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

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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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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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
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 12

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
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 19

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) y+ 12y
(iii) \(\frac { { y }^{ 3 } }{ 4 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 22
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 23

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
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 26

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 20 Substitution

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)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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.

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

IMPORTANT POINTS

  1. Substitution : The value of an expression depends on the value of its variable (s).
  2. 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
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 44
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 4

Question 2.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 6
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 7

Question 3.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 9

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:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 12

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
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 17

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:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 19

Question 10.
If a = 3, find the values of a2 and 2a.
Solution:
a2 = (3)2 = 3 x 3 = 9
2a = (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
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 21
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 22
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 23
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 24

Question 3.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 25
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 26
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 27

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:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 29
(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:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 31

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) xy
(ii) yx
(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
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 36

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 19 Fundamental Operations

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations (Related to Algebraic Expressions)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations (Related to Algebraic Expressions)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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.

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

IMPORTANT POINTS

  1. Fundamental Operations : In mathematics, the operations : addition (+), subtraction (-), multiplication (x) and division (÷) are called the four fundamental operations.
  2. Addition and Subtraction :
    • Addition of Like Terms :
      • When all the terms are positive, add their coefficients.
      • When all the terms are negative, add their coefficients without considering their negative signs and then prefix the minus sign to the sum.
    • Addition of Unlike Terms : As discussed above, the sum of two or more like terms is a single like term ; but the two unlike terms cannot be added together to get a single term.
    • Subtraction of Like Terms : The same rules, as those for subtraction of integers, are applied for the subraction of like terms. The result of subtraction of two like terms is also a like term.

Add the positive terms together and negative terms separately together. Then, find the result of two terms obtained.

Fundamental Operations Exercise 19A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
(i) 5 + 4 = ………… and 5x + 4x = ………….
(ii) 12 + 18 = ………… and 12x2y + 18x2y = ………….
(iii) 7 + 16 = ………….. and 7a + 16b = …………
(iv) 1 + 3 = ………… and x2y + 3xy2 = ………..
(v) 7 – 4 = …………… and 7ab – 4ab = …………..
(vi) 12 – 5 = ………… and 12x – 5y = ……………
(vii) 35 – 16 = ………….. and 35ab – 16ba = ………….
(viii) 28 – 13 = …………. and 28ax2 – 13a2x = ………….
Solution:
(i) 5 + 4 = 9 and 5x + 4x = 9x
(ii) 12 + 18 = 30 and 12x2y + 18x2y = 30x2y
(iii) 7 + 16 = 23 and 7a + 16 b = 7a + 16b
(iv) 1 + 3 = 4 and x2y + 3xy2 = x2y + 3xy2
(v) 7 – 4 = 3 and 7ab – 4ab = 3ab
(vi) 12 – 5 = 7 and 12x – 5y = 12x – 5y
(vii) 35 – 16 = 19 and 35ab – 16ba = 19ab
(viii) 28 – 13 = 15 and 28ax2 – 13a2x = 28ax2 – 13a2x

Question 2.
Fill in the blanks :
(i) The sum of – 2 and – 5 = …………. and the sum of – 2x and – 5x = …………….
(ii) The sum of 8 and – 3 = ………….. and the sum of 8ab and – 3ab = ………….
(iii) The sum of – 15 and – 4 = …………….. and the sum of – 15x and -4y = ………………
(iv) 15 + 8 + 3 = ……….. and 15x + 8y + 3x = …………….
(v) 12 – 9 + 15 = …………… and 12ab – 9ab + 15ba = ……………..
(vi) 25 – 7 – 9 = and 25xy – 7xy – 9yx = ……………
(vii) – 4 – 6 – 5 = …………. and – 4ax – 6ax – 5ay = …………….
Solution:
(i) The sum of – 2 and – 5 = – 7 and the sum of – 2x and – 5x = -7x
(ii) The sum of 8 and -3 = 5 and the sum of 8ab and – 3ab = 5ab
(iii) The sum of – 15 and – 4 = – 19 and the sum of – 15x and – 4y = – 15x – 4y
(iv) 15 + 8 + 3 = 26 and 15x + 8y + 3x = 18x + 8y
(v) 12 – 9 + 15 = 18 and 12ab – 9ab + 15ba = 18ab
(vi) 25 – 7 – 9 = 9 and 25xy – 7xy – 9yx = 9xy
(vii) – 4 – 6 – 5 = – 15 and – 4ax – 6ax – 5ay = – 10ax – 5ay

Question 3.
Add:
(i) 8xy and 3xy
(ii) 2xyz, xyz and 6xyz
(iii) 2a, 3a and 4b
(iv) 3x and 2y
(v) 5m, 3n and 4p
(vi) 6a, 3a and 9ab
(vii) 3p, 4q and 9q
(viii) 5ab, 4ba and 6b
(ix) 50pq, 30pq and 10pr
(x) – 2y, – y and – 3y
(xi) – 3b and – b
(xii) 5b, – 4b and – 10b
(xiii) – 2c, – c and – 5c
Solution:
(i) 8xy + 3xy = 11xy
(ii) 2xyz + xyz + 6xyz = (2 + 1 + 6) xyz = 9xyz
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 1
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 2

Question 4.
Evaluate :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 3
Solution:
(i) 6a – a – 5a – 2a = 6a – (1 + 5 + 2).a
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 4

Question 5.
Evaluate :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 6

Question 6.
Subtract the first term from the second :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 7
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 8

Question 7.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 9
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 12
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 13
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 14

Fundamental Operations Exercise 19B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the sum of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 15
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 16
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 17
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 18

Question 2.
Add the following expressions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 19
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 20
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 21

Question 3.
Evaluate :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 22
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 23
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 24

Question 4.
Subtract :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 25
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 26
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 27

Question 5.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 28
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 29

Question 6.
From the sum of x + y – 2z and 2x – y + z subtract x + y + z.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 30

Question 7.
From the sum of 3a – 2b + 4c and 3b – 2c subtract a – b – c.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 31

Question 8.
Subtract x – 2y – z from the sum of 3x – y + z and x + y – 3z.
Solution:
(3x – y + z) + (x + y – 3 z) – (x – 2y – z)
= 3x – y + z + x + y – 3z – x + 2y + z
= 3x + x – x – y + y + 2y + z + z – 3z
= 3x + 2y – z

Question 9.
Subtract the sum of x + y and x – z from the sum of x – 2z and x + y + z
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 32

Question 10.
By how much should x + 2y – 3z be increased to get 3x ?
Solution:
3x – (x + 2y – 3z)
= 3x – x – 2y + 3z
= 2x – 2y + 3z

Question 11.
The sum of two expressions is 5x2 – 3y2. If one of them is 3x2 + 4xy – y2, find the other.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 33

Question 12.
The sum of two expressions is 3a2 + 2ab – b2. If one of them is 2a2 + 3b2, find the other.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 34

Fundamental Operations Exercise 19C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 35
Solution:
(i) 6 x 3 = 18 and 6x x 3x = 6 x 3x x x x = 18x2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 36

Question 2.
Fill in the blanks :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 37
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 38
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 39

Question 3.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 40
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 41

Question 4.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 42
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 43
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 44

Question 5.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 45
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 46
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 47

Question 6.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 48
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 49

Question 7.
Multiply :
(i) x + 2 and x + 10
(ii) x + 5 and x – 3
(iii) x – 5 and x + 3
(iv) x – 5 and x – 3
(v) 2x+ y and x+ 3y
(vi) (3x – 5y) and (x + 6y)
(vii) (x + 9y) and (x – 5y)
(viii) (2x + 5y) and (2x + 5y)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 51

Question 8.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 52
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 53
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 54

Question 9.
Find the product of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 55
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 56
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 57
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 58

Fundamental Operations Exercise 19D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Divide :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 59
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 60

Question 2.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 61
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 62
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 63
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 64

Question 3.
Divide :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 65
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 66
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 67
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 68

Question 4.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 69
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 70

Question 5.
Divide :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 71
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 72
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 73
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 74

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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.

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

IMPORTANT POINTS

  1. Algebra : Algebra is a generalized form of Arithmetic. In Arithmetic, we use numbers, such as : 3, – 8, 0.63, etc., each of which has one definite value ; whereas in Algebra, we use letters along with numbers.
    For Example : 5x, 3x – 4, 7a + b, 3y – 5x, x + 3y – 9z, etc.
    The letters used in Algebra are called variables or literal numbers or simply literals.
  2. Signs and Symbols : In Algebra, the signs +, -, x and ÷ are used with the same meaning as in Arithmetic.
    Following sign and symbols are frequently used in algebra and have the same meanings as they have in any other branch of Mathematics.
    = means, “is equal to”
    ≠ means, “is not equal to”
    < means, “is less than” > means, “is greater than”
    \(\nless\) means, “is not less than”
    \(\ngtr\) means, “is not greater than”
    ∴ means, “therefore”
    ∵ means, “because” or “since”
    ~ means, “difference between”
    ⇒ means, “implies that”.
  3. To Write a Given Statement in Algebraic Form
    Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 1

Fundamental Concepts Exercise 18A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express each of the following statements in algebraic form :
(i) The sum of 8 and x is equal to y.
(ii) x decreased by 5 is equal to y.
(iii) The sum of 2 and x is greater than y.
(iv) The sum of x and y is less than 24.
(v) 15 multiplied by m gives 3n.
(vi) Product of 8 and y is equal to 3x.
(vii) 30 divided by b is equal to p.
(viii) z decreased by 3x is equal to y.
(ix) 12 times of x is equal to 5z.
(x) 12 times of x is greater than 5z.
(xi) 12 times of x is less than 5z.
(xii) 3z subtracted from 45 is equal to y.
(xiii) 8x divided by y is equal to 2z.
(xiv) 7y subtracted from 5x gives 8z.
(xv) 7y decreased by 5x gives 8z.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 2

Question 2.
For each of the following algebraic expressions, write a suitable statement in words:
(i) 3x + 8=15
(ii) 7 – y > x
(iii) 2y – x < 12
(iv) 5 ÷ z = 5
(v) a + 2b > 18
(vi) 2x – 3y= 16
(vii) 3a – 4b > 14
(viii) b + 7a < 21
(ix) (16 + 2a) – x > 25
(x) (3x + 12) – y < 3a
Solution:
(i) 3x plus 8 is equal to 15
(ii) 1 decreased by y is greater than x
(iii) 2y decreased by x is less than 12
(iv) 5 divided by z is equal to 5
(v) a increased by 2b is greater than 18
(vi) 2x decreased by 3y is equal to 16
(vii) 3a decreased by 4b is greater than 14
(viii) b increased la is less than 21
(ix) The sum of 16 and 2a decreased by x is greater than 25
(x) The sum of 3x and 12 decreased by y is less than 3a.

Fundamental Concepts Exercise 18B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Separate the constants and the variables from each of the following:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 4

Question 2.
Group the like terms together :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 6
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 7

Question 3.
State whether true or false :
(i) 16 is a constant and y is a variable but 16y is variable.
(ii) 5x has two terms 5 and x.
(iii) The expression 5 + x has two terms 5 and x
(iv) The expression 2x2 + x is a trinomial.
(v) ax2 + bx + c is a trinomial.
(vi) 8 x ab is a binomial.
(vii) 8 + ab is a binomial.
(viii) x3 – 5xy + 6x + 7 is a polynomial.
(ix) x3 – 5xy + 6x + 7 is a multinomial.
(x) The coefficient of x in 5x is 5x.
(xi) The coefficient of ab in – ab is – 1.
(xii) The coefficient of y in – 3xy is – 3
Solution:
(i) True
(ii) False
(iii) True
(iv) False
(v) True
(vi) False
(vii) True
(viii) True
(ix) True
(x) False
(xi) True
(xii) False

Question 4.
State the number of terms in each of the following expressions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 8
Solution:
(i) 2 terms
(ii) 2 terms
(iii) 2 terms
(iv) 2 terms
(v) 3 terms
(vi) 2 term
(vii) 2 terms
(viii) 3 terms
(ix) 3 terms

Question 5.
State whether true or false:
(i) xy and – yx are like terms.
(ii) x2y and – y2x are like terms.
(iii) a and – a are like terms.
(iv) – ba and 2ab are unlike terms.
(v) 5 and 5x are like terms.
(vi) 3xy and 4xyz are unlike terms.
Solution:
(i) True
(ii) False
(iii) True
(iv) False
(v) False
(vi) True

Question 6.
For each expression, given below, state whether it is a monomial, or a binomial or a trinomial.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 9
Solution:
(i) Monomial
(ii) Binomial
(iii) Monomial
(iv) Monomial
(v) Trinomial
(vi) Binomial
(vii) Trinomial
(viii) Binomial
(ix) Trinomial

Question 7.
Write down the coefficient of x in the following monomial :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 11

Question 8.
Write the coefficient of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 12
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 13

Question 9.
State the numeral coefficient of the following monomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 14
(vii) – 7x ÷ y
(viii) – 3x ÷ (2y)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 15

Question 10.
Write the degree of each of the following polynomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 16
Solution:
(i) 2
(ii) 2
(iii) 10
(iv) 20
(v) 3
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 17

Fundamental Concepts Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express each of the following statements in algebraic form :
(i) The sum of 3x and 4y is 8.
(ii) 5x decreased by 7 gives y.
(iii) 31 added to 4x gives 6x.
(iv) 3x subtracted from 89 gives 44.
Solution:
(i) 3x + 4y = 8
(ii) 5x – 7 = y
(iii) 4x + 37 = 6x
(iv) 89 – 3x = 44

Question 2.
Group the like terms :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 18
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 19

Question 3.
Write the number of terms in each of the following polynomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 20
Solution:
(i) 2 terms
(ii) 3 terms
(iii) 3 terms
(iv) 4 terms
(v) 3 terms

Question 4.
For each expression, given below, state whether it is a monomial, or a binomial or a trinomial:
(i) x + y
(ii) 5x – 4y
(iii) 7x2 + 5x + 8
(iv) 64 + 3 ÷ 6
(v) 9 ÷ a x b
(vi) 8a ÷ b
Solution:
(i) binomial
(ii) binomial
(iii) trinomial
(iv) 6a + 3 ÷ b = 6a + \(\frac { 3 }{ b }\)
It has two terms
It is binomial
(v) 9 ÷ a x b = \(\frac { 9b }{ a }\)
It has one term
It is monomial.
(vi) monomial

Question 5.
Write the coefficient of x2y in :
(i) -7x2yz
(ii) 8abx2y
(iii) – x2y
Solution:
(i) – 7z
(ii) 8ab
(iii) -1

Question 6.
Write the coefficient of :
(i) x2 in – 8x2y
(ii) y in -4y
(iii) x in – xy2
Solution:
(i) – 8y
(ii) – 4
(iii) – y2

Question 7.
Write the numeral coefficient in :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 21
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 22

Question 8.
Write the degree of each of the following polynomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 23
Solution:
(i) 8
(ii) 4
(iii) 2
(iv) 1
(v) 3
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 24
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 25

Question 9.
Write each statement, given below in algebraic form :
(i) 28 more than twice of x is equal to 45.
(ii) 3y reduced by 5z is greater than 8x.
(iii) 6x divided by 13y is less than 17.
(iv) 9 multiplied by 5x is equal to 2y.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 26

Question 10.
State whether true or false :
(i) If 23 is a constant and x is a variable, 23 + x is constant.
(ii) If 23 is a constant and x is a variable, 23x is a variable.
(iii) If y is a variable and 57 is a constant, y – 57 is a variable.
(iv) If 3x and 2y are variable; each of 3x + 2y, 3x – 2y, 3x ÷ 2y and 3x x 2y is a variable.
Solution:
(i) False
Sum of a constant and a variable is also variable.
(ii) True
Product of a constant and a variable is variable.
(iii) True
Constant subtracted from a variable is also variable.
(iv) True
Sum, difference product or quotient of two variables is also variable.

