Samacheer Kalvi Class 7 Maths Term 3 Chapter 3 Algebra Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 3 chapter 3 Algebra. Here students can easily find all the solutions for Algebra Exercise 3.1, 3.2 and 3.3. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Algebra Exercise 3.1 Solutions :
(1) Fill in the blanks :
(i) (p – q)2 = ____
(ii) The product of (x + 5) and (x − 5) is ____.
(iii) The factors of x2 – 4x k+ 4 are ____.
(iv) Express 24ab2c2 as product of its factors is ___.
Solution :
(ii) → x²–25.
(iii) → (x–2) and (x–2).
(iv) → 2 × 2 × 2 × 3 × a × b × b × c × c.
(2) Say whether the following statements are True or False.
(i) (7x + 3) (7x − 4) = 49x² – 7x – 12
(ii) (a – 1)² = a² – 1
(iii) (x² + y²) (y² + x²) = (x² + y²)²
(iv) 2p is the factor of 8pq.
Solution :
(i) → True
(ii) → False
(iii) → True
(iv) → True
(3) Express the following as the product of its factors.
(i) 24 ab²c²
(ii) 36 x³y²z
(iii) 56 mn²p²
Solution :
(i) 24 ab²c² = 24 × a × b × b × c × c
= 8 × 3 × a × b × b × c × c
= 2 × 2 × 2 × 3 × a ×b × b × c × c
(ii) 36 x³y²z = 36 × x × x × x × y × y × z
= 6 × 6 × x × x × x × y × y × z
= 2 × 3 × 2 × 3 × x × x × x × y × y × z
(iii) 56 mn²p² = 56 × m × n × n × p ×p
= 8 × 7 × m × n × n × p × p
= 2 × 2 × 2 × 7 × m × n × n × p × p
(4) Using the identity (x + a) (x + b) = x² + x (a + b) + ab following product
(i) (x + 3) (x + 7)
(ii) (6a + 9) (6a − 5)
(iii) (4x + 3y) (4x + 5y)
(iv) (8 + pq) (pq + 7)
Solution :
(i) (x + 3)(x + 7)
Since, (x +a)(x +b)= x² + x(a + b) + ab
= (x + 3)(x + 7) = x² + x (3+7) +3×7
= x² + x(10) +21
(ii) (6a + 9)(6a − 5)
= (6a + 9)(6a − 5) = (6a)² + 6a (9-5) + 9×(-5)
= 62a² + 6a(4) – 45
= 36a² + 24a – 45
(iii) (4x + 3y)(4x + 5y)
= (4x + 3y)(4x + 5y) = (4x)² + 4x(3y+5y) + 3y×5y
= 16x² + 4x×8y + 15y²
= 16x² + 32xy + 15y²
(iv) (8 + pq)(pq + 7)
= (8 + pq)(pq + 7) = (pq + 8)(pq + 7)
= (pq)² + pq(8+7) + 8×7
= p²q² + 15pq + 56
(5) Expand the following squares, using suitable identities
(i) (2x + 5)²
(ii) (b – 7)²
(iii) (mn + 3p)²
(iv) (xyz – 1)²
Solution :
(i) (2x + 5)²
We know, (a + b)2 = a² + 2ab + b²
= (2x + 5)² = (2x)² + 2×2a×5 + 5²
= 4x² +20a + 25
(ii) (b – 7)²
We know the identity, (a – b)² = a² – 2ab + b²
= (b – 7)² = b² – 2×b×7 + 7²
= b² – 14b + 49
(iii) (mn + 3p)²
We know the identity, (a + b)² = a² + 2ab + b²
= (mn + 3p)² = (mn)² + 2× mn × 3p + (3p)²
= m²n² + 6mnp + 9p²
(iv) (xyz – 1)²
We know the identity, (a – b)² = a² – 2ab + b²
= (xyz – 1)² = (xyz)² – 2 × xyz × 1 + 1²
