirty thousand four
(iv) 75, 132, 684 = Seventy five million, one hundred thirty two thousand, six hundred eighty four
(v) 815, 906 = Eight hundred fifteen thousand, nine hundred six
(vi) 5, 420, 700 = Five million, four hundred twenty thousand, seven hundred

Question 2.
Write the following numbers, placing the commas, according to Indian system :
(i) 835629 = …………
(ii) 35640254 = …………
(iii) 2826040 = ………
Solution:
(i) 835629 = 8, 35, 629
(ii) 35640254 = 3, 56, 40, 254
(iii) 2826040 = 28, 26, 041

Question 3.
Write the following numbers, by placing the commas, according to International system :
(i) 6509820 = ………..
(ii) 428140584 = …………
(iii) 63560981 = ……….
Solution:
(i) 6509820 = 6, 509, 820
(ii) 428140584 = 428, 140, 584
(iii) 63560981 = 63, 560, 981

Question 4.
Fill in the blanks :
(i) Four lakh sixty seven thousand three hundred six.
= ……………… (In numeral form)
= …………… (In International System)
= ……………. (In International numeration)
(ii) Thirteen lakh forty five.
= ……………. (In numeral form)
= ……………. (In International System)
= ……………. (In International numerals)
Solution:
(i) Four lakh sixty seven thousand three hundred six
= 4, 67, 306
= 467, 306
= Four hundred sixty seven thousand three hundred six
(ii) Thirteen lakh forty five
= 13, 00, 45
= 130, 045
= One hundred thirty thousand forty five

Question 5.
Fill in the blanks :
(i) Six hundred four thousand eight hundred forty seven.
= ……………. (In numeral form)
= ……………. (In Indian System)
= ……………… (Number name in Indian System)
(ii) Two million three hundred ten thousand one hundred four
= …………….. (In numeral form)
= …………… (In Indian System)
= ……………. (In Indian numeration)
= …………….. (In international numerals)
Solution:
(i) Six hundred four thousand eight hundred forty seven
= 6, 04, 847
= 604, 847
= Six lakh four thousand eight hundred forty seven
(ii) Two million three hundred ten thousand one hundred four
= 2, 310, 104
= 23, 10, 104
= Twenty three lakh ten thousand one hundred four
= 2, 310, 104

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 2 Estimation

November 10, 2023May 16, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 2 Estimation

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 2 Estimation

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Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Estimation Exercise 2A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Round off each of the following to the nearest ten :
(i) 62
(ii) 265
(iii) 543
(iv) 8261
(v) 6294
(vi) 3008
(vii) 72326
Solution:
If the digit at ones place is less than 5, replace ones digit by 0, and keep the other digit as they are :
And if the digit at ones place is 5 or more than 5, increase tens digit by 1, and replace ones digit by 0.
(i) 62 – 60
(ii) 265 – 270
(iii) 543 – 540
(iv) 8261 – 8260
(v) 6294 – 6290
(vi) 3008-3010

Question 2.
Round off each of the following to the nearest hundred :
(i) 748
(ii) 784
(iii) 2667
(iv) 5432
(v) 6388
(vi) 59237
Solution:
If the tens digit is less than 5, replace each one of tens and once digits by 0 and keep the other digits as they are :
(i) 748 – 700
(ii) 784 – 800
(iii) 2667 – 2700
(iv) 5432 – 5400
(v) 6388 – 6400
(vi) 59237 – 59200

Question 3.
Round off each of the following to the nearest thousand :
(i) 6475
(ii) 6732
(iii) 5352
(iv) 32568
(v) 9248
(vi) 83294
Solution:
If the tens digit is 5 or more than 5, increase the hundreds digit by 1 and replace each of tens digit and ones digit by 0.
(i) 6475 – 6000
(ii) 6732 – 7000
(iii) 25352 – 25000
(iv) 32568 – 33000
(v) 9248 – 9000
(vi) 83294 – 83000

