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) 9x^{3} × 2x^{4}

Answer:

9x^{3} × 2y^{4}

= 9 × 2 × x^{3} × y^{4}

= 18x^{3}y^{4}

(ii) – 6a^{2} × 5a^{7}

Answer:

– 6 a^{2} × 5a^{7}

= -6 × 5 × a^{2} × a^{7}

= – 30 a^{9}

(iii) – 8y^{9} × – 4y^{3}

Answer:

– 8y^{9} × – 4y^{3}

= – 8 × – 4 × y^{9} × y^{3}

= 32y^{12}

Question 2.

Multiply the monomials:

(i) 7pq and \(\frac{4}{3}\)p^{2}q^{3}

Answer:

7pq and \(\frac{4}{3}\)p^{2}q^{3}

= 7pq × \(\frac{4}{3}\)p^{2}q^{3}

= 7 × \(\frac{4}{3}\) × pq × p^{3}q^{3}

= \(\frac{28}{3}\)p^{3}q^{4}

(ii) 12a^{2}b^{6}c^{8} and -3a^{7}b^{4}c^{3}

Answer:

12a^{2}b^{6}c^{8} and -3a^{7}b^{4}c^{3}

= 12 × ( -3) × a^{2}b^{6}c^{8} × a^{7}b^{4}c^{3}

= -36 × a^{9}b^{10}c^{11} = -36a^{9}b^{10}c^{11}

(iii) \(\frac{2}{5}\)x^{2}y and \(\frac{5}{3}\)x^{3}y^{2}z^{2}

Answer:

\(\frac{2}{5}\)x^{2}y and \(\frac{5}{3}\)x^{3}y^{2}z^{2}

= \(\frac{2}{5}\) × \(\frac{5}{3}\)x^{2}y × x^{3}y^{2}z^{2}

= \(\frac{2}{3}\) x^{5}y^{3}z^{2}

Question 3.

Multiply the monomials:

(i) 3x^{7}, 4x^{2} and – 5x^{3}

Answer:

3x^{7} × 4x^{2} × (- 5x^{3})

= 3 × 4 × -5 × x^{7} × x^{2} × x^{3}

= – 60x^{12}

(ii) 1.2 a^{2}b^{2}, 5ab^{4}c^{2} and 1.1a^{5}bc^{7}

Answer:

1.2 a^{2}b^{2} × 5ab^{4}c^{2} × 1.1a^{5}bc^{7}

= 1.2 × 5 × 1.1 × a^{2}p^{2} × ab^{4}c^{2} × a^{5}bc^{7}

= 6.60 × a^{2} × a × a^{5} × b^{2} × b^{4} × b × c^{2} × c^{7}

= 6.6 a^{8}b^{7}c^{9}

(iii) \(\frac{3}{4}\)pq, \(\frac{1}{2}\)qr^{2},-5p^{2}r^{3} and-6r^{5}

Answer:

\(\frac{3}{4}\)pq × \(\frac{1}{2}\)qr^{2} × (-5p^{2}r^{3}) × -6r^{5}

= \(\frac{3}{4}\) × \(\frac{1}{2}\) × (- 5) × (- 6) × pq × qr^{2} × p^{2}r^{3} × r^{5}

=\(\frac{45}{4}\) p × p^{2} × q × q × r^{2} × r^{3} × r^{5}

= \(\frac{45}{4}\) p^{3}q^{2}r^{10}

Question 4.

Find the product of (\(\frac{1}{2}\)x^{3}) (- 10 x) (\(\frac{1}{5}\) x^{2}) and verify the result for x – 1.

Answer:

(\(\frac{1}{2}\)x^{3}) (- 10 x) (\(\frac{1}{5}\) x^{2})

= \(\frac{1}{2}\) × (-10) × \(\frac{1}{5}\)x^{3} × x × x^{2}

= (-1) × x^{6}

= -x^{6}

Verification:

L.H.S = (\(\frac{1}{2}\)x^{3}) (- 10 x) (\(\frac{1}{5}\) x^{2})

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 = – x^{6} = – (1)^{6} = – 1

Hence verified L.H.S. = R.H.S.

