RS Aggarwal Solutions for Class 6 Maths Chapter 8–Algebraic Expressions

Class 6: Maths Chapter 8 solutions. Complete Class 6 Maths Chapter 8 Notes.

RS Aggarwal Solutions for Class 6 Maths Chapter 8–Algebraic Expressions

RS Aggarwal 6th Maths Chapter 8, Class 6 Maths Chapter 8 solutions

Ex 8A Solutions

Question 1.
Solution:
(i) x + 12
(ii) y – 7
(iii) a – b
(iv) (x + y) + xy
(v) 13x (a + b)
(vi) 7y + 5x
(vii) x+y5
(viii) 4 – x
(ix) xy−2
(x) x²
(xi) 2x + y
(xii) y² + 3x 
(xiii) x – 2y
(xiv) y³ – x³
(xv) x8×y

Question 2.
Solution:
Marks scored in English = 80
Marks scored in Hindi = x
∴ Total score in the two subjects = 80 + x

Question 3.
Solution:
We can write :
(i) b × b × b ×….15 times = 6¹⁵
(ii) y × y × y ×…..20 times = y²⁰
(iii) 14 × a × a × a × a × b × b × b= 14a⁴ b³
(iv) 6 × x × x × y × y = 6x²y²
(v) 3 × z × z × z × y × y × x= 3z³y²x

Question 4.
Solution:
We can write :
(i) x²y⁴ = x × x × y × y × y × y
(ii) 6y⁵ = 6 × y × y × y × y × y
(iii) 9xy²z = 9 × x × y × y × z
(iv) 10a³b³c³ = 10 × a × a × a × b × b × b × c × c × c

Ex 8B Solutions

Question 1.
Solution:
(i) Substituting a = 2 and b = 3 in the , given expression, we get :
a + b = 2 + 3 = 5
(ii) Substituting a = 2 and b = 3 in the given expression, we get :
a² + ab = (2)² + 2 x 3
= 4 + 6 = 10
(iii) Substituting a = 2 and b = 3 in the given expression, we get :
ab – a² = 2 x 3 – (2)²
= 6 – 4 = 2
(iv) Substituting a = 2 and b = 3 in the given expression, we get :
2a – 3b = 2 x 2 – 3 x 3
= 4 – 9 = – 5
(v) Substituting a = 2 and b = 3 in the given expression, we get :
5a² – 2ab = 5 x (2)² – 2 x 2 x 3
= 5 x 4 – 4 x 3
= 20 – 12 = 8
(vi) Substituting a = 2 and b = 3 in the given expression, we get :
a³ – b³ = (2)³ – (3)³ = 2 x 2 x 2 – 3 x 3 x 3
= 8 – 27 = – 19

Question 2.
Solution:
(i) Substituting x = 1, y = 2 and z = 5 in the given expression, we get :
3x – 2y + 4z = 3 x 1 – 2 x 2 + 4 x 5
= 3 – 4 + 20 = 23 – 4 = 19

Question 3.
Solution:
(i) Substituting p = – 2, q = – 1 and r = 3
in the given expression, we get :

Question 4.
Solution:
(i) The coefficient of x in 13x is 13
(ii) The coefficient of y in – 5y is – 5
(iii) The coefficient of a in 6ab is 6b
(iv) The coefficient of z in – 7xz is – 7x
(v) The coefficient of p in – 2pqr is – 2qr
(vi) The coefficient of y² in 8xy²z is 8xz
(vii) The coefficient of x³ in x³ is 1
(viii) The coefficient of x² in – x² is -1

Question 5.
Solution:
(i) The numerical coefficient of ab is 1
(ii) The numerical coefficient of – 6bc is – 6
(iii) The numerical coefficient of 7xyz is 7
(iv) The numerical coefficient of – 2x³y²z is – 2.