Question 4.
Round off
(i) 578 to the nearest ten.
(ii) 578 to the nearest hundred.
(iii) 4327 to the nearest thousand.
(iv) 32974 to the nearest ten-thousand.
(v) 27487 to the nearest ten-thousand.
Solution:
(i) 578 – 580
(ii) 578 – 600
(iii) 4327 – 4000
(iv) 32974 – 30000
(v) 27487 – 30000

Question 5.
Round off each of the following to the nearest tens, hundreds and thousands.
(i) 864
(ii) 1249
(iii) 54, 547
(iv) 68, 076
(v) 56, 293
(vi) 7, 293
(vii) 89, 24, 379
Solution:
(i) 864 – 860
864 – 900
864 – 1000
(ii) 1249 – 1250
1249 – 1200
1249 – 1000
(iii) 54, 547 – 54550
54, 547 – 54500
54, 547 – 55000
(iv) 68, 076 – 68, 080
68, 076 – 68, 100
68, 076 – 68, 000
(v) 56, 293 – 56, 290
56, 293 – 56, 300
56, 293 – 56, 000
(vi) 7, 293 – 7290
7, 293 – 7, 300
7, 293 – 7000
(vii) 89, 24, 379 – 89, 24, 380
89, 24, 379 – 89, 24, 400
89, 24, 379 – 89, 24, 000

Question 6.
Round off the following to the nearest tens ;
(i) ₹ 562
(ii) 837 m
(iii) 545 cm
(iv) ₹ 27
Solution:
(i) ₹ 562
₹ 562 – ₹ 560
(ii) 837 m
837 – 840 m
(iii) 545 cm
545 – 550
(iv) ₹ 21
₹ 27 – ₹ 30

Question 7.
List all the numbers which can be round off to 30.
Solution:
The numbers that can be rounded off to 30 are:
26, 27, 28, 29, 31, 32, 33, 34

Question 8.
List all the numbers which can be rounded off to 50.
Solution:
The numbers that can be rounded off to 50 are:
46, 47, 48, 49, 51, 52, 53, 54

Question 9.
Write the smallest and the largest numbers which are rounded off to 80.
Solution:
The smallest number which is rounded off to 80 is 75
and the largest number which is rounded off to 80 is 84.

Question 10.
Write the smallest and the largest numbers which are rounded off to 130.
Solution:
The smallest number which is rounded off to 130 is 125
and The largest number which is rounded off to 130 is 134.

Estimation Exercise 2B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Estimate the sum of each pair of numbers to the nearest ten :
(i) 67 and 44
(ii) 34 and 87
(iii) 23 and 66
(iv) 78 and 18
(v) 96 and 55
(vi) 76 and 62
(vii) 457 and 175
(viii) 474 and 173
(ix) 527 and 267
Solution:
(i) 67 and 44
67 to the nearest ten 70 and, 44 to the nearest ten = 40
∴ Required sum (70 + 40) = 110
(ii) 34 and 87
34 to the nearest ten = 30 and, 87 to the nearest ten = 90
∴ Required sum = (30 + 90) = 120
(iii) 23 and 66
23 to the nearest ten = 20 and, 66 to the nearest ten = 70
∴ Required sum = (20 + 70) = 90
(iv) 78 and 18
78 to the nearest ten = 80 and, 18 to the nearest ten = 20
∴ Required sum = (80 + 20) = 100
(v) 96 and 55
96 to the nearest ten = 100 and, 55 to the nearest ten = 60
∴ Required sum = (100 + 60) = 160
(vi) 76 and 62
76 to the nearest ten = 80 and, 62 to the nearest ten = 60
∴ Required sum = (80 + 60) = 140
(vii) 457 and 175
457 to the nearest ten = 460 and, 175 to the nearest ten = 180
∴ Required sum = (460 + 180) = 640
(viii) 474 and 173
474 to the nearest ten = 470 and, 173 to the nearest ten = 170
∴ Required sum = (470 + 170) = 640
(ix) 527 and 267
527 to the nearest ten = 530 and, 267 to the nearest ten = 270
∴ Required sum = (530 + 270) = 800

