Samacheer Kalvi Class 9 Maths Chapter 3 Algebra Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students by providing Solutions for Samacheer Kalvi Class 9 Maths chapter 3 Algebra. Here students can easily find all the solutions for Algebra Exercise 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 3.10, 3.11, 3.12, 3.13, 3.14 and 3.15. 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 :
(2) Write the coefficient of x 2 and x in each of the following polynomials.
(i) 4 + 2/5x² – 3x
(ii) 6 – 2x² + 3x³ – √7x
(iii) πx² – x + 2
(iv) √3x² + √2x + 0.5
(v) x²- 7/2x + 8
Solution :
(i) Given, 4 + 2/5x² – 3x
Here, Co-efficient of x² = 2/5
Co-efficient of x = – 3
(ii) 6 – 2x² + 3x³ – √7x
Here, Co-efficient of x² = – 2
Co-efficient of x = √7
(iii) πx² – x + 2
Here, Co-efficient of x² = π
Co-efficient of x = -1
(iv) √3x² + √2x + 0.5
Here, Co-efficient of x² = √3
Co-efficient of x = √2
(v) x²- 7/2x + 8
Here, Co-efficient of x² = 1
Co-efficient of x = – 7/2
(4) Rewrite the following polynomial in standard form
(i) x – 9 + √7x³ + 6²
(ii) √2x² – 7/2x⁴ + x – 5x²
(iii) 7x³ – 6/5x² + 4x – 1
(iv) y² + √5y³ – 11 – 7/3y + y⁴
Solution :
(i) Standard form = √7x³ + 6x² + x -9
(ii) Standard form = -7/2 x⁴ – 5x³ + √2x² + x
(iii) Standard form = 7x³ – 6/5 x2 + 4x -1
(iv) Standard form = 9y⁴ + √5y³ + y2 – 7/3 – 11
(6) Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial
(i) p(x) = 7x² + 6x – 1
q(x) = 6x – 9
(ii) f(y) = 6y² – 7y + 2
g(y) = 7y + y³
(iii) h(z) = z⁵ – 6z⁴ + z
f(z) = 6z² + 10z – 7
Solution :
(i) 7x² + 6x – 1
+ 6x – 9
(-) (+)
————-
7x² + 8
∴ the degree of the resultant polynomial is 2.
(ii) 6y² – 7y + 2
Y³ + 0 + 7y + 0
(-) (-) (-) (-)
————————–
-y³ + 6y² – 14y + 2
∴ the degree of the resultant polynomial is 3
(iii) z⁵ – 6z⁴ + 0 + 0 + z
6z² + 10z + 7
(-) (-)
—————————
Z⁵ – 6z⁴ – 6z² – 9z + y
∴ the degree of the resultant polynomial is 5.
(10) The cost of a chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x and y. If x = 10, y =5 find the amount paid by him.
Solution :
Total cost = (x + y) (x + y) = (x + y)²
If x = 10, y = 5
So, (10 + 5)² = 225
Therefore, the amount paid by him is 225.
(11) The length of a rectangle is (3x+2) units and it’s breadth is (3x–2) units. Find its area in terms of x. What will be the area if x = 20 units.
Solution :
Total area = (3x – 2) (3x + 2)
= (3x)² – 2²
If x = 20 then,
= (60)² – 4
= 3600 – 4
= 3596
Therefore, the area will be 3596.
Algebra Exercise 3.2 Solutions :
(1) Find the value of the polynomial f(y) = 6y – 3y² + 3 at
(i) y = 1
(ii) y = –1
(iii) y = 0
Solution :
(i) f(y) = 6y – 3y² + 3
At y = 0, f(0) = 3
So, the value of the polynomial is 3.
(ii) f(y) = 6y – 3y² + 3
At, y = 1, f(1) = 6 – 3 + 3 = 6
So, the value of the polynomial is 6.
(iii) f(y) = 6y – 3y² + 3
At y = -1, f(-1)
= -6 -3 + 3
= – 6
Hence, the value of the polynomial is -6.
(2) If p(x) = x²– 2√2x + 1, find p(2√2)
Solution :
p(x) = x² – 2 √2x + 1
Then, p(2√2) = (2√2)2 – 2√2× 2√2 + 1
= 8 – 8 + 1
= 1
(3) Find the zeros of the polynomial in each of the following :
(i) p(x) = x – 3
(ii) p(x) = 2x + 5
(iii) q(y) = 2y – 3
(iv) f(z) = 8z
(v) p(x) = ax when a ≠ 0
(vi) h(x) = ax + b, a ≠ 0, a, b ∈ R
Solution :
(i) x – 3 = 0
= x = 3
Because, At x = 3 the polynomial be zero.
(ii) 2x + 5 = 0
= x = -5/2
At -5/2 the polynomial be zero.
(4) Find the roots of the polynomial equations
(i) 5x – 6 = 0
(ii) x +3 = 0
(iii) 10x + 9 = 0
(iv) 9x – 4 = 0
Solution :
(i) 5x – 6 = 0
∴ x = 6/5
(ii) x + 3 = 0
∴ x = -3
(iii) 10x + 9 = 0
∴ x = -9/10
(iv) 9x – 4 = 0
∴ x = 4/9
(5) Verify whether the following are zeros of the polynomial indicated against them, or not
(i) p(x) = 2x – 1, x = 1/2
(ii) p(x) = x³ – 1, x = 1
(iii) p(x) = ax + b, x = -b/a
(iv) p(x) = (x+3)(x-4), x = 4, x = -3
Solution :
(i) p(x) = 2x – 1
At x = 1/2, p(1/2) – 2 × 1/2 – 1 = 0
This is zero polynomial
(ii) P(x) = x³ – 1
At x = 1, p(1) = 1 – 1 = 0
This is zero polynomial
(iii) p(x) = ax + b
At x = -b/a, p(-b/a) = a × (-b/a) + b = 0
This is a zero polynomial
(iv) p(x) = (x + 3) (x – 4)
At x = -3, p(-3) = (3 + 3) (x – 4) = 0
At x = 4, p(4) = (x + 3) (4 – 4) = 0
This is zero polynomial.
