DAV Class 7 Maths Chapter 6 Worksheet 2 Solutions

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The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 6 Worksheet 2 Solutions of Algebraic Expressions offer comprehensive answers to textbook questions.

DAV Class 7 Maths Ch 6 WS 2 Solutions

Question 1. Find the product:
(i) (x2 + 3xy) (4x)
Answer:
(x2 + 3xy) (4x)
= x2 × 4x + 3xy × 4x
= 4x3 + 12 ×2y

(ii) 9 pqr (2 p2q – 3q2r)
Answer:
9pqr (2p2q – 3q2r)
= 9 pqr × 2 p2q – 9 pqr × 3q2r
= 18p3q2r – 27 pq3r2

(iii) \(\frac{3}{4}\)a2(\(\frac{2}{3}\)b2 + 8ab)
Answer:
\(\frac{3}{4}\)a2(\(\frac{2}{3}\)b2 + 8ab)
= \(\frac{3}{4}\)a2 × \(\frac{2}{3}\) b2 + \(\frac{3}{4}\) a2 × 8 ab
= \(\frac{1}{2}\)a2b2 + 6a3b

DAV Class 7 Maths Chapter 6 Worksheet 2 Solutions

Question 2.
Find the following products and then evaluate when x = 2, y = – 1.
(i) (7x + 9y2) (3xy2)
Solution:
(i)(7x + 9y2) (3xy2)
Put x = 2 and y = -1
= 7x × 3 xy2 + 9y2 × 3 xy2
= 21x2y2 + 27xy4
= 21 (2)2 (- 1)2 + 27(2) (- 1)4
= 21 × 4 + 27 × 2 × 1
= 84 + 54
= 138

(ii) -11x(2y5 – \(\frac{3}{11}\)x2y3)
Answer:
-11x(2y5 – \(\frac{3}{11}\)x2y3)
= -11x × 2y5 – 11x × –\(\frac{3}{11}\) x2y3
= -22xy5 + 3x3y3

Put x = 2, y = -1
= -22(2)(-1)5 + 3(2)3(-1)3
= -22 × 2 × -1 + 3 × 8 × – 1
= 44 – 24
= 20

(iii) 0.5x (2x2y2 + 1.5 xy3) = 0.5x × 2x2y2 + 0.5x × 1.5xy3
= 1.0 x3y2 + 0.75 x2y3

Put x = 2, y = – 1
= 1.0 (2)3 (-1)2 + 0.75(2)2 (-1)3
= 1.0 × 8 × 1-0.75 × 4 × -1
= 8 + 3.00 = 11

Question 3.
Find the product of 8s2(t2 – 2st) and verify the result when s = 1 and t = 5.
Answer:
8s2 (t2 – 2st)
= 8s2t2 – 8s2 x 2st
= 8s2t2 – 16s3t

Verification:
L.H.S. = 8s2 (t2 – 2st)
Put s = 1 and t = 5 = 8(1)2 [(5)2 – 2(1) (5)]
= 8 × 1 [25 -10]
= 8 × 15
= 120
R.H.S. = 8s2t2 -16 s3t
Put s = 1 and t = 5
= 8(1)2(5)2 – 16(1)3(5)
= 8 × 1 × 25 – 16 × 1 × 5
= 200 – 80
= 120
Hence Verified L.H.S. = R.H.S.

DAV Class 7 Maths Chapter 6 Worksheet 2 Solutions

Question 4.
Find the product of \(\frac{2}{7}\)x2 (7y + 14x) and verify the result when x = 2 and y = 3.
Answer:
\(\frac{2}{7}\) x2 (7y + 14x) = \(\frac{2}{7}\) x2 × 7y + \(\frac{2}{7}\) x2 × 14x
= 2x2y + 4x3

Verification:
L.H.S. = jx2(7y + 14x)
Put x = 2 and y = 3
= \(\frac{2}{7}\)(2)2 [7(3) + 14 (2)]
= \(\frac{2}{7}\) × 4 [21 + 48]
= \(\frac{2}{7}\) × 49 = 56
R.H.S. = 2 x2y + 4x3
= 2(2)2 (3) + 4 (2)3
= 2 × 4 × 3 + 4 × 8
= 24 + 32
= 56
Hence verified L.H.S. = R.H.S.

Question 5.
Find the product of 0.1xy (3x + 2y) and verify the result when x = 5 and y = – 1.
Answer:
0.2xy(3x + 2y) = 0.2 xy × 3x + 0.2 xy × 2y
= 0.6 x2y + 0.4 xy2

Verification:
L.H.S. = 0.2 xy (3x + 2y)
Put x = 5 and y = – 1
= 0.2(5)(- 1) [3(5) + 2(-1)] = -1.0 [15 – 2]
= -1 × 13
=-13
= 0.6x2y + 0.4 xy2
= 0.6(5)2(-1) + 0.4(5) (-1)2
= 0.6 × 25 × – 1 + 0.4 × 5 × 1
= – 15 + 2
= – 13
Hence Verified L.H.S. = R.H.S.

Question 6.
Simplify:
(i) ab (a2 – b2) + b3 (a- 2b)
Answer:
ab(a2 – b2) + b3 (a – 2b)
= ab × a2 – ab × b2 + b3 a – b3 × 2b
= a3b – ab3 + ab3 – 2b4
= a3b – 2b4

(ii) -6 x(xy + 2y2) – 3y2(2x2 + y)
Answer:
– 6x (xy + 2y2) – 3y2 (2x2 + y)
= – 6x × xy – 6x × 2y2 – 3y2 × 2x2 – 3y2 × y
= – 6x2y – 12xy2 – 6x2y2 – 3y3

(iii) 2x (x – y2) – 3y (xy + 2x) – xy (x + y)
Answer:
2x(x – y2) – 3y (xy + 2x) – xy (x + y)
= 2x × x – 2x × y2 – 3y × xy – 3y × 2x – xy × x – xy × y
= 2x2 – 2xy2 – 3xy2 – 6xy – x2y – xy2
= 2x2 – 6xy2 – x2y – 6xy