Question 2.
Estimate the sum of each pair of numbers to the nearest hundred :
(i) 336 and 782
(ii) 546 and 342
(iii) 270 and 495
(iv) 4280 and 5295
(v) 4230 and 2410
(vi) 30047 and 39287
Solution:
(i) 336 and 782
336 to the nearest hundred = 300 and, 782 to the nearest hundred = 800
∴ Required sum = (300 + 800) = 1100
(ii) 546 and 342 and, 342 to the nearest hundred = 300
∴ Required sum = (500 + 300) = 800
(iii) 270 and 495
270 to the nearest hundred = 300 and, 495 to the nearest hundred = 500
∴ Required sum = (300 + 500) = 800
(iv) 4280 and 5295
4280 to the nearest hundred = 4300 and, 5295 to the nearest hundred = 5300
∴ Required sum = (4300 + 5300) = 9600
(v) 4230 and 2410
4230 to the nearest hundred = 4200 and, 2410 to the nearest hundred = 2400
∴ Required number = (4200 + 2400) = 6600
(vi) 30047 and 39287
30047 to the nearest hundred = 30000 and, 39287 to the nearest hundred = 39300
∴ Required sum = (30000 + 39300) = 69, 300

Question 3.
Estimate the sum of the following pair of numbers to the nearest thousand:
(i) 53826 and 36455
(ii) 56802 and 22475
Solution:
(i) 53826 and 36455
53826 to the nearest hundred = 54000 and, 36455 to the nearest hundred = 36000
∴ Required sum = (54000 + 36000) = 90000
(ii) 56802 and 22475
56802 to the nearest thousand = 57000 and, 22475 to the nearest thousand = 22000
∴ Required sum = (57000 + 22000) = 79000

Question 4.
Estimate the following differences correct to nearest ten :
(i) 82 – 27
(ii) 96 – 36
(iii) 508 – 248
Solution:
(i) 82 – 27
82 to the nearest ten = 80 and, 27 to the nearest ten = 30
∴ Required difference = (80 – 30) = 50
(ii) 96 – 36
96 to the nearest ten = 100 and, 36 to the nearest ten = 40
∴ Required difference = (100 – 40) = 60
(iii) 508 – 248
508 to the nearest ten = 510 and, 248 to the nearest ten = 250
∴ Required difference = (510 – 250) = 260

Question 5.
Estimate each difference to the nearest hundred :
(i) 769 – 314
(ii) 856 – 687
(iii) 6352 – 2086
Solution:
(i) 769 – 314
769 to the nearest hundred = 800 and, 314 to the nearest hundred = 300
∴ Required difference = (800 – 300) = 500
(ii) 856 – 687
856 to the nearest hundred = 900 and, 687 to the nearest hundred = 700
∴ Required difference = (900 – 700) = 200
(iii) 6352 – 2086
6352 to the nearest hundred = 6400 and, 2086 to the nearest hundred = 2100
∴ Required difference = (6400 – 2100) = 4300

Question 6.
Estimate each difference to the nearest thousand :
(i) 45974 – 38766
(ii) 76003 – 48399
Solution:
(i) 45974 – 38766
45974 to the nearest thousand = 46000 and, 38760 to the nearest thousand = 39000
∴ Required difference = (46000 – 39000) = 7000
(ii) 76003 – 48399
76003 to the nearest thousand = 76000 and, 48399 to the nearest thousand = 48000
∴ Required difference = (76000 – 48000) = 28000

