Class 9 Maths Chapter 4 End of Chapter Solutions – Ganita Manjari
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 4 End of Chapter (EOC) Exercise. This review section combines all core algebraic identities, factorization techniques, polynomial reductions, and word problems to ensure you are fully prepared for your school examinations.
🔑 Chapter Overview & Key Formulas:
- Binomial & Trinomial Squares: \((a \pm b)^2 = a^2 \pm 2ab + b^2\) and \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)
- Difference of Squares & Cubes: \(a^2 - b^2 = (a - b)(a + b)\) and \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)
- Cubic Expansions: \((a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3\) and \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
NCERT Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercise Solutions
Question 1
Use suitable identities to find the following products:
Solution:
(i) \((-3x + 4)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = -3x\) and \(b = 4\):
\(= (-3x)^2 + 2(-3x)(4) + (4)^2\)
\(= \mathbf{9x^2 - 24x + 16}\)
(ii) \((2s + 7)(2s - 7)\)
Using \((a + b)(a - b) = a^2 - b^2\) with \(a = 2s\) and \(b = 7\):
\(= (2s)^2 - (7)^2\)
\(= \mathbf{4s^2 - 49}\)
(iii) \(\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)\)
Using the difference of squares identity with \(a = p^2\) and \(b = \frac{1}{2}\):
\(= (p^2)^2 - \left(\frac{1}{2}\right)^2\)
\(= \mathbf{p^4 - \frac{1}{4}}\)
(iv) \((2n + 7)(2n - 7)\)
Using \((a + b)(a - b) = a^2 - b^2\) with \(a = 2n\) and \(b = 7\):
\(= (2n)^2 - (7)^2\)
\(= \mathbf{4n^2 - 49}\)
(v) \((s - 2t)(s^2 + 2st + 4t^2)\)
Using the sum of cubes identity \((a - b)(a^2 + ab + b^2) = a^3 - b^3\) with \(a = s\) and \(b = 2t\):
\(= (s)^3 - (2t)^3\)
\(= \mathbf{s^3 - 8t^3}\)
(vi) \(\left(\frac{1}{2r} - 4r\right)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = \frac{1}{2r}\) and \(b = 4r\):
\(= \left(\frac{1}{2r}\right)^2 - 2\left(\frac{1}{2r}\right)(4r) + (4r)^2\)
\(= \mathbf{\frac{1}{4r^2} - 4 + 16r^2}\)
(vii) \((-3m + 4k - l)^2\)
Using \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\) with \(a = -3m\), \(b = 4k\), and \(c = -l\):
\(= (-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m)\)
\(= \mathbf{9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm}\)
(viii) \(\left(x - \frac{1}{3}y\right)^3\)
Using \((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\) with \(a = x\) and \(b = \frac{1}{3}y\):
\(= (x)^3 - 3(x)^2\left(\frac{1}{3}y\right) + 3(x)\left(\frac{1}{3}y\right)^2 - \left(\frac{1}{3}y\right)^3\)
\(= \mathbf{x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3}\)
(ix) \(\left(\frac{7}{2}k - \frac{2}{3}m\right)^3\)
Using the cubic identity with \(a = \frac{7}{2}k\) and \(b = \frac{2}{3}m\):
\(= \left(\frac{7}{2}k\right)^3 - 3\left(\frac{7}{2}k\right)^2\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{2}{3}m\right)^2 - \left(\frac{2}{3}m\right)^3\)
\(= \mathbf{\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3}\)
Question 2
Find the values using suitable identities:
