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) `a^2` + ab
(iii) ab − `a^2`
(iv) 2a − 3b
(v) 5`a^2` − 2ab
(vi) `a^3` − `b^3`
Solution:
(i) a + b
(By putting the value of a =2 and b = 3 )
= 2 + 3
= 5
(ii) `a^2` + ab
(By putting the value of a =2 and b = 3 )
= `(2)^2` + (2)(3)
= 4 + 6
= 10
(iii) ab − `a^2`
(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) 5`a^2` − 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) `a^3` − `b^3`
(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) 3`x` − 2y + 4z
(ii) `x^2` + `y^2` + `z^2`
(iii) 2`x^2` − 3`y^2` + `z^2`
(iv) `x`y + yz − z`x`
(v) 2`x^2`y − 5yz + `x``y^2`
(vi) `x^3` − `y^3`− `z^3`
Solution:
(i) 3`x` − 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) `x^2` + `y^2` + `z^2`
(By putting the value of `x` = 1, y = 2 and z = 5 )
= `(1)^2` + `(2)^2` + `(5)^2`
= 1 + 4 + 25
= 30
(iii) 2`x^2` − 3`y^2` + `z^2`
(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) `x`y + yz − z`x`
(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) 2`x^2`y − 5yz + `x``y^2`
(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) `x^3` − `y^3`− `z^3`
(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) `p^2` + `q^2` − `r^2`
(ii) 2`p^2` − `q^2` + 3`r^2`
(iii) p − q − r
(iv) `p^3` + `q^3` + `r^3` + 3pqr
(v) 3`p^2`q + 5p`q^2` + 2pqr
(vi) `p^4` + `q^4` − `r^4`
Solution:
(i) `p^2` + `q^2` − `r^2`
(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) 2`p^2` − `q^2` + 3`r^2`
(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) `p^3` + `q^3` + `r^3` + 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) 3`p^2`q + 5p`q^2` + 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) `p^4` + `q^4` − `r^4`
(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 13`x`
(ii) y in − 5y
(iii) a in 6ab
(iv) z in − 7`x`z
(v) p in − 2pqr
(vi) `y^2` in 8`x``y^2`z
(vii) `x^3` in `x^3`
(viii) `x^2` in − `x^2`
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 `y^2` in `8xy^2z` is 8xz.
(vii) Coefficient of `x^3` in `x^3` is 1.
(viii) Coefficient of `x^2` in `−x^2` is − 1.
5. Write the numerical coefficient of :
(i) ab
(ii) − 6bc
(iii) 7xyz
(iv) − 2`x^3``y^2`z
Solution:
(i) Numerical coefficient of ab is 1.
(ii) Numerical coefficient of − 6bc is − 6.
(iii) Numerical coefficient of `7 xyz` is 7.
(iv) Numerical coefficient of `−2x^3y^2z` is − 2.
6. Write the constant term of :
(i) 3`x^2` + 5`x` + 8
(ii) 2`x^2` − 9
(iii) 4`y^2` − 5y + `frac\{3}{5}`
(iv) `z^3` − 2`z^2` + z − `frac\{8}{3}`
Solution:
(i) The constant term of 3`x^2` + 5`x` + 8 is 8
(ii) The constant term of 2`x^2` − 9 is − 9
(iii) The constant term of 4`y^2` − 5y + `frac\{3}{5}` is `frac\{3}{5}`
(iv) The constant term of `z^3` − 2`z^2` + z − `frac\{8}{3}` is − `frac\{8}{3}`
7. Identify the monomial, binomials and trinomials in the following:
(i) − 2`x`yz
(ii) 5 + 7`x^3``y^3``z^3`
(iii) − 5`x^3`
(iv) a + b − 2c
(v) `x`y + yz − z`x`
(vi) `x^5`
(vii) a`x^3` + b`x^2` + c`x` + d
(viii) −14
(ix) 2`x` + 1
Solution:
(i) − 2`x`yz
It has only one term. So, it is monomial
(ii) 5 + 7`x^3``y^3``z^3`
It has only two term. So, it is binomials
(iii) − 5`x^3`
It has only one term. So, it is monomial
(iv) a + b − 2c
It has only three term. So, it is trinomials
(v) `x`y + yz − z`x`
It has only three term. So, it is trinomials
(vi) `x^5`
It has only one term. So, it is monomial
(vii) a`x^3` + b`x^2` + c`x` + 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) 2`x` + 1
It has only two term. So, it is binomials
8. Writ all the terms of the algebraic expressions :
(i) 4`x^5` − 6`y^4` + 7`x^2`y − 9
(ii) 9`x^3` − 5`z^4` + 7`x^3`y − `x`yz
Solution:
(i) All the terms of 4`x^5` − 6`y^4` + 7`x^2`y − 9 are
4`x^5`, − 6`y^4`, 7`x^2`y, − 9
(ii) All the terms of 9`x^3` − 5`z^4` + 7`x^3`y − `x`yz are
9`x^3`, − 5`z^4`, 7`x^3`y, − `x`yz
9. Identity like terms in the following:
(i) `a^2`, `b^2`, − 2`a^2`, `c^2`, 4a
(ii) 3`x`, 4`x`y, − yz, `frac\{1}{2}`zy
(iii) −2`x``y^2`, `x^2`y, 5`y^2``x`, `x^2`z
(iv) abc, a`b^2`c, ac`b^2`, `c^2`ab, `b^2`ac, `a^2`bc, ca`b^2`
Solution:
The terms that have same literals are called like terms.
(i) `a^2` and − 2`a^2` are like terms.
(ii) − yz and `frac\{1}{2}`zy are like terms.
(iii) −2`x``y^2` and 5`y^2``x` are like terms.
(iv) `ab^2c`, `acb^2`, `b^2ac` and `cab^2` are like terms.
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