Question 7.
Estimate each of the following products by rounding of each number to the nearest ten :
(i) 49 x 52
(ii) 63 x 38
(iii) 27 x 54
(iv) 53 x 85
(v) 74 x 67
(vi) 25 x 33
Solution:
(i) 49 x 52
49 to the nearest ten = 50 and, 52 to the nearest ten = 50
∴ Required product = (50 x 50) = 2500
(ii) 63 x 38
63 ta the nearest ten = 60 and, 38 to the nearest ten = 40
∴ Required product = (60 x 40) = 2400
(iii) 27 x 54
27 to the nearest ten = 30 and, 54 to the nearest ten = 50
∴ Required product = (30 x 50) = 1500
(iv) 53 x 85
53 to the nearest ten = 50 and, 85 to the nearest ten = 90
∴ Required product = (50 x 90) = 4500
(v) 74 x 67
74 to the nearest ten = 70 and, 67 to the nearest ten = 70
∴ Required product = (70 x 70) = 4900
(vi) 25 x 33
25 to the nearest ten = 30 and, 33 to the nearest ten = 30
∴ Required product = (30 x 30) = 900

Question 8.
Estimate each of the following products by rounding off each number to the nearest hundred :
(i) 477 x 213
(ii) 624 x 236
(iii) 333 x 247
(iv) 537 x 283
(v) 382 x 127
(vi) 427 x 328
Solution:
(i) 477 x 213
477 to the nearest hundred = 500 and, 213 to the nearest hundred = 200
∴ Required product = (500 x 200) = 100000
(ii) 624 x 236
624 to the nearest hundred = 600 and, 236 to the nearest hundred = 200
∴ Required product = (600 x 200) = 120000
(iii) 333 x 247
333 to the nearest hundred = 300 and, 247 to the nearest hundred = 200
∴ Required product = (300 x 200) = 60000
(iv) 537 x 283
537 to the nearest hundred = 500 and, 283 to the nearest hundred = 300
∴ Required product = (500 x 300) = 150000
(v) 382 x 127
382 to the nearest hundred = 400 and, 127 to the nearest hundred = 100
∴ Required product = (400 x 100) = 40000
(vi) 427 x 328
472 to the nearest hundred = 500 and, 328 to the nearest hundred = 300
∴ Required product = (500 x 300) = 150000

Question 9.
Estimate each of the following products by rounding off the first number correct to nearest ten and the other number correct to nearest hundred :
(i) 28 x 287
(ii) 432 x 128
(iii) 48 x 165
(iv) 72 x 258
(v) 83 x 664
(vi) 44 x 250
Solution:
(i) 28 x 287
28 to the nearest ten = 30 and, 287 to the nearest hundred = 300
∴ Required product = (30 x 300) = 9000
(ii) 432 x 128
432 to the nearest ten = 430 and, 128 to the nearest hundred = 100
∴ Required product = (430 x 100) = 43000
(iii) 48 x 165
48 to the nearest ten = 50 and, 165 to the nearest hundred = 200
∴ Required product = (50 x 200) = 10000
(iv) 72 x 258
72 to the nearest ten = 70 and, 258 to the nearest hundred = 300
∴ Required product = (70 x 300) = 21000
(v) 83 x 664
83 to the nearest ten = 80 and, 664 to the nearest hundred = 700
∴ Required product = (80 x 700) = 56000
(vi) 44 x 250
44 to the nearest ten = 40 and, 250 to the nearest hundred = 300
∴ Required product = (40 x 300) = 12000

Question 10.
Estimate each of the following quotients by converting each number to the nearest ten :
(i) 87 ÷ 28
(ii) 84 ÷ 23
(iii) 77 ÷ 22
(iv) 198 ÷ 24
(v) 355 ÷ 26
(vi) 444 ÷ 42
(vii) 843 ÷ 33
Solution:
(i) 87 ÷ 28
(87 ÷ 28) is (approximately to the nearest 10) equal to 90 ÷ 30 = 3
(ii) 84 ÷ 23
84 ÷ 23 is (approximately to the nearest 10) equal to 80 ÷ 20 = 4
(iii) 77 ÷ 22
77 ÷ 22 is (approximately to the nearest 10) equal to 80 ÷ 20 = 4
(iv) 198 ÷ 24
198 ÷ 24 is (approximately to the nearest 10) equal to 200 ÷ 20 = 10
(v) 355 ÷ 26
355 ÷ 26 is (approximately to the nearest 10) equal to 360 ÷ 30 = 12
(vi) 444 ÷ 42
444 ÷ 42 is (approximately to the nearest 10) equal to 440 ÷ 40 = 11
(vii) 843 ÷ 33
843 ÷ 33 is (approximately to the nearest 10) equal to 840 ÷ 30 = 28