Algebra Exercise 3.3 Solutions :
(1) Check whether p(x) is a multiple of g(x) or not.
p(x) = x³ – 5x² + 4x – 3; g(x) = x – 2
Solution :
Let, g(x) = 0
Then, x – z = 0
= x = 2
P(x) = x³ – 5x² + 4x – 3
= 23 – 5z² + 4.2 – 3
= -7 ≠ 0
This is not a multiple of g(x)
(2) By remainder theorem, find the remainder when, p(x) is divided by g(x) where
(i) p(x) = x³ – 2x² – 4x – 1 g(x) = x + 1
(ii) p(x) = 4x³ – 12x² + 14x – 3 g(x) = 2x – 1
(iii) p(x) = x³ – 3x³ + 4x + 50 g(x) = x – 3
Solution :
(i) g(x) = x + 1
= x + 1 = 0
= x = -1
P(-1) = (-1)³ – 2 (1)² – 4 (-1) – 1
= -1 – 2 + 4 – 1 = 0
Remainder = 0
(ii) 2x -1 = 0
= x = 1/2
P(1/2) = 4(1/2)³ – 12 × 1/2² + 14.1/2 – 3
= 4 1/8 – 12 × 1/4 + 14/2 – 3
= 1/2 – 3 + 7 -3
= 1/2 + 1 = 3/2
So, Remainder = 3/2
(iii) x – 3 = 0
= x = 3
P(3) = 3³ – 3.3² + 4.3 + 50
= 27 – 27 + 12 + 50
= 62
Remainder = 62
(5) For what value of k is the polynomial p(x) = 2x³ – kx² + 3x + 10 exactly divisible by (x – 2)
Solution :
x – 2 = 0 = x = 2
P(2) = 2.2³ – k2² + 3.2 + 10
= 16 – 4k + 6 + 10
= 32 – 4k = 0
= k = 8
So, the value of k is 8.
(6) If two polynomials 2x³ + ax² + 4x – 12 and x³ + x² –2x+ a leave the same remainder when divided by (x – 3), find the value of a and also find the remainder.
Solution :
Given, x – 3 = 0
= x = 3
F(3) = 2.83 + a3² + 4.3 – 12 = 54 + 9a
x-3 = 0 = x = 3 Since , remainder are same
g(3) = 3³ + 3² -2.3 + a = 27 – 9 – 6 + a
= a+30
= 9a + 54 = a + 30
= 8a = -24 = a = -3
Therefore, f(3) = 9(-3) + 54 = 27
(7) Determine whether (x – 1) is a factor of the following polynomials :
(i) x³ + 5x² – 10x + 4
(ii) x⁴ + 5x² – 5x + 1
Solution :
(i) p(1) = 0
= 1³ + 5.1² – 10 + 4
= 1 + 5 – 10 + 4 = 0
Therefore, (x-1) is a factor.
(ii) p(1) = 0
P(1) = 1⁴ + 5.1² – 5(1) + 1
= 1 + 5 – 5 + 1
= 2 ≠ 0
Therefore, (x-1) is not a factor.
(8) Using factor theorem, show that (x -5) is a factor of the polynomial 2x³ – 5x² – 28x + 15
Solution :
p(5) = 0
Because, p(5) = 2.5³ – 5.5² – 28.5 + 15
= 250 – 125 – 140 + 15
= 0
So, (x-5) is a factor of 2x³ – 5x² – 28x + 15
(9) Determine the value of m, if (x+3) is a factor of x³ – 3x² – mx + 24
Solution :
Since, x + 3 is a factor
So, p(-3) = (-3)³ – 3(-3)² – m (-3) + 24 = 0
Because, -27 – 9 + 3m + 24 = 0
= 3m – 30 = 0
= m = 10
Therefore, the value of m is 10.
(11) If (x-1) divides the polynomial kx³ – 2x² + 25x − 26 without remainder, then find the value of k
Solution :
Since, x – 1 is a factor
So, p(1) = K.1³ – 2.1² + 25.1 – 26 = 0
= K – 2 + 25 – 26 = 0
= K – 3 = 0
= K = 3
Therefore, the value of k is 3.
Algebra Exercise 3.4 Solutions :
(1) Expand the following :
(ii) (-p + 2q + 3r)²
(iii) (2p + 3) (2P – 4) (2P – 5)
(iv) (3a + 1) (3a – 2) (3a + 4)
Solution :
(ii) Given, (-p + 2q + 3r)²
= (-p)² + (2q)² + (3r)² + 2 × (-p) 2q + 2 × 2q × 3r + 2 × 3r (-p)
= p² + 4q² + 9r² – 4pq + 12qr – 6pr
(iii) Given, (2p + 3) (2P – 4) (2P – 5)
= (2P + 3) (4P² – 10P – 8P + 20)
= 8P³ – 36 P² + 40P + 12P² – 54 P + 60
= 3P³ – 24 – P² – 14P + 60
(iv) (3a + 1) (3a – 2) (3a + 4)
= (3a + 1) (9a² + 6a – 8)
= 27a³ + 18a² – 24a + 9a² + 6a – 8
= 27a³ + 27a² – 18a – 8
(2) Using algebraic identity, find the coefficients of x², x and constant term without actual expansion
(i) (x + 5)(x + 6)(x + 7)
(ii) (2x + 3)(2x – 5)(2x – 6)
Solution :
(i) (x + 5) (x + 6) (x +y)
= x³ + (a + b + c) x² (a b + b c + c a) x + a b c
Co-efficient of x² = 5 + 6 + 7 = 18
Coefficient of x = 5 × 6 + 6 × 7 + 5 × 7
= 30 + 42 + 35 = 107
Constant term = 5 × 6 × 7 = 210
(ii) Coefficient of x² = 4(3 – 5 – 6) = 4 × -8 = -32
Coefficient of x = 2(3 × -5 + (5× – 6) + (-6 × 3)]
= 2(-15 + 30 – 18) = -6
Constant term = 3 × -5x -6 = 90
(4) Expand :
(i) (3a – 4b)³
(ii) (x + 1/y)³
Solution :
(i) Given, (3a – 4b)³
= (3a)³ – 3(3a)² (4b) + 3(3a) (4b)² – (4b)³
= 27a³ – 108 a² b + 144 ab² – 64 b³
(ii) Given, (x + 1/y)³
= x³ + 3x². 1/y + 3x 1/y² + 1/y³
(5) Evaluate the following by using identities:
(i) 98³
(ii) 1001³
Solution :
(i) Given, 98³ = (100 – 2)³
= 100³ – 3100².2 + 3.100.2² – 2³
= 1000000 – 60000 + 1200 – 8
= 941192
(ii) Given, 1001³ – (1000 + 1)³
= 1000³ + 3 × 1000² × 1 + 3 × 1000 × 1 + 1
= 100,000000 + 3 × 000000 + 3000 + 1
= 1003003001
(6) If (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x² + y² + z²
Solution :
We know, (x + y + z)² = x² + y² + z² + 2xy + 2xy + 2xz + 2yz
= x² + y² + z² = (x + y + z)² -2(x y + x y + y z)
= 92 – 2 × 26
= 81 -52 = 29
So, the value is 29.
