Fundamental Concepts of Algebra Class 6 RS Aggarwal Exe-12D Goyal Brothers ICSE Maths Solutions Ch-12. We provide step by step Solutions of substitution in polynomial for ICSE Class-6 Foundation RS Aggarwal of Goyal Brothers Prakashan . Visit official Website CISCE for detail information about ICSE Board Class-6 Mathematics.
Fundamental Concepts of Algebra Class 6 RS Aggarwal Exe-12D Goyal Brothers ICSE Maths Solutions Ch-12
| Board | ICSE |
| Subject | Maths |
| Class | 6th |
| Ch-12 | Fundamental Concepts of Algebra |
| Writer | RS Aggrawal |
| Book Name | Foundation |
| Exe-12D | Substitution in Polynomial |
| Academic Session | 2024 – 2025 |
Substitution in Polynomial
the process of replacing each variable by a given value of it is called substitution
Exercise- 12D
(Fundamental Concepts of Algebra Class 6 RS Aggarwal Exe-12D Goyal Brothers ICSE Maths Solutions Ch-12)
Que-1: Find the values of :
(i) 2x+3y, when x=3 and y=4
(ii) 4x-5y, when x=5 and y=3
(iii) 3a-4b+5c, when a=10, b=7 and c=5
(iv) 8m-5n-3p, when m=15, n=-4 and p=20
(v) 5ab+7bc-3ca, when a=2, b=3 and c=4
Sol: (i) 2x+3y
2(3)+3(4) = 6+12
= 16
(ii) 4x-5y
4(5)-5(3) = 20-15
= 5
(iii) 3a-4b+5c
3(10)-4(7)+5(5)
= 30-28+25 = 27
(iv) 8m-5n-3p
8(15)-5(-4)-3(20)
120+20-60 = 80
(v) 5ab+7bc-3ca
5(2)(3)+7(3)(4)-3(4)(2)
30+84-24 = 90
Que-2: If x = -3 , find the values of :
(i) 5-2x (ii) x²+2 (iii) 6-x²
Sol: (i) 5-2x
x = -3
= 5-2(-3)
= 5+6 = 11
(ii) x²+2
x = -3
= (-3)²+2
= 9+3 = 11
(iii) 6-x²
x = -3
= 6-(-3)²
= 6-9 = -3
Que-3: If x = 5, find the values of :
(i) x²+x+3 (ii) 5+4x-x² (iii) 9x²-7x
Sol: (i) x²+x+3
x = 5
= (5)²+(5)+3
= 25+8 = 33
(ii) 5+4x-x²
x = 5
= 5+4(5)-(5)²
= 5+20-25 = 0
(iii) 9x²-7x
x = 5
= 9(5)²-7(5)
= 9(25)-35
= 225-35 = 190
Que-4: Find the values of :
(i) 6x²-5x+4, when x = 5 , (ii) 3x²-4x-2 when x = -4 , (iii) 2x³-3x²+5x+6, when x = 3 , (iv) x⁴-2x³+x²-7x-10, when x = -2
Sol: (i) 6x²-5x+4
x = 5
= 6(5)²-5(5)+4
= 6(25)-25+4
= 150-21 = 129
(ii) 3x²-4x-2
x = -4
= 3(-4)²-4(-4)-2
= 48+16-2 = 62
(iii) 2x³-3x²+5x+6
x = 3
= 2(3)³-3(3)²+5(3)+6
= 54-27+15+6
= 48
(iv) x⁴-2x³+x²-7x-10
x = -2
= (-2)⁴-2(-2)³+(-2)²-7(-2)-10
= 16-2(-8)+4+14-10
= 16+16+8
= 40
Que-5: If a= 4 and b= 5, find the values of :
(i) ab-a² (ii) 6a²-2ab (iii) 5a²-3b² (iv) 4b²-3ab
Sol: (i) ab-a²
a = 4, b = 5
= (4)(5)-(4)²
= 20-16 = 4
(ii) 6a²-2ab
a = 4, b= 5
= 6(4)²-2(4)(5)
= 6(16)-40
= 96-40 = 56
(iii) 5a²-3b²
a = 4, b = 5
= 5(4)²-3(5)²
= 5(16)-3(25)
= 80-75 = 5
(iv) 4b²-3ab
a = 4, b= 5
= 4(5)²-3(4)(5)
= 4(25)-60
= 100-60 = 40
Que-6: If x = 4, y = 3 and z = -2, find the value of :
(i) x-2y+3z (ii) xy+yz+zx (iii) -2xyz
Sol: (i) x-2y+3z
x = 4, y = 3 and z = -2
= 4-2(3)+3(-2)
= 4-6-6 = -8
(ii) xy+yz+zx
x = 4, y = 3 and z = -2
= (4)(3)+(3)(-2)+(-2)(4)
= 12-6-8 = -2
(iii) -2xyz
x = 4, y = 3 and z = -2
= -2(4)(3)(-2)
= 48.