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Selina Concise Chemistry Class 6 ICSE Solutions – Air and Atmosphere

November 10, 2023May 11, 2022 by Raju

Selina Concise Chemistry Class 6 ICSE Solutions – Air and Atmosphere

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

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Selina Class 6 Chemistry ICSE Solutions Physics Biology Maths Geography History & Civics

Selina Concise ICSE Solutions for Class 6 Chemistry Chapter 6 Air and Atmosphere

POINTS TO REMEMBER

Air cannot be seen as it is transparent we only feel its presence when it blows.
The thick layer of air around earth upto 320 km above the surface is called atmosphere.
Air is a mixture of various gases, water vapours, which protects us from harmful rays of Sun. It consists of nearly 80% nitrogen, 20% oxygen.
The components of air can be separated by simple physical means. Has no formula.
N₂ is inert gas, reduces the effect of O₂ present in air, N₂ is helpful in plant growth as it provides nitrates used by plant for formation of protein which is important nutrient for animals.
Oxygen is supporter of combustion no burning is possible without O₂ and no life is possible without it.
CO₂ is must for plant life, plants prepare their food taking CO₂ from air. CO₂ maintains temp, of earth and without CO₂ our planet would be too cold.
Ozone in the upper layers of atmosphere protects the earth from harmful ultraviolet rays of Sun.
Air is useful in many ways : Acts as medium for propagation of sound, birds, insects to fly in air, air ships, aeroplanes to fly only must due to presence of air. Air helps in movement of yachts, parachutes, aircrafts etc. Air is must for animals and plants on earth and air dissolved in water for aquatic life.
Respiration : Is a slow chemical process when O₂ present in the inhaled air reacts with the digested food material in the body to release energy, CO₂ and water. Sugar + Oxygen -> Carbon dioxide + Water + Energy.
Breathing : The complete process in which air is inhaled and exhaled is called breathing.
Combustion : Burning of substance in air completely producing CO₂, heat and light.
Photosynthesis : Preparation of food by green plants taking CO₂ from air, water from soil in presence of sun light.
Oxygen in air remains constant it is not depleted as plants during day time produce lot of O₂ in air.
Soil contains air trapped in between its particle and organisms living there breath this oxygen. Due to this earth worms come out of soil during heavy rains as water gets into the pores in the soil.
Common pollutant of air are smoke, dust, coal, emitted by industries and automobiles.
Gases like SO₂, NO₂ also pollute air as they form corresponding acids when they come in contact with water vapours in air and fall down in the form of acid rain with rain water.
These pollutants reduce the visibility which lead to accidents, cause global warming.

EXERCISE-I

Question 1.
What is atmosphere ?
Answer:
The earth is surrounded by a thick layer of air called the atmosphere that extends upto a height of about 320 kilometres above the surface of the earth.

Question 2.
Why can’t we see air ?
Answer:
We can’t see air because it is colourless, odourless and transparent gaseous matter.

Question 3.
What is wind ?
Answer:
Fast moving air is called wind.

Question 4.
What would have happened if there would have been no atmosphere around the earth ?
Answer:
Without atmosphere life would not be possible as atmosphere protects us from harmful gases. We could not live without
air present in atmosphere. In absence of the atmopshere, the earth would get so cold at night that we would not be able to survive. No CO₂ and N₂ for plants without atmosphere.
P.Q. When water is heated, we see bubbles rising up. Why ?
Answer:
Water has a lot of air dissolved on it. This is what allows breathing to fishes and other aquatic beings. The solubility of gases decreases when the temperature is raised, and that is why the dissolved air bubbles out from the water.