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage)

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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.

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

IMPORTANT POINTS

  1. Percent: Out of one hundred, is called percent and it is denoted as (%) e.g. 10%, 15% ….
    • Percent can be expressed in fraction and decimal such as given below :
      Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 1
    • A fraction or decimal can be expressed in percent.
    • We use percentage in profit and loss and also in finding interest etc.
  2. Percentage : When a quantity is expressed in the percent form, it is called percentage.
  3. To convert a given Fraction or Decimal into Percentage (Percent Form) : Multiply the given fraction or the given decimal by 100 and at the same time write the sign of percentage.
  4. To convert a given Percentage into a Fraction or Decimal: Remove the sign of percentage and at the same time divide by 100. Then reduce the fraction obtained to its lowest terms or decimal as required.
  5. To Express one Quantity (number) as a percentage of the other : Divide first quantity by the second and at the same time multiply the result by 100%.
    Keep in Mind :

    • Percent or percentage has no unit.
    • In order to express one quantity as a percentage of another quantity ; both the quantities must have same units.
  6. To find the Increase or Decrease Percent :
    Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 2

Percent Exercise 16A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express each of the following statements in the percentage form :
(i) 13 out of 20
(ii) 21 eggs out of 30 are good
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 3

Question 2.
Express the following fractions as percent :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 4
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 37
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 6

Question 3.
Express as percent:
(i) 0.10
(ii) 0.02
(iii) 0.7
(iv) 0.15
(v) 0.032
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 7

Question 4.
Convert into fractions in their lowest terms:
(i) 8%
(ii) 20%
(iii) 85%
(iv) 250%
(v) 12\(\frac { 1 }{ 2 }\) %
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 8

Question 5.
Express as decimal fractions :
(i) 25%
(ii) 108%
(iii) 95%
(iv) 4.5%
(v) 29.2%
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 9

Question 6.
Express each of the following natural numbers as percent :
(i) 7
(ii) 2
(iii) 19.5
(iv) 5.37
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 10

Percent Exercise 16B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express :
(i) Rs 5 as a percentage of Rs 25.
(ii) 80 paise as a percent of Rs 4.
(iii) 700 gm as a percentage of 2.8 kg.
(iv) 90 cm as a percent of 4.5 m.
Solution:
(i) \(\frac { 5 }{ 25 }\) x 100 = 20%
(ii) 80 paise as a percent of 400 paise (as/rupee = 100 paise)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 11

Question 2.
Express the first quantity as a percent of the second :
(i)) 40 P, ₹ 2
(ii) 500 gm, 6 kg
(iii) 42 seconds, 6 minutes
Solution:
40 p, ₹ 2 = 40 p to 200 p
(1 Rupee = 100 paise)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 12

Question 3.
Find the value of each of the following:
(i) 20% of ₹ 150
(ii) 90% of 130
(iii) 15% of 2 minutes
(iv) 7.5 % of 500 kg.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 13

Question 4.
If a man spends 70% of his income, what percent does he save ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 14

Question 5.
A girl gets 65 marks out of 80. What percent marks did she get ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 15

Question 6.
A class contains 25 children, of which 6 are girls. What percentage of the class are the boys.
Solution:
Total number of students = 25
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 16

Question 7.
A tin contains 20 litres of petrol. Due to leakage, 3 litres of petrol is lost. What percent is still present in the tin ?
Solution:
Total petrol in tin = 20 litres
last due to leakage = 3 litres
Balance petrol in tin = (20 – 3) = 17 litres
Percentage of petrol in tin = \(\frac { 17 }{ 20 }\) x 100 = 85%

Question 8.
An alloy of copper and zinc contains 45% copper and the rest is zinc. Find the weight of zinc in 20 kg of the alloy.
Solution:
Total weight of alloy = 20 kg
Weight copper = 20 x 45% = 20 x \(\frac { 45 }{ 100 }\) = 9 kg
Weight of zinc = (total weight of alloy – weight of copper) = 20 – 9 = 11 kg

Question 9.
A boy got 60 out of 80 in Hindi, 75 out of 100 in English and 65 out of 70 in Arithmetic. In which subject his percentage of marks the best ? Also, find his overall percentage.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 17
Now total marks he gets = 60 + 75 + 65 = 200
Total marks = 80 + 100 + 70 = 250
Percent marks obtained = \(\frac { 200 }{ 250 }\) x 100 = 80%

Question 10.
In a camp, there were 500 soldiers. 60 more soldiers joined them. What percent of the earlier (original) number have joined the camp.
Solution:
Number of soldiers = 500
More joined them = 60
Percentage to join the earlier = \(\frac { 60 }{ 500 }\) x 100 = 12%

Question 11.
In a plot of ground of area 6000 sq. m, only 4500 sq. m is allowed for construction. What percent is to be left without construction ?
Solution:
Total ground area = 6000 sq. m.
Allowed for construction = 4500 sq.m.
Area left without construction = 6,000 sq. m – 4500 sq. m = 1500 sq. m
Percentage of construction left = \(\frac { 1500 }{ 6000 }\) x 100 = 25%

Question 12.
Mr. Sharma has a monthly salary of ₹ 8,000. If he spends ₹ 6,400 every month; find :
(i) his monthly expenditure as percent.
(ii) his monthly savings as percent.
Solution:
Monthly salary of Mr. Sharma = ₹ 8000
He spends every month = ₹ 6400
His savings = ₹ 8000 – 6400 = ₹ 1600
(i) Percent expenditure = \(\frac { 6400}{ 8000 }\) x 100% = 80%
(ii) Percent savings = \(\frac { 1600}{ 8000 }\) x 100% = 20%

Question 13.
The monthly salary of Rohit is ₹ 24,000. If his salary increases by 12%, find his new monthly salary
Solution:
Salary = ₹ 24000
New salary = ₹ 24000 + 12% of 24000
= ₹ 24000 + \(\frac { 12 }{ 100 }\) x 24000
= ₹ 24000 + 2880 = ₹ 26880
New salary = ₹ 26880

Question 14.
In a sale, the price of an article is reduced by 30%. If the original price of the article is ₹ 1,800, find :
(i) the reduction in the price of the article
(ii) reduced price of the article.
Solution:
(i) Original price of article = ₹ 1800
Reduction = 30%
Reduction in price = 30% of 1800
= \(\frac { 30 }{ 100 }\) x 1800 = ₹ 540
(ii) Reduced price of the article = Original price – Reduction = ₹ 1800 – ₹ 540 = ₹ 1260

Question 15.
Evaluate :
(i) 30% of 200 + 20% of 450 – 25% of 600
(ii) 10% of ₹ 450 – 12% of ₹ 500 + 8% of ₹ 500.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 18
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 19

Percent Exercise 16C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The price of rice rises from Rs. 30 per kg to Rs. 36 per kg. Find the percentage rise in the price of rice.
Solution:
First price of rice = Rs. 30 per kg
Rised price = Rs. 36 per kg
Rise per kg = 36 – 30 = Rs. 6
Percent rise = \(\frac { 6 }{ 30 }\) x 100 = 20%

Question 2.
The population of a small locality was 4000 in 1979 and 4500 in 1981, By what percent had the population increase ?
Solution:
Year 1979 population = 4,000
Year 1981 population = 4,500
Increase in population = (4,500 – 4,000) = 500
percentage of increase in population = \(\frac { 500 }{ 4000 }\) x 100 = 12.5%

Question 3.
The price of a scooter was ₹ 8000 in 1975. It came down to ₹ 6000 in 1980. By what percent had the price of the scooter came down ?
Solution:
Original cost of scooter = ₹ 8,000
Reduced cost of scooter = ₹ 6000
Reduction in price of scooter = ₹ 8,000 – ₹ 6,000 = ₹ 2,000
Percentage of reduction = \(\frac { 2000 }{ 8000 }\) x 100 = 25%

Question 4.
Find the resulting quantity when :
(i) ₹ 400 is decreased by 8%.
(ii) 25 km is increased by 5%.
(iii) a speed of 600 km/h is increased by 12\(\frac { 1 }{ 2 }\) %
(iv) there is 2.5% increase in a salary of ₹ 62, 500.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 20
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 21
= ₹ 1562.50
Resulting quantity (salary) = ₹ 62500 + ₹ 1562.50 = ₹ 64062.50

Question 5.
The population of a village decreased by 12%. If the original population was 25,000, find the population after decrease ?
Solution:
Original population = 25,000
Decrease in population = 12% Population after decrease
= 25,000 – 12% of 25,000
= 25,000 – \(\frac { 12 }{ 100 }\) x 25,000
= 25,000 – 3,000 = 22,000

Question 6.
Out of a salary of Rs. 13,500,1 keep 1/3 as savings. Of the remaining money, I spend 50% on food and 20% on house rent. How much do I spend on food and house rent ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 22

Question 7.
A tank can hold 50 litres of water. At present, it is only 30% full. How many litres of water shall I put into the tank so that it becomes 50% full ?
Solution:
Capacity of tank = 50 litres
30% of capacity = 30% of 50 litres
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 23

Question 8.
In an election, there are a total of 80,000 voters and two candidates, A and B. 80% of the voters go to the polls out of which 60% vote for A. How many votes does B get.
Solution:
Member of voters = 80,000
Total vote polled = 80% of 80,000 = \(\frac { 80 }{ 100 }\) x 80,000 = 64,000
Vote polled to A = 60% of 64,000 = \(\frac { 60 }{ 100 }\) x 64000 = 38,400
Vote polled to B = Total vote polled – vote polled to A = 64,000 – 38,400 = 25,600

Question 9.
70% of our body weight is made up of water. Find the weight of water in the body of a person whose body weight is 56 kg.
Solution:
Water in human body = 70%
Weight of a man = 56 kg
Quantity of water in him = 70% of 56
= \(\frac { 70 }{ 100 }\) = x 56 = 39.2 kg

Question 10.
Only one-fifth of water is available in liquid form. This limited amount of water is replenished and used by man recurrently. Express this information as percent, showing :
(i) water available in liquid form.
(ii) water available in frozen form.
Solution:
Let total quantity of water = 1
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 24

Question 11.
By weight, 90% of tomato and 78% of potato is water. Find :
(i) the weight of water in 25 kg of tomato.
(ii) the total quantity, by weight, of water in 90 kg of potato and 30 kg of tomato
(iii) the weight of potato which contains 39 kg of water.
Solution:
Water in tomato = 90% and water in potato = 78%
(i) Weight of water in 25 kg of tomato
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 25

Percent Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Rohit’s age is 12 years and Geeta’s age is 15 years. Express :
(i) Rohit’s age as a percent of Geeta’s age.
(ii) Geeta’s age as a percent of Rohit’s age.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 26

Question 2.
A class has 30 boys and 20 girls. Find:
(i) the percentage of girls in the class
(ii) the percentage of boys in the class
(iii) percentage of number of boys as compared with number of girls.
Solution:
Total students in class = Number of boys + Number of girls = 30 + 20 = 50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 27

Question 3.
Mrs. Sharma went to the market with ₹ 800 in her purse. When she returned to her home, ₹ 240 were still left in her purse. What percent of her money did she spend in the market ?
Solution:
Money in her purse = ₹ 800
Balance in her purse = ₹ 240
Money spent = ₹ 800 – ₹ 240 = ₹ 560
Percentage of money spent = \(\frac { 560 }{ 800 }\) x 100 = 70%

Question 4.
In a mixture of two liquids A and B, 35% is liquid B. If the total quantity of the mixture is 20 kg, find the quantity of A, by weight.
Solution:
Total quantity of A and B = 20 kg
Quantity of B = 35% of 20 = \(\frac { 35 }{ 100 }\) x 20 = 7kg
Quantity of A = Total quantity – Quantity of B = 20 kg – 7 kg = 13 kg
Hence, quantity of A = 13 kg.