Question 6.
Solution:
(i) The constant term is 8
(ii) The constant term is – 9
(iii) The constant term is 35
(iv) The constant term is –83

Question 7.
Solution:
(i) The given expression contains only one term, so it is monimial.
(ii) The given expression contains only two terms, so it is binomial.
(iii) The given expression contains only one term, so it is monomial.
(iv) The given expression contains three terms, so it is trinomial.
(v) The given expression contains three terms, so it is trinomial.
(vi) The given expression contains only one term, so it is monomial.
(vii) The given expression contains four terms, so it is none of monomial, binomial and trinomial.
(viii) The given expression contains only one term so it is monomial.
(ix) The given expression contains two terms, so it is binomial.

Question 8.
Solution:
(i) The terms of the given expression 4x⁵ – 6y⁴ + 7x²y – 9 are :
4x⁵, – 6y⁴, 7x²y, – 9
(ii) The terms of the given expression 9x³ – 5z⁴ + 7x³y – xyz are :
9x³, – 5z⁴, 7x³y, – xyz.

Question 9.
Solution:
(i) We have : a², b², – 2a², c², 4a
Here like terms are a², – 2a²
(ii) We have : 3x, 4xy, – yz, 12 zy
Here like terms are – yz, 12 zy
(iii) We have : – 2xy², x²y, 5y²x, x²z
Here like terms are – 2xy², 5y²x
(iv) We have :
abc, ab²c, acb², c²ab, b²ac, a²bc, cab²
Here like terms are ab²c, acb², b²ac, cab².

Ex 8C Solutions

Question 1.
Solution:
(i) The required sum
= 3x + 7x
= (3 + 7) x
= 10x

Question 2.
Solution:
(i) Adding columnwise,
we get

Question 3.
Solution:
(i) Arranging the like terms column wise and adding, we get :

Question 4.
Solution:
(i) We have :
2x – 5x = (2 – 5)x = – 3x
(ii) We have :
6x – y – (- xy) = 6xy + xy = 7xy
(iii) We have : 5b – 3a
(iv) We have : 9y – ( – 7x) = 9y + 7x
(v) We have : – 7x² – 10x² = ( – 7 – 10)x²
= – 17x²
(vi) We have : b² – a² – (a² – b²)
= b² – a² – a² + b²
= b² + b² – a² – a²
= (1 + 1) b² + ( – 1 – 1) a²
= 2b² – 2a²

Question 5.
Solution:
(i) Arranging the like terms column wise, we get :

Question 6.
Solution:
(i) Rearranging and collecting the like terms, we get :

Question 7.
Solution:
We have:

Question 8.
Solution:
We have :
A = 7x² + 5xy – 9y²
B = – 4x² + xy + 5y²
C = 4y² – 3x² – 6xy

= 0+0+0 = 0
Hence the result

Question 9.
Solution:
Required expression

Question 10.
Solution:
Substituting the values of P, Q, R and S, we have :
P + Q + R + S = (a² – b² + 2ab)

Question 11.
Solution:
Required expression

Question 12.
Solution:
Required expression

Question 13.
Solution:
Required expression

Question 14.
Solution:
Required expression

Question 15.
Solution:
Sum of 5x – 4y + 6z and – 8x + y – 2z
= 5x – 4y + 6z – 8x + y – 2z
= 5x – 8x – 4y + y + 6z – 2z
= – 3x – 3y + 4z
Sum of 12x – y + 3z and – 3x + 5y – 8z
= 12x – y + 3z – 3x + 5y – 8z
= 12x – 3x – y + 5y + 3z – 8z
= 9x + 4y – 5z

Question 16.
Solution:
Required expression

Question 17.
Solution:
Required expression

Ex 8D Solutions

Simplify :

Question 1.
Solution:
We have : a – (b – 2a)
= a – b + 2a
= a + 2a – b
= (1 + 2) a – b
= 3a – b.

Question 2.
Solution:
We have : 4x – (3y – x + 2z)
= 4x – 3y + x – 2z
= 4x + x – 2y – 2z
= 5x – 3y – 2z

Question 3.
Solution:
We have :
(a² + b² + 2ab) – (a² + b² – 2ab)
= a² + b² + 2ab – a² – b² + 2ab
= a² – a² + b² – b² + 2ab + 2ab
= 0 + 0 + (2 + 2) ab
= 4 ab

Question 4.
Solution:
We have :
– 3 (a + b) + 4 (2a – 3b) – (2a – b)
= – 3a – 3b + 8a – 12b – 2a + b
= – 3a + 8a – 2a – 3b – 12b + b
= ( – 3 + 8 – 2) a + ( – 3 – 12 + 1) b
= 3a – 14 b.