Question 5.
Why is air called a mixture ? Give five facts in support of your answer.
Answer:
Air is a mixture because :

Air has no formula, a mixture has no formula whereas compound has a formula.
No energy changes are involved to form air from various gases.
When air is formed out of its constituents no change in mass and no change in volume takes place.
Properties of air vary from place to place and time to time. i.e. there is more CO₂ in towns as compared to villages where more oxygen prevails as compared to towns.
Components of air can be separated by simple physical methods.

Question 6.
What are the main components of air ? Write down the composition of three main gases present in air by volume.
Answer:
Main components of air are :

Question 7.
What do you observe when

(a) Ice cold water is filled in a glass tumbler.
(b) A burning candle is covered with an inverted jar.
(c) Carbon dioxide gas is passed through lime water.
(d) A beam of light is allowed to enter in a closed dark room through a small hole.

Answer:

(a) We will observe that fine water droplets get deposited on the outer wall of the glass tumbler.
(b) The candle burn more brightly because candle gets oxygen support in burning.
(c) When CO₂ gas is passed through lime water it forms calcium carbonate which is white precipitate (turbidity). This gives the milky white appearance to the solution.
(d) We will observe randomly moving dust particles in the beam of light. This confirms the presence of dust particles in the air.

Question 8.
Write the chemical name of –

(a) Lime water
(b) The white insoluble solid formed on reaction of carbon dioxide with lime water.

Answer:

(a) Calcium hydroxide Ca(OH)₂
(b) Calcium carbonate – CaCO₃

These droplets were present in air as moisture which got condensed on the cooler walls of glass tumbler. This shows the presence of water vapours in air.

EXERCISE-II

Question 1.
Name two important processes supported by oxygen present in air.
Answer:
Two processes supported by oxygen present in air are :

Combustion
Respiration

Question 2.
Give two uses of the following components present in air:

(a) oxygen
(b) nitrogen
(c) carbon dioxide
(d) water vapour

Answer:

(a) Oxygen: For respiration and breathing no life is possible without oxygen.
Oxygen is the most vital component which is responsible for two most important processes,
(i) Respiration and
(ii) Combustion.
(b) Nitrogen : Present in air gets fixed up in the soil as nitrates used by plants for their growth and for the formation of protein an important nutrient for animals. It is used to make fertilizers and various nitrogenous products.
(c) Carbon dioxide : Carbon dioxide is used by plants to prepare their food by photosynthesis. Dry ice (solid carbon-dioxide) is used as a refrigerant. Carbon dioxide is used to prepare fizzy drinks like soda water.
(d) Water vapour : Water vapors present in air provide moisture for both plants and animals. It also helps in predicting climatic conditions of a particular area as its amount varies from place to place and time to time.

Question 3.
Define the following:

(a) Respiration
(b) Photosynthesis
(c) Combustion.

Answer:

(a) Respiration : Respiration is a chemical process that takes place in all living beings (slow in plants). In this process, oxygen present in inhaled air reacts with the digested food material in the body. This results in release of energy, carbon-dioxide and water.
Sugar + Oxygen → CO₂ + Water + Energy.
(b) Photosynthesis : Process of producing food by green plants taking CO₂ from air, water from soil by roots and in presence of sun light and chlorophyll is called photosynthesis.
(c) Combustion : Combustion also called burning “is burning of substance in oxygen of air completely producing heat and light is called combustion.”

Question 4.
What are fuels ? Give two examples of modern fuels.
Answer:
Fuels : The substances which burn in air to produce large amount of energy in the form of heat and light are called fuels.
Examples :

1. CNG (Compressed natural gas)
2. LPG (Liquefied natural gas)

Question 5.
Give reasons :

(a) Aquatic animals and plants are able to survive in matter.
(b) A burning candle stops burning if covered with a glass tumbler.
(c) Mountaineers and divers carry oxygen cylinders with them.
(d) When water is heated, we see bubbles rising up.