Solution:
(i) \(17 \times 21\)
\(= (19 - 2)(19 + 2) = (19)^2 - (2)^2\)
\(= 361 - 4 = \mathbf{357}\)
(ii) \(104 \times 96\)
\(= (100 + 4)(100 - 4) = (100)^2 - (4)^2\)
\(= 10000 - 16 = \mathbf{9984}\)
(iii) \(24 \times 16\)
\(= (20 + 4)(20 - 4) = (20)^2 - (4)^2\)
\(= 400 - 16 = \mathbf{384}\)
(iv) \(147^3\)
\(= (150 - 3)^3 = (150)^3 - 3(150)^2(3) + 3(150)(3)^2 - (3)^3\)
\(= 3375000 - 202500 + 4050 - 27 = \mathbf{3176523}\)
(v) \(199^3\)
\(= (200 - 1)^3 = (200)^3 - 3(200)^2(1) + 3(200)(1)^2 - (1)^3\)
\(= 8000000 - 120000 + 600 - 1 = \mathbf{7880599}\)
(vi) \(127^3\)
\(= (130 - 3)^3 = (130)^3 - 3(130)^2(3) + 3(130)(3)^2 - (3)^3\)
\(= 2197000 - 152100 + 3510 - 27 = \mathbf{2048383}\)
(vii) \((-107)^3\)
\(= -(100 + 7)^3 = -[(100)^3 + 3(100)^2(7) + 3(100)(7)^2 + (7)^3]\)
\(= -(1000000 + 210000 + 14700 + 343) = \mathbf{-1225043}\)
(viii) \((-299)^3\)
\(= -(300 - 1)^3 = -[(300)^3 - 3(300)^2(1) + 3(300)(1)^2 - (1)^3]\)
\(= -(27000000 - 2700000 + 900 - 1) = \mathbf{-24300899}\)
Question 3
Factor the following algebraic expressions:
Solution:
(i) \(4y^2 + 1 + \frac{1}{16y^2}\)
\(= (2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2 = \mathbf{\left(2y + \frac{1}{4y}\right)^2}\)
(ii) \(9m^2 - \frac{1}{25n^2}\)
\(= (3m)^2 - \left(\frac{1}{5n}\right)^2 = \mathbf{\left(3m - \frac{1}{5n}\right)\left(3m + \frac{1}{5n}\right)}\)
(iii) \(27b^3 - \frac{1}{64b^3}\)
\(= (3b)^3 - \left(\frac{1}{4b}\right)^3 = \mathbf{\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)}\)
(iv) \(x^2 + \frac{5x}{6} + \frac{1}{6}\)
Splitting middle term \(\frac{5}{6}x\) into \(\frac{1}{2}x + \frac{1}{3}x\):
\(= x^2 + \frac{1}{2}x + \frac{1}{3}x + \frac{1}{6} = \mathbf{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)}\)
(v) \(27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}\)
\(= (3u)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3 = \mathbf{\left(3u - \frac{1}{5}\right)^3}\)
(vi) \(64y^3 + \frac{1}{125}z^3\)
\(= (4y)^3 + \left(\frac{1}{5}z\right)^3 = \mathbf{\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2\right)}\)
(vii) \(p^3 + 27q^3 + r^3 - 9pqr\)
\(= (p)^3 + (3q)^3 + (r)^3 - 3(p)(3q)(r) = \mathbf{(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)}\)
(viii) \(9m^2 - 12m + 4\)
\(= (3m)^2 - 2(3m)(2) + (2)^2 = \mathbf{(3m - 2)^2}\)
(ix) \(9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz\)
Taking out \(\frac{1}{3}\) common: \(\frac{1}{3}(27x^3 - 8y^3 + z^3 + 18xyz)\)
\(= \frac{1}{3}\left[(3x)^3 + (-2y)^3 + (z)^3 - 3(3x)(-2y)(z)\right]\)
\(= \mathbf{\frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3xz)}\)
(x) \(4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy\)
\(= (2x)^2 + (3y)^2 + (6z)^2 + 2(2x)(3y) + 2(3y)(6z) + 2(6z)(2x) = \mathbf{(2x + 3y + 6z)^2}\)
(xi) \(27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}\)
\(= (3u)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right)^3 = \mathbf{\left(3u - \frac{1}{6}\right)^3}\)
Question 4
Simplify the following:
Solution:
(i) \(\frac{4x^2 + 4x + 1}{4x^2 - 1}\)
