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RS Aggarwal Class 6 Maths Solutions Chapter 8- Algebraic Expression Exercise 8B

RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 8- Algebraic Expression

RS Aggarwal Class 6 Math Solution Chapter 8- Algebraic Expression Exercise 8B is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily.  RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.

 Rs Aggarwal Class 6 Math Solution Chapter 8- Algebraic Expression 


Exercise 8B

1. If a = 2 and b = 3, find the value of :

    (i) a + b

    (ii) a2 + ab

    (iii) ab − a2

    (iv) 2a − 3b

    (v) 5a2 − 2ab

    (vi) a3 −  b3

Solution: 

    (i)  a + b

    (By putting the value of a =2 and b = 3 )

    = 2 + 3     

    = 5

    (ii)  a2 + ab

    (By putting the value of a =2 and b = 3 )

    =  (2)2 + (2)(3)     

    = 4 + 6

    = 10

    (iii) ab − a2

    (By putting the value of a =2 and b = 3 )

    = (2)(3) (2)2    

    = 6 − 4

    = 2

    (iv)  2a − 3b

    (By putting the value of a =2 and b = 3 )

    =  2(2) − 3(3)     

    =  4 − 9

    =  − 5

    (v) 5a2 − 2ab

    (By putting the value of a =2 and b = 3 )

    = 5(2)2 − 2(2)(3)     

    = 5(4) − 2(6)

    = 20 − 12

    = 8

    (vi) a3 −  b3

    (By putting the value of a =2 and b = 3 )

    = (2)3 −  (3)3    

    = 8 − 27

    =  − 19


2. If x = 1, y = 2 and z = 5, find the value of :

    (i) 3x −  2y + 4z

    (ii) x2 + y2 + z2

    (iii) 2x2 − 3y2 + z2

    (iv) xy + yz − zx

    (v) 2x2y −  5yz + xy2

    (vi) x3 −  y3−  z3

Solution: 

    (i) 3x −  2y + 4z

     (By putting the value of  x = 1, y = 2 and z = 5 )

    =   3(1) −  2(2) + 4(5)   

    = 3 − 4 + 20

    = 23 − 4

    = 19

    (ii) x2 + y2 + z2

     (By putting the value of  x = 1, y = 2 and z = 5 )

    = (1)2 + (2)2 + (5)2   

    = 1 + 4 + 25

    = 30

    (iii) 2x2 − 3y2 + z2

     (By putting the value of  x = 1, y = 2 and z = 5 )

    = 2(1)2 − 3(2)2 + (5)2    

    = (2)(1) − 3(4) + 25

    = 2 − 12 + 25

    = 27 − 12

    = 15

    (iv) xy + yz − zx

     (By putting the value of  x = 1, y = 2 and z = 5 )

    = (1)(2) + (2)(5) − (5)(1)    

    =2 + 10 − 5

    = 12 − 5

    = 7

    (v) 2x2y −  5yz + xy2

     (By putting the value of  x = 1, y = 2 and z = 5 )

    = 2(1)2(2) −  5(2)(5) + (1)(2)2    

    = 2(1)(2) −  5(2)(5) + (1)(4)

    = 4 −  50 + 4

    = 8 − 50

    = − 42

    (vi) x3 −  y3−  z3

     (By putting the value of  x = 1, y = 2 and z = 5 )

    = (1)3 −  (2)3−  (5)3    

    = 1 −  8 − 125

    = 1 − 133

    = − 132


3. If p = − 2, q = − 1 and r = 3, find the value of :

    (i) p2 + q2r2

    (ii) 2p2q2 + 3r2

    (iii) p − q − r

    (iv) p3 + q3 + r3 + 3pqr

    (v) 3p2q + 5pq2 + 2pqr

    (vi) p4 + q4r4

Solution: 

    (i)  p2 + q2r2
    
    (By putting the value of  p = − 2, q = − 1 and r = 3 )

    = (2)2 + (1)2(3)2     
    
    = 4 + 1 − 9

    = 5 − 9

    = − 4

    (ii) 2p2q2 + 3r2

    (By putting the value of  p = − 2, q = − 1 and r = 3 )

    = 2(2)2(1)2 + 3(3)2    

    = 2(4) − (1) + 3(9)

    = 8 − 1 + 27

    = 35 − 1

    = 34

    (iii) p − q − r

    (By putting the value of  p = − 2, q = − 1 and r = 3 )

    =  (− 2) − (− 1) − 3    

    =  − 2 + 1 − 3 

    = − 5 + 1

    = − 4

    (iv) p3 + q3 + r3 + 3pqr

    (By putting the value of  p = − 2, q = − 1 and r = 3 )