Question 5.
A girl got 375 marks out of 500 in the first term examination, 560 marks out of 800 in the second term examination and 840 marks out of 1200 in the third term examination. Find:
(i) her percentage score in the first term examination.
(ii) her percentage score in the second term examination.
(iii) her percentage score in the third term examination.
(iv) the total marks secured in all the three examinations.
(v) the total marks scored in all the three examinations.
(vi) her percentage score on the whole in all the three examinations.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 28

Question 6.
Out of his monthly income of ₹ 2,500; a man spends ₹ 1,750. What percent of his income does he save every month ?
Solution:
Monthly income = ₹ 2,500
Spending = ₹ 1,750
Saving = monthly income – spending = 2,500- 1,750 = ₹ 750
Percentage of income he saves = \(\frac { 750 }{ 2500 }\) x 100 = 30%

Question 7.
Mr. Singh’s monthly salary is ₹ 15,000. This month he was promoted with an increment of ₹ 3,000 in his salary. Express his increment as a percent of his original salary.
Solution:
Monthly salary = ₹ 15,000
Increment on promotion = ₹ 3,000
Percentage of increment to monthly salary = \(\frac { 3000 }{ 15000 }\) x 100 = 20%

Question 8.
(i) The price of an article increased from ₹ 16 to ₹ 20 ; find the percentage increase.
(ii) The price of an article decreased from Rs 20 to Rs 16 ; find the percentage decrease.
Solution:
(i) Original price = ₹ 16
Increased price = ₹ 20
Amount of increase = 20 – 16 = ₹ 4
Percentage of increase = \(\frac { 4 }{ 16 }\) x 100 = 25%
(ii) Original price = ₹ 20
Decrease price = ₹ 16
Amount of decrease = 20 – 16 = ₹ 4
Percentage of decrease = \(\frac { 4 }{ 20 }\) x 100 = 20%

Question 9.
(i)) The salary of a man is ₹ 7,200 per month, which is now increased by 8%. Find his new salary per month.
(ii) The salary of Mr. Sahni is ₹ 8,400 per month, which is now decreased by 8%. Find his new salary per month.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 29

Question 10.
Find the percentage change from the first quantity to the second :
(i) ₹ 80, ₹ 120
(ii) 75 kg, 60 kg
(iii) 50 cm, 45 cm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 30

Question 11.
The original price of an article is ₹ 640. Find its new price when its price is :
(i) increased by 30%
(ii) decreased by 20%
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 31

Question 12.
Find the number that is :
(i) 50% more than 48
(ii) 30% less than 70
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 32

Question 13.
Evaluate :
(i) 8% of 900 – 12% of 750 + 20% of 165.
(ii) 70% of 70 + 90% of 90 – 120% of 120.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 33

Question 14.
Approximately 97.3% water on the earth is not fit for drinking. Find :
(i) the percentage of water on the earth that is fit for drinking.
(ii) The total volume of water available in certain part of the earth where there is 21,600 m3 of drinking water.
Solution:
Approximately water on earth which is not fit for drinking = 97.3%
(i) Water fit for drinking = 100 – 97.3 = 2.7%
(ii) At a certain place, the water which is fit for drinking = 21600 m3
Volume of total water on that place
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 34

Question 15.
Air is an important inexhaustible natural resource. It is essential for the survival of human beings, microbes, plants and animals. The following table shows the percentage of various gases in air.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 35
(i) In 800 m3 of air, calculate the approximate quantities of nitrogen, oxygen and other gases.
(ii) If a certain quantity (by volume) of air contains 4,200 litres of oxygen, find the total quantity of air taken and the amount of nitrogen in it.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 36

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time

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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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Idea of Speed, Distance and Time Exercise 17A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
A train covers 51 km in 3 hours. Calculate its speed. How far does the train go in 30 minutes?
Solution:
Given : Distance = 51 km
Time = 3 hours
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 1

Question 2.
A motorist travelled the distance between two towns, which is 65 km, in 2 hours and 10 minutes. Find his speed in metre per minute.
Solution:
Distance between two towns = 65 km
Time taken = 2 hr 10 min
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 3

Question 3.
A train travels 700 metres in 35 seconds. What is its speed in km/h?
Solution:
Distance = 700 m
Time taken = 35 sec
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 4

Question 4.
A racing car covered 600 km in 3 hours 20 minutes. Find its speed in metre per second. How much distance will the car cover in 50 sec?
Solution:
Distance covered = 600 km
Time taken = 3 hr 20 min
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 5
= Speed x Time
= 50 x 50 m = 2500 m or 2.50 km

Question 5.
Rohit goes 350 km in 5 hours. Find :
(i) his speed
(ii) the distance covered by Rohit in 6.2 hours
(iii) the time taken by him to cover 210 km.
Solution:
Distance covered = 350 km
Time taken = 5 hours
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 6

Question 6.
A boy drives his scooter with a uniform speed of 45 km/h. Find :
(i) the distance covered by him in 1 hour 20 min.
(ii) the time taken by him to cover 108 km.
(iii) the time taken to cover 900 m.
Solution:
Speed of the scooter = 45 km/h
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 7
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 8

Question 7.
I travel a distance of 10 km and come back in 2\(\frac { 1 }{ 2 }\) hours. What is my speed?
Solution:
Total distance covered = 10 km + 10 km = 20 km
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 9

Question 8.
A man walks a distance of 5 km in 2 hours. Then he goes in a bus to a nearby town, which is 40 km, in further 2 hours. From there, he goes to his office in an autorickshaw, a distance of 5 km, in \(\frac { 1 }{ 2 }\) hour. What was his average speed during the whole journey?
Solution:
Distance of 5 km travelled on foot in 2 hours
Distance of 40 km travelled by bus in 2 hours
Distance of 5 km travelled by Rickshaw in \(\frac { 1 }{ 2 }\) hour
Total distance covered = 5 + 40 + 5 = 50 km
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 10

Question 9.
Jagan went to another town such that he covered 240 km by a car going at 60 kmh-1. Then he covered 80 km by a train, going at 100 kmh-1 and the rest 200 km, he covered by a bus, going at 50 kmh-1. What was his average speed during the whole journey?
Solution:
Distance covered 240 km by car with speed 60 km/h
Distance covered 80 km by train with speed 100 km/h
and rest distance covered 200 km by bus with speed 50 km/h
Total distance covered = (240 + 80 + 200) km = 520 km
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 12

Question 10.
The speed of sound in air is about 330 ms-1. Express this speed in kmh-1. How long will the sound take to travel 99 km?
Solution:
Speed of sound in air = 330 m/sec
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 13

Idea of Speed, Distance and Time Exercise 17B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
A train 180 m long is running at a speed of 90 km/h. How long will it take to pass a railway signal?
Solution:
Distance = 180 m
Speed = 90 km/h
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 14

Question 2.
A train whose length is 150 m, passes a telegraph pole in 10 sec. Find the speed of the train in km/h.
Solution:
Distance = 150 m
Time taken = 10 sec
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 15

Question 3.
A train 120 m long passes a railway platform 160 m long in 14 sec. How long will it take to pass another platform which is 100 m long?
Solution:
Distance covered = 120 m + 160 m = 280 m
Time taken = 14 seconds
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 16

Question 4.
Mr. Amit can walk 8 km in 1 hour 20 minutes.
(a) How far does he go in :
(i) 10 minutes ?
(ii) 30 seconds ?
(b) How long will it take him to walk :
(i) 2500 m ?
(ii) 6.5 km ?
Solution:
Amit walks 8 km in 1 hour 20 min
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 17
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 18

Question 5.
Which is greater : a speed of 45 km/h or a speed of 12.25 m/sec?
How much is the distance travelled by each in 2 seconds?
Solution:
First speed = 45 km/h
Second = 12.25 m/sec
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 19

Question 6.
A and B start from the same point and at the same time with speeds 15 km/h and 12 km/h respectively, find the distance between A and B after 6 hours if both move in :
(i) same direction
(ii) the opposite directions.
Solution:
A’s speed = 15 km/h
B’s speed = 12 km/h
Distance covered by A in 6 hours = 15 x 6 = 90 km
and Distance covered by B in 6 hours = 12 x 6 = 72 km
(i) Distance between A and B when they move in the same direction = 90 – 72 = 18 km
(ii) Distance between A and B, when they move in the opposite directions = 90 + 72 = 162 km

Question 7.
A and B start from the same place, in the same direction and at the same time with speeds 6 km/h and 2 m/sec respectively. After 5 hours who will be ahead and by how much?
Solution:
A’s speed = 6 km/h
B’s speed = 2 m/sec
Distance covered by A in 5 hours = 6 x 5 = 30 km
and distance covered by B in 5 hours = 5 x 60 x 60 x 2 m = 36000 m
= \(\frac { 3600 }{ 1000 }\) = 36 km
B will be ahead and 36 – 30 = 6 km ahead.

Question 8.
Mohit covers a certain distance in 6 hrs by his scooter at a speed of 40 kmh-1.
(i) Find the time taken by Manjoor to cover the same distance by his car at the speed of 60 kmh-1.
(ii) Find the speed of Joseph, if he takes 8 hrs to complete the same distance.
Solution:
Mohit’s speed = 40 km/h or kmh-1
Distance covered in = 6 hours
Distance = 40 x 6 = 240 km
(i) Manjoor car’s speed = 60 kmh-1
He will cover the distance of 240 km in = \(\frac { 240 }{ 60 }\) = 4 hours
(ii) Joseph covered that distance in 8 hours
His speed = \(\frac { 240 }{ 8 }\) = 30 kmh-1

Question 9.
A boy swims 200 m in still water and then returns back to the point of start in total 10 minutes. Find the speed of his swim in
(i) ms-1
(ii) kmh-1.
Solution:
Distance swimed by a boys of 200 m + 200 m = 400 m
Time taken = 10 minutes
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 20

Question 10.
A distance of 14.4 km is covered in 2 horus 40 minutes. Find the speed in ms-1. With this speed Sakshi goes to her school, 240 m away from her house and then returns back. How much time, in all, will Sakshi take?
Solution:
Distance = 14.4 km
Time taken to cover = 2 hrs 40 min
= \(2\frac { 2 }{ 3 } =\frac { 8 }{ 3 }\) hrs
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 21
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 17 Idea of Speed, Distance and Time image - 22

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions

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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
Decimal Fraction : A fraction, whose denominator is 10 or a higher power of 10 e.g. 100, 1,000, 10,000 etc. is known as decimal fraction.
Number of Decimal Places: The number of digits in the decimal part of a number is the number of decimal places in it.
When the given number has only decimal part in it. It is always written 0 before it as 0.7, 0.55 are written as 0.7, 0.55.

Conversion of a Fraction into a Decimal Fraction :

    1. When the denominator is 10,100,1000, 10,000 etc. : Counting from right to left of the numerator of the given fraction, mark the decimal point after as many digits as the number of zeroes in it denominator
      Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 1
  1. When the denominator is not, 10, 100, 1000, 10,000 etc.
    Multiply both, the numerator and denominator of the given fraction, by a suitable number to get the denominator 10 or a power of 10 and then proceed as above, e.g.
    Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 2
  2. Conversion of a given Decimal Fraction into a Non-Decimal Fraction : Remove the decimal point and at the same time write 1 in the denominator, as many zeroes to the right of 1 as there are digits in the decimal part e.g.,
    Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 4
    Zero or zeores written at the right of a decimal number does not change its value, e.g. 3.4 is the same as 3.40, 3.400, 3.4000 etc.

Decimal Fractions Exercise 15A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write the number of decimal places in each of the following :
(i) 7.03
(ii) 0.509
(iii) 146.2
(iv) 0.0065
(v) 8.03207
Solution:
(i) 7.03, the decimal part is .03 which contains two digits.
Number 7.03 has 2 decimal places.
(ii) 0.509, the decimal part is 0.509 which contains three digits.
Number 0.509 has 3 decimal places
(iii) 146.2, the decimal part is .2 which contains one digits.
Number 146.2 has 1 decimal places.
(iv) 0.0065, the decimal part is .0065 which contains four digits.
Number 0.0065 has 4 decimal places
(v) 8.03207, the decimal part is .03207 which contains five digits.
Number 8.03207 has 5 decimal places.

Question 2.
Convert the given unlike decimal fractions into like decimal fractions:
(i) 1.36, 239.8 and 47.008
(ii) 507.0752, 8.52073 and 0.808
(iii) 459.22, 7.03093 and 0.200037
Solution:
(i) 1.36 = 1.360
239:8 = 239.800
47.008 = 47.008
(ii) 507.0752 = 507.07520
8.52073 = 8.52073
0.808 = 0.80800
(iii) 459.22 = 459.220000
7.03093 = 7.030930
0.200037 = 0.200037

Question 3.
Change each of following fractions to a decimal fraction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 6.

Question 4.
Convert into a decimal fraction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 7
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 53
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 9

Question 5.
Change the given decimals fractions to fractions in their lowest terms :
(i) 0.05
(ii) 3.95
(iii) 4.005
(iv) 0.876
(v) 50.06
(vi) 0.01075
(vii) 4.8806
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 10

Decimal Fractions Exercise 15B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Add the following :
(i) 0.243, 2.47 and 3.009
(ii) 0.0736, 0.6095 and 0.9107
(iii) 1.01, 257 and 0.200
(iv) 18, 200.35, 11.72 and 2.3
(v) 0.586, 0.0586 and 0.00586
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 12

Question 2.
Find the value of :
(i) 6.8 – 2.64
(ii) 2 – 1.0304
(iii) 0.1 – 0.08
(iv) 0.83 – 0.342
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 13

Question 3.
Subtract :
(i) 0.43 from 0.97
(ii) 2.008 from 22.1058
(iii) 0.18 from 0.6
(iv) 1.002 from 17
(v) 83 from 92.05
Solution:
(i) 0.43 from 0.97
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 14

Question 4.
Simplify :
(i) 3.5 – 2.43 + 0.075
(ii) 7.84 + 0.3 – 4.016
(iii) 2.987 – 1.25 – 0.54
(iv) 52.9 – 231.666 + 204
(v) 8.57 – 6.4432 – 1.70 + 0.683
Solution:
(i) 3.5 – 2.43 + 0.075
= 3.500 + 0.075 – 2.43
= 3.575 – 2.430 = 1.145
(ii) 7.84 + 0.3 – 4.016
= 7.840 + 0.300 – 4.016
= 8.140 – 4.016
= 4.124
(iii) 2.987 – 1.25 – 0.54
= 2.987 – 1.79
= 2.987 – 1.790
= 1.197
(iv) 52.9 – 231.666 + 204
= 52.9 – 231.666 + 204.0
= 256.9 – 231.666
= 256.900 – 231.666
= 25.234

Question 5.
From the sum of 75.75 and 4.9 subtract 28.465.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 15

Question 6.
Subtract the sum of 8.14 and 12.9 from 32.7.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 16

Question 7.
Subtract the sum of 34.27 and 159.8 from the sum of 20.937 and 200.6.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 17

Question 8.
From the sum of 2.43 and 4.349 subtract the sum of 0.8 and 3.15.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 18

Question 9.
By how much does the sum of 18.0495 and 34.9644 exceed the sum of 7.6752 and 24.876 ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 19

Question 10.
What least number must be added to 89.376 to get 1000?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 20

Decimal Fractions Exercise 15C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Multiply :
(i) 5.6 and 8
(ii) 38.46 and 9
(iii) 0.943 and 62
(iv) 0.0453 and 35
(v) 7.5 and 2.5
(vi) 4.23 and 0.8
(vii) 83.54 and 0.07
(viii) 0.636 and 1.83
(ix) 6.4564 and 1000
(x) 0.076 and 100
Solution:
(i) 5.6 x 8 = 44.8
(ii) 38.46 x 9 = 346.14
(iii) 0.943 and 62
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 21
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 22

Question 2.
Evaluate :
(i) 0.0008 x 26
(ii) 0.038 x 95
(iii) 1.2 x 2.4 x 3.6
(iv) 0.9 x 1.8 x 0.27
(v) 1.5 x 1.5 x 1.5
(vi) 0.025 x 0.025
(vii) 0.2 x 0.002 x 0.001
Solution:
(i) 0.0008 – 26
Since, 8 x 26 = 208
0.0008 x 26 = 0.0208
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 23
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 24
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 25

Question 3.
Multiply each of the following numbers by 10, 100 and 1000 :
(i) 3.9
(ii) 2.89
(in) 0.0829
(iv) 40.3
(v) 0.3725
Solution:
(i) 3.9 x 10 = 39
3.9 x 100 = 390.0 = 390
3.9 x 1000 = 3900.0 =3900
(ii) 2.89 x 10 = 28.9
2.89 x 100 = 289
2.89 x 1000 = 2890.00 = 2890
(iii) 0.0829 x 10 = 0.829
0.0829 x 100 = 8.29
0.0829 x 1000 = 82.9
(iv) 40.3 x 10 = 403
40.3 x 100 = 4030
40.3 x 1000 = 40300
(v) 0.3725 x 10 = 3.725
0.3725 x 100 = 37.25
0.3725 x 1000 = 372.5

