# Number Line Prime Class 6 Solutions Chapter 10

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## Number Line Prime Class 6 Solutions Chapter 10 Introduction to Algebra

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Number Line Prime Class 6 Math Book, Chapter 10, Introduction to Algebra. Here students can easily find step by step solutions of all the problems for Introduction to Algebra, Exercise 10A, 10B and 10C 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. Here in this post students will get chapter 10 solutions. Here in this post all the solutions are based on latest Syllabus.

Introduction to Algebra Exercise 10A Solution :

Question no – (1)

Solution :

Let the number be x

(a) X + 3

(b) X – 5

(c) X – 2

(d) 2x + 1

(e) 3x – 2

(f) 2/3 x

(g) 5/6x +2

(h) 3x

(i) 2y

(j) X = y+9

Question no – (2)

Solution :

(a) A number taken away from 18

(b) 4 added with thrice of a number.

(c) One fourth of a number

(d) 3 less than a number.

Question no – (3)

Solution :

(a) X and 1

(b) X and -6

(c) 2x and 5

(d) 4x and -3

(e) X and – 1/2

(f) 4/3 x and 2

Question no – (4)

Solution :

Numerical Coefficient

(a) 2

(b) 4

(c) 10

(d) -1

(e) 1/3

(f) -3

Introduction to Algebra Exercise 10B Solution :

Question no – (1)

Solution :

(a) X + 1

(b) X + 2y

(c) 2a + 3b

(d) 10 n + 100

(e) 6x +7

(f) a + b + c + d

(g) m2n + nm

(h) x2 + Y2 + 2xy

(i) X – 2

(j) 3x – 14

(k) 10x – 100

Question no – (2)

Solution :

(a) a × b

= ab

(b) x × y × z

= xyz

(c) pq × r

= pqr

(d) m2 n × nm2

= m4 n2

(e) ab × a3 b2

= a4 b3

(f) 2abc × 4a2 b2c2

= 8a3b3c3

(g) 5x × 6y × 7z

= 210 xyz

(h) x × x2 × x3

= x6

(i) 2a ÷ a

= 2a/a = 2

(j) 10 xy × 2Y

= 20x2y

(k) 50 p2q2r2 ÷ 5 pqr

= 50 p2q2r2/5pqr

= 10 pqr

(l) –a3b3 ÷ 4a

= – a3b3/4a

= – 1/4 a2b3

Question no – (3)

Solution :

 No. Like Terms : Unlike Terms : (a) a, 2a ab (b) 3a2b2, a2b2 2a2b, a (c) (10xy, 6yx) (2yz 7=8zy) 5zx (d) (2pqr, – 9pqr, rpq) 4p3q2, 5r2p3

Question no – (4)

Solution :

Like terms

(a) 2a, – a

(b) – x, 4x

(c) – mn, 2mn

(d) (a3 and 2a3) (+ 3ab2, ab2)

(e) No like terms

Introduction to Algebra Exercise 10C Solution :

Question no – (1)

Solution :

(a) 3

= Monomial

(b) x – 2y + 10

= Trinomial

(c) 2x + 5y

=  Binomial

(d) x² – y² – z² + 144

=  Polynomial

(e) 2xyz

=  Monomial

(f) -9p²q²

= Monomial

Question no – (2)

Solution :

If P is the Perimeter and ‘a’ is Side.

So, P = 5 × a

Question no – (3)

Solution :

The equation become

= Pqr/100

Question no – (4)

Solution :

Cost of each Pencil will be,

= 150/x

Question no – (5)

Solution :

(a) 2n for n

= 1, 2, 3, 4, 5) or 1 m ≤ mn, ≤ 5

(b) 10 n -9 for 1 ≤ n, ≤ 5

(c) 5n + 2 for 1 ≤ n, ≤ 5

(d) 3n – 1

= 1 ≤ n ≤ 6

Question no – (6)

Solution :

6x + 2

If x = 100 then

6 × 100 + 2

= 602

Question no – (7)

Solution :

10x – 4

For x = 35,

= 10 × 35 – 4

= 350 -4

= 346

Question no – (8)

Solution :

(a) 2x + 5

= 2 × 2 + 5

= 4 + 5

= 9

(b) – 4x

= – 4 x2

= – 8

(c) 5x – 3

= 5 × 2 – 3

= 10 – 3

= 7

(d) 25/3 × 6x

= 25/3 × 6 × 2

= 100

(e) 5x/4

= 5 × 2/4

= 5/2

(f) 30x + 2

= 30 × 2 + 2

= 60 +2

= 62

Question no – (9)

Solution :

(a) – 3xyz

= – 3 × x × Y × z

(b) 4a2b

= 2 × 2 × a × a × b

(c) 27 P

= 3 × 3 × 3 × p

(d) -12 cd

= -2 × 2 × 3 × c × d

Question no – (11)

Solution :

Side = s

Perimeter

= s + s + s+ s + s + s + s + s

= 4s

Question no – (12)

Solution :

Given Diameter = 2 × radius

Also Diameter = d

and radius = r

Therefore, d = 2r

Next Chapter Solution :

Updated: June 22, 2023 — 4:51 am