SEBA Class 10 Maths Chapterwise Solutions (2026–27) – Updated HSLC Answers PDF & Important Questions

Q1(i). Find the roots of the quadratic equation by factorisation:
\(x^2 - 3x - 10 = 0\)

Answer:
Given equation:
\[ x^2 - 3x - 10 = 0 \]
We need two numbers whose: Sum = −3 and Product = −10 These numbers are −5 and 2 Split the middle term: \[ x^2 - 5x + 2x - 10 = 0 \] Group terms: \[ x(x - 5) + 2(x - 5) = 0 \] Take common factor: \[ (x - 5)(x + 2) = 0 \] So, \[ x - 5 = 0 \quad \text{or} \quad x + 2 = 0 \]
\[ x = 5 \quad \text{or} \quad x = -2 \]
Therefore, the roots are 5 and −2.

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(ii) \(2x^2 + x - 6 = 0\)

Answer:
Given equation:
\[ 2x^2 + x - 6 = 0 \]
Multiply 2 × (−6) = −12
We need two numbers whose sum = 1 and product = −12
These numbers are 4 and −3

Split the middle term:
\[ 2x^2 + 4x - 3x - 6 = 0 \]
Group terms:
\[ 2x(x + 2) - 3(x + 2) = 0 \]
Take common factor:
\[ (2x - 3)(x + 2) = 0 \]
So,
\[ 2x - 3 = 0 \quad \text{or} \quad x + 2 = 0 \] \[ x = \frac{3}{2} \quad \text{or} \quad x = -2 \]
Therefore, the roots are \( \frac{3}{2} \) and −2.

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(iii) \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)

Answer:
Given:
\[ \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \]
Multiply \(\sqrt{2} \times 5\sqrt{2} = 10\)
Numbers with sum = 7 and product = 10 → 5 and 2

Split the middle term:
\[ \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \]
Group terms:
\[ x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \]
\[ (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \]
\[ x = -\frac{5}{\sqrt{2}},\; x = -\sqrt{2} \]

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(iv) \(2x^2 - x + \frac{1}{8} = 0\)

Answer:
Given:
\[ 2x^2 - x + \frac{1}{8} = 0 \]
Multiply the whole equation by 8 to remove fraction:
\[ 16x^2 - 8x + 1 = 0 \]
Multiply 16 × 1 = 16
We need two numbers whose sum = −8 and product = 16
These numbers are −4 and −4

Split the middle term:
\[ 16x^2 - 4x - 4x + 1 = 0 \]
Group terms:
\[ 4x(4x - 1) -1(4x - 1) = 0 \]
Take common factor:
\[ (4x - 1)(4x - 1) = 0 \]
\[ 4x - 1 = 0 \Rightarrow x = \frac{1}{4} \]
Therefore, both roots are equal.

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(v) \(100x^2 - 20x + 1 = 0\)

Answer:
Given:
\[ 100x^2 - 20x + 1 = 0 \]
Multiply 100 × 1 = 100
We need two numbers whose sum = −20 and product = 100
These numbers are −10 and −10

Split the middle term:
\[ 100x^2 - 10x - 10x + 1 = 0 \]
Group terms:
\[ 10x(10x - 1) -1(10x - 1) = 0 \]
Take common factor:
\[ (10x - 1)(10x - 1) = 0 \]
\[ 10x - 1 = 0 \Rightarrow x = \frac{1}{10} \]
Therefore, both roots are equal.

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(vi) \(2x^2 - 7x + 6 = 0\)

Answer:
Given:
\[ 2x^2 - 7x + 6 = 0 \]
Multiply 2 × 6 = 12
We need two numbers whose sum = −7 and product = 12
These numbers are −3 and −4

Split the middle term:
\[ 2x^2 - 3x - 4x + 6 = 0 \]
Group terms:
\[ x(2x - 3) - 2(2x - 3) = 0 \]
Take common factor:
\[ (2x - 3)(x - 2) = 0 \]
\[ x = \frac{3}{2},\; x = 2 \]

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(vii) \(x^2 - 10x - 96 = 0\)

Answer:
Given:
\[ x^2 - 10x - 96 = 0 \]
Multiply 1 × (−96) = −96
We need two numbers whose sum = −10 and product = −96
These numbers are −16 and 6

Split the middle term:
\[ x^2 - 16x + 6x - 96 = 0 \]
Group terms:
\[ x(x - 16) + 6(x - 16) = 0 \]
Take common factor:
\[ (x - 16)(x + 6) = 0 \]
\[ x = 16,\; x = -6 \]

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(viii) \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)

Answer:
Given:
\[ \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \]
Multiply:
\[ \sqrt{3} \times 7\sqrt{3} = 21 \]
We need two numbers whose sum = 10 and product = 21
These numbers are 7 and 3