Que-7: If p = 3, q = -5 and r = 4, find the value of :
(i) p²-2qr (ii) q²-3pr (iii) r²-pq (iv) pq+qr+rp (v) 3pq + 10pr + 2qr (vi) p²+q²+r²-pq-qr-pr
Sol: (i) p²-2qr
p = 3, q = -5 and r = 4
= (3)²-2(-5)(4)
= 9+40 = 49
(ii) q²-3pr
p = 3, q = -5 and r = 4
= (-5)²-3(3)(4)
= 25-36 = -11
(iii) r²-pq
p = 3, q = -5 and r = 4
= (4)²-(3)(-5)
= 16+15 = 31
(iv) pq+qr+rp
p = 3, q = -5 and r = 4
= (3)(-5)+(-5)(4)+(4)(3)
= -15-20+12
= -23
(v) 3pq+10pr+2qr
p = 3, q = -5 and r = 4
= 3(3)(-5)+10(3)(4)+2(-5)(4)
= -45+120-40
= 35
(vi) p²+q²+r²-pq-qr-pr
p = 3, q = -5 and r = 4
= (3)²+(-5)²+(4)²-(3)(-5)-(-5)(4)-(3)(4)
= 9+25+16+15+20-12
= 73
Que-8: If a = 3 and b = -4, find the value of :
(i) (a+b)/(a-b) (ii) ab/(a²+b²) (iii) (a²+b²)/(b²-a²)
Sol: (i) (a+b)/(a-b)
a = 3, b = -4
= (3+(-4))/(3-(-4))
= -1/7
(ii) ab/(a²+b²)
a = 3, b = -4
= [(3)(-4)]/[(3)²+(-4)²]
= -12/(9+16)
= -12/25
(iii) (a²+b²)/(b²-a²)
a = 3, b = -4
= [(3)²+(-4)²]/[(-4)²-(3)²]
= (9+16)/(16-9)
= 25/7
Que-9: If a = 8, b = 2 and c = 3, find the value of [16bc-5ab]/[abc-3b²]
Sol: [16bc-5ab]/[abc-3b²]
a = 8, b = 2 and c = 3
= [16(2)(3)-5(8)(2)]/[(8)(2)(3)-3(2)²]
= [96-80]/[48-12]
= 16/36
= 4/9
Que-10: If x = 4, y = 3 and z = -2, find the value of [x²+y²-z²]/[xy+yz-zx]
Sol: [x²+y²-z²]/[xy+yz-zx]
x = 4, y = 3 and z = -2
= [(4)²+(3)²-(-2)²]/[(4)(3)+(3)(-2)-(-2)(4)]
= [16+9-4]/[12-6+8]
= 21/14
= 3/2
Que-11: State true or false : (i) x²+y² = z², when x = 5, y = 12 and z = 13 (ii) The value of 5x-3 is 2, when x = 0 (iii) The value of x^y is 8, when x = 3 and y = 2 (iv) When a = 3, then a² and 2a have equal values.
Sol: – (i) True
(ii) False
(iii) False
(iv) False
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