Question 5.
Solution:
We have :
– 4x² + {(2x² – 3) – (4 – 3x²)}
= – 4x² + {2x² – 3 – 4 + 3x²}
[removing grouping symbol]
= – 4x² + {5x² – 7)
= – 4x² + 5x² – 7
(removing grouping symbol {})
= x² – 7

Question 6.
Solution:
We have :
– 2 (x² – y² + xy) – 3 (x² + y² – xy)
= – 2x² + 2y² – 2xy – 3x² – 3y² + 3xy
= – 2x² – 3x² + 2y² – 3y² – 2xy + 3xy
= ( – 2 – 3)x² + (2 – 3) y² + ( – 2 + 3)xy
= – 5x² – y² + xy

Question 7.
Solution:
a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
[removing grouping symbol( )]
= a – [2b – 3a + 2b – 3c]
(removing grouping symbol {})
= a – [4b – 3a – 3c]
= a – 4b + 3a + 3c
(removing grouping symbol [ ])
= 4a – 4b + 3c

Question 8.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
– x + [5y – {x – (5y – 2x)}]
= – x + [5y – {x – 5y + 2x}]
= – x + [5y – {3x – 5y}]
= – x + [5y – 3x + 5y]
= – x + [ 10y – 3x]
= – x + 10y – 3x
= – x – 3x + 10y
= – 4x + 10y

Question 9.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
86 – [15x – 7 (6x – 9) – 2 {10x – 5(2 – 3x)}]
= 86 – [15x – 42x + 63 – 2 {10x – 10 + 15x}
= 86 – [ 15x – 42x + 63 – 2 {25x – 10}]
= 86 – [15x – 42x + 63 – 50x + 20]
= 86 – 15x + 42x – 63 + 50x – 20
= (86 – 63 – 20) – 15x + 42x + 50x
= (86 – 83) + (- 15 + 42 + 50) x
= 3 + 77x

Question 10.
Solution:
Removing the innermost grouping ‘ symbol () first, then { } and then [ ], we have :
12x – [3x³ + 5x² – {7x² – (4 – 3x – x³) + 6x³} – 3x]
= 12x – [3x³ – 5x² – {7x² – 4 + 3x + x³ + 6x³} – 3x]
= 12x – [3x³ + 5x² – {7x² – 4 + 3x + 7x³} – 3x]
= 12x – [3x³ + 5x² – 7x² + 4 – 3x – 7x³ – 3x]
= 12x – [3x³ – 7x³ + 5x² – 7x² + 4 – 3x – 3x]
= 12x – [ – 4x³ + 2x² + 4 – 6x]
= 12x + 4x³ + 2x² – 4 + 6x
= 12x + 6x + 4x³ + 2x² – 4
= 18x + 4x³ + 2x² – 4
= 4x³ + 2x² + 18x – 4

Question 11.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
5a – [a² – {2a (1 – a + 4a²) – 3a (a² – 5a – 3)}] – 8a
= 5a – [a² – {2a – 2a² + 8a³ – 3a³ + 15a² + 9a}] – 8a
= 5a – [a² – {2a + 9a – 2a² + 15a² + 8a³ – 3a³}] – 8a
= 5a – [a² – {11a + 13a² + 5a³}] – 8a
= 5a – [a² – 11a – 13a² – 5a³] – 8a
= 5a – a² + 11a + 13a² + 5a³ – 8a
= 5a + 11a – 8a – a² + 13a² + 5a³
= 8a + 12a² + 5a³
= 5a3 + 12a² + 8a.