Answer:

(a) Aquatic animals and plants use oxygen dissolved in water and survive.
(b) Oxygen is necessary for combustion when covered with a glass tumbler supply of oxygen stops.
(c) For artificial respiration mountaineers carry oxygen cylinders as at high altitudes, the air is thin and breathing becomes difficult. Also divers carry oxygen cylinder for artificial respiration as there is less oxygen dissolved in water (less dense) and breathing becomes difficult.
(d) These bubbles come from the air dissolved in water. The marine life uses air dissolved in water.

Question 6.
Name the processes which maintain the balance between oxygen and carbon dioxide in the air. How is it done?
Answer:
Respiration and combustion are the processes which maintain the balance between oxygen and CO₂ in the air.

Respiration : Respiration is a chemical process that takes place in all living beings. In this process, oxygen present in the inhaled air reacts with the digested food material in the body. This results in the release of energy, carbon dioxide and water.
Combustion : Burning or combustion is a process in which a substance reacts chemically with oxygen and gets oxidised, with the release of energy in the form of heat and light. It is a fast process. During the process of burning, along with energy, carbon dioxide and water vapour are also produced.

Question 7.
State two similarities and two differences between respiration and burning.
Answer:
Similarities:
Burning

Oxygen is needed to combine with carbon and hydrogen in compound.
CO₂ and H₂O are formed with release of energy.

Respiration

Oxygen is needed tocombine with C and H2 of food.
CO₂ and H₂O are formed with release of energy.

Differences :
Burning

It occurs at higher temperature.
Is fast process.
A natural and continous process.

Respiration

It occurs at body temperature.
Is slow process.
An artificial & discont inuous process.

Question 8.
Define rusting ? What are the two necessary conditions for rusting of iron. Give the chemical name of rust.
Answer:
Rusting : Slow conversion of iron into its hydrated oxide in the presence of moisture and air is called rusting.
Conditions for rusting:

Presence of moisture (water).
Presence of oxygen (air).

Chemical name of rust is hydrated iron oxide [Fe20rxH20]

Question 9.
How is air useful to :

(a) water boats
(b) agriculture
(c) windmills
(d) scooters and cars.

Answer:

(a) Air helps movement of water boats.
(b) Air speeds up drying up of agricultural products like grains, pulses fruits etc. Air helps in pollination of flowers and dispersel of seeds.
(c) Windmills work where there is sufficient movement of air.
(d) Air filled tyres of cars move smoothly on road as there is less friction.

Question 10.
State the full form of LPG and CNG ? How are the two different in their composition ?
Answer:
LPG (Liquefied Petroleum Gas): It is obtained from crude petroleum oil. It mainly contains gaseous compounds known as isobutane and butane. Popularly it is known as cooking gas. It is the best fuel for domestic purposes and in laboratories. It is available in cylinders. It is also supplied through pipes in big cities.
CNG (Compressed Natural Gas): It is produced along with crude oil. It mainly contains methane gas. It has become a popular fuel for vehicles like three wheeler scooters, cars and buses. It is a cheap fuel as well as pollution free. It is used as a substitute of petrol.
Difference in composition
LPG is obtained from crude petroleum oil. It mainly contains gaseous compounds known as isobutane and butane. While
CNG is produced along with crude oil. It mainly contains methane gas.

Question 11.

(a) Why is nitrogen important to all living beings ?
(b) What is nitrogen fixation ?

Answer:

(a) Nitrogen constitutes 78% of air by volume. It is of vital importance to the plants, animals and human beings as it is needed to prepare vital nutrient ‘protein’ to every living being which is necessary for their growth.
(b) Nitrogen cannot be absorbed directly by plants. It is first fixed up in the soil as nitrites and nitrates and then absorbed by the plants in soluble forms. This phenomenon is called nitrogen fixation.

EXERCISE-III

Question 1.
What is air pollution ?
Answer:
Air Pollution : “Mixing of UNWANTED and HARMFUL SUBSTANCES in air is called AIR POLLUTION.”

Question 2.
Mention five causes of air pollution.
Answer:
CAUSES OF AIR POLLUTION ARE:

Burning of FOSSIL FUELS and FIBRES.
Cutting of forests.
Erruption of volcanoes.
Increase in POPULATION.
Agricultural activities like use of chemical fertilisers, insecticides, pesticides and burning of husks etc.

