# DAV Class 7 Maths Chapter 6 Worksheet 1 Solutions

The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 6 Worksheet 1 Solutions of Algebraic Expressions offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 6 WS 1 Solutions

Question 1.
Find the product:
(i) 9x3 × 2x4
9x3 × 2y4
= 9 × 2 × x3 × y4
= 18x3y4

(ii) – 6a2 × 5a7
– 6 a2 × 5a7
= -6 × 5 × a2 × a7
= – 30 a9

(iii) – 8y9 × – 4y3
– 8y9 × – 4y3
= – 8 × – 4 × y9 × y3
= 32y12

Question 2.
Multiply the monomials:
(i) 7pq and $$\frac{4}{3}$$p2q3
7pq and $$\frac{4}{3}$$p2q3
= 7pq × $$\frac{4}{3}$$p2q3
= 7 × $$\frac{4}{3}$$ × pq × p3q3
= $$\frac{28}{3}$$p3q4

(ii) 12a2b6c8 and -3a7b4c3
12a2b6c8 and -3a7b4c3
= 12 × ( -3) × a2b6c8 × a7b4c3
= -36 × a9b10c11 = -36a9b10c11

(iii) $$\frac{2}{5}$$x2y and $$\frac{5}{3}$$x3y2z2
$$\frac{2}{5}$$x2y and $$\frac{5}{3}$$x3y2z2
= $$\frac{2}{5}$$ × $$\frac{5}{3}$$x2y × x3y2z2
= $$\frac{2}{3}$$ x5y3z2

Question 3.
Multiply the monomials:
(i) 3x7, 4x2 and – 5x3
3x7 × 4x2 × (- 5x3)
= 3 × 4 × -5 × x7 × x2 × x3
= – 60x12

(ii) 1.2 a2b2, 5ab4c2 and 1.1a5bc7
1.2 a2b2 × 5ab4c2 × 1.1a5bc7
= 1.2 × 5 × 1.1 × a2p2 × ab4c2 × a5bc7
= 6.60 × a2 × a × a5 × b2 × b4 × b × c2 × c7
= 6.6 a8b7c9

(iii) $$\frac{3}{4}$$pq, $$\frac{1}{2}$$qr2,-5p2r3 and-6r5
$$\frac{3}{4}$$pq × $$\frac{1}{2}$$qr2 × (-5p2r3) × -6r5
= $$\frac{3}{4}$$ × $$\frac{1}{2}$$ × (- 5) × (- 6) × pq × qr2 × p2r3 × r5
=$$\frac{45}{4}$$ p × p2 × q × q × r2 × r3 × r5
= $$\frac{45}{4}$$ p3q2r10

Question 4.
Find the product of ($$\frac{1}{2}$$x3) (- 10 x) ($$\frac{1}{5}$$ x2) and verify the result for x – 1.
($$\frac{1}{2}$$x3) (- 10 x) ($$\frac{1}{5}$$ x2)
= $$\frac{1}{2}$$ × (-10) × $$\frac{1}{5}$$x3 × x × x2
= (-1) × x6
= -x6

Verification:
L.H.S = ($$\frac{1}{2}$$x3) (- 10 x) ($$\frac{1}{5}$$ x2)
Put x = 1, = [$$\frac{1}{2}$$ × (1)3] [-10 × 1] [$$\frac{1}{5}$$ × (1)2]
= $$\frac{1}{2}$$ × (-10) × $$\frac{1}{5}$$
= -1
R.H.S = – x6 = – (1)6 = – 1
Hence verified L.H.S. = R.H.S.

Question 5.
Find the product of (- 3xyz) ($$\frac{4}{9}$$x2z)($$\frac{-27}{2}$$xy2z) and verify the result for x = y = 3 and z = -1.
Solution:
(- 3 xyz) ($$\frac{4}{9}$$yx2z) ($$\frac{-27}{2}$$xy2z)
= -3 × $$\frac{4}{9}$$ × x × x2 × x × y × y2 × z

Verification:
L.H.S. = (- 3 xyz) ($$\frac{4}{9}$$yx2 z) ($$\frac{-27}{2}$$xy2z)
Put x = 2, y = 3 and z = – 1
= ($$\frac{4}{9}$$ × 2 × 3 × -1) ($$\frac{-27}{2}$$× 22 ×- -1)
= (18) ($$\frac{-16}{9}$$)(243)
= (18) (- 16) (27)
= – 7776
= 18x4y3z3
= 18 (2)4 (3)3 (- 1)3
= 18 × 16 × 27 × (- 1)
= – 7776
Hence Verified L.H.S. = R.H.S.