= x²y²z² – 2xyz + 1
(6) Using the identity (a + b)(a − b) = a² – b², find the following product
(i) (p + 2) (p − 2)
(ii) (1 + 3b) (3b − 1)
(iii) (4 − mn) (mn + 4)
(iv) (6x + 7y) (6x – 7y)
Solution :
(i) (p + 2)(p − 2)
Since the identity (a + b)(a − b) = a² – b²
Thus, (p + 2)(p − 2)
= p² – 2²
= p² – 4
(ii) (1 + 3b)(3b − 1)
Since the identity (a + b)(a − b) = a² – b²
Thus, (1 + 3b)(3b − 1)
= (3b + 1 )(3b − 1)
= 3²b² – 12 = 9b² – 1
(iii) (4 − mn)(mn + 4)
Since the identity (a + b)(a − b) = a² – b²
Thus, (4 − mn)(mn + 4)
= (4 − mn)(4 + mn )
= 4² – m² n² = 16 – m²n²
(iv) (6x + 7y)(6x – 7y)
Since the identity (a + b)(a − b) = a² – b²
Thus, (6x + 7y)(6x – 7y)
= (6x)² – (7y)²
= 36 x² – 49y²
(7) Evaluate the following, using suitable identity
(i) 51²
(ii) 103²
(iii) 998²
(iv) 47²
(v) 297 × 303
(vi) 990 × 1010
(vii) 51 × 52
Solution :
(i) 51² = (50+1)²
= 50² + 2×50×1 + 1² …. ( since, (a+b)² = a²+ 2ab + b²)
= 2500 + 100 +1
= 2601
(ii) 103² = (100+3)²
= 100² + 2×100×3 + 3² …. ( since, (a+b)² = a² + 2ab + b²)
= 10000 + 600 + 9
= 10609
(iii) 9982 = (1000-2)²
= 1000² – 2×1000 × 2 + 2² …. ( since, (a-b)² = a² – 2ab + b²)
= 1000000 – 4000 +4
= 1000000 – 3996
= 9,96,004
(iv) 47² = (50 – 3)²
= 502 – 2×50×3 + 32 …. ( since, (a-b)2 = a² – 2ab + b² )
= 2500 – 300 + 9
= 2200 +9
= 2209
(v) 297 × 303 = (300 – 3) ×(300+3)
= 300² – 3² … { since, a² – b² = (a – b) ×(a+b) }
= 90000 – 9
= 89991
(vi) 990 × 1010 = (1000-10) ×(1000 +10)
= 1000² – 10² … { since, a² – b² = (a – b) ×(a+b) }
= 1000000 – 100
= 999900
(vii) 51 × 52 = (50 + 1) × (50 + 2)
= 50² + 50 (1 + 2) + 1 × 2 …. (since, (x + a) × (x + b) = x² + x(a + b) + ab)
= 2500 + 50 × 3 + 2
= 2500 + 152
= 2652
(8) Simplify (a + b)² – 4ab
Solution :
Given, (a + b)² – 4ab
= (a+b)² – 4ab = a² + 2ab + b² – 4ab …. ( since, (a+b)² = a² + 2ab + b²)
= a² – 2ab + b²
= (a – b)² …… ( since, (a-b)² = a² – 2ab + b² )
(9) Show that, (m – n)² + (m + n)² = 2(m² + n²)
Solution :
To show that, ( m – n)² + ( m + n)² = 2(m² + n²)
LHS = ( m – n)² + ( m + n)²
= m² – 2mn + n² + ( m + n)² …. ( since, (a-b)² = a² – 2ab + b² )
= m² – 2mn + n² + m² + 2mn + n² …. ( since, (a+b)² = a² + 2ab + b²)
= 2m² + 2n²
= 2(m² + n²)
= RHS
Hence, ( m – n)² + ( m + n)² = 2(m² + n²)
(10) If a + b = 10 and ab = 18, find the value of a² + b²
Solution :
Given, a + b = 10 and ab = 18
We know, (a + b)² = a² + 2ab + b²
10² = a² + 2 × 18 + b²
100 = a² + b² + 36
a² + b² = 100 – 36
a² + b² = 64
Hence, the value of a² + b² is 64.