Numerator \(= (2x + 1)^2\), Denominator \(= (2x - 1)(2x + 1)\)
\(= \frac{(2x + 1)^2}{(2x - 1)(2x + 1)} = \mathbf{\frac{2x + 1}{2x - 1}}\)
(ii) \(\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}\)
\(= \frac{27(a^3 - (2b)^3)}{9(a^2 - (2b)^2)} = \frac{3(a - 2b)(a^2 + 2ab + 4b^2)}{(a - 2b)(a + 2b)} = \mathbf{\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}}\)
(iii) \(\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}\)
Numerator \(= (s + 5t)(s^2 - 5st + 25t^2)\), Denominator \(= (s - 7t)(s + 5t)\)
\(= \frac{(s + 5t)(s^2 - 5st + 25t^2)}{(s - 7t)(s + 5t)} = \mathbf{\frac{s^2 - 5st + 25t^2}{s - 7t}}\)
Question 5
Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units:
Solution:
(i) \(25a^2 - 30ab + 9b^2\)
Factorizing the area expression: \((5a - 3b)^2 = (5a - 3b)(5a - 3b)\)
Therefore, possible expressions are:
\(\mathbf{\text{Length} = 5a - 3b, \quad \text{Breadth} = 5a - 3b}\)
(ii) \(36s^2 - 49t^2\)
Factorizing the area expression: \((6s)^2 - (7t)^2 = (6s - 7t)(6s + 7t)\)
Therefore, possible expressions are:
\(\mathbf{\text{Length} = 6s + 7t, \quad \text{Breadth} = 6s - 7t}\)
Question 6
Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units:
Solution:
(i) \(6a^2 - 24b^2\)
\(= 6(a^2 - 4b^2) = 6(a - 2b)(a + 2b)\)
Possible expressions for dimensions are: \(\mathbf{6, \, (a - 2b), \, (a + 2b)}\)
(ii) \(3ps^2 - 15ps + 12p\)
\(= 3p(s^2 - 5s + 4) = 3p(s - 1)(s - 4)\)
Possible expressions for dimensions are: \(\mathbf{3p, \, (s - 1), \, (s - 4)}\)
Question 7
The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Solution:
Side of inner square playground \(= 40\) m.
Area of inner playground \(= 40^2 = 1600\text{ m}^2\).
Since a path of width \(s\) is created all around, the side of the outer square \(= (40 + 2s)\) m.
Area of outer square \(= (40 + 2s)^2 = 1600 + 160s + 4s^2\text{ m}^2\).
Area of the path \(= \text{Area of outer square} - \text{Area of inner square}\)
\(= (1600 + 160s + 4s^2) - 1600\)
\(= \mathbf{4s^2 + 160s}\text{ square metres}\).
Question 8
If a number plus its reciprocal equals \(\frac{10}{3}\), find the number.
Solution:
Let the number be \(x\). According to the given condition:
\(x + \frac{1}{x} = \frac{10}{3}\)
\(\frac{x^2 + 1}{x} = \frac{10}{3}\)
\(3(x^2 + 1) = 10x\)
\(3x^2 - 10x + 3 = 0\)
Splitting middle term \(-10x\) into \(-9x - x\):
\(3x^2 - 9x - x + 3 = 0\)
\(3x(x - 3) - 1(x - 3) = 0\)
\((3x - 1)(x - 3) = 0\)
Therefore, \(x = 3\) or \(x = \frac{1}{3}\).
The number is \(\mathbf{3}\) (or its reciprocal \(\mathbf{\frac{1}{3}}\)).
Question 9
A rectangular pool has area \(2x^2 + 7x + 3\) square hastas. If its width is \(2x + 1\) hastas, find its length.
Solution:
\(\text{Area} = \text{Length} \times \text{Width}\)
\(\text{Area} = 2x^2 + 7x + 3\)
Factorizing the area by splitting the middle term (\(6x\) and \(x\)):
\(= 2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)\)
Since \(\text{Width} = 2x + 1\):
\(\text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{(2x + 1)(x + 3)}{2x + 1} = \mathbf{x + 3}\text{ hastas}\).