Question 4.
Evaluate :
(i) 8.64 ÷ 8
(ii) 0.0072 ÷ 6
(iii) 20.64 ÷ 16
(iv) 1.602 ÷ 15
(v) 13.08 ÷ 4
(vi) 3.204 ÷ 9
(vii) 3.024 ÷ 12
(viii) 5.15 ÷ 5
(ix) 3 ÷ 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 26
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 27

Question 5.
Divide each of the following numbers by 10,100 and 1000 :
(i) 49.79
(ii) 0.923
(iii) 0.0704
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 28

Question 6.
Evaluate :
(i) 9.4 ÷ 0.47
(ii) 6.3 ÷ 0.09
(iii) 2.88 ÷ 1.2
(iv) 8.64 ÷ 1.6
(v) 37.188 ÷ 3.6
(vi) 16.5 ÷ 0.15
(vii) 3.2 ÷ 0.005
(viii) 3.24 ÷ 0.0016
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 29
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 30

Question 7.
Fill in the blanks with 10,100,1000, or 10000 etc.:
(i) 7.85 x ………….. = 78.5
(ii) 0.442 x ………. = 442
(in) 0.0924 x ………… = 9.24
(iv) 0.00187 x …………. = 18-7
(v) 2.6 x …………. = 2600
(vi) 0.08 x ………… = 80
(vii) 96.7 ÷ ………….. = 0.967
(viii) 5.2 ÷ ……………. = 0.52
(ix) 33.15 ÷ ……………….. = 0.03315
(x) 0.7 ÷ ………… = 0.007
(xi) 0.00672 x ……… = 67.2
Solution:
(i) 7.85 x 10 = 78.5
(ii) 0.442 x 1000 = 442
(iii) 0.0924 x 100 = 9.24
(iv) 0.00187 x 10000 = 18.7
(v) 2.6 x 1000 = 2600
(vi) 0.08 x 1000 = 80
(vii) 96.7 ÷ 100 = 0.967
(viii) 5.2 ÷ 10 = 0.52
(ix) 33.15 ÷ 1000 = 0.03315
(x) 0.7 ÷ 100 = 0.007
(xi) 0.00672 x 10000 = 67.2

Question 8.
Evaluate :
(i) 9.32 – 28.54 ÷ 10
(ii) 0.234 x 10 + 62.8
(iii) 3.06 x 100 – 889.4 ÷ 100
(iv) 2.86 x 7.5 + 45.4 ÷ 0.2
(v) 97.82 x 0.03 – 0.54 ÷ 0.3
Solution:
(i) 9.32 – 28.54 ÷ 10
= 9.32 – 2.854
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 31
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 32

Decimal Fractions Exercise 15D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express in paise :
(i) Rs. 8.40
(ii) Rs. 0.97
(iii) Rs. 0.09
(iv) Rs. 62.35
Solution:
(i) Rs. 8.40 = 8.40 x 100 paise [1Rs. = 100 Paise]
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 33

Question 2.
Express in rupees :
(i) 55 P
(ii) 8 P
(iii) 695 P
(iv) 3279 P
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 34

Question 3.
Express in centimetre (cm) :
(i) 6 m
(ii) 8.54 m
(iii) 3.08 m
(iv) 0.87 m
(v) 0.03 m
(vi) 25.04 m
Solution:
(i) 6 x 100 = 600 cm
(ii) 8.54 x 100 = 854 cm
(iii) 3.08 x 100 = 308 cm
(iv) 0.87 x 100 = 87 cm
(v) 0.03 x 100 = 3 cm
(vi) 25.04 x 100 = 2504 cm

Question 4.
Express in metre (m) :
(i) 250 cm
(ii) 2328 cm
(iii) 86 cm
(iv) 4 cm
(v) 107 cm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 35
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 36

Question 5.
Express in gramme (gm) :
(i) 6 kg
(ii) 5.543 kg
(iii) 0.078 kg
(iv) 3.62 kg
(v) 4.5 kg
Solution:
(i) 6 x 1000 = 6000 gm
(ii) 5.543 x 1000 = 5543 gm
(iii) 0. 078 kg = 0.078 x 1000 g = 78 g (1 kg = 1000 g)
(iv) 3.62 x 1000 = 3620 gm
(v) 4.5 x 1000 = 4500 gm

Question 6.
Express in kilogramme (kg) :
(i) 7000 gm
(ii) 6839 gm
(iii) 445 gm
(iv) 8 gm
(iv) 93 gm
(vi) 13545 gm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 37

Question 7.
Add (giving answer in rupees) :
(i) Rs. 5.37 and Rs. 12
(ii) Rs. 24.03 and 532 paise
(iii) 73 paise and Rs. 208
(iv) 8 paise and Rs. 1536
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 38

Question 8.
Subtract :
(i) Rs. 35.74 from Rs. 63.22
(ii) 286 paise from Rs. 7.02
(iii) Rs. 0.55 from 121 paise
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 39
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 40

Question 9.
Add (giving answer in metre) :
(i) 2.4 m and 1.78 m
(ii) 848 cm and 2.9 m
(iii) 0.93 m and 64 cm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 41

Question 10.
Subtract (giving answer in metre) :
(i) 5.03 m from 19.6 m
(ii) 428 cm from 1033 m
(iii) 0.84 m from 122 cm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 42

Question 11.
Add (giving answer in kg) :
(i) 2.06 kg and 57.864 kg
(ii) 778 gm and 1.939 kg
(iii) 0.065 kg and 4023 gm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 43

Question 12.
Subtract (giving answer in kg) :
(i) 9.462 kg from 15.6 kg
(ii) 4317 gm from 23 kg
(iii) 0.798 kg from 4169 gm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 44

Decimal Fractions Exercise 15E – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The cost of a fountain pen is Rs. 13.25. Find the cost of 8 such pens.
Solution:
Cost of 1 fountain Pen = Rs. 13.25
Cost of 8 fountain Pen = 13.25 x 8 = 106.00 = Rs. 106

Question 2.
The cost of 25 identical articles is Rs. 218.25. Find the cost of one article.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 45

Question 3.
The length of an iron rod is 10.32 m. The rod is divided into 4 pieces of equal lengths. Find the length of each piece.
Solution:
The length of iron rod = 10.32 m
Dividing in 4 equal parts = \(\frac { 10.32 }{ 4 }\) = 2.58 m

Question 4.
What will be the total length of cloth required to make 5 shirts, if 2.15 m of cloth is needed for each shirt ?
Solution:
Cloth required for each shirt = 2.15 m
Cloth required for 5 shirts = 2.15 x 5 m = 10.75 m

Question 5.
Find the distance walked by a boy in 1\(\frac { 1 }{ 2 }\) hours, if he walks at 2.150 km every hour.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 47

Question 6.
83 note-books are sold at Rs. 15.25 each. Find the total money (in rupees) obtained by selling these note-books.
Solution:
Sale price of 1 note-book = Rs. 15.25
Sale of 83 books = Rs. 15.25 x 83 = Rs. 1265.75 paise
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 48

Question 7.
If length of one bed-cover is 2.1 m, find the total length of 17 bed-covers.
Solution:
Length of one bed-cover = 2.1 m
Length of 17 bed-cover = 17 x 2.1 = 35.7 m

Question 8.
A piece of rope is 10 m 67 cm long. Another rope is 16 m 32 cm long. By how much is the second rope longer than the first one ?
Solution:
Length of one rope = 10 m 67 cm
Length of another rope = 16 m 32 cm
Difference in length = 16 m \(\frac { 32 }{ 100 }\) cm – 10 m \(\frac { 67 }{ 100 }\) cm
= 16.32 m – 10.67 m
= 5.65 m or 5 m 65 cm.

Question 9.
12 cakes of soap together weigh 5 kg and 604 gm. Find the weight of
(i) One cake in both kg and gramme
(ii) 5 cakes in kg.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 49

Question 10.
Three strings of lengths 50 m 75 cm; 68 m 58 cm and 121 m 3 cm, respectively, are joined together to get a single string of greatest length, And the length of the single string obtained.
If this single string is then divided into 12 equal pieces ; find the length of each piece.
Solution:
1st string 50 m 75 cm = 50.75 m
2nd string 68 m 58 cm = 68.58 m
3rd string 121 m3 cm= 121.03 m
On joining three total length = 240.36 m
Now, one string = 240.36 m
Dividing 12 parts = \(\frac { 240.36 }{ 12 }\) = 20.3 m.

Decimal Fractions Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write th& following decimal numbers in ascending order of value
(i) 5.054, 5.250, 5.245 and 5.0543
(ii) 62.443, 62.434, 62.344 and 62.444
Solution:
(i) 5.054, 5.250, 5.245 and 5.0543
Writing them in like decimals :
5.0540, 5.2500, 5.2450, 5.0543
Now arranging in ascending order :
5.0540, 5.0543, 5.2450, 5.2500
=> 5.054 < 5.0543 < 5.245 < 5.250
(ii) 62.443, 62.434, 62.344 and 62.444
There are in like decimals :
Now writing in ascending order.
62.344, 62.434, 62.443, 62.444
or 62.344 < 62.434 < 62.443 < 62.444

Question 2.
What number added to 0.805 gives 1 ?
Solution:
The required number will be formed by subtracting 0.805 from 1
Required number = 1 – 0.805 = 1.000 – 0.805 = 0.195

Question 3.
What must be subtracted from 3 to get 2.462 ?
Solution:
The required number can be formed by subtracting 2.462 from 3
Required number = 3 – 2.462 = 3.000 – 2.462 = 0.538

Question 4.
By how much should 83.407 be decreased to get 27.78 ?
Solution:
The required number can be formed by subtracting 27.78 from 83.407
Required number = 83.407 – 27.78 = 83.407 – 27.780 = 55.627

Question 5.
Two articles weigh 32.674 kg and 40.038 kg respectively. Find :
(i) the total weight of both the articles.
(ii) the difference in the weights of both the articles.
Solution:
Weight of first article = 32.674 kg
Weight of second article = 40.038 kg
(i) Total weight of both the articles = (32.674 + 40.038) kg = 72.712 kg
(ii) Difference between the weights of the articles = (40.038 – 32.674) kg = 7.364 kg

Question 6.
By how much does the sum of 34.07 and 15.239 exceed the sum of 16.40 and 27.08?
Solution:
Sum of 34.07 and 15.239 = 34.070 + 15.239 = 49.309
and sum of 16.40 and 27.08 = 16.40 + 27.08 = 43.48
Difference between their sums = 49.309 – 43.48 = 49.309 – 43.480 = 5.829

Question 7.
The cost of 1 kg of fruit is Rs. 27.50. What is the cost of 3.6 kg of fruit ?
Solution:
Cost of 1 kg fruit = Rs. 27.50
Cost of 3.6 kg fruit = Rs. 27.50 x 3.6 = Rs. 99.00

Question 8.
Evaluate :
(i) 0.8 x 0.8 x 0.8
(ii) 0.8 ÷ 0.8 x 0.8
(iii) 0.8 x 0.8 ÷ 0.8
(iv) 0.8 ÷ 0.8 of 0.8
(v) 0.8 of 0.8 ÷ 0.8
Solution:
(i) 0.8 x 0.8 x 0.8 = 0.512
(ii) 0.8 ÷ 0.8 x 0.8
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 50

Question 9.
Evaluate :
(i) 3.5 x (4.2 + 2.6)
(ii) 3.5 x 4.2 + 3.5 x 2.6
Are (i) and (ii) equal ?
Solution:
(i) 3.5 x (4.2 + 2.6) = 3.5 x (6.8) = 23.8
(ii) 3.5 x 4.2 + 3.5 x 2.6 = 14.7 + 9.1 = 23.8
Yes results of (i) and (ii) are equal.

Question 10.
Evaluate :
(i) (3.87 – 2.09) x 2.4
(ii) 3.87 x 2.4 – 2.09 x 2.4
Are (i) and (ii) equal ?
Solution:
(i) (3.87 – 2.09) x 2.4 = 1.78 x 2.4 = 4.272
(ii) 3.87 x 2.4 – 2.09 x 2.4 = 9.288 – 5.016 = 4.272
Yes, results of (i) and (ii) are equal.

Question 11.
A 4.85 m long pole is divided into 5 equal parts. Find the length of each part.
Solution:
Length of pole = 4.85 m
It is divided into 5 equal parts Length of each part = 4.85 ÷ 5 m = 0.97 m
Hence length of each part = 0.97 m

Question 12.
A car can run 16.8 km consuming one litre of petrol. How many kilometres will it run on 3.7 litres of petrol ?
Solution:
A car can go in one litre = 16.8 km
It will go in 3.7 litres of petrol = 16.8 x 3.7 km = 62.16 km

Question 13.
A certain amount of money is distributed among 28 persons. If each person gets Rs. 62.45 and Rs. 5.78 is left, find the original amount of money.
Solution:
Number of persons = 28
Each person gets = RS. 62.45
Total amount distributed to 28 persons = Rs. 62.45 x 28 = Rs. 1748.60
Amount left undistributed = Rs. 5.78
Total amount = Rs. 1748.60 + 5.78 = Rs. 1754.38

Question 14.
Complete the following table :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 51
Solution:
The given table has been completed as follows:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 15 Decimal Fractions image - 52

Question 15.
The difference between two numbers is 47.364. If the smaller number is 31.855 ; find the bigger one.
Solution:
Difference of two number = 47.364
Smaller number = 31.855
Bigger number = 47.364 + 31.855 = 79.219

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 1
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 3

Fractions Exercise 14A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
For each expression, given below, write a fraction :
(i) 2 out of 7 = ………..
(ii) 5 out of 17 = ………..
(iii) three-fifths = ………
Solution:
(i) 2 out of 7 = \(\frac { 2 }{ 7 }\)
(ii) 5 out of 17 = \(\frac { 5 }{ 17 }\)
(iii) three-fifths = \(\frac { 3 }{ 5 }\)

Question 2.
Fill in the blanks :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 4
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 5
Solution:
(i) Proper
(ii) Improper
(iii) Improper
(iv) 1
(v) -1
(vi) Mixed
(vii) Like
(viii) Unlike fraction
(ix) Equal fraction
(x) Like
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 6

Question 3.
From the following fractions, separate :
(i) Proper fractions
(ii)Improper fractions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 7
Solution:
We know that proper fraction is a fraction whose numerator is less than its denominator and improper fraction is the fraction whose numerator is greater them its denominator :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 8