Split the middle term:
\[ \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \]
Group terms:
\[ x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \]
Take common factor:
\[ (\sqrt{3}x + 7)(x + \sqrt{3}) = 0 \]
\[ x = -\frac{7}{\sqrt{3}},\; x = -\sqrt{3} \]

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(ix) \(x^2 + 2\sqrt{2}x + 2 = 0\)

Answer:
Given:
\[ x^2 + 2\sqrt{2}x + 2 = 0 \]
Multiply 1 × 2 = 2
We need two numbers whose sum = \(2\sqrt{2}\) and product = 2
These numbers are \(\sqrt{2}\) and \(\sqrt{2}\)

Split the middle term:
\[ x^2 + \sqrt{2}x + \sqrt{2}x + 2 = 0 \]
Group terms:
\[ x(x + \sqrt{2}) + \sqrt{2}(x + \sqrt{2}) = 0 \]
Take common factor:
\[ (x + \sqrt{2})(x + \sqrt{2}) = 0 \]
\[ x = -\sqrt{2} \]
Therefore, equal roots.

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(x) \(14x + 5 - 3x^2 = 0\)

Answer:
Rearrange:
\[ 3x^2 - 14x - 5 = 0 \]
Multiply 3 × (−5) = −15
We need two numbers whose sum = −14 and product = −15
These numbers are −15 and 1

Split the middle term:
\[ 3x^2 - 15x + x - 5 = 0 \]
Group terms:
\[ 3x(x - 5) + 1(x - 5) = 0 \]
Take common factor:
\[ (3x + 1)(x - 5) = 0 \]
\[ x = -\frac{1}{3},\; x = 5 \]

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Q3. Find two numbers whose sum is 27 and product is 182.

Answer:
Let the two numbers be x and y.
Given:
\[ x + y = 27 \quad ...(1) \] \[ xy = 182 \quad ...(2) \]
From (1),
\[ y = 27 - x \quad ...(3) \]
Substituting (3) in (2), we get:
\[ x(27 - x) = 182 \] \[ 27x - x^2 = 182 \] \[ x^2 - 27x + 182 = 0 \]
Now factorising:
We need two numbers whose sum = −27 and product = 182
These numbers are −13 and −14

\[ x^2 - 13x - 14x + 182 = 0 \] \[ x(x - 13) - 14(x - 13) = 0 \] \[ (x - 13)(x - 14) = 0 \]
\[ x = 13 \quad \text{or} \quad x = 14 \]
So from (3):
\[ y = 14 \quad \text{or} \quad y = 13 \]
Therefore, the two numbers are 13 and 14.

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Q4. Find two consecutive positive integers, sum of whose squares is 365.

Answer:
Let the first integer be x.
Then the next integer will be = x + 1
Given:
\[ x^2 + (x+1)^2 = 365 \]
\[ x^2 + x^2 + 2x + 1 = 365 \] \[ 2x^2 + 2x + 1 = 365 \] \[ 2x^2 + 2x - 364 = 0 \]
Divide by 2:
\[ x^2 + x - 182 = 0 \]
Now factorising:
Numbers whose sum = 1 and product = −182 → 14 and −13

\[ x^2 + 14x - 13x - 182 = 0 \] \[ x(x + 14) - 13(x + 14) = 0 \] \[ (x + 14)(x - 13) = 0 \]
\[ x = 13 \quad \text{or} \quad x = -14 \]
Since integers are positive:
\(x = 13\)
Next integer = 14

Therefore, the numbers are 13 and 14.

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Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer:
Let base of triangle be x cm.
Then its altitude will be = x − 7 cm
Using Pythagoras theorem:
\[ (\text{base})^2 + (\text{altitude})^2 = (\text{hypotenuse})^2 \] \[ x^2 + (x - 7)^2 = 13^2 \]
\[ x^2 + x^2 - 14x + 49 = 169 \] \[ 2x^2 - 14x + 49 = 169 \] \[ 2x^2 - 14x - 120 = 0 \]
Divide by 2:
\[ x^2 - 7x - 60 = 0 \]
factorising:
Numbers whose sum = −7 and product = −60 → −12 and 5

\[ x^2 - 12x + 5x - 60 = 0 \] \[ x(x - 12) + 5(x - 12) = 0 \] \[ (x - 12)(x + 5) = 0 \]
\[ x = 12 \quad \text{or} \quad x = -5 \]
Rejecting negative value.
Base of triangle = 12 cm
Altitude of triangle = 5 cm

Therefore, the sides are 12 cm and 5 cm.