(7) Find 27a³ + 64b³, if 3a + 4b = 10 and ab = 2
Solution :
Given, (3a + 4b)³
= (3a)³ + 3(3a)² 4b + 3(3a) (4b)² + (4b)³
= 27a³ + 16b³ = (3a + 4b)³ – 3 × 3a × 4b × (3a + 4b)
= (10)³ – 36 × 2 × 10
= 1000-720
= 280
(8) Find x³ – y³, if x – y = 5 and xy = 14
Solution :
x³ – y³ = (x-y)³ + 3xy (x-y)
= 5³ + 3 × 14 × 5
= 125 + 210
= 335
(9) If a + 1/a = 6, then find the value of a³ + 1/a³
Solution :
a³ + 1/a³ = (a + 1/a³ – 3a × 1/a (a + 1/a)
= 6³ – 3 × 6
= 216 -18
= 198
So, the value of a³ + 1/a³ is 198.
(10) If x² + 1/x² = 23, then find the value of x +1/x and x³ + 1/x³
Solution :
(x + 1/x)² = x² + 1/x² + 2 × 1/x × x
= (x + 1/x)² = 23 + 2
= x + 1/x √25
= 5
X + 1/x³ = (x + 1/x)³ – 3x × 1/x (x + 1/x)
= 53 – 3 × 5
= 125 – 15
= 110
(11) If (y – 1/y)³ = 27, then find the value of y³- 1/y³
Solution :
y³ – 1/y³ = (y – 1/y)³ + 3y. 1/y (y – 1/y)
= 27 + 3.3 √27
= 27 + 3 × 3
= 36
Therefore, the value of y³- 1/y³ is 36.
(12) Simplify :
(i) (2a + 3b + 4c)(4a² + 9b²+ 16c² – 6ab -12bc – 8ca)
Solution :
Given, (2a + 3b + 4c) (4a² + 9b² + 16c² – 6ab – 12 c – 18 ac)
= (2a)³ + (3b)³ + (4c)³ – 3 × 2a × 3b × 4c
= 8a³ + 27b³ + 64c³– 72abc
(13) By using identity evaluate the following
(i) 7³ – 10³ + 3³
(ii) 1 + 1/8 – 27/8
Solution :
7³ – 10³ + 3³
If (a + b + c) (a² + b² + c² – a b – b c – ca) = a³ + b³ + c³ – 3 a b c
If a + b + c = 0
Then, a³ + b³ + c³ = 3 a b c
Because 7 – 10 + 3 = 0
= 7³ – 10³ + 3³ = 3 × 7 × (-10) × 3
= – 630
(14) If 2x – 3y – 4z = 0, then find 8x³ – 27y³ – 64z³
Solution :
x + y + z = 0 then x³ + y³ + z³ = 3 x y z
8x³– 21 y³ – 64 z³ = (2x)³ + (-3y)³ + (-4z)³
= 3 × 2x (-3y) × (-4z)
= 72xyz
Algebra Exercise 3.5 Solutions :
(1) Factorise the following expressions :
(i) 2a² + 4a²b + 8a²c
(ii) ab – ac – mb + mc
Solution :
(i) 2a² + 4a²b + 8a²c
= 2a² (1 + 2b + 4c)
(ii) ab – ac – mb + mc
= a (b – c) – m (b – c)
= (b – c) (a – m)
(2) Factorise the following :
(i) x² + 4x + 4
(ii) 3a² – 24ab + 48b²
(iii) x⁵ – 16x
(iv) m² + 1/m² – 23
(v) 6 – 216x²
(vi) a² + 1/a² – 18
Solution :
(i) x² + 4x + 4
= x² + 2.x.2 + 2²
= (x + 2)²
(ii) 3a² – 24ab + 48b²
= 3 (a² – 8ab + 16b²)
= 3 [a² – 2.a.4b + (4b)²]
= 3 [a – 4b)²
(iii) x⁵ – 16x
= x [(x)²)² – 4²]
= x (x² – 4) (x² + 4)
= x (x² + 4) (x + 2) (x – 2)
(iv) m² + 1/m² – 23
= (m + 1/m)² – 2.m.1/m – 23
= (m + 1/m)² – 25
= (m + 1/m)² – 5²
= (m + 1/m + 5) (m + 1/m – 5)
(v) 6 – 216x²
= 6 [1 – (6x)²]
= 6 (1 + 6x) (1 – 6x)
(vi) a² + 1/a² – 18
= (a – 1/a)² + 2.a². 1/a² – 18
= (a – 1/a)² – 16
= (a – 1/a)² – 4²
= (a – 1/a + 4) (a – 1/a – 4)
(3) Factorise the following :
(i) 4x² + 9y² + 25z² + 12xy + 30yz + 20xz
(ii) 25x² + 4y² – 9z² – 20xy + 12yz – 30xz
Solution :
(i) 4x² + 9y² + 25z² + 12xy + 30yz + 20xz
= (2x)² + (3y)² + (5z)² + 2. (2x) (3y) + 2 (3y) (5z) + 2(2x) (5z)
= (2x + 3y + 5z)²
(ii) 25x² + 4y² – 9z² – 20xy + 12yz – 30xz
= (5x)² + (-2y)² + (-3z)² – 2.(5x) (-2y) + 2 (-2y) (-3z) – 2 (-3z) (5x)
= (5x – 2y – 3z)²
(4) Factorise the following :
(i) 8x³ + 125y³
(ii) 27x³ – 8y³
(iii) a⁶ – 64
Solution :
(i) 8x³ + 125y³
= (2x)³ + (5y)³
= (2x + 5y) (4x² – 10xy + 25y²)
(ii) 27x³ – 8y³
= (3x)³ – (2y)³
= (3x – 2y) (9x² + 6xy + 4y²)
(iii) a⁶ – 64
= (a²)³ – 43
= (a² – 4) (a² + 4a² + 4²)