Question 4.
Change the following mixed fractions to improper fractions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 9
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 10

Question 5.
Change the following improper fractions to mixed fractions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 11
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 12

Question 6.
Change the following groups of fractions to like fractions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 13
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 14
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 15
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 16

Fractions Exercise 14B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Reduce the given fractions to their lowest terms :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 17
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 18

Question 2.
State, whether true or false ?
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 19
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 20
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 21

Question 3.
Which fraction is greater ?
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 22
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 23

Question 4.
Which fraction is smaller ?
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 24
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 25

Question 5.
Arrange the given fractions in descending order of magnitude :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 26
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 27
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 28
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 29

Question 6.
Arrange the given fractions in ascending order of magnitude :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 30
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 31
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 32
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 33

Question 7.
I bought one dozen bananas and ate five of them. What fraction of the total number of bananas was left ?
Solution:
Number of bananas bought = 1
Dozen = 12
Number of bananas eaten by me = 5
Number of bananas left = 12 – 5 = 7
Fraction = \(\frac { 7 }{ 12 }\)

Question 8.
Insert the symbol ‘=’ or ‘>’ or ‘<’ between each of the pairs of fractions, given below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 34
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 107
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 35
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 36
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 37

Question 9.
Out of 50 identical articles, 36 are broken. Find the fraction of :
(i) The total number of articles and the articles broken.
(ii) The remaining articles and total number of articles.
Solution:
Total number of articles = 50
Number of articles broken = 36
Remaining articles = 50 – 36 = 14
Now (i) the fraction of the total number of articles and articles broken = \(\frac { 50 }{ 36 }\)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 38

Fractions Exercise 14C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Add the following fractions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 39
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 40
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 41
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 42
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 43

Question 2.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 44
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 45
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 46
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 47
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 48
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 49
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 50

Fractions Exercise 14D – Selina Concise Mathematics Class 6 ICSE Solutions

Point to Remember :
BODMAS :- While simplifying an expressions we can involve six operation in following orders.
B Stands for “BRACKET”
O Stands for “OF”
D Stands for “DIVISION”
M Stands for “MULTIPLICATION”
A Stands for “ADDITION”
S Stands for “SUBTRACTION”

Question 1.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 51
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 52
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 53

Question 2.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 54
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 55

Question 3.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 56
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 57
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 58
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 59

Question 4.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 60
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 61
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 62
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 63
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 64

Question 5.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 65
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 66
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 67
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 68
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 69
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 70
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 71
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 72

Fractions Exercise 14E – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
From a rope of 10\(\frac { 1 }{ 2 }\) m long, 4\(\frac { 5 }{ 8 }\) m is cut off. Find the length of the remaining rope.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 73

Question 2.
A piece of cloth is 5 metre long. After washing, it shrinks by \(\frac { 1 }{ 25 }\) of its length. What is the length of the cloth after washing ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 74

Question 3.
I bought wheat worth Rs. 12\(\frac { 1 }{ 2 }\), rice worth Rs. 25\(\frac { 3 }{ 4 }\) and vegetables worth
Rs. 10\(\frac { 1 }{ 4 }\). If I gave a hundred-rupee note to the shopkeeper ; how much did he return to me
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 75

Question 4.
Out of 500 oranges in a box, \(\frac { 3 }{ 25 }\) are rotten and \(\frac { 1 }{ 5 }\) are kept for some guests. How many oranges are left in the box?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 76

Question 5.
An ornament piece is made of gold and copper. Its total weight is 96g. If \(\frac { 1 }{ 12 }\) of the ornament hi copper, find the weight of gold in it.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 77

Question 6.
A girl did half of some work on Monday and one-third of it on Tuesday. How much will she have to do on Wednesday in order to complete the work ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 78

Question 7.
A man spends \(\frac { 3 }{ 8 }\) of his money and 8 still has Rs. 720 left with him. How much money did he have at first ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 79
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 80

Question 8.
In a school, \(\frac { 4 }{ 5 }\) of the students are boys, and the number of girls is 100. Find the number of boys.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 81

Question 9.
After finishing \(\frac { 3 }{ 4 }\) of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 82

Question 10.
When Ajit travelled 15 km, he found that one-fourth of his journey was still left. What was the full length of the journey ?
Solution:
matics Class 6 ICSE Solutions Chapter 14 Fractions image - 83

Question 11.
In a particular month, a man earns Rs. 7,200. Out of this income, he spends \(\frac { 3 }{ 10 }\) on food, \(\frac { 1 }{ 4 }\) on house rent, \(\frac { 1 }{ 10 }\) on insurance and \(\frac { 2 }{ 25 }\) on holidays. How much did he save in that month ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 84
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 85

Fractions Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 86
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 87

Question 2.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 88
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 89

Question 3.
Evaluate :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 90
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 91
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 92
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 93
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 94
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 95
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 106

Question 4.
Mr. Mehra gave one-third of his money to his son, one-fifth of his money to his daughter and the remaining amount to his wife. If his wife got Rs. 91,000, how much money did Mr. Mehra have originally?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 96

Question 5.
A sum of Rs. 84,000 is divided among three persons A, B and C. If A gets one-fourth of it and B gets one-fifth of it; how much did C get ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 97
= Rs. 84,000 – (37,800) = Rs. 46,200

Question 6.
In one hour Rohit walks 3\(\frac { 2 }{ 5 }\) km. How much distance will he cover in 2\(\frac { 1 }{ 2 }\) hours?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 98

Question 7.
An 84 m long string is cut into pieces each of length 5\(\frac { 1 }{ 4 }\) m. How many pieces are obtained ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 99

Question 8.
In buying a ready made shirt-two-fifths of my pocket money is spent If Rs. 540 is still left with me, find :
(i) The money I had before I bought the shirt.
(ii) The emit of the shirt
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 100
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 101

Question 9.
Mohan leaves Rs. 1,20,000 to his wife and three children such that two-fifths of this money is given to his wife and the remaining is distributed equally among the children. Find, how much each child gets ?
Solution:
Total amount = Rs. 12,0,000
Amount given to his wife = \(\frac { 2}{ 5 }\) of Rs. 1,20,000
= Rs. 2 x 24,000 = Rs. 48,000
Remaining amount = Rs. 120000 – Rs. 48000 = Rs. 72000
This amount is distributed among three children equally.
Each’s share = Rs. 72,000 x \(\frac { 1 }{ 3 }\) = Rs. 24,000

Question 10.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 102
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 103
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 104
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 14 Fractions image - 105

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method

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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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Unitary Method Exercise 13A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The price of 25 identical articles is ₹ 1,750. Find the price of :
(i) one article
(ii) 13 articles
Solution:
The price of 25 articles = ₹ 1,750
(i) Price of one article = ₹ \(\frac { 1750 }{ 25 }\) = ₹ 70
(ii) Now, Price of 13 articles = 13 x ₹ 70 = ₹ 910

Question 2.
A motorbike travels 330 km in 5 litres of petrol. How much distance will it cover in :
(i) one litre of petrol ?
(ii) 2.5 litres of petrol ?
Solution:
(i) Consuming 5 litres petrol in 30 km
Consuming 1 litre petrol, motorbike covers = \(\frac { 330 }{ 5 }\) km = 66 km
(ii) Consuming 2.5 litres petrol = 66 x 2.5 = 165 km

Question 3.
If the cost of a dozen soaps is ₹ 460.80, what will the cost of:
(i) each soap ?
(ii) 15 soaps ?
(iii) 3 dozen soaps ?
Solution:
(i) Cost of one dozen soap = ₹ 460.80
In one dozen = 12 soaps
Cost of each soap = ₹ \(\frac { 460.80 }{ 12 }\) = ₹ 38.4
(ii) Cost of 15 soaps = 15 x ₹ 38.4 = ₹ 576
(iii) Cost of 3 dozen soaps = (12 x 3 = 36) = 36 x ₹ 38.4 = ₹ 1382.4

Question 4.
The cost of 35 envelops is ₹ 105. How many envelops can be bought for ₹ 90 ?
Solution:
Envelops purchased by ₹ 105 = 35
Envelopes purchased by ₹ 1 = \(\frac { 35 }{ 105 }\)
In ₹ 90, the envelop will be bought = \(\frac { 35 }{ 105 }\) x 90 = 30

Question 5.
If the cost of 8 cans of juice is ₹ 280, then what will be the cost of 6 cans of juice ?
Solution:
Cost of 8 cans of juice = ₹ 280
Cost of 1 can of juice = \(\frac { 280 }{ 8 }\) = ₹ 35
then, cost of 6 cans of juice = 6 x ₹ 35 = ₹ 210

Question 6.
For ₹ 378, 9 cans of juice can be bought, then how many cans of juice can be bought for ₹ 504?
Solution:
In ₹ 378, the juice can bought = 9 cans
In ₹ 504, the cans of juice will be bought = \(\frac { 9 }{ 378 }\)
12 cans of juice can be bought in ₹ 504.

Question 7.
A motorbike travels 425 km in 5 hours. How much distance will be covered by it in 3.2 hours?
Solution:
Distance covered by motorbike = 425 km
Time taken = 5 hours
Distance covered by motorbike in 1 hour = \(\frac { 425 }{ 5 }\) km/hr = 85 km/hr
Then, distance covered in 3.2 hours = 85 x 3.2 = 272 km/hr

Question 8.
If the cost of a dozen identical articles is ₹ 672, what will be the cost of 18 such articles?
Solution:
Cost of one dozen articles = ₹ 672
Cost of one article = ₹ \(\frac { 672 }{ 12 }\) = ₹ 56
Cost of 18 articles = ₹ 56 x 18 = ₹ 1008

Question 9.
A car covers a distance of 180 km in 5 hours.
(i) How much distance will the car cover in 3 hours with the same speed ?
(ii) How much time will the car take to cover 54 km with the same speed?
Solution:
Distance covered by car 180 km in 5 hours
(i) Distance covered in 1 hour = \(\frac { 180 }{ 5 }\) = 36 km
Distance covered in 3 hours = 3 x 36 = 108 km
(ii) To cover a distance of 180 km, time taken = 5 hours
To cover a distance of 1 km, time taken = \(\frac { 5 }{ 180 }\)
To cover a distance of 54 km, time taken = \(\frac { 5 }{ 180}\) x 54 = 1.5 hours

Question 10.
If it has rained 276 cm in the last 3 days, how many cm of rain will fall in one week (7 days) ?
Assume that the rain continues to fall at the same rate.
Solution:
Rate of rainfall in 3 days = 276 cm
Rainfall in one day = \(\frac { 276 }{ 3 }\) = 92 cm
Rainfall in one week = 92 x 7 = 644 cm

Question 11.
Cost of 10 kg of wheat is ₹ 180.
(i) What is the cost of 18 kg of wheat ?
(ii) What quantity of wheat can be purchased in ₹ 432 ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method image - 1

Question 12.
Rohit buys 10 pens for ₹ 150 and Manoj buys 14 pens for ₹ 168. Who got the pens cheaper?
Solution:
Rohit buys 10 pens = ₹ 150
Cost of one pen = \(\frac { 150 }{ 10 }\) = ₹ 15
Manoj buys 14 pens = ₹ 168
Cost of one pen = \(\frac { 168 }{ 14 }\) = ₹ 12
Manoj buys cheaper pen.

Question 13.
A tree 24 m high casts a shadow of 15 m. At the same time, the length of the shadow casted by some other tree is 6 m. Find the height of the tree.
Solution:
Height of a tree which casts a shadow of 15 m = 24 m
Height of a tree which casts a shadow 24 of 1 m = \(\frac { 24 }{ 15 }\) m
Height of a tree which casts a shadow of 6 m = \(\frac { 24 }{ 15 }\) x 6 = 9.6 m

Question 14.
A loaded truck travels 18 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
A loaded truck travels in 25 minutes a distance of = 18 km
A loaded truck travels in 1 min a distance of = \(\frac { 18 }{ 25 }\) km
A loaded truck travels in shows or 300 minutes, a distance of = \(\frac { 18 }{ 25 }\) x 300 = 216 km

Unitary Method Exercise 13B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Weight of 15 books is 6 kg. What is the weight of 45 such books?
Solution:
Weight of 15 books = 6 kg
Weight of 1 book = \(\frac { 6 }{ 15 }\) kg
Weight of 45 such books = \(\frac { 15 }{ 6 }\) x 45 = 112.5 kg

Question 2.
A made 84 runs in 6 overs and B made 126 runs in 7 overs. Who made more runs per over?
Solution:
Runs scored by A in 6 overs = 84 runs
Runs scored by A in one over = \(\frac { 84 }{ 6 }\) = 14 runs
Runs scored by B in 7 overs = 126 runs
Runs scored by B in one over = \(\frac { 126 }{ 7 }\) = 18 runs
B score more runs per over than A.

