ML Aggarwal Algebraic Expression and Identities Check Your Progress Class 8 ICSE Ch-10 Maths Solutions. We Provide Step by Step Answer of Check Your Progress Questions for Algebraic Expression and Identities as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.
ML Aggarwal Algebraic Expression and Identities Check Your Progress Class 8 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 8th |
| Chapter-10 | Algebraic Expression and Identities |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Check Your Progress Questions |
| Edition | 2023-2024 |
Algebraic Expression and Identities Check Your Progress
ML Aggarwal Class 8 ICSE Maths Solutions
Page-189
Question 1. Add the following expressions:
(i) -5x2y + 3xy2 – 7xy + 8, 12x2y – 5xy2 + 3xy – 2
(ii) 9xy + 3yz – 5zx, 4yz + 9zx – 5y, -5xz + 2x – 5xy
Answer:
(i) (-5x2y + 3xy2 – 7xy + 8) + (12x2y – 5xy2 + 3xy – 2)
= 7x2y – 2xy2 – 4xy + 6
(ii) (9xy + 3yz – 5zx) + (4yz + 9zx – 5y, -5xz + 2x – 5xy)
= 4xy + 7yz – zx + 2x – 5y
Question 2. Subtract:
(i) 5a + 3b + 11c – 2 from 3a + 5b – 9c + 3
(ii) 10x2 – 8y2 + 5y – 3 from 8x2 – 5xy + 2y2 + 5x – 3y
Answer:
(i) 5a – 3b + 11c – 2 from 3a + 5b – 9c + 3
= (3a + 5b – 9c + 3) – (5a – 3b + 11c – 2)
= 3a + 5b – 9c + 3 – 5a + 3b – 11c + 2
= -2a + 8b – 20c + 5
(ii) 10x2 – 8y2 + 5y – 3 from 8x2 – 5xy + 2y2 + 5x – 3y
= (8x2 – 5xy + 2y2 + 5x – 3y) – (10x2 – 8y2 + 5y – 3)
= 8x2 – 5xy + 2y2 + 5x – 3y – 10x2 + 8y2 – 5y + 3
= – 2x2 – 5xy + 10y2 + 5x – 8y – 3
Question 3. What must be added to 5x2 – 3x + 1 to get 3x3 – 7x2 + 8?
Answer:
= (3x3 – 7x2 + 8) – (5x2 – 3x + 1)
= 3x3 – 7x2 + 8 – 5x2 + 3x – 1
= 3x3 – 12x2 + 3x + 7
Question 4. Find the product of
(i) 3x2y and -4xy2
(ii) –4⁄5xy, 5⁄7 yz and –14⁄9 zx
Answer:
(i) 3x2y and -4xy2
= 3x2 × (-4xy2)
= -12x2+1 y1+2
= 12x3y3
(ii) –(4/5)xy, (5/7)yz and –(14/9)zx
= –(4/5)xy × (5/7)yz × –(14/9)zx
= –(4/5) × (5/7) × –(14/9) x2y2z2
= (8/9)x2y2z2
Question 5. Multiply:
(i) (3pq – 4p2 + 5q2 + 7) by -7pq
(ii) (3⁄4x2y – 4⁄5xy + 5⁄6xy2) by – 15xyz
Answer:
(i) (3pq – 4p2 + 5q2 + 7) × (-7pq)
= -7pq × 3pq – 7pq × (-4p2) + (-7pq) (5q2) – 7pq × 7
= -21p2q2 + 28p3q – 35pq3 – 49pq
(ii) (3/4x2y – 4/5xy + 5/6xy2) × (– 15xyz)
Question 6. Multiply:
(i) (5x2 + 4x – 2) by (3 – x – 4x2)
(ii) (7x2 + 12xy – 9y2) by (3x2 – 5xy + 3y2)
Answer:
(i) (5x2 + 4x – 2) × (3 – x – 4x2)
= 5x2(3 – x – 4x2) + 4x(3 – x – 4x2) – 2(3x – x – 4x2)
= 15x2 – 5x3 – 20x4 + 12x – 4x2 – 16x3 – 6x + 2x + 8x2
= -20x4 – 21x3 + 19x2 + 14x – 6
(ii) (7x2 + 12xy – 9y2) x (3x2 – 5xy + 3y2)
= 7x2(3x2 – 5xy + 3y2) + 12xy(3x2 – 5xy + 3y2) – 9y2(3x2 – 5xy + 3y2)
= 21x4 – 35x3y + 21x2y2 + 36x3y – 60x2y2 + 36xy3 – 27x2y2 + 45xy3 – 27y4
= 21x4 + x3y + 81xy3 – 66x2y2 – 27y4
Question 7. Simplify the following expressions and evaluate them as directed:
(i) (3ab – 2a2 + 5b2) x (2b2 – 5ab + 3a2) + 8a3b – 7b4 for a = 1, b = -1
(ii) (1.7x – 2.5y) (2y + 3x + 4) – 7.8x2 – 10y for x = 0, y = 1.