Question 5.

Find the product of (- 3xyz) (\(\frac{4}{9}\)x^{2}z)(\(\frac{-27}{2}\)xy^{2}z) and verify the result for x = y = 3 and z = -1.

Solution:

(- 3 xyz) (\(\frac{4}{9}\)yx^{2}z) (\(\frac{-27}{2}\)xy^{2}z)

= -3 × \(\frac{4}{9}\) × x × x^{2} × x × y × y^{2} × z

Verification:

L.H.S. = (- 3 xyz) (\(\frac{4}{9}\)yx^{2} z) (\(\frac{-27}{2}\)xy^{2}z)

Put x = 2, y = 3 and z = – 1

= (\(\frac{4}{9}\) × 2 × 3 × -1) (\(\frac{-27}{2}\)× 2^{2} ×- -1)

= (18) (\(\frac{-16}{9}\))(243)

= (18) (- 16) (27)

= – 7776

= 18x^{4}y^{3}z^{3}

= 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 (a^{2}bc^{2}) (9ab^{2}c) (- 4ab^{2}c^{4}) and verify the result for a = b = -1, c = 1.

Answer:

(a^{2}bc^{2}) (9 ab^{2}c^{2}) (- 4 ab^{2}c^{4})

= 1 × 9 × -4 × a^{2}bc^{2} × ab^{2}c^{2} × ab^{2}c^{4}c^{4}

= – 36 × a^{2} × a × a × b × b^{2} × b^{2} × c^{2} × c^{2} × c^{4}

= – 36a^{4}b^{5}c^{8}

Verification:

L.H.S. = (a^{2}bc^{2}) (9ab^{2}c^{2}) (-4ab^{2}c^{4})

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 = -36a^{4} b^{5} c^{8}

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.

Answer:

The sides of the rectangle are 2a and 3a

Area of the rectangle = 2a× 3a

= 6a^{2} square units

Question 8.

Find the area of a rectangle whose length is thrice its breadth whereas breadth is 4x.

Answer:

Here breadth = 4x

Length = 3 × 4x = 12x

Area of the rectangle = length × breadth

= 12x × 4x

= 12 × 4 × x × x

= 48 x^{2} square units

Question 9.

Find the area of a rectangle whose breadth is b and length is square of breadth.

Answer:

Here breadth = b

Length = b^{2}

Area of rectangle = length × breadth

= b^{2} × b

= b^{3} 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., 2x
^{2}y, 3, 30zx^{2}etc. - Expression having two terms is called Binomial. e.g. 2x + y, 3 + x
^{2}, 4x^{3}+ 3y^{3}etc. - Expression having three terms is called Trinomial. e.g. 3x
^{2}+ x + 5, 2x^{2}+ xy + 3y^{2}, 5 + 6x + 7x^{2}etc. - Expression having many terms is called polynomial. e.g. x
^{5}– 3x^{4}+ x^{2}+ 3x + 5, 3 + 4x^{6}+ 3x^{5}+ 8x – 7x^{2}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 5x^{2}+ 6x and 4x + 7 is 5x^{2}+ 10x + 7; the like terms 6x and 4x add to 10x; the unlike terms 5x^{2}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:

x^{a}x^{b}= x^{a + 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 5x^{2} – 6y^{2} + 3xy – 5 and x^{2} + y^{2} + 7

Solution:

Example 2:

Subtract (- 3m^{2} + 5mn – 6n^{2}) from (2m^{2} – mn + 3n^{2})

Solution:

Example 3:

Find the product (3p^{2}q) × (\(\frac{2}{3}\) pq^{3}) × (2pq)

Solution:

(3p^{2}q) × (\(\frac{2}{3}\) pq^{3}) × (2pq)

= 3 × \(\frac{- 2}{3}\) × 2 × p^{2} × p × p × q × q^{3} × q

= – 4p^{4}q^{5}.