Question 12.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
3 – [x – {2y – (5x + y – 3) + 2x²} – (x² – 3y)]
= 3 – [x – {2y – 5x – y + 3 + 2x²} – x² + 3y]
= 3 – [x – {y – 5x + 3 + 2x²} – x² + 3y]
= 3 – [x – y + 5x – 3 – 2x² – x² + 3y]
= 3 – [6x + 2y – 3 – 3x²]
= 3 – 6x – 2y + 3 + 3x²
= 6 – 6x – 2y + 3x²

Question 13.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
xy – [yz – zx – {yx – (3y – xz} – (xy – zy)}]
= xy – [yz – zx – {yx – 3y + xz – xy + zy}]
= xv – [yz – zx – {- 3y + xz + zy}]
= xy – [yz – zx + 3y – xz – zy]
= xy – [ – 2xz + 3y]
= xy + 2xz – 3y

Question 14.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
2a – 3b – [3a – 2b – {a – c – (a – 2b)}]
= 2a – 3b – [3a – 2b – {a – c – a + 2b}]
= 2a – 3b – [3a – 2b – { – c + 2b}]
= 2a – 3b – [3a – 2b + c – 2b]
= 2a – 3b – 3a + 2b – c + 2b
= 2a – 3a – 3b + 2b + 2b – c
= – a + b – c

Question 15.
Solution:
Removing the innermost grouping symbol () first, then { } and ten [ ], we have:
– a – [a + {a + b – 2a – (a – 2b)} – b]
= – a – [a + {a + b – 2a – a + 2b} – b]
= – a – [a + { – 2a + 3b} – b]
= – a – [a – 2a + 3b – b]
= – a – a + 2a – 3b + b
= – 2a + 2a – 2b
= – 2 b

Question 16.
Solution:
Removing the innermost grouping symbol ‘—’ first, then ( ), then { } and then [ ], we have
2a – [4b – {4a – (3b – 2a+2b¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯)}]
= 2a – [4b – {4a – (3b – 2a – 2b)}]
= 2a – [4b – {4a – (b – 2a)}]
= 2a – [4b – {4a – b + 2a}]
= 2a – [4b – {6a – b}]
= 2a – [4b – 6a + b]
= 2a – [5b – 6a]
= 2a – 5b + 6a
= 8a – 5b.

Question 17.
Solution:
Removing the innermost grouping < symbol ( ) first, then { } and then [ ], we have :
5x – [4y – {7x – (3z – 2y) + 4z – 3(x + 3y – 2z)}]
= 5x – [4y – {7x – 3z + 2y + 4z – 3x – 9y + 6z}]
= 5x – [4y – {4x + 7z – 7y}]
= 5x – [4y – 4x – 7z + 7y]
= 5x – [11y – 4x – 7z]
= 5x – 11y + 4x + 7z
= 9x – 11y + 7z

RS Aggarwal Solutions for Class 6 Maths Chapter 8: Download PDF

RS Aggarwal Solutions for Class 6 Maths Chapter 8–Algebraic Expressions

Download PDF: RS Aggarwal Solutions for Class 6 Maths Chapter 8–Algebraic Expressions PDF

Chapterwise RS Aggarwal Solutions for Class 6 Maths :

• Chapter 1–Number System
• Chapter 2–Factors and Multiples
• Chapter 3–Whole Numbers
• Chapter 4–Integers
• Chapter 5–Fractions
• Chapter 6–Simplification
• Chapter 7–Decimals
• Chapter 8–Algebraic Expressions
• Chapter 9–Linear Equations in One Variable
• Chapter 10–Ratio, Proportion and Unitary Method
• Chapter 11–Line Segment, Ray and Line
• Chapter 12–Parallel Lines
• Chapter 13–Angles and Their Measurement
• Chapter 14–Constructions (Using Ruler and a Pairs of Compasses)
• Chapter 15–Polygons
• Chapter 16–Triangles
• Chapter 17–Quadrilaterals
• Chapter 18–Circles
• Chapter 19–Three-Dimensional Shapes
• Chapter 20–Two-Dimensional Reflection Symmetry (Linear Symmetry)
• Chapter 21–Concept of Perimeter and Area
• Chapter 22–Data Handling
• Chapter 23–Pictograph
• Chapter 24–Bar Graph

About RS Aggarwal Class 6 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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