= (a + 2) (a – 2) (5a² + 16)
(5) Factorise the following :
(i) x³ + 8y³ + 6xy – 1
(ii) l³ – 8m³ – 27n³ – 18lmn
Solution :
(i) x³ + 8y³ + 6xy – 1
= x³ + (2y)³ + (-1)³ – 3 (x) (2y) (-1)
= (x + 2y – 1) (x² + 4y² + 1 – 2xy + 2y + x)
(ii) l³ – 8m³ – 27n³ – 18lmn
= l³ – (2m)³ – (3n)³ – 3(l) (-2m) ( -3n)
= (l³ – 2m – 3n) [l² + (-2m) ² + (-3n) ² – 1× (-2m) – (- 2m) – 3n] – (3n × 1)]
= (l – 2m – 3n) (l² + 4m² + 9n² + 2lm – 6mn + 3nl)
Algebra Exercise 3.6 Solutions :
(1) Factorise the following :
(i) x² + 10x + 24
(ii) z² + 4z – 12
(iii) P² – 6p – 16
(iv) t² + 72 – 17t
(v) y² – 16y – 80
(vi) a² + 10a – 600
Solution :
(i) x² + 10x + 24
= x² + 4x + 6x + 24
= (x + 4) (x + 6)
(ii) z² + 4z – 12
= z² + 6z – 2z – 12
= z (z + 60 – 2 (z + 6)
= (z + 6) (z²)
(iii) P² – 6p – 16
= p² – 8p + 2p – 16
= (p – 8) (p + 2)
(iv) t² + 72 – 17t
= t² – 17t + 72
= t² – 8t – 9t + 72
= (t – 8) (t – 9)
(v) y² – 16y – 80
= y² – 20y + 4y – 80
= (y – 20) (y + 4)
(vi) a² + 10a – 600
= a² + 30a – 20a – 600
= a (a + 30) – 20 (a + 30)
= (a + 30) (a – 20)
(2) Factorise the following :
(i) 2a² + 9a + 10
(ii) 5x² – 29xy – 42y²
(iii) 9 – 18x + 8x²
(iv) 6x² + 16xy + 8y²
(v) (a + b)² + 9 (a + b) + 18
Solution :
(i) 2a² + 9a + 10
= 2a² + 4a + 5a + 10
= 2a (a + 2) + 5 (a + 2)
= (a + 2) (2a + 5)
(ii) 5x² – 29xy – 42y²
= 5x² – 35xy + 6xy – 42y²
= 5x (x – 7y) + 6y (x – 7y)
= (x – 7y) (5x + 6y)
(iii) 9 – 18x + 8x²
= 8x² – 6x – 12x + 9
= 2x (4x – 3) – 3 (4x – 3)
= (4x – 3) (2x – 3)
(iv) 6x² + 16xy + 8y²
= 2 (3x² + 8xy + 4y²)
= 2 (3x² + 6xy + 2xy + 4y²)
= 2 [3x (x + 2y) + 2y (x + 2y)]
= 2 [(x + 2y) (3x + 2y)]
(vi) (a + b)² + 9 (a + b) + 18
= (a + b)² + 6 (a + b) + 3 (a + b) + 18
= (a + b) [a + b] + 3 [(a + b) + 6]
= [(a + b) + 6] [(a + b) + 3]
(3) Factorise the following :
(i) (p – q)² – 6 (p – q) – 16
(ii) m² + 2mn – 24n²
(iii) √5a² + 2a – 3√5
(iv) a⁴ – 3a² + 2
(v) 8m³ – 2m²n – 15mn²
(vi) 1/x² + 1/y² + 2/xy
Solution :
(i) (p – q)² – 6 (p – q) – 16
= (p – q)² – 8 (p – q) + 2 (pq) – 16
= (p – q) [(p – q) – 8] + 2 [(p – q) – 8]
= [(p – q) – 8] [(p – q) + 2]
(ii) m² + 2mn – 24n²
= m² + 6mn – 4mn – 24n²
= m (m + 6n) – 4n (m + 6n)
= (m + 6n) (m – 4n)
(iii) √5a² + 2a – 3√5
= √5a² + 5a – 3a – 3√5
= √5a (a + √5) – 3 (a + √5)
= (a + √5) (√5a – 3)
(iv) a⁴ – 3a² + 2
= a⁴ – 2a² – a² + 2
= a² (a² – 2) – 1 (a² – 2)
= (a² – 2) (a² – 1)
(v) 8m³ – 2m²n – 15mn²
= m [8m² – 2mn – 15n²]
= m [8m² – 12mn + 10mn – 15n²]
= m [4m (2m – 3n) + 5n (2m – 3n)]
= m (2m – 3n) (4m + 5n)
(vi) 1/x² + 1/y² + 2/xy
= (1/x)² + 1/xy + 1/xy + 1/y²
= 1/x (1/x + 1/y) + 1/y (1/x + 1/y)
= (1/x + 1/y) (1/x + 1/y)
= (1/x + 1/y)²
Algebra Exercise 3.7 Solutions :
(1) Find the quotient and remainder of the following :
(i) (4x³ + 6x² – 23x + 18) ÷ (x+3)
(ii) (8y³ – 16y² + 16y – 15) ÷ (2y – 1)
(iii) (8x³ – 1) ÷ (2x -1)
(iv) (-18z + 14z² + 24z³ + 18) ÷ (3z + 4)
Solution :
(i) (4x³ + 6x² – 23x + 18) ÷ (x+3)
Here, Quotient = 4x² – 6x – 5
Remainder = +3
(ii) (8y³ – 16y² + 16y – 15) ÷ (2y – 1)
Here, Quotient = 4y² – 6y + 5
Remainder = – 10
(iii) (8x³ – 1) ÷ (2x -1)
Here, Quotient = 8z² – 6z + 2
Remainder = 10
(2) The area of a rectangle is x² + 7x + 12. if its breadth is (x+3), then find its length.