Answer:
(i) (3ab – 2a2 + 5b2) × (2b2 – 5ab + 3a2) + 8a3b – 7b4
= 3ab(2b2 – 5ab + 3a2) – 2a2(2b2 – 5ab + 3a2) + 5b2(2b2 – 5ab + 3a2) + 8a3b – 7b4
= 6ab32 – 15a2b2 + 9a3b – 4a2b2 + 10a3b – 6a4 + 10b4 – 25ab3 + 15a2b2 + 8a3b – 7b4
= 27a3b – 4a2b2 – 19ab3 – 6a4 + 3b4
Putting, a = 1 and b = (-1)
= 27(1 )3 (-1) – 4(1)2 (-1)2 – 19 (1) (-1)3 – 6(1)4 + 3(-1)4
= -27 – 4 + 19 – 6 + 3
= -37 + 22
= -15
(ii) (1.7x – 2.5y) (2y + 3x + 4) – 7.8x2 – 10y
1.7x(2y + 3x + 4) – 2.5y(2y + 3x + 4) – 7.8x2 – 10y
= 3.4xy + 5.1x2 + 6.8x – 5y2 – 7.5xy – 10y – 7.8x2 – 10y
= -2.7x2 – 4.1xy – 5y2 + 6.8x – 20y
Putting, x = 0 and y = 1
= -2.7 × 0 – 4.1 × 0 × 1 – 5(1)2 + 6.8 × 0 – 20 × 1
= 0 + 0 – 5 + 0 – 20
= -25
(ML Aggarwal Algebraic Expression and Identities Check Your Progress Class 8)
Question 8. Carry out the following divisions:
(i) 66pq2r3 ÷ 11qr2
(ii) (x3 + 2x2 + 3x) ÷ 2x
Answer:
(i) 66pq2r3/ 11qr2
= 6pq2-1r3-2
= 6pqr
(ii) (x3 + 2x2 + 3x)/ 2x
= x3/2x + 2x2/2x + 3x/2x
= ½ x2 + x + 3/2
Question 9. Divide 10x4 – 19x3 + 17x2 + 15x – 42 by 2x2 – 3x + 5.