(11) Factorise the following algebraic expressions by using the identity a²- b² = (a+b) (a – b)
(i) z² – 16
(ii) 9 – 4y²
(iii) 25a² – 49b²
(iv) x⁴ – y⁴
Solution :
(i) z² – 16
Using the identity a² – b² = (a + b) (a – b)
z² – 16 = z² – 4²
= (z – 4) (z + 4)
(ii) 9 – 4y²
Using the identity a² – b² = (a + b) (a – b)
9 – 4y² = 3² – 2²y²
= 3² – (2y)²
= ( 3-2y) (3+2y)
(iii) 25a² – 49b²
Using the identity a² – b² = (a + b) (a – b)
25a² – 49b² = (5a)² – (7b)²
= (5a – 7b) (5a + 7b)
(iv) x⁴ – y⁴
Using the identity a² – b² = (a + b) (a – b)
x⁴ – y⁴ = (x²)² – (y²)²
= (x² + y²) (x² – y²)
= (x² + y²) (x + b) (x – b)
(12) Factorise the following using suitable identity
(i) x² – 8x + 16
(ii) y² + 20y + 100
(iii) 36m² + 60m + 25
(iv) 64x² – 112xy + 49y²
(v) a² + 6ab + 9b² – c²
Solution :
(i) x² – 8x + 16
x² – 8x + 16 = x² – 2×x×4 + 4²
= ( x – 4 )² …. ( since, (a-b)² = a² – 2ab + b² )
= ( x – 4 )( x – 4 )
(ii) y² + 20y + 100
y² + 20y + 100 = y² + 2 × y × 10 + 10²
= ( y + 10)² ….. ( since, (a+b)²= a² + 2ab + b² )
= ( y + 10)( y + 10)
(iii) 36m² + 60m + 25
36m² + 60m + 25 = 6²m² + 2 ×5 ×6m + 5²
= (6m)² + 2 × 5 × 6m + 5² … ( since, (a+b)² = a² + 2ab + b² ).
= ( 6m + 5)²
= ( 6m + 5)( 6m + 5)
(iv) 64x² – 112xy + 49y²
64x² – 112xy + 49y² = 8²x² – 2 × 8x ×7y + 7²y²
= (8x)² – 2 × 8x ×7y + (7y)²
= ( 8x – 7y)² …. ( since, (a-b)² = a² – 2ab + b² )
= ( 8x – 7y)( 8x – 7y)
(v) a² + 6ab + 9b² – c²
a² + 6ab +9b² – c² = a² + 2× a × 3b + 3²b² – c²
= (a + 3b)² – c² ….. ( since, (a + b)² = a² + 2ab + b²)
= (a+3b – c) (a+3b + c) ……{since, a² – b² = (a-b) (a + b)}
Objective Type Question Solutions :
(14) If a + b = 5 and a² + b² = 13, then ab = ?
(i) 12
(ii) 6
(iii) 5
(iv) 13
Solution :
Correct Option → (ii)
If a + b = 5 and a² + b² = 13, then ab = 6
(14) (5 + 20) (- 20 – 5) = ?
(i) −425
(ii) 375
(iii) −625
(iv) 0
Solution :
Correct Option → (iii)
= (5 + 20) (- 20 – 5)
= −625
(15) The factors of x² – 6x + 9 are
(i) (x − 3)(x − 3)
(ii) (x − 3)(x + 3)
(iii) (x + 3)(x + 3)
(iv) (x − 6)(x + 9)
Solution :
Correct Option → (i)
The factors of x² – 6x + 9 are (x − 3)(x − 3)
(16) The common factors of the algebraic expressions ax²y, bxy² and cxyz is
(i) x²y
(ii) xy²
(iii) xyz
(iv) xy
Solution :
Correct Option → (iv)
xy is the correct answer.
Algebra Exercise 3.2 Solutions :
(1) Given that x ≥ y. fill in the blanks with suitable inequality signs
(i) y __ x
(ii) x + 6 __ y + 6
(iii) x2 __ xy
(iv) -xy __ -y²
(v) x − y __ 0
Solution :
(i) → y ≤ x
(ii) → x + 6 ≥ y + 6
(iii) → x2 ≥ xy
(iv) → −xy ≤ – y²
(v) → x − y ≥ 0
(2) Say True or False.