Question 10
If both \(x - 2\) and \(x - \frac{1}{2}\) are factors of \(px^2 + 5x + r\), show that \(p = r\).
Solution:
Let \(f(x) = px^2 + 5x + r\).
Since \(x - 2\) is a factor, \(f(2) = 0\):
\(p(2)^2 + 5(2) + r = 0 \implies 4p + 10 + r = 0 \implies 4p + r = -10 \quad \text{--- (Equation 1)}\)
Since \(x - \frac{1}{2}\) is a factor, \(f\left(\frac{1}{2}\right) = 0\):
\(p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r = 0 \implies \frac{p}{4} + \frac{5}{2} + r = 0\)
Multiplying by 4: \(p + 10 + 4r = 0 \implies p + 4r = -10 \quad \text{--- (Equation 2)}\)
Comparing Equation 1 and Equation 2:
\(4p + r = p + 4r\)
\(4p - p = 4r - r\)
\(3p = 3r \implies \mathbf{p = r}\) (Hence proved).
Question 11
If \(a + b + c = 5\) and \(ab + bc + ca = 10\), then prove that \(a^3 + b^3 + c^3 - 3abc = -25\).
Solution:
We know the standard identity:
\(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
First, find \(a^2 + b^2 + c^2\) using the identity \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\):
\(5^2 = a^2 + b^2 + c^2 + 2(10)\)
\(25 = a^2 + b^2 + c^2 + 20 \implies a^2 + b^2 + c^2 = 5\)
Now substitute into the cubic expression:
\(a^3 + b^3 + c^3 - 3abc = (5) \cdot [(a^2 + b^2 + c^2) - (ab + bc + ca)]\)
\(= 5 \cdot [5 - 10] = 5 \cdot (-5) = \mathbf{-25}\) (Hence proved).
Question 12
By factoring the expression, check that \(n^3 - n\) is always divisible by 6 for all natural numbers n. Give reasons.
Solution:
Factorizing the expression completely:
\(n^3 - n = n(n^2 - 1) = \mathbf{(n - 1)n(n + 1)}\)
Reasoning:
\((n - 1)n(n + 1)\) represents the product of three consecutive natural numbers. Among any three consecutive integers:
1. At least one number is always an even number (divisible by 2).
2. Exactly one number is always a multiple of 3.
Since both 2 and 3 are prime numbers and divide the product, their product \(2 \times 3 = 6\) must also divide the expression completely for all natural numbers \(n\).
Question 13
Find the value of:
Solution:
(i) \(x^3 + y^3 - 12xy + 64\), when \(x + y = -4\)
Rewrite the expression as: \(x^3 + y^3 + 4^3 - 3(x)(y)(4)\)
Using the identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\) with \(a = x, b = y, c = 4\):
Since \(x + y = -4\), we have \(x + y + 4 = 0\).
Therefore, \((x + y + 4)(\dots) = 0 \times (\dots) = \mathbf{0}\).
(ii) \(x^3 - 8y^3 - 36xy - 216\), when \(x = 2y + 6\)
Given \(x = 2y + 6 \implies x - 2y - 6 = 0\).
Rewrite the expression as: \(x^3 + (-2y)^3 + (-6)^3 - 3(x)(-2y)(-6)\)
Using the cubic identity with \(a = x, b = -2y, c = -6\):
Since \(a + b + c = x - 2y - 6 = 0\), the entire expression equals \(\mathbf{0}\).
Frequently Asked Questions (FAQs)
Q1: How do you handle complex cubic expansions in EOC problems?
A: Apply standard formulas like \((a \pm b)^3\) or group terms carefully to match the trinomial sum-of-cubes identity.
Q2: How is divisibility proven algebraically, like in Question 12?
A: Factorize the expression into consecutive integers and apply basic number theory properties regarding factors of 2 and 3.
Q3: What is the shortcut for problems like Question 11?
A: Use the expanded square identity to find \(a^2 + b^2 + c^2\) first, then substitute it directly into the expanded factored form of \(a^3 + b^3 + c^3 - 3abc\).
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