Answer:
(10x4 – 19x3 + 17x2 + 15x – 42) ÷ (2x2 – 3x + 5)
Thus, Quotient = 5x2 – 2x – 7 and Remainder = 4x – 7
Algebraic Expression and Identities Check Your Progress
ML Aggarwal Class 8 ICSE Maths Solutions
Page-190
Question 10. Using identities, find the following products:
(i) (3x + 4y) (3x + 4y)
(ii) (5⁄2𝑎−𝑏)(5⁄2𝑎−𝑏)
(iii) (3.5m – 1.5n) (3.5m + 1.5n)
(iv) (7xy – 2)(7xy + 7)
Answer:
(i) (3x + 4y) (3x + 4y)
= (3x + 4y)2
= (3x)2 + 2 × 3x × 4y + (4y)2 [Using, (a + b)2 = a2 + 2ab + b2]
= 9x2 + 24xy + 16y2
(ii) (5a/2 – b) (5a/2 – b)
= (5a/2 – b)2
= (5a/2)2 + 2 × 5a/2 × (-b) + (b)2 [Using, (a – b)2 = a2 – 2ab + b2]
= 25a2/4 – 5ab + b2
(iii) (3.5m – 1.5n) (3.5m + 1.5n)
= (3.5m)2 – (1.5n)2 [Using, (a – b)(a + b) = a2 – b2]
= 12.25m2 – 2.25n2
(iv) (7xy – 2)(7xy + 7)
= (7xy)2 + (-2 + 7) × (7xy) + (-2) × 7 [Using, (x + a)(x + b) = x2 + (a + b)x + ab]
= 49x2y2 + 35xy – 14
Question 11. Using suitable identities, evaluate the following:
(i) 1052
(ii) 972
(iii) 201 × 199
(iv) 872 – 132
(v) 105 × 107
Answer:
(i) (105)2 = (100 + 5)2
= (100)2 + 2 × 100 × 5 + (5)2 [Using, (a + b)2 = a2 + 2ab + b2]
= 10000 + 1000 + 25
= 11025
(ii) (97)2 = (100 – 3)2
= (100)2 – 2 × 100 × 3 + (3)2 [Using, (a – b)2 = a2 – 2ab + b2]
= 10000 – 600 + 9
= 10009 – 600
= 9409
(iii) 201 × 199 = (200 + 1) (200 – 1)
= (200)2 – (1)2 [Using, (a + b) (a – b) = a2 – b2]
= 40000 – 1
= 39999
(iv) 872 – 132
= (87 + 13) (87- 13) [Using, a2 – b2 = (a + b)(a – b)]
= 100 × 74
= 7400
(v) 105 × 107
= (100 + 5) (100 + 7)
= (100)2 + (5 + 7) × 100 + 5 × 7 [Using, (x + a)(x – b) = x2 + (a + b)x + ab]
= 10000 + 1200 + 35
= 11235
Question 12. Prove that following:
(i) (a + b)2 – (a – b)2 + 4ab
(ii) (2a + 3b)2 + (2a – 3b)2 = 8a2 + 18b2
Answer:
(i) Taking the RHS, we have
RHS = (a – b)2 + 4ab
= a2– 2ab + b2 + 4ab
= a2 + 2ab + b2
= (a + b)2 = L.H.S.
(ii) Taking the LHS, we have
LHS = (2a + 3b)2 + (1a – 3b)2
= (2a)2 + 2 × 2a × 3b + (3b)2 + (2a)2 – 2 × 2a × 3b + (3b)2
= 4a2 + 12ab + 9b2 + 4a2 – 12ab + 9b2
= 8a2 + 18b2 = RHS
Question 13. If x + 1/𝑥 = 5, evaluate
(i) x2 + 1/x2
(ii) x4 + 1/x4
Answer:
(i) We have, x + 1/x = 5
On squaring on both sides, we get
(x + 1/x)2 = 52
x2 + 1/x2 + 2 × x × 1/x = 25
x2 + 2 + 1/x2 = 25
x2 + 1/x2 = 25 – 2
x2 + 1/x2 = 23
(ii) Again, squaring x2 + 1/x2 = 23 on both sides, we get
(x2 + 1/x2)2 = 232
x4 + 1/x4 + 2 × x4 × 1/x4 = 529
x4 + 1/x4 + 2 = 529
x4+ 1/x4 = 529 – 2
x4 + 1/x4 = 527
Question 14. If a + b = 5 and a2 + b2 = 13, find ab.
Answer:
a + b = 5 and a2 + b2 = 13
On squaring a + b = 5 both sides, we get
(a + b)2 = (5)2
a2 + b2 + 2ab = 25
13 + 2ab = 25 ⇒ 2ab = 25 – 13 = 12
⇒ ab = 12/2 = 6
∴ ab = 6
— End of Algebraic Expression and Identities Check Your Progress Class 8 ICSE Maths Solutions :–
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