(i) Linear inequation has almost one solution.
(ii) When x is an integer, the solutions for x ≤ 0 are −1, −2,…
(iii) An inequation, −3 < x < −1, where x is an integer, cannot be represented in the number line.
(iv) x < −y can be rewritten as −y < x
Solution :
(i) → False
(ii) → False
(iii) → True
(iv) → False
(3) Solve the following inequations.
(i) x ≤ 7, where x is a natural number.
(ii) x − 6 < 1, where x is a natural number.
(iii) 2a + 3 ≤ 13, where a is a whole number.
(iv) 6x − 7 ≥ 35, where x is an integer.
(v) 4x − 9 > −33, where x is a negative integer.
Solution :
(i) Given, x ≤ 7
Since the solution belongs to the set of natural numbers, the solutions are 1, 2, 3, 4, 5, 6 and 7.
(ii) Given, x − 6 < 1
x − 6 + 6 < 1 + 6 …. ( adding 6 from the inequation)
(iii) Given, 4x − 9 > −33
4x − 9 + 9 > −33 + 9 …. ( adding 9 from the inequation)
4x > -24
4x/4 > -24/4 … ( dividing 4 from the inequation )
x > -6
Since the solution belongs to the set of a negative integer numbers, the solutions are -5 , -4, -3, -2 and -1.
Objective Type Question Solution :
(6) The solutions of the inequation 3 ≤ p ≤ 6 are (where p is a natural number)
(i) 4, 5 and 6
(ii) 3, 4 and 5
(iii) 4 and 5
(iv) 3, 4, 5 and 6
Solution :
Correct Option → (iv)
The solutions of the inequation 3 ≤ p ≤ 6 are 3, 4, 5 and 6
(7) The solution of the inequation 5x + 5 ≤ 15 are (where x is a natural number)
(i) 1 and 2
(ii) 0, 1 and 2
(iii) 2, 1, 0, −1, −2..
(iv) 1, 2, 3..
Solution :
Correct Option → (i)
The solution of the inequation 5x + 5 ≤ 15 are 1 and 2.
(8) The cost of one pen is ₹ 8 and it is available in a sealed pack of 10 pens. If Swetha has only ₹ 500, how many packs of pens can she buy at the maximum?
(i) 10
(ii) 5
(iii) 6
(iv) 8
Solution :
Correct Option → (iii)
She can buy maximum 6 packets of pen.
(9) The inequation that is represented on the number line as shown below is ___
(i) −4 < x < 0
(ii) −4 ≤ x ≤ 0
(iii) −4 < x ≤ 0
(iv) −4 ≤ x < 0
Solution :
Correct Option → (iii)
The inequation that is represented on the number line as shown below is −4 < x ≤ 0.
Algebra Exercise 3.3 Solutions :
(1) Using identity, find the value of
(i) (4.9)²
(ii) (100.1)²
(iii) (1.9) × (2.1)
Solution :
(i) (4.9)² = (5.0 – 0.1)²
= 5² – 2×5×0.1 + 0.1² … ( since, (a-b)² = a² – 2ab + b²)
= 25 – 1 +0.01
= 24 +0.01
= 24.01
(ii) (100.1)² = ( 100 + 0.1)²
= 100² + 2×100×0.1 + 0.1² … ( since, (a + b)² = a² + 2ab + b ²)
= 10000 + 20 + 0.01
= 10020.01
(iii) (1.9) × (2.1) = (2.0 – 0.1) × (2.0 + 0.1)
= (2)² – (O.1)² … (Since, (a – b) (a +b) = a² – b² )
= 4 – 0.01
= 3.99
(2) Factorise 4x² – 9y²
Solution :
4×2 – 9y2
= (2x)2 – (3y)2
= (2x + 3y) (2x – 3y)
(4) Show that (x + 2y)² – (x – 2y)² = 8xy
Solution :
LHS = (x + 2y)² – (x – 2y)²