Example 4:

Find the product 2x^{2}y (3xy^{2} – 61x^{2}y) and evaluate for x = 1 and y = – 1.

Solution:

2x^{2}y (3xy^{2} – 61x^{2}y) = 2x^{2}y × 3xy^{2} – 2x^{2}y × 61x^{2}y

= 2 × 3 × x^{2}y × xy^{2} – 2 61 x^{2}y x^{2}y

= 6x^{3}y^{3} – 122x^{4}y^{2}

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 (5x^{2} – 3y^{2}) – 5y (2x^{2} + y^{2})

Solution:

2x (5x^{2} – 3y^{2}) – 5y (2x^{2} + y^{2})

= 2x × 5x^{2} – 2x × 3y^{2} – 5y × 2x^{2} – 5y × y^{2}

= 2 × 5 × x × x^{2} – 2× 3 × x × y^{2} – 5 × 2 × y × x^{2} – 5 × 1 × y × y^{2}

= 10x^{3} – 6xy^{2} – 10x^{2}y – 5y^{3}

Example 6:

Find the product \(\frac{1}{3}\) ab^{2}(6b^{2} – 7ab) and verify the result for a = 1 and b = – 1.

Solution:

\(\frac{1}{3}\) ab^{2} (6b^{2} – 7ah) = \(\frac{1}{3}\) ab^{2} × 6b^{2} – \(\frac{1}{3}\) ab^{2} × 7ab

= \(\frac{1}{3}\) × 6ab^{2} × b^{2} – \(\frac{1}{3}\) × 7 × ab^{2} × ab

= 2ab^{4} – \(\frac{7}{3}\) a^{2}b^{3}

Verification:

L.H.S. = \(\frac{1}{3}\) ab^{2} (6b^{2} – 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. = 2ab^{4} – a^{2}b^{3}

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 (3x^{2}y – 2xy^{2}) (2xy^{2} + 3x^{2}y) and verify the result for x = 2 and y = 3.

Solution:

(3x^{2}y – 2xy^{2}) (y^{2} + 3x^{2}y)

= 3x^{2}y (2xy^{2} + 3x^{2}y) – 2xy^{2} (2xy^{2} + 3x^{2}y)

= 3x^{2}y × 2xy^{2} + 3x^{2}y × 3x^{2}y – 2xy^{2} × 2xy^{2} – 2xy^{2} × 3x^{2}y

= 6x^{3}y^{3} + 9x^{4}y^{2} – 4x^{2}y^{4} – 6x^{3}y^{3}

= 9x^{4}y^{2} – 4x^{2}y^{4}

Verification:

LH.S. = (3x^{2}y – 2xy^{2}) (2xy^{2} + 3x^{2}y)

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. = 9x^{4}y^{2} – 4x^{2}y^{4}

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) (5x^{2} – 6xy + 4y^{2}) and verify for x = 2 and y = 1.

Solution:

(3x + 2y) (5x^{2} – 6xy + 4y^{2}) = 3x (5x^{2} – 6xy + 4y^{2}) + 2y (5x^{2} – 6xy + 4y^{2})

= 3x × 5x^{2} – 3x × 6xy + 3x × 4y^{2} + 2y × 5x^{2} – 2y × 6xy + 2y × 4y^{2}

= 15x^{3} – 18x^{2}y + 12xy^{2} + 10x^{2}y – 12xy^{2} + 8y^{3}

= 15x^{3} – 8x^{2}y + 8y^{3}

Verification:

L.H.S. (3x + 2y) (5x^{2} – 6xy + 4y^{2})

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. = 15x^{3} – 8x^{2}y + 8y^{3}

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 25x^{3}y^{2} and 30x^{2}y^{3}.

Solution:

25x^{3}y^{2} and 30x^{2}y^{3}

H.C.F. of 25 and 30 is 5

H.C.F. of x^{3}y^{2} and x^{2}y^{3} = x^{2}y^{2}

So, H.CF.= 5x^{2}y^{2}