Question 6.
Find the product of (a2bc2) (9ab2c) (- 4ab2c4) and verify the result for a = b = -1, c = 1.
(a2bc2) (9 ab2c2) (- 4 ab2c4)
= 1 × 9 × -4 × a2bc2 × ab2c2 × ab2c4c4
= – 36 × a2 × a × a × b × b2 × b2 × c2 × c2 × c4
= – 36a4b5c8

Verification:
L.H.S. = (a2bc2) (9ab2c2) (-4ab2c4)
Put a = $$\frac{1}{2}$$, b = -1, c = 1
= [($$\frac{1}{2}$$)2 (-1)(1)2] [9($$\frac{1}{2}$$) (-1)2 (1)2] [-4($$\frac{1}{2}$$) (-1)2 (1)4]
= –$$\frac{1}{4} \times \frac{9}{2}$$ × (-2) = $$\frac{9}{4}$$
R.H.S = -36a4 b5 c8
Put a = , b = -1, c = 1
= -36($$\frac{1}{2}$$)4 (-1)5 (1)8
= -36 × $$\frac{1}{16}$$ × (-1) × 1 = $$\frac{9}{4}$$
Hence Verified L.H.S = R.H.S

Question 7.
Find the area of a rectangle whose sides are 2a and 3a.
The sides of the rectangle are 2a and 3a
Area of the rectangle = 2a× 3a
= 6a2 square units

Question 8.
Find the area of a rectangle whose length is thrice its breadth whereas breadth is 4x.
Length = 3 × 4x = 12x
Area of the rectangle = length × breadth
= 12x × 4x
= 12 × 4 × x × x
= 48 x2 square units

Question 9.
Find the area of a rectangle whose breadth is b and length is square of breadth.
Length = b2
Area of rectangle = length × breadth
= b2 × b
= b3 square units

### DAV Class 8 Maths Chapter 6 Worksheet 1 Notes

• Fundamental operations are (+, -, × and ÷).
• Expressions are made of terms. Terms are added to make an expression. For example, the addition of the terms 5xy and 8 gives the expression 5xy + 8.
• A term is a product of factors. The term 5xy in the expression 5xy + 8 is a product of factors x, y and 5. Factors containing variables are said to be algebraic factors.
• The coefficient is the numerical factor in the term.
• Terms which have the same algebraic factors are like terms. For example, 7xy and – 2 xy are like terms.
• Terms which have different algebraic factors are unlike terms. For example, 7xy and – 2y are not like terms.
• Combination of constants and variables separated by the four operations is called algebraic expression.
• Expression having only one term is called Monomial e.g., 2x2y, 3, 30zx2 etc.
• Expression having two terms is called Binomial. e.g. 2x + y, 3 + x2, 4x3 + 3y3 etc.
• Expression having three terms is called Trinomial. e.g. 3x2 + x + 5, 2x2 + xy + 3y2, 5 + 6x + 7x2 etc.
• Expression having many terms is called polynomial. e.g. x5 – 3x4 + x2 + 3x + 5, 3 + 4x6 + 3x5 + 8x – 7x2 etc.

• The sum (or difference) of two like terms is a like term with coefficient equal to the sum (or
difference) of the coefficients of the two like terms.
For example, 9xy – 4xy (9 – 4) xy i.e.; 5xy.
• When we add two algebraic expressions, the like terms are added as given above; the unlike
terms are left as they are. Thus, the sum of 5x2 + 6x and 4x + 7 is 5x2 + 10x + 7; the like terms 6x and 4x add to 10x; the unlike terms 5x2 and 7 are left as they are.
• In subtraction of algebraic expression from another, change the signs of all the terms of the expression which is to be subtracted and then the two expressions are added.
• Product of variables follows the rules of exponents:
xa xb = xa + b
• Product of monomials is the product of constants of the given monomials and their variables.
• To multiply a monomial and a binomial, the monomial is multiplied with each term of the binomial and the products are then added.
• To multiply two binomials, each term of one binomial is multiplied with each term of the second binomials and the products are then added.
• To multiply a binomial and a trinomial, each term of the binomial is multiplied with each term of the trinomial and the products are then added.
• Factorisation is the process of writing a given algebraic expression as a product of two or more algebraic factors.
• Factors which are common to both monomials are common factors.
• The product of HCF of numerical coefficients and the highest common powers of the variables gives the HCF of the monomials.
• Use the distribution property of multiplication over addition, to express the given algebraic expression as product of HCF and the quotient of the given algebraic expression divided by HCF.
• To factorize an algebraic expression containing a binomial as a common factor, we write the expression as a product of the binomial and quotient obtained by dividing the given expression by its binomial.
• Factorisation by regrouping the terms can be done in such a way that a factor can be taken out from each group and the expression can be factorised.

Example 1:
Add 5x2 – 6y2 + 3xy – 5 and x2 + y2 + 7
Solution:

Example 2:
Subtract (- 3m2 + 5mn – 6n2) from (2m2 – mn + 3n2)
Solution:

Example 3:
Find the product (3p2q) × ($$\frac{2}{3}$$ pq3) × (2pq)
Solution:
(3p2q) × ($$\frac{2}{3}$$ pq3) × (2pq)
= 3 × $$\frac{- 2}{3}$$ × 2 × p2 × p × p × q × q3 × q
= – 4p4q5.

