# Brilliant’s Composite Mathematics Class 6 Solutions Chapter 6

## Brilliant’s Composite Mathematics Class 6 Solutions Chapter 6 Algebraic Expressions

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 6, Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions, Exercise 6.1, 6.2, 6.3 and 6.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

Algebraic Expressions Exercise 6.1 Solution

Question no – (1)

Solution :

(i) The sum of literal number x and a number 10,

In algebraic expression = x + 10

(ii) x less than a number two times y

Algebraic expression = 2y – x

(iii) One-four of the sum of number a and b

Algebraic expression = (a + b) × 1/4 a + b/4

(iv) Product of x and y added to their sum

Algebraic expression = x + y + xy

(v) Four times x added 6 times y

Algebraic expression = 4x + 6y

(vi) One-half of p multiplied by the difference of x and y

Algebraic expression = p/2 (x – y)

(vii)  4 takes away from y

Algebraic expression = y – 4

Question no – (2)

Solution :

(i) Perimeter of a square is the product of 4 and length of its side.

Perimeter,

= 4 × length of its side

= P = 4l

(ii) Profit equal to the difference of selling price and cost price.

In numeral, = R = d/2

(iii) The radius of a circle is half of its diameter.

In numeral,

= R = d/2

= 1/2 × diameter of the circle

Algebraic Expressions Exercise 6.2 Solution

Question no – (1)

Solution :

(i) 4a3

= 4 × a × a × a

(ii) x8

= x × x × x × x × x × x × x × x

(iii) 7p2q2

= 7 × p × p × q × q

(iv) a2b3

= a × a × b × b × b

(v) 10x3y2z4

= 10 × x × x × x × y × y × z × z × z × z

Question no – (2)

Solution :

(i) x × x × x × x …..10 times

Exponential from = x10

(ii) 8 × a × a × b × b × b

Exponential from = 8a3b3

(iii) a × a × a × a × …..40 times

Exponential from = 4xy2z

(iv) 4 × x × y × y × z × z × z

Exponential from = 4xy2z2

(v) 7 × x × x × y × y × z × z

Exponential from = 7x2y3z2

Question no – (3)

Solution :

According to the question,

Length = 3b cm

height = 1/2 3b = 3/2b

Volume,

=  l × h × b

= 3/2 b × 3b × b

= 9/2b3 cm3

Hence, volume of the cuboid is 9/2b3 cm3

Question no – (4)

Solution :

(i) 8x4y – 7x3yz + 4/3x2yz2 – 2.5xyz

All the terms = 8x4y, – 7x3yz, 4/3x2yz2, – 2.5xyz

(ii) 3ab2 – 7ab2 + 8ab3c3 – 5abc

∴ All the terms = 3ab2,  -7ab2,  + 8ab3c3,  -5abc

(iii) 3x5 + 5y4 – 7x2y + 7

All the terms = 3x5,  + 5y4,  – 7x2y,  + 7

(iv) a– 3ab – b2 + c

All the terms =  a2 – 3ab,  – b2,  + c

Question no – (5)

Solution :

(i) -4

= Monomials

(ii) x + y

= Binomials

(iii) 4.5a

= Monomials

(iv) a3 – b3

= Binomials

(v) a2 + 2ab + b2

= Trinomials

(v) 2a + 3b

= Binomials

(vii) ax + by + c

= Trinomials

(viii) 5xy

= Monomials

Question no – (6)

Solution :

(i) Given, 3x

= 3

Coefficient of x is = 3

(ii) 4ax

= – 4a

Coefficient of x is = – 4a

(iii) 5xy²

= – 5y²

Coefficient of x is = – 5y²

(iv) xyz

= yz

Coefficient of x is = yz

Question no – (7)

Solution :

(i) 5x

= 5

Numerical Coefficient = 5

(ii) xy

= 1

Numerical Coefficient = 1

(iii) -6xy2

= -6

Numerical Coefficient = – 6

(iv) 7y2z

= 7

Numerical Coefficient =  7

(v) -5x2y2z

= -5

Numerical Coefficient = -5

Question no – (8)

Solution :

Given expression = 3x2 – 2xy + 4y2x3

∴ Power of x =  2, 1, 3

Question no – (9)

Solution :

(i) 3x – 7x

= Like terms

(ii) 16x, 16y

= Unlike terms

(iii) 9ab, -6b

= Unlike terms

(iv) 2xy , yz, 3, 2y/2

= Unlike terms

Question no – (10)

Solution :