Question 3.
Geeta types 108 words in 6 minutes. How many words would she type in half an hour?
Solution:
Words typed by Geeta in 6 minutes = 108
Words typed by Geeta in 1 minute = \(\frac { 108 }{ 6 }\)
Words typed by Geeta in half hour or 30 minutes = \(\frac { 108 }{ 6 }\) x 30 = 540 words

Question 4.
The temperature dropped 18 degree Celsius in the last 24 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next 18 days?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method image - 2

Question 5.
Mr. Chopra pays ₹ 12,000 as rent for 3 months. How much does he has to pay for a year if the rent per month remains same?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method image - 3

Question 6.
A truck requires 108 litres of diesel for covering a distance of 1188 km. How much diesel will be required by the truck to cover a distance of 3300 km?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method image - 4

Question 7.
If a deposit of ₹ 2,000 earns an interest of ₹ 500 in 3 years, how much interest would a deposit of ₹ 36,000 earn in 3 years with the same rate of simple interest?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method image - 5

Question 8.
If John walks 250 steps to cover a distance of 200 metres, find the distance covered by him in 350 steps.
Solution:
Distance covered with 250 steps = 200 m
Distance covered with 1 step = \(\frac { 200 }{ 250 }\) m
Distance covered with 350 steps = \(\frac { 200 }{ 250 }\) x 350 = 280 m

Question 9.
25 metres of cloth costs ₹ 1,012.50.
(i) What will be the cost of 20 metres of cloth of the same type?
(ii) How many metres of the same kind can be bought for ₹ 1,620?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method image - 6

Question 10.
In a particular week, a man works for 48 hours and earns ₹ 4,320. But in the next week he worked 6 hours less, how much has he earned in this week?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 13 Unitary Method image - 7

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion (Including Word Problems)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion (Including Word Problems)

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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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Proportion Exercise 12A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
In each of the following, check whether or not the given ratios form a proportion :
(i) 8 : 16 and 12 : 15
(ii) 16 : 28 and 24 : 42
(iii) 12 ÷ 3 and 8 ÷ 2
(iv) 25 : 40 and 20 : 32
(v) \(\frac { 15 }{ 18 }  and \frac { 10 }{ 12 }\)
(vi) \(\frac { 7 }{ 8 }\) and 14 : 16
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 1
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 2

Question 2.
Find the value of x in .each of the following proportions :
(i) x : 4 = 6 : 8
(ii) 14 : x = 7 : 9
(iii) 4 : 6 = x : 18
(iv) 8 : 10 = x : 25
(v) 5 : 15 = 4 : x
(vi) 16 : 24 = 6 : x
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 3
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 4

Question 3.
Find the value of x so that the given four numbers are in proportion :
(i) x, 6, 10 and 15
(ii) x, 4, 15 and 30
(iii) 2, x, 10 and 25
(iv) 4, x, 6 and 18
(v) 9, 12, x and 8
(vi) 4, 10, 36 and x
(vii) 7, 21, x and 45
(viii) 6, 8, 12 and x.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 5
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 6

Question 4.
The first, second and the fourth terms of a proportion are 6, 18 and 75, respectively. Find its third term.
Solution:
Let the third term = x
6 : 18 : : x : 75
= 18 x x = 6 x 75
x = \(\frac { 6\times 75 }{ 18 } =\frac { 75 }{ 3 }\) = 25
The third term of proportion is 25

Question 5.
Find the second term of the proportion whose first, third and fourth terms are 9, 8 and 24 respectively.
Solution:
Let the second term = x
9 : x : : 8 : 24
=> x x 8 = 24 x 9
x = \(\frac { 24\times 9 }{ 8 }\) = 3 x 9 = 27
The second term of proportion = 27

Question 6.
Find the fourth term of the proportion whose first, second and third terms are 18, 27, and 32 respectively.
Solution:
Let the fourth term = x
18 : 27 : : 32 : x
=> 18 x x = 27 x 32
=> x = \(\frac { 27\times 32 }{ 18 }\) = 3 x 16 = 48
Fourth term = 48

Question 7.
The ratio of the length and the width of a school ground is 5 : 2. Find the length, if the width is 40 metres.
Solution:
Let the length = x m,
width = 40 m
The ratio of length to width = x : 40
as per given statement 5 : 2 = x : 40
=> 2 x x = 40 x 5
x = \(\frac { 40\times 5 }{ 2 }\) = 20 x 5 = 100 m

Question 8.
The ratio of the sale of eggs on a Sunday and that of the whole week at a grocery shop was 2 : 9. If the total value of the sale of eggs in the same week was Rs 360, find the value of the sale of eggs that Sunday.
Solution:
Let, the sale of eggs on Sunday = x
Sale in week = Rs 360
According to question, 2 : 9 = x : 360
=> 9 x x = 360 x 2
x = \(\frac { 360\times 2 }{ 9 }\) = Rs 80
Sale on Sunday = Rs 80

Question 9.
The ratio of copper and zinc in an alloy is 9 : 8. If the weight of zinc, in the alloy, is 9.6 kg ; find the weight of copper in the alloy.
Solution:
Let the weight of copper = x kg
Weight of zinc = 9.6 kg.
According to question,
9 : 8 = x : 9.6
=> 8 x x = 9 x 9.6
=> x = \(\frac { 9\times 9.6 }{ 8 }\) = 9 x 1.2 = 10.8 kg.
Weight of cooper in alloy = 10.8

Question 10.
The ratio of the number of girls to the number of boys in a school is 2 : 5. If the number of boys is 225 ; find:
(i) the number of girls in the school.
(ii) the number of students in the school.
Solution:
Let, the number of girls in school = x
Number of boys in school = 225
According to question 2 : 5 = x : 225
=> 5 x x = 2 x 225
x = \(\frac { 2\times 225 }{ 5 }\) = 2 x 45 = 90
Number of girls in school = 90
Total number of student in the school = (number of boys + number of girls) = (225 + 90) = 315

Question 11.
In a class, one out of every 5 students pass. If there are 225 students in all the sections of a class, find how many pass ?
Solution:
Total number of students in all sections = 225
Given, One of every five students pass
Total students pass = 225 x \(\frac { 1 }{ 5 }\) = 45 studetns

Question 12.
Make set of all possible proportions from the numbers 15, 18, 35 and 42.
Solution:
The possible proportions that can be made from the numbers 15, 18, 35 and 42 are
(i) 15 : 35 :: 18 : 42
(ii) 42 : 18 :: 35 : 15
(iii) 42 : 35 :: 18 : 15
(iv) 15 : 18 :: 35 : 42

Proportion Exercise 12B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
If x, y and z are in continued proportion, then which of the following is true :
(i) x : y = x : z
(ii) x : x = z : y
(iii) x : y = y : z
(iv) y : x = y : z
Solution:
(iii) x : y = y : z

Question 2.
Which of the following numbers are in continued proportion :
(i) 3, 6 and 15
(ii) 15, 45 and 48
(iii) 6, 12 and 24
(iv) 12, 18 and 27
Solution:
(iii) and (iv)

Question 3.
Find the mean proportion between
(i) 3 and 27
(ii) 0.06 and 0.96
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 7

Question 4.
Find the third proportional to :
(i) 36, 18
(ii) 5.25, 7
(iii) ₹ 1.60, ₹ 0.40
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 8
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 9
=> x = 0.1

Question 5.
The ratio between 7 and 5 is same as the ratio between ₹ x and ₹ 20.50 ; find the value of x.
Solution:
Since, It is given that the ratio between 7 and 5 is same as the ratio between ₹ x and ₹ 20.50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 10

Question 6.
If (4x + 3y) : (3x + 5y) = 6 : 7, find :
(i) x : y
(ii) x, if y = 10
(iii) y, if x = 27
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 12

Question 7.
If \(\frac { 2y+5x }{ 3y-5x } =2\frac { 1 }{ 2 }\), find:
(i) x : y
(ii) x, if y = 70
(iii) y, if x = 33
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 13
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 14
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 15

Proportion Exercise 12C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Are the following numbers in proportion:
(i) 32, 40, 48 and 60 ?
(ii) 12,15,18 and 20 ?
Solution:
(i) 32, 40, 48 and 60 are in proportion
if 32 : 40 = 48 : 60
if 32 x 60 = 40 x 48
\(\left\{ \frac { a }{ b } =\frac { c }{ d } \Longrightarrow \quad ad=bc \right\}\)
if 1920 = 1920
Which is true.
32, 40, 48 and 60 are in proportion
(ii) 12, 15, 18 and 20 are in proportion
if 12 : 15 = 18 : 20
if 12 x 20 = 15 x 18 {ad = bc}
if 240 = 270
which is not true.
12, 15, 18 and 20 are not in proportion.

Question 2.
Find the value of x in each of the following such that the given numbers are in proportion.
(i) 14, 42, x and 75
(ii) 45, 135, 90 and x
Solution:
14, 42, x and 75 are in proportion
\(\frac { 14 }{ 42 } =\frac { x }{ 75 }\)
=> 14 x 75 =x x 42
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 16

Question 3.
The costs of two articles are in the ratio 7 : 4. If the cost of the first article is Rs. 2,800 ; find the cost of the second article.
Solution:
Ratio in the cost of two articles = 7 : 4
Cost of first article = Rs. 2800
Let cost of the second article = x
7 : 4 = 2800 : x
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 17

Question 4.
The ratio of the length and the width of a rectangular sheet of paper is 8 : 5. If the width of the sheet is 17.5 cm; find the length.
Solution:
Let length of sheet = x cm
Ratio in length and breadth = 8 : 5
and width = 17.5 cm
8 : 5 = x : 17.5
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 18
Length of sheet = 28 cm

Question 5.
The ages of A and B are in the ratio 6 : 5. If A’s age is 18 years, find the age of B.
Solution:
Ratio in the ages of A and B = 6 : 5
A’s age = 18 years
Let B’s age = x years
6 : 5 = 18 : x
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 19

Question 6.
A sum of Rs. 10, 500 is divided among A, B and C in the ratio 5 : 6 : 4. Find the share of each.
Solution:
Total amount = Rs. 10, 500
Ratio in A, B, and C = 5 : 6 : 4
Sum of ratio = 5 + 6 + 4 = 15
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 20

Question 7.
Do the ratios 15 cm to 2 m and 10 sec to 3 minutes form a proportion ?
Solution:
15 cm : 2 m : : 10 sec : 3 min
15 cm : 2 x 100 cm :: 10 sec : 30 x 60 sec
15 : 200 :: 10 : 1800
3 : 40 :: 1 : 180
No, they donot form a proportion

Question 8.
Do the ratios 2 kg : 80 kg and 25 g : 625 g form a proportion ?
Solution:
2 kg : 80 kg : : 25 g : 625 g
2 : 80 :: 25 : 625
1 : 40 :: 1 : 25
No, they donot form a proportion.

Question 9.
10 kg sugar cost ₹ 350. If x kg sugar of the same kind costs ₹ 175, find the value of x
Solution:
10 kg of sugar costs = ₹ 350
and x kg of sugar cost = ₹ 175
A.T.Q.
10 kg : x kg :: 350 : 175
=> 10 x 175 = 350 x x
=> 350x= 1750
=> x = \(\frac { 1750 }{ 350 }\) = 5
Hence, 5 kg of sugar costs ₹ 175

Question 10.
The length of two ropes are in the ratio 7 : 5. Find the length of:
(i) shorter rope, if the longer one is 22.5 ni
(ii) longer rope, if the shorter is 9.8 m.
Solution:
Length of the ropes are in the ratio = 7 : 5
(i) Let the length of shorter rope = x
Length of longer rope = 22.5 m
A.T.Q.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 21
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion image - 22

Question 11.
If 4, x and 9 are in continued proportion, find the value of x.
Solution:
4, x and 9 are in continued proportion
=> 4 : x = x : 9
=> x2 = 9 x 4
=> x = √36
x = 6

Question 12.
If 25, 35 and x are in continued proportion, find the value of x.
Solution:
25, 35 and x are in continued proportion
=> 25 : 35 = 35 : x
=> 25 x x = 35 x 35
=> x = \(\frac { 35\times 35 }{ 25 }\)
=> x = 49

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio

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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

1. Ratio : The relation between two quantities (both of the same kind and in the same unit) obtained on dividing one quantity by the other, is called the ratio.
Keep in Mind :
(i) The ratio between two numbers or quantities is denoted by the colon “ : ”.
Thus, the ratio between two quantities p and q = p : q
(ii) The ratio between two quantities of same kind and in the same units is obtained on dividing one quantity by the other. Thus, the ratio between 20 kg and 80 kg = \(\frac { 20 }{ 80 } =\frac { 1 }{ 4 }\) = 1 : 4
(iii) The first term of a ratio is called antecedent and its second term is called consequent. In the ratio 1:4; antecendent = 1 and consequent = 4.
(iv) A ratio must always by expressed in its lowest terms.
2. Proportion : A proportion is an expression which states that the two ratios are equal.
Keep in Mind :
In a proportion, its first and the fourth terms are called extremes whereas its second and the third terms are called means.
Thus, in 8 : 12 = 18 : 27; the terms 12 and 18 are means whereas 8 and 27 are extremes.
Also, Product of extremes = Product of Means

Ratio Exercise 11A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express each of the following ratios in its simplest form :
(a) (i) 4 : 6
(ii) 48 : 54
(iii) 200 : 250
(b) (i) 5 kg : 800 gm
(ii) 30 cm : 2 m
(iii) 3 m : 90 cm
(iv) 2 years : 9 months
(v) 1 hour : 45 minutes
(vi) 4 min : 45 sec
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 1
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 3
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 4

Question 2.
A field is 80 m long and 60 m wide. Find the ratio of its width to its length.
Solution:
Width of field = 60 m
Length of field = 80 m
Ratio between width and length = \(\frac { 60 }{ 80 } =\frac { 3 }{ 4 }\) = 3 : 4

Question 3.
State, true or false :
(i) A ratio equivalent to 7 : 9 is 27 : 21.
(ii) A ratio equivalent to 5 : 4 is 240 : 192.
(iii) A ratio of 250 gm and 3 kg is 1 : 12.
Solution:
(i) False.
Correct: A ratio equivalent to 7 : 9 is 9 : 7
(ii) True
(iii) True

Question 4.
Is the ratio of 15 kg and 35 kg same as the ratio of 6 years and 14 years ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 5

Question 5.
Is the ratio of 6 g and 15 g same as the ratio of 36 cm and 90 cm ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 6

Question 6.
Find the ratio between 3.5 m, 475 cm and 2.8 m.
Solution:
The given values = 3.5 m, 475 cm and 2.8 m
= 3.5 x 100 cm : 475 cm : 2.8 x 100 cm
= 350 cm : 475 cm : 280 cm
= 70 cm : 95 cm : 56 cm
The ratio is 70 : 95 : 56

Question 7.
Find the ratio between 5 dozen and 2 scores. [1 score = 20]
Solution:
Ratio between 5 dozen and 2 scores
Given = 1 score = 20
then, 2 scores = 2 x 20 = 40
and 1 dozen = 12,
5 dozen = 12 x 5 = 60
Then, ratio = 60 : 40 = 3 : 2

Ratio Exercise 11B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The monthly salary of a person is Rs. 12,000 and his monthly expenditure is Rs 8,500. Find the ratio of his:
(i) salary to expenditure
(ii) expenditure to savings
(ii) savings to salary
Solution:
Monthly salary of a person = Rs 12,000
Monthly expenditure = 8,500
Saving of the person = (12,000 – 8500) = Rs 3,500
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 7

Question 2.
The strength of a class is 65, including 30 girls. Find the ratio of the number of:
(i) girls to boys
(ii) boys to the whole class
(iii) the whole class to girls.
Solution:
Total strength of class (including boys and girls) = 65
Number of girls = 30
Number of boys = (65 – 30) = 35
Ratio between girls and boys = 30 : 35 = \(\frac { 30 }{ 35 } =\frac { 6 }{ 7 }\) = 6 : 7
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 8

Question 3.
The weekly expenses of a boy have increased from ₹ 1,500 to ₹ 2,250. Find the ratio of:
(i) increase in expenses to original expenses.
(ii) original expenses to increased expenses.
(iii) increased expenses to increase in expenses.
Solution:
Original expenses = ₹ 1500
Increased expenses = ₹ 2250
Increase in expenses = ₹ 2250 – ₹ 1500 = ₹ 750
Now,
(i) Ratio in increase in expenses to original expenses = ₹ 750 : ₹ 1500 = 1 : 2
(ii) Original expenses to increased expenses = ₹ 1500 : ₹ 2250
\(\frac { 1500 }{ 750 } =\frac { 2250 }{ 750 }\) = 2 : 3
(iii) Increased expenses to increased in expenses = ₹ 2250 : ₹ 750 = 3 : 1 (Dividing by 750)