Solution :
Here, Area of the rectangle is given,
x² + 7x + 12
Here, breadth = (x + 3)
And we know area of rectangle = length * breadth
So, If we divide area/breadth
x² + 7x + 12/(x + 3)
= x + 4
Hence, the length is x + 4
(3) The base of a parallelogram is (5x+4). find its height, if the area is 25x² – 16
Solution :
Here in question given,
Area = 25x² – 16
And base = (5x+4)
We know,
Area = base × height
Therefore,
25x² – 16/(5x+4)
= 5x – 4
Therefore, its height is 5x – 4
(6) If the quotient obtained on dividing (8x⁴ – 2x² + 6x – 7) by (2x + 1) is (4x³ + px² – qx + 3), then find p, q and also the remainder.
Solution :
Quotient 4x³ – 2x³ + 3 is compared
Coefficient of x² of p = -2
Coefficient of x of q = 0
Therefore, remainder = 3
(7) If the quotient obtained on dividing 3x³ + 11x² + 34x + 106 by x – 3 is 3x² + ax + b, then find a, b and also the remainder.
Solution :
Quotient 3x² + 20x + 94 compered
Coefficient of x of a = 20
Coefficient of x of b = 94
Remainder = 388
Algebra Exercise 3.8 Solutions :
(1) Factorise each of following polynomials using synthetic division
(i) x³ – 3x² – 10 + 24
(ii) p(x) = 2x³ – 3x² – 3x + 2
(iii) p(x) = – 7x + 3 + 4x³
(iv) p(x) = x³ – 7x + 6
(v) p(x) = x³ – 10x² – x + 10
Solution :
(i) x³ – 3x² – 10 + 24
Let (p(x) = x³ – 3x² – 10x + 24
Put x = 1
P(1) = 1 – 3 – 10 + 24 ≠ 10
(x – 1) is not factor
Again put x = 1
P (-1) = -1 – 3 + 24 ≠ 0
(x + 1) not factor
Let x = 2
P(2) = 8 – 12 – 20 + 24 = 0
So (x – 2) is a factor
1 -3 – 10 24 ÷ 2
= 1 -1 – 12 0
= 1 -1 – 12 0 ÷ -3
= 1 -4 0
So, x³ – 3x² – 10x + 24 = (x – 2) (x + 3) (x – 4)
(ii) p(x) = 2x³ – 3x² – 3x + 2
Put x = 1, p (x) = 1 – 3 – 3 + 2 ≠0
Put x = – 1, p (-1) = – 2 – 3 + 3 + 2 = 0
(x + 1) is a factor
2 -3 -3 2 ÷ -1
= 2 -5 2
∴ 2x³ – 3x² – 3x + 2 = (x + 1) (x – 2) (2x + 1)
(iii) p(x) = – 7x + 3 + 4x³
Putting x = 1 p (x) = 4 + 3 – 7 = 0
(x + 1) is factor
4 0 -7 3 ÷ 1
= 4 4 – 3 0
Quotient 4x² + 4x – 3
= 4x² + 6x – 2x – 3
= 2x (2x + 3) – 1 (2x + 3)
= (2x + 3) (2x – 1)
So, 4x³ + 3 – 7x = (2x + 3 (2x – 1) (x – 1)
(iv) p(x) = x³ – 7x + 6
Put x = 1, p(1) = 1 – 7 + 6 = 0
x – 1 is a factor
1 0 -7 6 ÷ -1
= 1 1 -6 0
1 1 -6 0 ÷ 2
= 1 3 0
The factors are (x – 1), (x – 2), (x + 3)
So, x³ – 7x + 6 = (x – 10 (x – 2) (x = 3)
(v) p(x) = x³ – 10x² – x + 10
Putting x = 1, p(x) = 1 – 10 – 1 + 10 = 0
(x – 1) is a factor
1 -10 -1 10 ÷ -1
= 1 -9 -10 0
1 -9 -10 0 ÷ -1
= 1 -10 0
Therefore, x³ – 10x² – x + 10 = (x – 1) (x + 1) (x – 10)
Algebra Exercise 3.9 Solutions :
(1) Find the GCD for the following :
(i) p⁵, p^11, p⁹
(ii) 4x³, y³, z³
(iii) 9a²b²c³, 15a³b²c³
(iv) 64x⁸, 240x⁶
(v) ab²c³, a²b³c, a³bc²
(vi) 35x⁵y³z⁴ , 49x²yz³ ,14xy²
(vii) 25ab³c, 100a²bc,125ab
(viii) 3abc, 5xyz, 7pqr
Solution :
(i) p⁵, p^11, p⁹
= p⁵ = p⁵
= p^11 = p⁵. p⁶
= p⁹= p⁵.p⁴
∴ GCD = p⁵
(ii) 4x³ ,y³, z³
= y³ = y³
= z³= z³
∴ GCD = 1
(iii) 9a²b²c³, 15a³b²c³
= 9a²b²c = 3 × 3a²b² c³
= 15a³b²c⁴ = 3 × 5 × a² × b² × c³ × c
∴ GCD = 3a²b²c³