= x² +2×x×2y + (2y)² – (x – 2y)² … ( since, (a + b)² = a² + 2ab + b²)
= x² + 4xy + 4y² – [x² – 2×x×2y + (2y)²] … (since, (a – b)² = a² – 2ab + b²)
= x² + 4xy + 4y² – x² + 4xy – 4y²
= 8xy
= RHS
Hence, (x + 2y)² – (x – 2y)² = 8xy
Challenge Problem Solutions :
(6) If X = a² – 1 and Y = 1 – b², then find X + Y and factorize the same
Solution :
Given, X = a² – 1 and Y = 1 – b²
Thus, X + Y = a² – 1 + 1 – b²
= a² – b²
= (a + b) (a – b)
(7) Find the value of (x – y) (x + y) (x² + y²)
Solution :
Given, (x – y) (x + y) (x² + y²)
= (x² – y²) (x² + y²)
= (x²)² – (y²)²
= x⁴ – y⁴
Thus, the value will be x⁴ – y⁴
(8) Simplify (5x – 3y)² – (5x + 3y)²
Solution :
(5x – 3y)² – ( 5x + 3y)² = 25x² – 2×5x×3y + 9y² – ( 5x + 3y)²
= 25x² – 30xy + 9y² – [ 25x² + 2×5x×3y + 9y² ]
= 25x² – 30xy + 9y² – 25x² – 30xy – 9y²
= -60xy ….(Simplified)
(9) Simplify (i) (a + b)² – (a – b)² (ii) (a + b)² + (a – b)²
Solution :
(i) (a + b)²– (a – b)²
= a² + 2ab + b² – [ a² – 2ab + b² ]
= a² + 2ab + b² – a² + 2ab – b²
= 4ab….(Simplified)
(ii) (a + b)² + (a – b)²
= a² + 2ab + b² – [a² – 2ab + b²]
= a² + 2ab + b² + a² – 2ab + b²
= 2a² + 2b²
= 2 (a² + b²) ….(Simplified)
(10) A square lawn has a 2m wide path surrounding it. if the area of the path is 136 m², find the area of lawn
Solution :
Given, a square lawn has a 2m wide path surrounding it.
Since, the area of the path is 136 sq.m.
Let x be side of square.
Thus, sides of outer square is x + 4
Now, the area of the path = area of outer square – area of inner square
136 = (x + 4)² – x²
136 = x² + 2 × x × 4 + 16 – x²
136 = 8x + 16
8x = 136 – 16
8x = 120
x = 120/8
x = 15 m
Thus, area of square = x² = 15² = 225 sq. m
(11) Solve the following inequalities.
(i) 4n + 7 ≥ 3n + 10, n is an integer.
(ii) 6(x + 6) ≥ 5(x − 3), x is a whole number.
(iii) −13 ≤ 5x + 2 ≤ 32, x is an integer.
Solution :
(i) Given, 4n + 7 ≥ 3n + 10
4n + 7 – 3n ≥ 3n + 10 – 3n … [Subtracting 3n from the inequation]
n + 7 – 7 ≥ 10 – 7 … [Subtracting 7 from the inequation]
n ≥ 3
Since, solution belongs to the set of integers, that are grater than or equal to 3, we take the values of x as 3, 4, 5,…
Therefore, the solutions are 3, 4, 5, …
(ii) Given, 6(x + 6) ≥ 5(x − 3)
6x + 36 ≥ 5x – 15
6x + 36 – 5x ≥ 5x – 15 – 5x … [Subtracting 5x from the inequation]
x + 36 – 36 ≥ -15 – 36 … [Subtracting 36 from the inequation]
x ≥ – 51
Since the solution belongs to the set of whole numbers, the solutions are 0,1, 2, 3 , 4 …
(iii) Given, −13 ≤ 5x + 2 ≤ 32
−13 -2 ≤ 5x + 2 – 2 ≤ 32 – 2 … [Subtracting 2 from the inequation]
-15 ≤ 5x ≤ 30
-15/5 ≤ 5x/5 ≤ 30/5 … [ dividing 5 from the inequation]
-3 ≤ x ≤ 6
Since the solution belongs to the set of an integers numbers, the solutions are -3, – 2, -1, 0, 1, 2, 3, 4, 5 and 6.
Next Chapter Solution :
