RD Sharma Solutions for Class 8 Maths Exercise 6.7 of Chapter 6 Algebraic Expressions and Identities are provided in PDF here. From the exam point of view, BYJU’S experts have prepared these solutions, which will help students gain mathematical skills to excel in their final exams. By practising RD Sharma Solutions regularly, students can score high marks in their exams. The PDFs can be downloaded easily from the links given below. In Exercise 6.7 of Chapter 6, Algebraic Expressions and Identities, we will study problems based on a special product.
RD Sharma Solutions for Class 8 Maths Exercise 6.7 Chapter 6 Algebraic Expressions and Identities
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EXERCISE 6.7 PAGE NO: 6.47
1. Find the following products:
(i) (x + 4) (x + 7)
(ii) (x – 11) (x + 4)
(iii) (x + 7) (x – 5)
(iv) (x – 3) (x – 2)
(v) (y2 – 4) (y2 – 3)
(vi) (x + 4/3) (x + 3/4)
(vii) (3x + 5) (3x + 11)
(viii) (2x2 – 3) (2x2 + 5)
(ix) (z2 + 2) (z2 – 3)
(x) (3x – 4y) (2x – 4y)
(xi) (3x2 – 4xy) (3x2 – 3xy)
(xii) (x + 1/5) (x + 5)
(xiii) (z + 3/4) (z + 4/3)
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
(xvi) (y2 + 5/7) (y2 – 14/5)
(xvii) (p2 + 16) (p2 – 1/4)
Solution:
(i) (x + 4) (x + 7)
Let us simplify the given expression
x (x + 7) + 4 (x + 7)
x2 + 7x + 4x + 28
x2 + 11x + 28
(ii) (x – 11) (x + 4)
Let us simplify the given expression
x (x + 4) – 11 (x + 4)
x2 + 4x – 11x – 44
x2 – 7x – 44
(iii) (x + 7) (x – 5)
Let us simplify the given expression
x (x – 5) + 7 (x – 5)
x2 – 5x + 7x – 35
x2 + 2x – 35
(iv) (x – 3) (x – 2)
Let us simplify the given expression
x (x – 2) – 3 (x – 2)
x2 – 2x – 3x + 6
x2 – 5x + 6
(v) (y2 – 4) (y2 – 3)
Let us simplify the given expression
y2 (y2 – 3) – 4 (y2 – 3)
y4 – 3y2 – 4y2 + 12
y4 – 7y2 + 12
(vi) (x + 4/3) (x + 3/4)
Let us simplify the given expression
x (x + 3/4) + 4/3 (x + 3/4)
x2 + 3x/4 + 4x/3 + 12/12
x2 + 3x/4 + 4x/3 + 1
x2 + 25x/12 + 1
(vii) (3x + 5) (3x + 11)
Let us simplify the given expression
3x (3x + 11) + 5 (3x + 11)
9x2 + 33x + 15x + 55
9x2 + 48x + 55
(viii) (2x2 – 3) (2x2 + 5)
Let us simplify the given expression
2x2 (2x2 + 5) – 3 (2x2 + 5)
4x4 + 10x2 – 6x2 – 15
4x4 + 4x2 – 15
(ix) (z2 + 2) (z2 – 3)
Let us simplify the given expression
z2 (z2 – 3) + 2 (z2 – 3)
z4 – 3z2 + 2z2 – 6
z4 – z2 – 6
(x) (3x – 4y) (2x – 4y)
Let us simplify the given expression
3x (2x – 4y) – 4y (2x – 4y)
6x2 – 12xy – 8xy + 16y2
6x2 – 20xy + 16y2
(xi) (3x2 – 4xy) (3x2 – 3xy)
Let us simplify the given expression
3x2 (3x2 – 3xy) – 4xy (3x2 – 3xy)
9x4 – 9x3y – 12x3y + 12x2y2
9x4 – 21x3y + 12x2y2
(xii) (x + 1/5) (x + 5)
Let us simplify the given expression
x (x + 1/5) + 5 (x + 1/5)
x2 + x/5 + 5x + 1
x2 + 26/5x + 1
(xiii) (z + 3/4) (z + 4/3)
Let us simplify the given expression
z (z + 4/3) + 3/4 (z + 4/3)
z2 + 4/3z + 3/4z + 12/12
z2 + 4/3z + 3/4z + 1
z2 + 25/12z + 1
(xiv) (x2 + 4) (x2 + 9)
Let us simplify the given expression
x2 (x2 + 9) + 4 (x2 + 9)
x4 + 9x2 + 4x2 + 36
x4 + 13x2 + 36
(xv) (y2 + 12) (y2 + 6)
Let us simplify the given expression
y2 (y2 + 6) + 12 (y2 + 6)
y4 + 6y2 + 12y2 + 72
y4 + 18y2 + 72
(xvi) (y2 + 5/7) (y2 – 14/5)
Let us simplify the given expression
y2 (y2 – 14/5) + 5/7 (y2 – 14/5)
y4 – 14/5y2 + 5/7y2 – 2
y4 – 73/35y2 – 2
(xvii) (p2 + 16) (p2 – 1/4)
Let us simplify the given expression
p2 (p2 – 1/4) + 16 (p2 – 1/4)
p4 – 1/4p2 + 16p2 – 4
p4 + 63/4p2 – 4
2. Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006
Solution:
(i) 102 × 106
We can express 102 as 100 + 2 and 106 as 100 + 6
Now, let us simplify
102 × 106 = (100 + 2) (100 + 6)
= 100 (100 + 6) + 2 (100 + 6)
= 10000 + 600 + 200 + 12
= 10812
(ii) 109 × 107
We can express 109 as 100 + 9 and 107 as 100 + 7
Now, let us simplify
109 × 107 = (100 + 9) (100 + 7)
= 100 (100 + 7) + 9 (100 + 7)
= 10000 + 700 + 900 + 63
= 11663
(iii) 35 × 37
We can express 35 as 30 + 5 and 37 as 30 + 7
Now, let us simplify
35 × 37 = (30 + 5) (30 + 7)
= 30 (30 + 7) + 5 (30 + 7)
= 900 + 210 + 150 + 35
= 1295
(iv) 53 × 55
We can express 53 as 50 + 3 and 55 as 50 + 5
Now, let us simplify
53 × 55 = (50 + 3) (50 + 5)
= 50 (50 + 5) + 3 (50 + 5)
= 2500 + 250 + 150 + 15
= 2915
(v) 103 × 96
We can express 103 as 100 + 3 and 96 as 100 – 4
Now, let us simplify
103 × 96 = (100 + 3) (100 – 4)
= 100 (100 – 4) + 3 (100 – 4)
= 10000 – 400 + 300 – 12
= 10000 – 112
= 9888
(vi) 34 × 36
We can express 34 as 30 + 4 and 36 as 30 + 6
Now, let us simplify
34 × 36 = (30 + 4) (30 + 6)
= 30 (30 + 6) + 4 (30 + 6)
= 900 + 180 + 120 + 24
= 1224
(vii) 994 × 1006
We can express 994 as 1000 – 6 and 1006 as 1000 + 6
Now, let us simplify
994 × 1006 = (1000 – 6) (1000 + 6)
= 1000 (1000 + 6) – 6 (1000 + 6)
= 1000000 + 6000 – 6000 – 36
= 999964
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