(iv) 64x⁸ , 240x⁶
= 64x⁸ = 16×4x⁸
= 240x⁶=15×16x⁶
∴ GCD =16x⁶
(v) ab²c³, a²b³c, a³bc²
= ab²c³, a²b³c, a³bc²
GCD = abc
(vi) 35x⁵y³z⁴, 49x²yz³, 14xy²
= 35x⁵y³z⁴ = 7×5×x⁵y³z⁴
= 49x²yz³ = 7×7x²yz³
=14xy²z² = 7×2xy²z²
∴ GCD = 7xyz²
(vii) 25ab³c, 100a²bc,125ab
= 25ab³c = 25ab³c
= 100a²bc = 25×4a²bc
= 125ab = 5 × 25ab
∴ GCD = 25ab
(viii) 3abc ,5xyz, 7pqr
= 3abc, 5xyz, 7pqr
∴ GCD = 1
(2) Find the GCD of the following :
(i) (2x + 5 ), (5x + 2)
(ii) a^m + 1, a^m + 2, a^m + 3
(iii) 2a² + a, 4a² -1
(iv) 3a², 5b³, 7c⁴
(v) x⁴ – 1, x² – 1
(vi) a³ – 9axc, (a – 3x)²
Solution :
(i) (2x + 5 ), (5x + 2)
= 2x + 5, 5x + 2
∴ GCD = 1
(ii) a^m + 1, a^m + 2,a^m + 3
= a^m + 1, a^m + 2, a^m + 3
∴ GCD = a^m + 1
(iii) 2a² + a,4a² – 1
= a(2a + 1),(2a + 1)(2a – 1)
∴ GCD = (2a + 1)
(iv) 3a²,5b³ ,7c⁴
= 3a²,56³,7c⁴
∴ GCD = 1
(v) x⁴ – 1, x² – 1
= (x² – 1) (x² + 1), (x² – ¬¬1)
∴ GCD = x² – 1
(vi) a³ – 9axc, (a – 3x) ²
= a(a² – 9ax²), (a – 3x) ²
= a(a – 3x) (a + 3x) , (a – 3x) (a – 3x)
∴ GCD = (a – 3x)
Algebra Exercise 3.11 Solutions :
(1) Solve, using the method of substitution.
(i) 2x – 3y = 7; 5x + y = 9
(ii) 1.5x + 0.1y = 6.2; 3x – 0.4y = 11.2
(iii) 10% of x + 20% of y = 24; 3x – y = 20
(iv) √2x – √3y = 1; √3x – √8y = 0
Solution :
(i) 2x – 3y = 7; 5x + y = 9
2x-3y = 7- (i)
5x + y = 9- (ii)
= 5x + y = 9
= y = 9 – 5x
Then, we get form (i)
2x – 3(9 – 5x)=7
= 2x – 27 = 15x = 7
= 17x = 34
= x = 34/17 = 2
So, y = 9 – 34/17 × 5
9 – 10 = -1
X = 2
Y = -1
(ii) 1.5x + 0.1y = 6.2; 3x – 0.4y = 11.2
1.5x + 0.1y = 6.2 …… (i)
3x – 0.4y = 11.2 …… (ii)
y = 0.62 – 1.5x
Now, 3x – 0.4 (6.2 – 1.5x) = 11.2
= 3x + 6x – 24.8 = 11.2
= + 9x = 11.2 + 24.8
= x = 36/9 = 4
0.1y = 6.2 – 1.5×4
20.1y = 6.2 – 6
y = 2
x = 4
y = 2
(iii) 10% of x + 20% of y = 24; 3x – y = 20
10/100 x + 20/100 y = 24
= 10x + 20y = 24×100
= x + 2y = 240
3x – y = 20
= y = 3x – 20
So, x + 2 (3x – 20) = 240
= x + 6x + 40 = 240
= 7x = 280
= x = 40
So, y = 3×40 – 20
= 120 – 20
= 100
x = 40
y = 100
(iv) √2x – √3y = 1; √3x – √8y = 0
√2x – √3y = 1 …… (i)
√3x – √8y = 0 …… (ii)
∵ √3x = √8y
= y = √3/√8 x
So, √2x – √3× √3/√8x = 1
= 4x – 3x/√8 = 1
= x = √8
So, y = √3/√8 ×√8
= √3
So, x = √8
y = √3
(2) Raman’s age is three times the sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman.
Solution :
Let, Raman’s age be = a
Sum of age of his sons are = b
So a = 3b
So, a + 5 = 2 (b+10)
= a + 5 = 2b + 20
= a – 2b = 15
So, 3b – 2b = 15
b = 15
So, a = 3×15 = 45
∴ Raman’s age is 45
(3) The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. If the digits are reversed, the number so formed exceeds the original number by 495. Find the number.
Solution :
Let, the number be x and y
∴ and x + y = 13 ……. (i)
So, 100y + x (100x + y) = 495
100y + x – 100x – y = 495
= – 99x + 99y = 495 ……. (ii)
Multiplying (i) by 99 and subtract from (ii)
– 99x + 99y = 495
+ 99x + 99y = 1287
——————————-
198y = 792
y = 792/798 = 4
So, y = 4 and x = 13 – 4 = 9
Therefore, the number is 409.