Question 4.
Reduce each of the following ratios to their lowest terms :
(i) 1 hour 20 min : 2 hours
(ii) 4 weeks : 49 days
(iii) 3 years 4 months : 5 years 5 months.
(iv) 2 m 40 cm : 1 m 44 cm
(v) 5 kg 500 gm : 2 kg 750 gm
Solution:
(i) 1 hour 20 min : 2 hour
= (1 x 60 + 20) minutes : 2 x 60 minutes
= 80 minutes : 120 minutes
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 9
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 10

Question 5.
Two numbers are in the ratio 9 : 2. If the smaller number is 320, find the larger number.
Solution:
Let the larger number = 9x
and smaller number = 2x
If smaller number is 320,
then, larger number will be = \(\frac { 9x\times 320 }{ 2x }\) = 1440

Question 6.
A bus travels 180 km in 3 hours and a train travels 450 km in 5 hours. Find the ratio of speed of train to speed of bus.
Solution:
Distance travelled by bus = 180 km
Time taken = 3 hours
Speed = \frac { Distance }{ Time } =\frac { 180 }{ 3 } = 60 km/hr
Distance travelled by train = 450 km
Time taken = 5 hours
Speed = \frac { Distance }{ Time } =\frac { 450 }{ 5 } = 90 km/hr
Ratio of speed of train to speed of bus = 90 : 60 = 3 : 2

Question 7.
In winters, a school opens at 10 a.m. and closes at 3.30 p.m. If the lunch interval is of 30 minutes, find the ratio of lunch interval to total lime of the class periods.
Solution:
Timing of a school (10 a.m. to 3.30 p.m) = 5 hours 30 minutes
Timing for lunch interval = 30 minutes
Total time of the class periods = 5 hours 30 minutes – 30 minutes
= 5 hours = 60 x 5 = 300 minutes
Ratio of lunch interval to total time of the class period = 30 minutes : 300 minutes = 1 : 10

Question 8.
Rohit goes to school by car at 60 km per hour and Manoj goes to school by scooty at 40 km per hour. If they both live in the same locality, find the ratio between the time taken by Rohit and Manoj to reach school.
Solution:
Rohit travel by car, speed of the car = 60 km/hr
Manoj travel by scooty, speed of the scooty = 40 km/hr
Since, It is given that, they live in the same locality
Hence, let the distance be k Time taken by Rohit to reach school =
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 11

Question 9.
In a club having 360 members, 40 play carrom, 96 play table tennis, 144 play badminton and remaining members play volley-ball. If no member plays two or more games, find the ratio of members who play :
(i) carrom to the number of those who play badminton.
(ii) badminton to the number of those who play table-tennis.
(iii) table-tennis to the number of those who play volley-ball.
(iv) volleyball to the n umber of those who play other games.
Solution:
Total members in a club = 360 members
Members who play carrom = 40
Members who play table tennis = 96
Members who play badminton = 144
Members who play volleyball = 360 – (40 + 96 + 144) = 360 – 280 = 80
(i) Ratio between the members who play carrom to the number of those who play badminton = 40 : 144 => 5 : 18
(ii) Ratio of members who play badminton to the number of those who play table- tennis = 144 : 96 => 6 : 4 = 3 : 2
(iii) Ratio of members who play table tennis to the number of those who play volley-ball = 96 : 80 = 6 : 5
(iv) Ratio of members who play volley-ball to the number of those who play other games = 80 : 280 =>4 : 14 = 2 : 7

Question 10.
The length of a pencil is 18 cm and its radius is 4 cm. Find the ratio of its length to its diameter.
Solution:
Length of a pencil = 18 cm
Radius of a pencil = 4 cm
Diameter of a pencil = 2r = 2 x 4 = 8 cm
Ratio of length of a pencil to its diameter = 18 : 8 = 19 : 2

Question 11.
Ratio of distance of the school from A’s home to the distance of the school from B’s home is 2 : 1.
(i) Who lives nearer to the school ?
(ii) Complete the following table :

Solution:
Ratio of distance of the school from A’s home to the distance of the school from B’s hence = 2 : 1
(i) B lives nearest to the school
(ii) Let A’s home is 2x km from school and B’s home is x km
Distance of school from A’s home = 2 x Distance of school from B’s home
=> If A lives at a distance of 4 km, then B lives at a distance of = \(\frac { 1 }{ 2 }\) x 4 = 2 km
=> If B lives at a distance of 9 km then A lives at a distance of = 2 x 9 = 18 km
=> If A lives at a distance of 8 km then B lives at a distance = \(\frac { 1 }{ 2 }\) x 8 = 4 km
=> If B lives at a distance of 8 km, then A lives at a distance = 2 x 8 = 16 km
=> If A lives at a distance of 6 km, then B lives at a distance = \(\frac { 1 }{ 2 }\) x 6 = 3 km

Question 12.
The student-teacher ratio in a school is 45 : 2. If there are 4050 students in the school, how many teachers must be there?
Solution:
Total number of students = 4050
Let total number of teachers = x
The student-teacher ratio in a school = 45 : 2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 12

Ratio Exercise 11C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
₹ 120 is to be divided between Hari and Gopi in the ratio 5 : 3. How much does each get ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 13

Question 2.
Divide 72 in the ratio \(2\frac { 1 }{ 2 } :1\frac { 1 }{ 2 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 14
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 15

Question 3.
Divide 81 into three parts in the ratio 2 : 3 : 4.
Solution:
Given ratio = 2 : 3 : 4
Sum of ratio = 2 + 3 + 4 = 9
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 16

Question 4.
Divide Rs 10,400 among A, B and C in the ratio \(\frac { 1 }{ 2 } :\frac { 1 }{ 3 } :\frac { 1 }{ 4 } \)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 17

Question 5.
A profit of Rs 2,500 is to be shared among three persons in the ratio 6 : 9 : 10. How much does each person get?
Solution:
Total profit = Rs 2,500
Given ratio = 6 : 9 : 10
Sum of ratio = 6 + 9 + 10 = 25
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 18

Question 6.
The angles of a triangle are in the ratio 3 : 7 : 8. Find the greatest and the smallest angles.
Solution:
Sum of angles of triangle = 180°
Given ratio 3 : 7 : 8
Sum of ratio = 3 + 7 + 8= 18
Smallest angle = \(\frac { 3 }{ 18 }\) x 180° = 30°
Greatest angle = \(\frac { 8 }{ 18 }\) x 180° = 80°

Question 7.
The sides of a triangle are in the ratio 3 : 2 : 4. If the perimeter of the triangle is 27 cm, find the length of each side.
Solution:
Ratio in the sides of a triangle = 3 : 2 : 4
Sum of ratios = 3 + 2 + 4 = 9
Perimeter of triangle = 27 cm
Length of first side = \(\frac { 27\times 3 }{ 9 }\) = 9 cm
Length of second side = \(\frac { 27\times 2 }{ 9 }\) = 6 cm
Length of third side = \(\frac { 27\times 4 }{ 9 }\) = 12 cm

Question 8.
An alloy of zinc and copper weighs 12\(\frac { 1 }{ 2 }\) kg. if in the alloy, the ratio of zinc and copper is 1 : 4, find the weight of copper in it.
Solution:
Weight of alloy = 12\(\frac { 1 }{ 2 }\) kg = \(\frac { 25 }{ 2 }\) kg.
Given ratio =1 : 4
Sum of ratio =1 + 4 = 5
Weight of copper = \(\frac { 4 }{ 5 } \times \frac { 25 }{ 2 }\) kg = 2 x 5 = 10 kg

Question 9.
How will Rs 31500 be shared between A, B and C ; if A gets the double of what B gets, and B gets the double of what C gets ?
Solution:
Let the share of C = 1
Share of B = double of C = 2 x 1 = 2
Share of A = double of B = 2 x 2 = 4
Given ratio (A : B : C) = 4 : 2 : 1
Sum of ratio = 4 + 2 + 1 = 7
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 19

Question 10.
Mr. Gupta divides Rs 81000 among his three children Ashok, Mohit and Geeta in such a way that Ashok gets four times what Mohit gets and Mohit gets 2.5 times what Geeta gets. Find the share of each of them.
Solution:
Let the share of Geeta = 1
Share of Mohit is (2-5 times of Geeta) = 2.5
Share of Ashok is (4 times of Mohit) = 4 x 2.5 = 10
Ratio = 1 : 2.5 : 10 = 1 x 2 :2.5 x 2 : 10 x 2 = 2 : 5 : 20
Sum of ratio = 2 + 5 + 20 = 27
Share of Geeta = \(\frac { 2 }{ 27 }\) x Rs 81000 = 2 x Rs 3000 = Rs 6000
Share of Mohit = \(\frac { 5 }{ 27 }\) x Rs 81000 = 5 x Rs 3000 = Rs 15000
Share of Ashok = \(\frac { 20 }{ 27 }\) x Rs 81000 = 20 x Rs 3000 = Rs 60000

Ratio Exercise 11D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Which ratio is greater:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 20
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 21

Question 2.
Which ratio is smaller :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 22
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 11 Ratio image - 23

Question 3.
Increase 95 in the ratio 5 : 8.
Solution:
Given ratio = 5 : 8
The increased quantity = \(\frac { 8 }{ 5 }\) x given quantity
= \(\frac { 8 }{ 5 }\) x 95 = 152

Question 4.
Decrease 275 in the ratio 11 : 7.
Solution:
Given ratio = 11 : 7
The decrease quantity= \(\frac { 11 }{ 7 }\) x given quantity.
= \(\frac { 11 }{ 7 }\) x 275 = 175

Question 5.
Decrease 850 in the ratio 17 : 6 and then increase the result in the ratio 5 : 9.
Solution:
The given quantity = 850 and decrease in the ratio =17 : 6
The decreased quantity = \(\frac { 6 }{ 17 }\) x 850 = 300
Now, the quantity is increased in the ratio 5 : 9
The resulting quantity = \(\frac { 9 }{ 5 }\) x 300 = 540

Question 6.
Decrease 850 in the ratio 17 : 6 and then decrease the resulting number again in 4 : 3.
Solution:
The given quantity = 850 and decrease in the ratio = 17:6
The decreased quantity = \(\frac { 6 }{ 17 }\) x 850 = 300
Now, the quantity is decreased again in the ratio = 4 : 3
The resulting quantity = \(\frac { 3 }{ 4 }\) x 300 = 225

Question 7.
Increase 1200 in the ratio 2 : 3 and then decrease the resulting number in the ratio 10 : 3.
Solution:
The given quantity = 1200 and increase in the ratio is 2 : 3
The increased quantity = \(\frac { 3 }{ 2 }\) x 1200 = 1800
Now, the quantity is decrease in the ratio 10 : 3
= \(\frac { 3 }{ 10 }\) x 1800 = 540

Question 8.
Increase 1200 in the ratio 3 : 7 and then increase the resulting number again in the ratio 4 : 7.
Solution:
The given quantity = 1200 increase in the ratio is 3 : 7
The increased quantity = \(\frac { 7 }{ 3 }\) x 1200 = 2800
Now, the quantity is increased again in the ratio 4 : 7
The resulting quantity = \(\frac { 7 }{ 4 }\) x 2800 = 4900

Question 9.
The number 650 is decreased to 500 in the ratio a : b, find the ratio a : b.
Solution:
The given quantity = 650
and the decreased quantity = 500
The ratio (a : b) by which 650 is decreased to 500 = \(\frac { 650 }{ 500 } =\frac { 13 }{ 10 }\)
The resulting ratio =13 : 10

Question 10.
The number 800 is increased to 960 in the ratio a : b, find the ratio a : b.
Solution:
The given quantity = 800
and the increased quantity = 960
The ratio (a : b) by which 800 is increased to 960 = \(\frac { 800 }{ 960 } =\frac { 5 }{ 6 }\)
The resulting ratio = 5 : 6

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 10 Sets

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 10 Sets

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 10 Sets

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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

In our day-to-day life we often speak or hear about different types of collections.
Such as :
(i) A collection of stamps.
(ii) A collection of toys :
(iii) A collection of books, etc.
In the same way, we have different types of groups made for different activities.
Such as :
(i) A group of boys playing hockey.
(ii) A group of girls playing badminton.
(iii) A group of students going for picnic, etc.
In mathematics, a collection of particular things or a group of particular objects is called a set.

  1. Definition of a Set : A set is a collection of well-defined objects.
  2. Meaning of “Well-Defined” : Well- defined means, it must be absolutely clear that which object belongs to the set and which does not.
  3. Elements (or, members) of a set : The objects used to form a set are called its elements or its members.
    Keep in Mind :

    • The pair of curly braces { } denotes a set.
    • The Greek letter Epsilon ‘ ε ’ is used for the words “belongs to”, “is an element of, etc.
      p ε A will be be read as “p belongs to set A” or “p is an element of set A”.
      In the same way ; q ε A, r ε A, s ε a. and t ε A.
      The symbol ‘ε not’ stands for “does not belong to” also for “is not an element of.
      a ε not A will be read as “a does not belong to set A” or “a is not an element of set A”.
  4. Properties of a set :
    • The change in order of writing the elements does not make any change in the set.
    • If one or many elements of a set are repeated, the set remains the same.
  5. Notation (Representation) of a set :
    • Description method : In this method, a well-defined description about the elements of the set is made.
    • Roster or Tabular Method : In this method, the elements (members) of a set are written inside a pair of curly braces and are separated by commas.
    • Rule or Set Builder Method : In this method, actual elements of the set are not listed but a rule or a statement or a formula, in the briefest possible way, is written inside a pair of curly braces.

Sets Exercise 10A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State whether or not the following elements form a set; if not, give reason:
(i) All easy problems in your text book.
(ii) All three sided figures.
(iii) The first five counting numbers.
(iv) All the tall boys of your class.
(v) The last three days of the week.
(vi) All triangles that are difficult to draw.
(vii) The first three letters of the English alphabet.
(viii) All tasty fruits.
(ix) All clever boys of class 6.
(x) All good schools in Delhi.
(xi) All the girls in your class, whose heights are less than your height.
(xii) All the boys in your class, whose heights are more than your height.
(xiii) All the problems in your Mathematics book, which are difficult for Amit.
Solution:
(i) No; some problems may be easy for one person but may be difficult to some other person. Objects are not well- defined.
(ii) Yes.
(iii) Yes.
(iv) No; it is not mentioned that the boys must be taller than which boy. If we consider three boys A, B and C; boy B can be taller than A but not necessarily taller than C.
(v) Yes
(vi) No; it may be difficult for one boy to draw a given triangle but to some other boy it may be easy to draw the same triangle.
(vii) Yes
(viii) No; a fruit may be tasty for one person and may not be tasty to other person / persons.
(ix) No; clever in what respect and from whom out of six ?
(x) No; all the people can not find the same schools as good as others said. So, the objects are not well-defined.
(xi) Yes
(xii) Yes
(xiii) Yes.