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Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Answer:
Let number of articles be x.
Then cost per article will be = 2x + 3
so, the total cost will be = number × cost per article
\[ x(2x + 3) = 90 \]
\[ 2x^2 + 3x = 90 \] \[ 2x^2 + 3x - 90 = 0 \]
Factorising:
Multiply 2 × (−90) = −180
Numbers whose sum = 3 and product = −180 → 15 and −12

\[ 2x^2 + 15x - 12x - 90 = 0 \] \[ x(2x + 15) - 6(2x + 15) = 0 \] \[ (2x + 15)(x - 6) = 0 \]
\[ x = 6 \quad \text{or} \quad x = -\frac{15}{2} \]
Rejecting the negative value.
Number of articles = 6
Cost per article = \(2(6) + 3 = 15\)

Therefore, number of articles = 6 and cost per article = Rs 15.

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Q7. If \(x^2 - 2px + p^2 = 0\), then the value of \( \frac{p}{x} \) is:

(a) 0
(b) -1
(c) 1
(d) 2

Answer: (c) 1

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Q8. A student has done the following steps to find the roots of the equation \(x^2 - 3x - 10 = 0\). Identify the first mistake.

Step 1: \(x^2 - 3x - 10 = 0\)
Step 2: \(x^2 - 5x - 2x - 10 = 0\)
Step 3: \(x(x - 5) - 2(x - 5) = 0\)
Step 4: \((x - 5)(x - 2) = 0\)
Step 5: \(x = 5\) and \(x = 2\)

(a) Step 2
(b) Step 3
(c) Step 4
(d) Step 5

Answer: (a) Step 2

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Q9. Find the sum and product of the roots of the quadratic equation \(2x^2 - 9x + 4 = 0\).

Answer:
For equation \(ax^2 + bx + c = 0\):
Sum of roots = \(-\frac{b}{a}\)
Product of roots = \(\frac{c}{a}\)

Here,
\(a = 2,\; b = -9,\; c = 4\)

Sum = \(\frac{9}{2}\)
Product = \(2\)

Therefore, sum = \( \frac{9}{2} \), product = 2.

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Q10. If one root of the quadratic equation \(2x^2 + kx - 6 = 0\) is 2, find the value of k and the other root.

Answer:
Given that one root = 2
Substitute \(x = 2\) in equation:
\[ 2(2)^2 + k(2) - 6 = 0 \] \[ 8 + 2k - 6 = 0 \] \[ 2k + 2 = 0 \Rightarrow k = -1 \]
Now equation becomes:
\[ 2x^2 - x - 6 = 0 \]
Factorise:
Multiply 2 × (−6) = −12
Numbers: −4 and 3

\[ 2x^2 - 4x + 3x - 6 = 0 \] \[ 2x(x - 2) + 3(x - 2) = 0 \] \[ (2x + 3)(x - 2) = 0 \]
Other root = \(-\frac{3}{2}\)

Therefore, \(k = -1\) and other root = \( -\frac{3}{2} \).

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Q11. A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway, shown in the figure.

The total area of the lawn and the walkway is 360 square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are 12 metres by 10 metres.

Based on the information given above, answer the following questions:

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(i) Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway = x m.

Answer:
Length of lawn = 12 m
Breadth of lawn = 10 m

Walkway is of width x on all sides.
So, total length = \(12 + 2x\)
Total breadth = \(10 + 2x\)

Total area = 360 m²
\[ (12 + 2x)(10 + 2x) = 360 \] \[ 120 + 24x + 20x + 4x^2 = 360 \] \[ 4x^2 + 44x + 120 = 360 \] \[ 4x^2 + 44x - 240 = 0 \] Divide by 4:
\[ x^2 + 11x - 60 = 0 \]
Therefore, required quadratic equation is \[ x^2 + 11x - 60 = 0 \]

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(ii) Solve the quadratic equation to find the width of the walkway x.

Answer:
\[ x^2 + 11x - 60 = 0 \]
Numbers whose sum = 11 and product = −60 → 15 and −4

\[ x^2 + 15x - 4x - 60 = 0 \] \[ x(x + 15) - 4(x + 15) = 0 \] \[ (x + 15)(x - 4) = 0 \]
\[ x = -15 \quad \text{or} \quad x = 4 \]
Reject negative value.
Therefore, width of walkway = 4 m.

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(iii) If the cost of paving the walkway at the rate of Rs. 50 per square metre is Rs. 12,000, calculate the area of the walkway.

Answer:
Cost per m² = Rs. 50
Total cost = Rs. 12,000

\[ \text{Area of walkway} = \frac{12000}{50} = 240 \text{ m}^2 \]
Therefore, area of walkway = 240 m².

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(iv) Find the perimeter of the lawn.

Answer:
Length = 12 m, Breadth = 10 m
\[ \text{Perimeter} = 2(12 + 10) = 44 \text{ m} \]
Therefore, perimeter of the lawn = 44 m.

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