Algebra Exercise 3.12 Solutions :
(1) Solve by the method of elimination
(i) 2x – y = 3; 3x + y = 7
(ii) x – y = 5; 3x + 2y = 25
(iii) x/10 + y/5 = 14; x/8 + y/6 = 15
(iv) 3 (2x+ y) = 7xy; 3 (x+3y) = 11xy
(v) 4/x + 5y = 7; 3/x + 4y = 5
(vi) 13x + 11y = 70; 11x + 13y = 74
Solution :
(i) 2x – y = 3; 3x + y = 7
2x – y = 3
3x + y = 7
(+) (+) (-)
———————————-
5x = 10
∴ x = 2
y = 2x – 3
= 4 – 3 = 1
x = 2
y = 1
(ii) x – y = 5; 3x + 2y = 25
x – y = 5 …… (i)
3x + 2y = 25
3x – 3y = 15
(+) (-) (-)
———————————-
5y = 10
∴ y = 2
x = 5+2 = 3
x = 7
y = 2
(iii) x/10 + y/5 = 14; x/8 + y/6 = 15
x/10 + y/5 = 14
= x + 2y = 140
x/8 + y/6 = 15
= 3x + 4y/24 = 15
= 3x + 4y = 360
So, 3x + 4y = 360
3x + 6y = 420
(-) (-) (-)
———————————-
-2y = -60
∴ y = 30
x = 140 – 60
= 80
x = 80
y = 30
(iv) 3 (2x+ y) = 7xy; 3 (x+3y) = 11xy
3 (2x + y) = 7xy
= 6x + 3y = 7xy ……. (i)
3 (x + 3y) = 11xy
= 3x + 9y = 11xy ……. (ii)
Dividing (i) and (ii) by xy we get and put 1/x = a, 1/y = b
= 3a + 6b= 7 ……. (iii)
And 3/y + 9/x = 11
= 9a + 3b = 11 ……. (iv)
9a + 3b = 11
9a + 18b = 21
(-) (-) (-)
———————————-
15b = 10
= b = 10/15 = 2/3
And 3a = 7 – 6× 2/3
= a = 5/3 = 1
∴ 1/x = 1
= x = 1
And 1/y = 2/3
y = 3/2
x = 1
y = 3/2
(v) 4/x + 5y = 7; 3/x + 4y = 5
4/x + 5y = 7
3/x + 4y = 5
Putting 1/x = a and get
4a + 5y = 7 …… (i)
3a + 4y = 5 ……. (ii)
So, 12a + 15y = 21
12a + 16y = 20
(-) (-) (-)
———————————-
– y = 1
= y = – 1
So, 4a = 7+5
= a = 3
So, 1/x = a = 3
= x = 1/3
x = 1/3
y = – 1
(vi) 13x + 11y = 70; 11x + 13y = 74
13x + 11y = 70 ……… (i)
11x + 13y = 74 …….. (ii)
143x + 121y = 770
143x + 169y = 962
———————————-
– 48y = – 192
= y = 4
∴ 13x = 70 – 11×4
= x = 70-44/13
= x = 26/13 = 2
x = 2
y = 4
(2) The monthly income of A and B are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 5,000 per month, find the monthly income of each.
Solution :
Let the monthly income be 3x and 4x
So and the expenditures be 5y d 7y
So, 3x – 5y = 5000 ….(i)
4x – 7y = 5000 ….(ii)
12x – 20y = 20000
12x – 21y = 15000
(-) (+) (-)
———————————-
y = 5000
So, 3x = 5000 + 25000
= x = 10000
Monthly of A = 30000 Rs
Monthly of B = 40000 Rs
(3) Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.
Solution :
Let, the age of son x year
Therefore, 5 years ago son’s = (x-5) year
Now we have,
3(x + 5) = 7(x-5) + 10
= 3x + 15 = 7x – 35 + 10
= 3x – 7x = – 35 + 10 – 10
= -4x = – 40
= x = 10
Son’s age 10 year
Present age of father = 3(x + 5) – 5
= 3(10 + 5) – 5
= (45 – 5)
= 40 year
Algebra Exercise 3.13 Solutions :
(1) Solve by cross-multiplication method :
(i) 8x – 3y = 12; 5x = 2y + 7
(ii) 6x + 7y – 11 = 0; 5x + 2y = 13
(iii) 2/x + 3/y = 5; 3/x – 1/y + 9 = 0
Solution :
(i) 8x – 3y = 12; 5x = 2y + 7
8x – 3y = 12
5x – 2y = 7
From, cross-multiplication method we get
x/22 = b/-33 = 1/-11
= x/22 = 1/-33
∴ x = 22/-11
b/-33 = 1/-11
= b = 3
(ii) 6x + 7y – 11 = 0; 5x + 2y = 13
6x + 7y – 11 = 0
5x + 2y – 13 = 0
x/-91 + 22 = y/78 – 55 = 1/12 – 35
x/-69 = y/23 = 1/23
x/-69 = 1/23
= x = 3
y/23 = 1/23
= y = 1
(iii) 2/x + 3/y = 5; 3/x – 1/y + 9 = 0
2/x + 3/y = 5, 3/x – 1/y + 9 = 0
Put, 1/x = a, 1/y = b
2a + 3b – 5 = 0
3a – b + 9 = 0
a/27-5 = b/18+15 = 1/-2-9
= a/22 = b/33 = 1/-11
a/22 = 1/-11
= a = -2
b/33 = 1/-11
= b = -3
So, x = – 1/2
y = – 1/3
(2) Akshaya has 2 rupee coins and 5 rupee coins in her purse. If in all she has 80 coins totalling ₹ 220, how many coins of each kind does she have.
Solution :
Let, number of 2 rupee coins be x
Number of 5 rupee coins be y
x + y = 80 = x + y – 80 = 0
2x + 5y = 220 = 2x + 5y – 220 = 0
x/-220 + 400 = y/-220 + 160 = 1/5 – 2
= x/180 = y/60 = 1/3
= x = 180/3
= 60
y/60 = 1/3
= y = 20
Thus, number of 2 rupee coins will be 60
And number of 5 rupee coins will be 20
(3) It takes 24 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 8 hours and the pipe of the smaller diameter is used for 18 hours. Only half of the pool is filled. How long would each pipe take to fill the swimming pool.
Solution :
Let the larger pipe take time hours
The smaller pipe take time y hours
∴ 1/x + 1/y = 1/24
In 1 hour the larger pipe can fill = 1/x
In 1 hour the smaller pipe can fill = 1/y
8/x + 18/y = 1/2
Put 1/x = a, 1/y = b
a + b = 1/24
= 24a + 24b – 1 = 0
8a + 18b = 1/2
= 16a + 36b – 1 = 6
a/-24+36 = b/-16 + 24 = 1/264 – 384
= a/12 = b/8 = 1/480
a = 12/480 = 1/40
So, x = 40
b = 8/480 = 1/60
y = 60
Algebra Exercise 3.15 Solutions :
(1) If x³ + 6x² + kx + 6 is exactly divisible by (x + 2), then k = ?
(1) -6
(2) -7
(3) -8
(4) 11
Solution :
(-2)³ + 6(-2)² + k(-2) + 6 = 0
= – 8 – 24 – 2k + 6 = 0
= k = 11
Hence, alternative (4) 11 is the correct answer of this question.