Sets Exercise 10B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
If set A = {2, 3, 4, 5, 6}, state which of the following statements arc true and which are false :
(i) 2 ∈ A
(ii) 5, 6 ∈ A
(iii) 3, 4, 7 ∈ A
(iv) 2, 8 ∈ A
Solution:
(i) True
(ii) True
(iii) False
(iv) False

Question 2.
If set B = {4, 6, 8, 10, 12, 14}. State, which of the following statements is correct and which is wrong :
(i) 5 ∈ B
(ii) 12 ∈ B
(iii) 14 ∈ B
(iv) 9 ∈ B
(v) B is a set of even numbers between 2 and 16.
(vi) 4,6 and 10 are die members of the set B. Also, write the wrong statements correctly.
Solution:
(i) Wrong ; 5 \(\notin \) B
(ii) Correct
(iii) Correct
(iv) Wrong ; 9 \(\notin \) B
(v) Correct
(vi) Correct.

Question 3.
State, whether true or false :
(i) Sets {4, 9, 6,2} and {6, 2, 4, 9} are not the same.
(ii) Sets {0, 1, 3, 9, 4} and {4, 0, 1, 3, 9} are the same.
(iii) Sets {5, 4} and {5, 4, 4, 5} are not the same.
(iv) Sets {8, 3} and {3, 3, 8} are the same.
(v) Collection of vowels used in the word ‘ALLAHABAD’ forms a set.
(vi) If P is the set of letters in the word ‘ROOP’; then P = (p, o, r)
(vii) If M is the set of letters used in the word ‘MUMBAI’, then: M = {m, u, b, a, i}
Solution:
(i) False.
(ii) True.
(iii) False.
(iv) True.
(v) True.
(vi) True.
(vii) True.

Question 4.
Write the set containing :
(i) the first five counting numbers.
(ii) the three types of angles.
(iii) the three types of triangles.
(iv) the members of your family.
(v) the first six consonants of the English Alphabet.
(vi) the first four vOWels of the English Alphabet.
(vii) the names of any three Prime-Ministers of India.
Solution:
(i) {1, 2, 3, 4, 5}
(ii) {acute-angle, obtuse-angle, right-angle}.
(iii) {scalene triangle, isosceles triangles, equilateral triangle}.
(iv) { Write the name of family member}.
(v) {b, c, d, f, g, h}
(vi) {a, e, i, o}
(vii) {J.L. Nehru, A.B. Vajpayee, Dr. Manmohan Singh}.

Question 5.
(a) Write the members (elements) of each set given below :
(i) {3, 8, 5, 15, 12, 7}
(ii) {c, m, n, o, s}
(b) Write the sets whose elements are :
(i) 2, 4, 8, 16, 64 and 128
(ii) 3, 5, 15, 45, 75 and 90
Solution:
(a) (i) 3, 8, 5, 15, 12 and 7
(ii) c, m, n, o and s
(b) (i) {2, 4, 8, 16, 64, 128}
(ii) {3, 5, 15, 45, 75, 90}

Question 6.
(i) Write the set of letters used in the word ‘BHOPAL’.
(ii) Write the set of vowels used im the word ‘BENGAL’.
(iii) Write the set of consonants used in the word ‘HONG KONG’.
Solution:
(i) {b, h, o, p, a, 1}
(ii) {e, a}
(iii) {h, o, n, g, k}

Sets Exercise 10C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write each of the following sets in the Roster Form :
(i) The set of five numbers each of which is divisible by 3.
(ii) The set of integers between – 4 and 4.
(iii) {x: x is a letter in the word ‘ SCHOOL’}
(iv) {x: x is an odd natural number between 10 and 20}
(v) {Vowels used in the word ‘AMERICA’}
(vi) {Consonants used in the * word ‘MADRAS’}
Solution:
(i) {3, 6, 9, 12, 15}
(ii) {-3, -2, -1, 0, 1, 2, 3}
(iii) {s, c, h, o, 1}
(iv) {11, 13, 15, 17, 19}
(v) (a, e, i)
(vi) {m, d, r, s}

Question 2.
Write each given set in the Roster Form :
(i) All prime numbers between one and twenty.
(ii) The squares of first four natural numbers.
(iii) Even numbers between 1 and 9.
(iv) First eight letters of the English alphabet.
(v) The letters of the word ‘BASKET’.
(vi) Four cities of India whose names start with the letter J.
(vii) Any four closed geometrical figures.
(viii) Vowels used in the word ‘MONDAY’.
(ix) Single digit numbers that are perfect squares as well.
Solution:
(i) {2, 3, 5, 7, 11, 13, 17, 19}
(ii) {12, 22, 32, 42} = {1, 4, 9, 16}
(iii) {2, 4, 6, 8}
(iv) {a, b, c, d, e, f, g, h}
(v) {b, a, s, k, e, t}
(vi) {Jaipur, Jodhpur, Jalandhar, Jaunpur}
(vii) {∆, O, □, O}
(viii) {o, a}
(ix) {0, 1, 4, 9}

Question 3.
Write each given set in the Set- Builder Form :
(i) {2, 4, 6, 8, 10}
(ii) {2, 3, 5, 7, 11}
(iii) {January, June, July}
(iv) {a, e, i, o, u}
(v) {Tuesday, Thursday}
(vi) {1,4,9, 16, 25}
(vii) {5,10,15,20,25,30}
Solution:
(i) {x : x is an even natural number less than 12}
(ii) {x : x is a prime number less than 12}
(iii) {x : x is a months of the year whose name starts with letter J}
(iv) {x : x is a vowel in English alphabets}
(v) {x : x is a day of the week whose name starts with letter T}
(vi) {x: x is a perfect square natural number upto 25}
(vii) {x : x is a natural number upto 30 and divisible by 5}.

Question 4.
Write each of the following sets in Roster (tabular) Form and also in Set-Builder Form.
(i) Set of all natural numbers that can divide 24 completely.
(ii) Set of odd numbers between 20 and 35.
(iii) Set of letters used in the word ‘CALCUTTA’.
(iv) Set of names of the first five months of a year.
(v) Set of all two digit numbers that are perfect square as well.
Solution:
(i) {1,-2, 3, 4, 6, 8, 12, 24}; {x : x is a natural number which divides 24 completely}
(ii) {21, 23, 25, 27, 29, 31, 33}; {x:x is an odd number between 20 and 35}
(iii) {c, a, 1, u, t}; {x: x is a letter used in the word ‘CALCUTTA’}
(iv) {January, February, March,April, May}; {x : x is name of first five months of a year}
(v) {16, 25, 36, 49, 64, 81}; {x : x is a perfect square two digit number}.

Question 5.
Write, in Roster Form, the set of :
(i) the first four odd natural numbers each divisible by 5.
(ii) th counting numbers between 15 and 35; each of which is divisible by 6.
(iii) the names of the last three days of a week.
(iv) the names of the last four months of a year.
Solution:
(i) {5, 15, 25, 35}
(ii) {18, 24, 30}
(iii) {Friday, Saturday, Sunday}
(iv) {September, October, November, December}.

Sets Exercise 10D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State, whether the given set is infinite or finite :
(i) {3, 5, 7, ………}
(ii) {1, 2, 3, 4}
(iii) {………., -3, -2, -1, 0, 1, 2}
(iv) {20, 30, 40, 50, …, 200}
(v) {7, 14, 21, ……….., 2401}
Solution:
(i) Infinite
(ii) Finite
(iii) Infinite
(iv) Finite
(v) Finite

Question 2.
(i) Which of the following sets is empty ?
(ii) Set of counting numbers between 5 and 6.
(iii) Set of odd numbers between 7 and 19. Set of odd numbers between 7 and 9.
(iv) Set of even numbers which are not divisible by 2.
(v) {0}.
Solution:
(i), (iii) and (iv)

Question 3.
State, which pair of sets, given below, are equal sets or equivalent sets:
(i) {3, 5, 7} and {5, 3, 7}
(ii) {8, 6, 10, 12} and {3, 2, 4, 6}
(iii) {7, 7, 2, 1,2} and {1, 2, 7}
(iv) {2, 4, 6, 8, 10} and {a, b, d, e, m}
(v) {5, 5, 2, 4} and {5, 4, 2, 2}
Solution:
(i) Equal
(ii) Equivalent
(iii) Equal
(iv) Equivalent
(v) Equal

Question 4.
State, which of the following are finite or infinite sets :
(i) Set of integers
(ii) {Multiples of 5}
(iii) {Fractions between 1 and 2}
(iv) {Number of people in India}
(v) Set of trees in the world
(vi) Set of leaves on a tree
(vii) Set of children in all the schools of Delhi
(viii) { …….., -4, -2, 0, 2, 4, 6, 8}
(ix) {- 12, – 9, – 6, – 3, 0, 3, 6, …….}
(x) {Number of points in a line segment 4 cm long}.
Solution:
(i) Infinite
(ii) Infinite
(iii) Infinite
(iv) Finite
(v) Infinite
(vi) Finite
(vii) Finite
(viii) Infinite
(ix) Infinite
(x) Infinite

Question 5.
State, whether or not the following sets are empty :
(i) {Prime numbers divisible by 2}
(ii) {Negative natural numbers}
(iii) {Women with height 5 metre}
(iv) {Integers less than 5}
(v) {Prime numbers between 17 and 23}
(vi) Set of even numbers, not divisible by 2
(vii) Set of multiples of 3, which are more than 9 and less than 15.
Solution:
(i) Not empty
(ii) Empty
(iii) Empty
(iv) Not empty
(v) Not empty
(vi) Empty
(vii) Not empty

Question 6.
State, if the given pairs of sets are equal sets or equivalent sets :
(i) {Natural numbers less than five} and {Letters of the word ‘BOAT’}.
(ii) {2, 4, 6, 8, 10} and {even natural numbers less than 12}
(iii) {1, 3, 5, 7, ………..} and set of odd natural numbers.
(iv) {Letters of the word MEMBER} and {Letters of the word ‘REMEMBER’}.
(v) {Negative natural numbers} and {50th day of a month}
(vi) {Even natural numbers} and {odd natural numbers}.
Solution:
(i) Equivalent
(ii) Equal
(iii) Equal
(iv) Equal
(v) Equal
(vi) Equivalent

Question 7.
State, whether the following are finite or infinite sets :
(i) {2, 4, 6, 8, …… 800}
(ii) {……, -5, -4, -3, -2}
(iii) {x : x is an integer between – 60 and 60}
(iv) {No. of electrical appliances working in your house}
(v) {x : x is a whole number greater than 20}
(vi) {x : x is a whole number less than 20}
Solution:
(i) Finite
(ii) Infinite
(iii) Finite
(iv) Finite
(v) Infinite
(vi) Finite

Question 8.
For each statement, given below, write True or False :
(i) {…., -8, -4, 0, 4, 8} is a finite set.
(ii) { – 32, – 28, – 24, – 20, ……….. , 0, 4, 8, 16} is an infinite set.
(iii) {x : x is a natural number less than 1} is the empty set.
(iv) {Whole numbers between 15 and 16} = {Natural numbers between 5 and 6}
(v) {Odd numbers divisible by 2} is the empty set.
(vi) {Even natural numbers divisible by 3} is the empty set.
(vii) {x : x is positive and x < 0} is the empty set.
(viii) {.., -5, -3, -1, 1, 3, 5, ..} is a finite set.
Solution:
(i) False
(ii) False
(iii) True
(iv) True (each set is the empty set)
(v) True
(vi) False (6 is an even natural number which is divisible by 3)
(vii) True (no positive number can be less than 0)
(viii) False

Question 9.
State, giving reasons, which of the following pairs of sets are disjoint sets and which are or overlapping sets :
(i) A = {Girls with ages below 15 years} and B = {Girls with ages above 15 years}
(ii) C = {Boys with ages above 20 years} and D = {Boys with ages above 27 years}
(iii) A = {Natural numbers between 35 and 60} and B = {Natural numbers between 50 and 80}
(iv) P = {Students of class IX studying in I.C.S.E. Board} and Q = {Students of class IX}
(v) A = {Natural numbers multiples of 3 and less than 30} and B = {Natural numbers divisible by 4 and between 20 and 45}
(vi) P = {Letters in the word ‘ALLAHABAD’} and Q = {Letters in the word ‘MUSSOORIE’}
Solution:
(i) Disjoint sets; as no girl can be of age below 15 years and also above 15 years
(ii) Overlapping sets; as boys above 27 years are also above 20 years.
(iii) Overlapping sets; as natural numbers from 50 to 59 are common to both the sets.
(iv) Overlapping sets; as students of class IX studying in I,C.S.E. board are common.
(v) Overlapping sets; as natural number 24 is common to both the sets.
(vi) Disjoint sets; as no letter is common to both the sets.

Sets Exercise 10E – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write the cardinal number of each of the following sets :
(i) A = {0, 1, 2, 4}
(ii) B = {-3, -1, 1, 3, 5, 7}
(iii) C = { }
(iv) D= {3, 2, 2, 1, 3, 1, 2}
(v) E = {Natural numbers between 15 and 20}
(vi) F = {Whole numbers from 8 to 14}.
Solution:
(i) A = {0, 1, 2, 4} i.e. n (A) = 4
(ii) B = {-3, -1, 1, 3, 5, 7} i.e. n(B) = 6
(iii) C = { } i.e. n(C) = 0
(iv) D = {3, 2, 2, 1,3, 1, 2} => D = {3, 2, 1,} i.e. n(D) = 3
(v) E = {16, 17, 18 19} i.e. n(E) = 4
(vi) F={8, 9, 10, 11, 12, 13, 14} i.e. n(F) = 7

Question 2.
Given :
(i) A = {Natural numbers less than 10}
B = {Letters of the word ‘PUPPET’}
C = {Squares of first four whole numbers}
D = {Odd numbers divisible by 2}. Find :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 10 Sets image - 1
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 10 Sets image - 2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 10 Sets image - 3

Question 3.
State true or false for each of the following. Correct the wrong statement.
(i) If A = {0}, then n(A) = 0.
(ii) n(φ) = 1.
(iii) If T ={a, l, a, h, b, d, h), then n(T) = 5.
(iv) If B ={1, 5, 51, 15, 5, l},then n(B) = 6.
Solution:
(i) A = {0} then n(A) = 0 which is false.
True statement is n (A) = 1
(ii) n(φ) = 1, which is false.
i.e. n (φ) = 0
(iii) T = {a, l, a, h, b, d, h} i.e. T = {a, l, h, b, d)
i.e. n(T) = 5 which is true.
(iv) B = {1, 5, 51, 15, 5, 1} n(B) = 6 which is false.
i.e. B = {1, 5, 51, 15} => n(B) = 4

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