    = (2)3+(1)3+(3)3+3(2)(1)(3)

    = − 8 + (− 1) + 27 + 18

    = − 8 − 1 + 45

    = − 9 + 45

    = 36

    (v)  3p2q + 5pq2 + 2pqr

    (By putting the value of  p = − 2, q = − 1 and r = 3 )

    =  3(2)2(1)+5(2)(1)2+2(2)(1)(3)

    = 3(4)(−1) + 5(−2)(1) + 2(2)(3)

    = −12 −10 + 12

    = −22 + 12

    = −10

    (vi) p4 + q4r4

    (By putting the value of  p = − 2, q = − 1 and r = 3 )

    = (2)4 + (1)4(3)4

    = 16 + 1 − 81

    = 17 − 81

    = − 64


4. Write the coefficient of :

    (i) x in 13x

    (ii) y in − 5y

    (iii) a in 6ab

    (iv) z in − 7xz

    (v) p in − 2pqr

    (vi) y2 in 8xy2z

    (vii) x3 in x3

    (viii) x2 in − x2

Solution: 

    (i) Coefficient of x in 13x is 13.

    (ii) Coefficient of y in − 5y is − 5.

    (iii) Coefficient of a in 6ab is 6b.

    (iv) Coefficient of z in − 7xz is − 7x.

    (v) Coefficient of p in − 2pqr is − 2qr.

    (vi) Coefficient of y2 in 8xy2z is 8xz.

    (vii) Coefficient of x3 in  x3 is 1.

    (viii) Coefficient of x2 in x2 is − 1.


5. Write the numerical coefficient of :

    (i) ab

    (ii) − 6bc

    (iii) 7xyz

    (iv) − 2x3y2z

Solution: 

    (i) Numerical coefficient of ab is 1.

    (ii) Numerical coefficient of − 6bc is − 6.

    (iii) Numerical coefficient of 7xyz is 7.

    (iv) Numerical coefficient of 2x3y2z is − 2.



6. Write the constant term of :

    (i) 3x2 + 5x + 8

    (ii) 2x2 − 9
    
    (iii) 4y2 − 5y + 35

    (iv) z3 − 2z2 + z − 83

Solution: 

    (i) The constant term of   3x2 + 5x + 8 is 8

    (ii) The constant term of  2x2 − 9 is − 9

    (iii) The constant term of  4y2 − 5y + 35 is 35

    (iv) The constant term of   z3 − 2z2 + z − 83 is − 83


7. Identify the monomial, binomials and trinomials in the following:

    (i) − 2xyz

    (ii) 5 + 7x3y3z3

    (iii) − 5x3

    (iv) a + b − 2c

    (v) xy + yz − zx

    (vi) x5

    (vii) ax3 + bx2 + cx + d

    (viii) −14

    (ix) 2x + 1

Solution: 

    (i) − 2xyz  

        It has only one term. So, it is monomial

    (ii) 5 + 7x3y3z3

        It has only two term. So, it is binomials
    
    (iii) − 5x3

        It has only one term. So, it is monomial

    (iv) a + b − 2c

        It has only three term. So, it is trinomials 

    (v) xy + yz − zx

        It has only three term. So, it is trinomials

    (vi) x5

        It has only one term. So, it is monomial

    (vii) ax3 + bx2 + cx + d

        It has four term. So, it is not a monomial, or, a binomials or, a trinomials

    (viii) −14

        It has only one term. So, it is monomial

    (ix) 2x + 1

        It has only two term. So, it is binomials



8. Writ all the terms of the algebraic expressions :

    (i) 4x5 − 6y4 + 7x2y − 9 

    (ii) 9x3 − 5z4 + 7x3y − xyz

Solution: 

    (i) All the terms of 4x5 − 6y4 + 7x2y − 9 are

        4x5,  − 6y4,   7x2y,  − 9 

    (ii) All the terms of 9x3 − 5z4 + 7x3y − xyz are

        9x3,  − 5z4,  7x3y,  − xyz


9. Identity like terms in the following:

    (i) a2, b2, − 2a2, c2, 4a

    (ii) 3x, 4xy, − yz, 12zy

    (iii) −2xy2, x2y, 5y2x, x2z

    (iv) abc, ab2c, acb2, c2ab, b2ac, a2bc, cab2

Solution: 

    The terms that have same literals are called like terms.

    (i) a2 and − 2a2 are like terms.

    (ii) − yz and 12zy are like terms.

    (iii) −2xy2 and 5y2x are like terms.

    (iv) ab2c, acb2, b2ac and cab2 are like terms.





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