(2) The root of the polynomial equation 2x + 3 = 0 is
(1) 1/3
(2) – 1/3
(3) – 3/2
(4) – 2/3
Solution :
Correct option – (3)
The root of the polynomial equation 2x + 3 = 0 is -3/2
(3) The type of the polynomial 4 – 3x³ is
(1) Constant polynomial
(2) Linear polynomial
(3) Quadratic polynomial
(4) Cubic polynomial
Solution :
Correct option – (4)
The polynomial 4 – 3x³ is a type of Cubic polynomial.
(4) If x^51 + 51 is divided by x + 1, then the remainder is
(1) 0
(2) 1
(3) 49
(4) 50
Solution :
Correct option – (4)
If x^51 + 51 is divided by x + 1, then the remainder is 50.
(5) The zero of the polynomial 2x + 5 is
(1) 5/2
(2) -5/2
(3) 2/5
(4) -2/5
Solution :
Correct option – (2)
The zero of the polynomial 2x+5 is -5/2.
(6) The sum of the polynomials p(x) = x³ – x² – 2, q(x) = x² – 3x + 1
(1) x³ – 3x – 1
(2) x³ + 2x² – 1
(3) x³ – 2x² – 3x
(4) x³ – 2x² + 3x – 1
Solution :
Correct option – (1)
The Sum will be x³ – 3x – 1
(7) Degree of the polynomials (y³-2) (y³+1) is
(1) 9
(2) 2
(3) 3
(4) 6
Solution :
Correct option – (4)
Degree of the polynomials (y³-2) (y³+1) is 6.
(8) Let the polynomials be
(A) – 13q⁵ + 4q² + 12q
(B) (x² + 4) (x² + 9)
(C) 4q⁸ – q⁶ + q²
(D) -5/7 y^12 + y³ + y⁵
Then ascending order of their degree is
(1) A, B, D, C
(2) A, B, C, D
(3) B, C, D, A
(4) B, A, C, D
Solution :
Correct option – (4)
B, A, C, D is the correct order.
(9) If p(a) = 0 then (x – a) is a _____ of p(x)
(1) Divisor
(2) Quotient
(3) Remainder
(4) Factor
Solution :
Correct option – (4)
If p(a) = 0 then (x – a) is a “Factor” of p(x).
(10) Zeros of (2 – 3x) is ___
(1) 3
(2) 2
(3) 2/3
(4) 3/2
Solution :
Correct option – (3)
Zeros of (2 – 3x) is 2/3.
(12) If x – 3 is a factor of p(x), then the remainder is
(1) 3
(2) -3
(3) p(3)
(4) p(-3)
Solution :
Correct option – (3)
If x – 3 is a factor of p(x), then the remainder is p(3).
(13) (x + y) (x² – xy + y²) is equal to
(1) (x + y)³
(2) (x – y)³
(3) x³ + y³
(4) x³ – y³
Solution :
Correct option – (3)
(x + y) (x² – xy + y²) is equal to x³ + y³.
(14) (a + b – c)² is equal to
(1) (a – b + c)²
(2) (-a – b + c)²
(3) (a + b + c)²
(4) (a – b – c)²
Solution :
Correct option – (2)
(a + b – c)² is equal to (-a – b + c)²
(15) If (x + 5) and (x – 3) are the factors of ax²+bx+c,then values of a, b and c are
(1) 1,2,3
(2) 1,2,15
(3) 1,2,-15
(4) 1, -2,15
Solution :
Correct option – (3)
The value of A = 1
The value of B = 2
The value of C = -15
(16) Cubic polynomial may have maximum of ___ linear factors
(1) 2
(2) 2
(3) 3
(4) 4
Solution :
Correct option – (3)
Cubic polynomial may have maximum of 3 linear factors.
(17) Degree of the constant polynomial is ___
(1) 3
(2) 2
(3) 1
(4) 0
Solution :
Correct option – (4)
Degree of the constant polynomial is 0.
(19) Find the value of m form the equation 2x + 3y =m. if its one solution is x = 2 and y = – 2
(1) 3
(2) -2
(3) 10
(4) 0
Solution :
Correct option – (2)
Therefore, the value of m will be -2.
(20) Which of the following is a linear equation
(1) x + 1/x = 2
(2) x (x – 1) = 2
(3) 3x + 5 = 2/3
(4) x³ – x = 5
Solution :
Correct option – (3)
Here, 3x + 5 = 2/3 is a linear equation.
(22) If (2, 3) is a solution of linear equation 2x + 3y = k then, the value of k is
(1) 12
(2) 6
(3) 0
(4) 13
Solution :
Correct option – (4)
So, the value of k is = 13
(23) Which condition does not satisfy the linear equation ax + by + c = 0
(1) a ≠ 0, b = 0
(2) a = 0, b ≠ 0
(3) a = 0, b = 0, c ≠ 0
(4) a ≠ 0, b ≠ 0
Solution :
Correct option – (3)
a = 0, b = 0, c ≠ 0 does not satisfy the linear equation ax + by + c = 0
(24) Which of the following is not a linear equation in two variable
(1) ax + by + c = 0
(2) 0x + 0y + c = 0
(3) 0x + by + c = 0
(4) ax + 0y + c = 0
Solution :
Correct option – (2)
Here, 0x + 0y + c = 0 is not a linear equation in two variable.
(25) The value of k for which the pair of linear equations 4x + 6y -1 = 0 and 2x + ky – 7 = 0 represents parallel lines is
(1) k = 3
(2) k = 2
(3) k = 4
(4) k = -3
Solution :
Correct option – (1)
k = 3 is the correct answer to this question.
(29) GCD of any two prime numbers is ____
(1) -1
(2) 0
(3) 1
(4) 2
Solution :
Correct option – (3)
GCD of any two prime numbers is 1.
(30) The GCD of x⁴ – y⁴ and x² – y² is
(1) x⁴ – y⁴
(2) x² – y²
(3) (x + y) ²
(4) (x + y)⁴
Solution :
Correct option – (2)
The GCD of x⁴ – y⁴ and x² – y² is x² – y².
Next Chapter Solution :
👉 Geometry