Question 3.
Name two air pollutants which

(a) affect our health
(b) cause acid rain
(c) cause global warming.

Answer:

(a) Two air pollutants that affect our health are :

DUST
SMOKE

(b) Two air pollutants that cause acid rain are :

SULPHUR DIOXIDE (SO₂) and
NITROGEN DIOXIDE (NO₂)

(c) Two air pollutants that cause global warming are :

CARBON DIOXIDE (CO₂)
METHANE (CH₄)

Question 4.
What is meant by ozone depletion?
Answer:
Ozone is present in the upper layer of atomosphere called stratosphere. It protects the earth from harmful ultraviolet rays of the sun.
Some air pollutants like chlorofluorocarbon (CFC) react with ozone present in the atmosphere. These pollutants reduce the density of the air. The ultra-violet rays from the sun reach directly on the earth which affect human health, causing skin diseases and cancer and also increase the earth’s temperature.

Question 5.
State four steps to be taken to control air pollution.
Answer:
Four steps taken to control air pollution :

Tall chimneys be installed in factories and power houses so that smoke and gases rise high and get diluted.
GROW more tall trees to absorb CO₂ Also plants help in bringing rains.
Use unleaded petrol and CNG in all public transport vehicles
Industries should be located far away from residential areas.

Question 6.
Name three greenhouse gases.
Answer:
Carbon dioxide, methane, nitrous oxide, etc.

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks

(a) The layer of air around the earth is called the atmosphere.
(b) Although we cannot see air, we can feel it.
(c) Air is a mixture of gases.
(d) Plants and animals maintain the balance of carbon- dioxide and oxygen in air.
(e) Polluted air is harmful for health.
(f) The supporter of combustion in air is oxygen.
(g) Green plants need CO₂ water and light to prepare food.
(h) Oxygen is used in burning the food to get energy.
(i) Aquatic plants and animals use dissolved oxygen in water.
(j) The reddish brown powder on the surface of iron nails exposed to air and moisture is called rust.

2. Write ‘true’ or ‘false for the following

(a) Air is a compound.
Answer. False
Correct: Air is a mixture.

(b) Carbon dioxide is given out during photosynthesis.
Answer. False
Correct : Oxygen is given out during photosynthesis.

(c) Respiration needs nitrogen.
Answer. False
Correct: Respiration needs oxygen.

(d) The composition of air was discovered by Lavoisier.
Answer. True

(e) The major component of air is oxygen.
Answer. False
Correct: The major component of air is nitrogen.

MULTIPLE CHOICE QUESTIONS

Tick (√) the correct alternative from the choice given for the following statements

1. Air consists of

only oxygen
only nitrogen
only carbon dioxide
all of these

2. Air pollution is due to the

cutting of green plants
gases like carbon monoxide, sulphur dioxide etc.
smoke given out by factories
all of the above

3. The gases which cause acid rain are

sulphur dioxide and oxygen
nitrogen and oxygen
carbon dioxide and water vapour
nitrogen dioxide and sulphur dioxide

4. Rust is

hydrated iron oxide
hydrated copper sulphate
anhydrous iron oxide
none of the above

5. Photosynthesis is a process in which plants

take in oxygen and give out carbon dioxide
take in carbondioxide and give out oxygen
take in nitrogen and give out oxygen .
none of the above.

6. Fuels which do not leave any residue on burning are

coal and wood
coal and LPG
wood and CNG
LPG and CNG

7. Respiration

is a slow process
is a natural and continuous process
takes place at body temperature
all of the above

8. Which of the following is common in combustion and respiration

oxygen
release of heat and light
natural process
nitrogen

9. Which of the following is not a green house gas ?

carbondioxide
sulphur dioxide
methane
nitrogen

10. The substance which accelerates the speed of a reaction without itself undergoing any change is called

catalyst
pollutant
fuel
none of the above.