From the question we get,  x = 1, y = 2 and z = -3

(i) 2xy4 – 15x2y + z

= 2xy4 – 15x2y + z

= 2.1.24 – 15.12. 2 (- 3)

= 32 – 30 – 3

= – 1

(ii) x3 + y3 + z3 + 3xyz

= 13 + 23 + (- 3)3 + 3.1.2. (- 3)

= 1 + 8 – 27 – 18

= -36

Question no – (11)

Solution :

According to the question, b = 6, c = 4 and x = 2

(i) 3b – 2 (x + x)

= 3b – 2(c + x)

= 3.6 – 2 (4 + 2)

= 18 – 12

= 6

Thus, the value will be 6

(ii) (2b + c – 3)

= 2b + c – 3

= 2 × 6 + 4 – 3

= 12 + 1

= 13

Hence, the value will be 13

(iii) 3(b + x + c)

= 3 (6 + 2 + 4)

= 3 × 12

= 36

Therefore, the value will be 36

Question no – (12)

Solution :

According to question the value of,

a = 2,

b = 3

c = 4

(i) a2bc

= a2bc

= 22 × 3 × 4

= 4 × 4 × 3

= 48

(ii) abc

= 2 × 3 × 4

= 24

(iii) ab3c

= ab3c

= 2 × 33 × 4

= 8 × 27

= 216

(iv) abc3

= 2 × 3 × 43

= 6 × 64

= 384

(v) a2b2c2

= 22 × 32 × 42

= 4 × 9 × 16

= 36 × 16

= 576

Algebraic Expressions Exercise 6.3 Solution

Question no – (1)

Solution :

(i) 4x2y, 8x2y and – 2x2y

Now, 4x2y + 8x2y + (- 2x2y)

= 12x2y – 2x2y

= 10x2y

(ii) x3, – 3x3, 2x3, – 4x3

Now, x– (3x3) + 2x3 + (-4x3)

= 3x3 – 3x3 + 2x3 – 4x3

= – 4x3

(iii) +2xy, – 3xy, + 6xy, – 4xy

Now, 2xy + (- 2xy) + 6xy + (- 4xy)

= 8xy – 3xy – 4xy

= xy

Question no – (2)

Solution :

(i) x – 2y – 3z and 3y – 2x – 5x

Now, (x – 2y – 3z) + (3y – 2x – 5z)

= x – 2y – 3z + 3y – 2x – 5z

= – x + y – 8z

(ii) 4x2 + 5y – 6x2, 6y + 3x2, 5x2 -7y and 9x2 – 10y + 6z2

Now, 4x2 + 5y – 6x2, 6y + 3x2, 5x2 – 7y, 9x2 – 10y + 6z2

= 21x2 – 6y

(iii) 3x2 + 4y – 5x3, 5y + 2x2, 7x2 – 8y and 4x2 – 9y – 5x3

Now, 3x2 + 4y – 5 z3 + 5y + 2x2 + 7x2 – 8y + 4x2 – 9y – 5z3

= 16x2 + y – 9y – 10z3

= 16x2 – 8y – 10z3

(iv) x + 2x + z, x + y + 2z and x – y – z.

Now, x + 2x + z, x + y + 2z + x – y – z.

= 4x + 3z – z + 4

= 5x + 2z

(v) x2 – 6x + 7, 8x – 15 + 2x2 and 4x2 – 6x – 10.

Now, x2 – 6x + 7, 8x – 15 + 2x+ 4x2 – 6x – 10.

= 7x2+ 2x – 6x – 8 – 10

= 7x2 – 4x – 18

Question no – (3)

Solution :

According to the question,

(i) 7x – 7y + 8z

– 9x + 7y – 9z

7y – 2z

x – y
———————————-
– x + 6 – 3z

(ii) – 2x2 + 4y2 + 6

– x2 – 5y2 – 8

– 3x2 – 2y2 + 4
———————————-
-2x2 – 3y2 + 2

Question no – (4)

Solution :

(i) 2x – 4y from 4x + 3y

= (4x + 3y) – (2x – 4y) (according to the question)

= 4x + 3y – 2x + 4y

= 2x + 7y

(ii) – x2 – 3z from 5x2 – y + z + 7

5x2 – y + z + 7 + x2 + 3z

= 6x2 – y + 4z + 7

Question no – (5)

Solution :

= (4ab + b2) – (a2 + 2ab + b2) …(according to the question)

= 4ab + b2 – a2 – 2ab – b2

= 2ab – a2

Therefore, 2ab – ashould be added.

