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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
= Radius circle
= 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 = 4xy2z3
(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,
Breadth = b cm
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) a2 – 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, x3 – (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 + 2x2 + 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 – a2 should 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
= a2 – 2ab…(Simplified)
(ii) (a2 + b2 + 2ab) – (a2 + b2 – 2ab)
Now,(a2 + b2 + 2ab) – (a2 + b2 – 2ab)
= a2 + b2 + 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)
= 2a2 + 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)
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