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
Answer:
9x³ × 2y⁴
= 9 × 2 × x³ × y⁴
= 18x³y⁴

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

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

Question 2.
Multiply the monomials:
(i) 7pq and \(\frac{4}{3}\)p²q³
Answer:
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
Answer:
12a²b⁶c⁸ and -3a⁷b⁴c³
= 12 × ( -3) × a²b⁶c⁸ × a⁷b⁴c³
= -36 × a⁹b¹⁰c¹¹ = -36a⁹b¹⁰c¹¹

(iii) \(\frac{2}{5}\)x²y and \(\frac{5}{3}\)x³y²z²
Answer:
\(\frac{2}{5}\)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
Answer:
3x⁷ × 4x² × (- 5x³)
= 3 × 4 × -5 × x⁷ × x² × x³
= – 60x¹²

(ii) 1.2 a²b², 5ab⁴c² and 1.1a⁵bc⁷
Answer:
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
Answer:
\(\frac{3}{4}\)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.
Answer:
(\(\frac{1}{2}\)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.
Answer:
(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.
Answer:
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.
Answer:
Here breadth = 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.
Answer:
Here breadth = b
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:
  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 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