Question no – (6)

Solution :

= 3x + 2y + 3z + 3x – 4y + 5z – (6x + 7y – 2z) …(according to the question)

= 6x – 2y + 8z – 6x – 7y + 22

= -7y + 10z

Question no – (7)

Solution :

As per the given question,

= (4x4 – 3x3 + 6x2 + 4x3 + 4x – 3 – 3x4 – 5x2 + 2x) – (5x4 – 7x3 – 3x + 4)

= x4 + x3 + x2 + 6x – 3 – 5x4 + 7x3 + 3x – 4

= -4x4 + 8x3 + x2 + 9x – 7

Question no – (8)

Solution :

In the given question we get,

a = 2x2 + 5x – 7

b = x2 – 3x – 3

c = -3x2 – 2x + 10

∴ a + b + c = 2x2 + 5x – 7 + x2 – 3x – 3 – 3x2 – 2x + 10

= (5x – 3x – 2x) + (-7 – 3 + 10)

= 0 (Proved)

Algebraic Expressions Exercise 6.4 Solution

Question no – (1)

Solution :

(i) (2a2 + b2) – (a2 + 2ab + b2)

2a2 + b2 – a2 – 2ab + b2

= a– 2ab…(Simplified)

(ii) (a2 + b2 + 2ab) – (a2 + b2 – 2ab)

Now,(a2 + b2 + 2ab) – (a2 + b2 – 2ab)

= a2 + b+ 2ab – a2 – b2 + 2ab

= 4ab…(Simplified)

(iii) -3p2 + 4pq – q2, 4p2 + q2, – 3pq + 4q2 and – 3p2 + 4pq

Now, -3p2 + 4pq – q2, 4p2 + q2, – 3pq + 4q2 and – 3p2 + 4pq

= (-3p2 + 4p2 – 3p2) + (4pq – 3pq + 4pq) + (-q2 + q2 + 4q2)

= – 2p2 + 5pq + 4q2(Simplified)

(iv) (a2 + b2 – ab) + (a2 + b2 + ab) – (2a2 + 2ab)

Now, (a2 + b2 – ab) + (a2 + b2 + ab) – (2a2 + 2ab)

= 2a+ 2b2 – 2a2 – 2ab

= 2b2 – 2ab…(Simplified)

(v) -5 (a + b) + 2(2a – b) + 4a – 7

Now, -5 (a + b) + 2(2a – b) + 4a – 7

= -5a – 5b + 4a – 2b + 4a – 7

= 3a – 7b – 7…(Simplified)

(vi) 4x3 – [9x2 – {- 5x3 – (2 – 7x2) + 6x}]

Now, 4x3 – [9x2 – {- 5x3 – (2 – 7x2) + 6x}]

= 4x3 – [9x2 – {- 5x3 – 2 + 7x2 + 6x}]

= 4x3 – [9x2 + 5x3 + 2 – 7x2 – 6x]

= 4x3 – 9x2 – 5x3 – 2 + 7x2 + 6x

= – x3 – 2x2 + 6x – 2...(Simplified)

(vii) -m – [m + {m + n – 2m – (m – 2n)} – n]

Now, – m – [m + {m + n – 2m – (m – 2n)} – n]

= – m – [m + {m + n – 2m – m + 2n} – n]

= – m – [m + { -2m + 3n} – n]

= – m – [m – 2m + 3n – n]

= – m – m + 2m – 3n + n

= – 2n…(Simplified)

(viii) 5 + [x – {2y – (6x + y – 4) + 2x2} – (x2 – 2y)]

Now, 5 + [x – {2y – (6x + y – 4) + 2x2} – (x2 – 2y)]

= 5 + [x – {2y – 6x – y + 4 + 2x2} – x2 + 2y]

= 5 + [x – {y – 6x + 2x2 + 4} – x2 + 2y]

= 5 + [x – y + 6x – 2x2 – 4 – x2 + 2y]

= 5 + 7x – 3x2 + y – 4

= 7x + y – 3x2 + 1

= 1 + 7x + y – 3x2(Simplified)

Question no – (2)

Solution :

(i) x + y – 3z + 7

= x + y – (3z – 7)

(ii) xy² + yz² + zx²

= xy² – (- yz² – zx²)

(iii) – y + z + x² – y² – a²

= – y + z + x² – (x² + a²)

(iv) 9a + 5xy – 7x² + 8y – 6

= 9a + 5xy – 7x² – (- 8y + 6)

Next Chapter Solution :

👉 Chapter 7 👈

Updated: June 10, 2023 — 10:55 am