Example 4:
Find the product 2x2y (3xy2 – 61x2y) and evaluate for x = 1 and y = – 1.
Solution:
2x2y (3xy2 – 61x2y) = 2x2y × 3xy2 – 2x2y × 61x2y
= 2 × 3 × x2y × xy2 – 2 61 x2y x2y
= 6x3y3 – 122x4y2
Put x = 1 and y = – 1
= 6(1)3(- 1)3 – 122 (1)4 (- 1)2
= 6 × 1 – 1 – 122 × 1 × 1
= – 6 – 122
= – 128.

Example 5:
Simplify: 2x (5x2 – 3y2) – 5y (2x2 + y2)
Solution:
2x (5x2 – 3y2) – 5y (2x2 + y2)
= 2x × 5x2 – 2x × 3y2 – 5y × 2x2 – 5y × y2
= 2 × 5 × x × x2 – 2× 3 × x × y2 – 5 × 2 × y × x2 – 5 × 1 × y × y2
= 10x3 – 6xy2 – 10x2y – 5y3

Example 6:
Find the product $$\frac{1}{3}$$ ab2(6b2 – 7ab) and verify the result for a = 1 and b = – 1.
Solution:
$$\frac{1}{3}$$ ab2 (6b2 – 7ah) = $$\frac{1}{3}$$ ab2 × 6b2 – $$\frac{1}{3}$$ ab2 × 7ab
= $$\frac{1}{3}$$ × 6ab2 × b2 – $$\frac{1}{3}$$ × 7 × ab2 × ab
= 2ab4 – $$\frac{7}{3}$$ a2b3

Verification:
L.H.S. = $$\frac{1}{3}$$ ab2 (6b2 – 7ab)
Put a = 1 and b = – 1
= $$\frac{1}{3}$$ × 1 × (- 1)2 [6 × (- 1)2 – 7 × 1 × (- 1)]
= $$\frac{1}{3}$$ × 1 × 1 [6 × 1 + 7]
= $$\frac{1}{3}$$ [6 + 7]
= $$\frac{1}{3}$$ × 13
= $$\frac{13}{3}$$

R.H.S. = 2ab4 – a2b3
Put a = 1 and b = – 1
= 2 × 1 × (- 1)4 – $$\frac{7}{3}$$ × (1)2 × (- 1)3
= 2 × 1 × 1 – $$\frac{7}{3}$$ × 1 × (- 1)
= 2 + $$\frac{7}{3}$$
= $$\frac{13}{3}$$
Hence verified LH.S. = R.H.S.

Example 7:
Find the product (3x2y – 2xy2) (2xy2 + 3x2y) and verify the result for x = 2 and y = 3.
Solution:
(3x2y – 2xy2) (y2 + 3x2y)
= 3x2y (2xy2 + 3x2y) – 2xy2 (2xy2 + 3x2y)
= 3x2y × 2xy2 + 3x2y × 3x2y – 2xy2 × 2xy2 – 2xy2 × 3x2y
= 6x3y3 + 9x4y2 – 4x2y4 – 6x3y3
= 9x4y2 – 4x2y4

Verification:
LH.S. = (3x2y – 2xy2) (2xy2 + 3x2y)
Put x = 2 and y = 3
= [3 × (2)2 × 3 – 2 × 2 × (3)2] [2 × 2 × (3)2 + 3 × (2) × 3]
= [3 × 4 × 3 – 2× 2 × 9] [2 × 2 × 9 + 3 × 4 × 3]
= [36 – 36] [36 + 36]
= 0 × 72 = 0

R.H.S. = 9x4y2 – 4x2y4
Put x = 2, y = 3
= 9 × (2)4 (3)2 – 4 × (2)2 (3)4
= 9 × 16 × 9 – 4 × 4 × 81
= 1296 – 1296 = 0
Hence verified LH.S. = R.H.S.

Example 8:
Find the product (3x + 2y) (5x2 – 6xy + 4y2) and verify for x = 2 and y = 1.
Solution:
(3x + 2y) (5x2 – 6xy + 4y2) = 3x (5x2 – 6xy + 4y2) + 2y (5x2 – 6xy + 4y2)
= 3x × 5x2 – 3x × 6xy + 3x × 4y2 + 2y × 5x2 – 2y × 6xy + 2y × 4y2
= 15x3 – 18x2y + 12xy2 + 10x2y – 12xy2 + 8y3
= 15x3 – 8x2y + 8y3

Verification:
L.H.S. (3x + 2y) (5x2 – 6xy + 4y2)
Put x = 2 and y = 1
= (3 × 2 + 2 × 1) [5(2)2 – 6 × 2 × 1 + 4(1)2]
= (6 + 2) [5 × 4 – 12 + 4 × 1]
= 8[20 – 12 + 4]
= 8 × 12 = 96
R.H.S. = 15x3 – 8x2y + 8y3
Put x = 2, and y = 1
= 15 (2)3 – 8 (2)2 (1) + 8 (1)3
= 15 × 8 – 8 × 4 × 1 + 8 × 1
= 120 – 32 + 8 = 96
Hence verified LH.S. = R.H.S.

Example 9:
Find the H.C.F of 25x3y2 and 30x2y3.
Solution:
25x3y2 and 30x2y3
H.C.F. of 25 and 30 is 5
H.C.F. of x3y2 and x2y3 = x2y2
So, H.CF.= 5x2y2